Genius Physics (Class 11) - For IIT-JEE and CBSE [1]

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genius Physics class XI

2012 Physical World

•

by Pradeep Kshetrapal

Science means organized knowledge.

It is human nature to observe things and happenings around in the nature and then to relate them. This knowledge is organized so that it become well connected and logical. Then it is known as Science. It is a systematic attempt to understand natural phenomenon and use this knowledge to predict, modify and control phenomena. Scientific Method Scientific methods are used to observe things and natural phenomena. It includes several steps : • Observations • Controlled experiments, • Qualitative and quantitative reasoning, • Mathematical modeling, • Prediction and • Verification or falsification of theories. There is no ‘final’ theory in science and no unquestioned authority in science. • Observations and experiments need theories to support them. Sometimes the existing theory is unable to explain the new observations, hence either new theories are formed or modification is done in the existing theories. • For example to explain different phenomena in light, theories are changed. To explain bending of light a new Wave-theory was formed, and then to explain photoelectric effect help of quantum mechanics was taken. Natural Sciences can be broadly divided in three branches namely Physics, Chemistry and biology • Physics is a study of basic laws of nature and their manifestation in different phenomenas. Principal thrusts in Physics • There are two principal thrusts in Physics; • 1.Unification 2. reduction Unification • Efforts are made to explain different phenomena in nature on the basis of one or minimum laws. This is principle of Unification.

Example: Phenomena of apple falling to ground, moon revolving around earth and weightlessness in the rocket, all these phenomena are explained with help of one Law that is, Newtons Law of Gravitation. Reductionism • To understand or to derive the properties of a bigger or more complex system the properties of its simpler constituents are taken into account. This approach is called reductionism. It is supposed to be the heart of Physics. For example a complex thermo dynamical system can be understood by the properties of its constituent like kinetic energy of molecules and atoms.

• •

• •

The scope of Physics can be divided in to two domains; Macroscopic and Microscopic. Macroscopic domain includes phenomena at the level of Laboratory, terrestrial and astronomical scales. Microscopic domain I ncludes atomic, molecular and nuclear phenomena. Recently third domain in between is also thought of with name Mesoscopic Physics. This deals with group of Hundreds of atoms • Scope of physics is very wide and exciting because it deals with objects of size as large as Universe (1025m) and as small as 10-14 m, the size of a nucleus. The excitement of Physics is experienced in many fields Like: • Live transmissions through television. • Computers with high speed and memory, • Use of Robots, • Lasers and their applications

Physics in relation to other branches of Science Physics in relation to Chemistry. • Chemical bonding, atomic number and complex structure can be explained by physics phenomena of Electrostatic forces, • taking help of X-ray diffraction.

genius Physics class XI

2012 by Pradeep Kshetrapal

• Physics in relation to other Science • Physics in relation to Biological Sciences: Physics helps in study of Biology through its inventions. Optical microscope helps to study bio-samples, electron microscope helps to study biological cells. X-rays have many applications in biological sciences. Radio isotopes are used in cancer. • Physics in relation with Astronomy: • Giant astronomical telescope developed in physics are used for observing planets. Radio telescopes have enabled astronomers to observe distant limits of universe. • Physics related to other sciences: Laws of Physics are used to study different phenomenas in other sciences like Biophysics, oceanography, seismology etc. Fundamental Forces in Nature There is a large number of forces experienced or applied. These may be macroscopic forces like gravitation, friction, contact forces and microscopic forces like electromagnetic and inter-atomic forces. But all these forces arise from some basic forces called Fundamental Forces. Fundamental Forces in Nature.. 1. Gravitational force. • It is due to Mass of the two bodies. • It is always attractive. • It operates in all objects of universe. • Its range is infinite It’s a weak force. 10-38 times compared to strong Nuclear force 2.Electromagnetic Forces: • It’s due to stationery or moving Electrical charge • It may be attractive or repulsive. • It operates on charged particles • Its range is infinite • Its stronger 1036 times than gravitational force but 10-2 times of strong Nuclear force. 3. Strong nuclear force: • Operate between Nucleons • It may be attractive or repulsive

•

Its range is very short, within nuclear size (10-15 m). Its strongest force in nature

4.Weak Nuclear force: • Operate within nucleons I.e. elementary particles like electron and neutrino. • It appears during radioactive b decay. • Has very short range 10-15m. • 10-13 times than Strong nuclear force.

Conservation Laws • In any physical phenomenon governed by different forces, several quantities do not change with time. These special quantities are conserved quantities of nature. 1.For motion under conservative force, the total mechanical Energy of a body is constant. 2.Total energy of a system is conserved, and it is valid across all domains of nature from microscopic to macroscopic. Total energy of the universe is believed to be constant. 3.Conservation of Mass was considered another conservation law, till advent of Einstein. Then it was converted to law of conservation of mass plus energy. Because mass is converted into energy and vice-versa according to equation E = mc2 The examples are annihilation and pair production. 4.Momentum is another quantity which is preserved. Similar is angular momentum of an isolated system. 5.Conservation of Electric charge is a fundamental law of nature. 6. Later there was development of law of conservation of attributes called baryon number, lepton number and so on. The laws of nature do not change with change of space and time. This is known as symmetry of space and time. This and some other symmetries play a central role in modern physics. Conservation laws are connected to this.

genius Physics

Classes XI

2011 by Pradeep Kshetrapal

Laws of Physics related to technology : Principal of Physics

Technology

Electromagnetic Induction

Electricity Generation

Laws of Thermodynamics

Steam, petrol, or diesel Engine

Electromagnetic Waves propagation

Radio, TV, Phones

Nuclear chain reaction

Nuclear reactor for power

Newtons Second & Third Law

Rocket propulsion

Bernoulli’s theorem

Aero planes

Population inversion

Lasers

X-rays

Medical Diagnosis

Ultra high magnetic fields

Superconductors

Digital electronics

Computers and calculators

Electromagnetic Induction

Electricity Generation

Physicist and their contributions Name

Contribution

country

Isaac Newton

Law of Gravitation, Laws of Motion, Reflecting telescope

U.K.

Galileo Galilei

Law of Inertia

Italy

Archimedes

Principle of Buoyancy, Principle of Lever

Greece

James Clerk Maxwell

Electromagnetic theory, light is an e/m wave.

U.K.

W.K.Roentgen

X-rays

Germany

Marie S. Curie

Discovery of Radium, Polonium, study of Radioactivity

Poland

Albert Einstein

Law of Photo electricity, Theory of Relativity

Germany

S.N.Bose

Quantum Statistics

India

James Chadwick

Neutron

U.K.

Niels Bohr

Quantum model of Hydrogen atom

Denmark

Earnest Rutherford

Nuclear model of Atom

New Zealand

C.V.Raman

Inelastic Scattering of light by molecules

India

Christian Huygens

Wave theory of Light

Holland

Michael Faraday

Laws of Electromagnetic Induction

U.K.

Edvin Hubble

Expanding Universe

U.S.A.

H.J.Bhabha

Cascade process in cosmic radiation

India

Abdus Salam

Unification of week and e/m interactions

Pakistan

R.A.Milikan

Measurement of Electronic Charge

U.S.A.

E.O.Lawrence

Cyclotron

U.S.A.

genius Physics

Classes XI

2011 by Pradeep Kshetrapal

Wolfgong Pauli

Quantum Exclusion principle

Austria

Louis de Broglie

Wave nature of matter

France

J.J.Thomson

Electron

U.K.

S.Chandrashekhar

Chandrashekhar limit, structure of stars

India

Christian Huygens

Wave theory of Light

Holland

Michael Faraday

Laws of Electromagnetic Induction

U.K.

Edvin Hubble

Expanding Universe

U.S.A.

Henrick Hertz

Electromagnetic Waves

Germany

J.C.Bose

Ultra short radio waves

India

Hideki Yukava

Theory of Nuclear Forces

Japan

W.Heisenberg

Quantum mechanics, Uncertainty principle

Germany

M.N.Saha

Thermal Ionization

India

G.N.Ramachandran

Triple Helical structure of proteins

india

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 1

Notes and exercises by

Page

1

Pradeep Kshetrapal

For students of Pradeep Kshetrapal only at Jamanipali, Ghantaghar Korba and Kusmunda.

genius PHYSICS by Pradeep Kshetrapal

Page

2

Units, Dimensions and Measurement 2

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 3

Physical World •

Science means organized knowledge.

It is human nature to observe things and happenings around in the nature and then to relate them. This knowledge is organized so that it become well connected and logical. Then it is known as Science. It is a systematic attempt to understand natural phenomenon and use this knowledge to predict, modify and control phenomena. Scientific Method Scientific methods are used to observe things and natural phenomena. It includes several steps : • Observations • Controlled experiments, • Qualitative and quantitative reasoning, • Mathematical modeling, • Prediction and • Verification or falsification of theories. There is no ‘final’ theory in science and no unquestioned authority in science. • Observations and experiments need theories to support them. Sometimes the existing theory is unable to explain the new observations, hence either new theories are formed or modification is done in the existing theories. • For example to explain different phenomena in light, theories are changed. To explain bending of light a new Wave-theory was formed, and then to explain photoelectric effect help of quantum mechanics was taken. Natural Sciences can be broadly divided in three branches namely Physics, Chemistry and biology •

Physics is a study of basic laws of nature and their manifestation in different phenomenas.

Principal thrusts in Physics • There are two principal thrusts in Physics; • 1.Unification 2. reduction Unification • Efforts are made to explain different phenomena in nature on the basis of one or minimum laws. This is principle of Unification. Example: Phenomena of apple falling to ground, moon revolving around earth and weightlessness in the rocket, all these phenomena are explained with help of one Law that is, Newtons Law of Gravitation.

Page

3

Reductionism • To understand or to derive the properties of a bigger or more complex system the properties of its simpler constituents are taken into account. This approach is called reductionism. It is supposed to be the heart of Physics. For example a complex thermo dynamical system can be understood by the properties of its constituent like kinetic energy of molecules and atoms.

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 4 • • • •

The scope of Physics can be divided in to two domains; Macroscopic and Microscopic. Macroscopic domain includes phenomena at the level of Laboratory, terrestrial and astronomical scales. Microscopic domain I ncludes atomic, molecular and nuclear phenomena. Recently third domain in between is also thought of with name Mesoscopic Physics. This deals with group of Hundreds of atoms • Scope of physics is very wide and exciting because it deals with objects of size as large as Universe (1025m) and as small as 10-14 m, the size of a nucleus. The excitement of Physics is experienced in many fields Like: • Live transmissions through television. • Computers with high speed and memory, • Use of Robots, • Lasers and their applications

Physics in relation to other branches of Science Physics in relation to Chemistry. • Chemical bonding, atomic number and complex structure can be explained by physics phenomena of Electrostatic forces, • taking help of X-ray diffraction. Physics in relation to other Science • Physics in relation to Biological Sciences: Physics helps in study of Biology through its inventions. Optical microscope helps to study bio-samples, electron microscope helps to study biological cells. X-rays have many applications in biological sciences. Radio isotopes are used in cancer. • Physics in relation with Astronomy: • Giant astronomical telescope developed in physics are used for observing planets. Radio telescopes have enabled astronomers to observe distant limits of universe. • Physics related to other sciences: Laws of Physics are used to study different phenomenas in other sciences like Biophysics, oceanography, seismology etc.

Page

2.Electromagnetic Forces: • It’s due to stationery or moving Electrical charge • It may be attractive or repulsive. • It operates on charged particles • Its range is infinite • Its stronger 1036 times than gravitational force but 10-2 times of strong Nuclear force.

4

Fundamental Forces in Nature There is a large number of forces experienced or applied. These may be macroscopic forces like gravitation, friction, contact forces and microscopic forces like electromagnetic and inter-atomic forces. But all these forces arise from some basic forces called Fundamental Forces. Fundamental Forces in Nature.. 1. Gravitational force. • It is due to Mass of the two bodies. • It is always attractive. • It operates in all objects of universe. • Its range is infinite It’s a weak force. 10-38 times compared to strong Nuclear force

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 5 3. Strong nuclear force: • Operate between Nucleons • It may be attractive or repulsive • Its range is very short, within nuclear size (10-15 m). • Its strongest force in nature 4.Weak Nuclear force: • Operate within nucleons I.e. elementary particles like electron and neutrino. • It appears during radioactive b decay. • Has very short range 10-15m. • 10-13 times than Strong nuclear force. Conservation Laws • In any physical phenomenon governed by different forces, several quantities do not change with time. These special quantities are conserved quantities of nature. 1. For motion under conservative force, the total mechanical Energy of a body is constant. 2. Total energy of a system is conserved, and it is valid across all domains of nature from microscopic to macroscopic. Total energy of the universe is believed to be constant. 3. Conservation of Mass was considered another conservation law, till advent of Einstein. Then it was converted to law of conservation of mass plus energy. Because mass is converted into energy and vice-versa according to equation E = mc2 The examples are annihilation and pair production. 4. Momentum is another quantity which is preserved. Similar is angular momentum of an isolated system. 5. Conservation of Electric charge is a fundamental law of nature. 6. Later there was development of law of conservation of attributes called baryon number, lepton number and so on. The laws of nature do not change with change of space and time. This is known as symmetry of space and time. This and some other symmetries play a central role in modern physics. Conservation laws are connected to this.

Technology

Electromagnetic Induction

Electricity Generation

Laws of Thermodynamics

Steam, petrol, or diesel Engine

Electromagnetic Waves propagation

Radio, TV, Phones

Nuclear chain reaction

Nuclear reactor for power

Newtons Second & Third Law

Rocket propulsion

Bernoulli’s theorem

Aero planes

Population inversion

Lasers

X-rays

Medical Diagnosis

Ultra high magnetic fields

Superconductors

Digital electronics

Computers and calculators

Electromagnetic Induction

Electricity Generation Page

Principal of Physics

5

Laws of Physics related to technology :

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 6

Physicist and their contributions country

Isaac Newton

Law of Gravitation, Laws of Motion, Reflecting telescope

U.K.

Galileo Galilei

Law of Inertia

Italy

Archimedes

Principle of Buoyancy, Principle of Lever

Greece

James Clerk Maxwell

Electromagnetic theory, light is an e/m wave.

U.K.

W.K.Roentgen

X-rays

Germany

Marie S. Curie

Discovery of Radium, Polonium, study of Radioactivity

Poland

Albert Einstein

Law of Photo electricity, Theory of Relativity

Germany

S.N.Bose

Quantum Statistics

India

James Chadwick

Neutron

U.K.

Niels Bohr

Quantum model of Hydrogen atom

Denmark

Earnest Rutherford

Nuclear model of Atom

New Zealand

C.V.Raman

Inelastic Scattering of light by molecules

India

Christian Huygens

Wave theory of Light

Holland

Michael Faraday

Laws of Electromagnetic Induction

U.K.

Edvin Hubble

Expanding Universe

U.S.A.

H.J.Bhabha

Cascade process in cosmic radiation

India

Abdus Salam

Unification of week and e/m interactions

Pakistan

R.A.Milikan

Measurement of Electronic Charge

U.S.A.

E.O.Lawrence

Cyclotron

U.S.A.

Wolfgong Pauli

Quantum Exclusion principle

Austria

Louis de Broglie

Wave nature of matter

France

J.J.Thomson

Electron

U.K.

S.Chandrashekhar

Chandrashekhar limit, structure of stars

India

Christian Huygens

Wave theory of Light

Holland

Michael Faraday

Laws of Electromagnetic Induction

U.K.

Edvin Hubble

Expanding Universe

U.S.A.

Henrick Hertz

Electromagnetic Waves

Germany

J.C.Bose

Ultra short radio waves

India

Hideki Yukava

Theory of Nuclear Forces

Japan

W.Heisenberg

Quantum mechanics, Uncertainty principle

Germany

M.N.Saha

Thermal Ionization

India

G.N.Ramachandran

Triple Helical structure of proteins

india

6

Contribution

Page

Name

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 7

1.1 Physical Quantity. A quantity which can be measured and by which various physical happenings can be explained and expressed in form of laws is called a physical quantity. For example length, mass, time, force etc. On the other hand various happenings in life e.g., happiness, sorrow etc. are not physical quantities because these can not be measured. Measurement is necessary to determine magnitude of a physical quantity, to compare two similar physical quantities and to prove physical laws or equations. A physical quantity is represented completely by its magnitude and unit. For example, 10 metre means a length which is ten times the unit of length 1 kg. Here 10 represents the numerical value of the given quantity and metre represents the unit of quantity under consideration. Thus in expressing a physical quantity we choose a unit and then find that how many times that unit is contained in the given physical quantity, i.e. Physical quantity (Q) = Magnitude × Unit = n × u Where, n represents the numerical value and u represents the unit. Thus while expressing definite amount of physical quantity, it is clear that as the unit(u) changes, the magnitude(n) will also change but product ‘nu’ will remain same. i.e.

n u = constant,

or

n1u1  n 2 u 2  constant ;

 n

1 u

i.e. magnitude of a physical quantity and units are inversely proportional to each other .Larger the unit, smaller will be the magnitude.

1.2 Types of Physical Quantity. (1) Ratio (numerical value only) : When a physical quantity is a ratio of two similar quantities, it has no unit. e.g. Relative density = Density of object/Density of water at 4oC Refractive index = Velocity of light in air/Velocity of light in medium Strain = Change in dimension/Original dimension

Note : 

Angle is exceptional physical quantity, which though is a ratio of two similar physical

quantities (angle = arc / radius) but still requires a unit (degrees or radians) to specify it along with its numerical value. (2) Scalar (Magnitude only) : These quantities do not have any direction e.g. Length, time, work, energy etc. Magnitude of a physical quantity can be negative. In that case negative sign indicates that the numerical value of the quantity under consideration is negative. It does not specify the direction. Scalar quantities can be added or subtracted with the help of following ordinary laws of addition or subtraction.

Page

Vector physical quantities can be added or subtracted according to vector laws of addition. These laws are different from laws of ordinary addition.

7

(3) Vector (magnitude and direction) : e.g. displacement, velocity, acceleration, force etc.

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 8

Note : 

There are certain physical quantities which behave neither as scalar nor as vector. For

example, moment of inertia is not a vector as by changing the sense of rotation its value is not changed. It is also not a scalar as it has different values in different directions (i.e. about different axes). Such physical quantities are called Tensors.

1.3 Fundamental and Derived Quantities. (1) Fundamental quantities : Out of large number of physical quantities which exist in nature, there are only few quantities which are independent of all other quantities and do not require the help of any other physical quantity for their definition, therefore these are called absolute quantities. These quantities are also called fundamental or base quantities, as all other quantities are based upon and can be expressed in terms of these quantities. (2) Derived quantities : All other physical quantities can be derived by suitable multiplication or division of different powers of fundamental quantities. These are therefore called derived quantities. If length is defined as a fundamental quantity then area and volume are derived from length and are expressed in term of length with power 2 and 3 over the term of length.

Note : 

In mechanics Length, Mass and time are arbitrarily chosen as fundamental

quantities. However this set of fundamental quantities is not a unique choice. In fact any three quantities in mechanics can be termed as fundamental as all other quantities in mechanics can be expressed in terms of these. e.g. if speed and time are taken as fundamental quantities, length will become a derived quantity because then length will be expressed as Speed  Time. and if force and acceleration are taken as fundamental quantities, then mass will be defined as Force / acceleration and will be termed as a derived quantity.

1.4 Fundamental and Derived Units. Normally each physical quantity requires a unit or standard for its specification so it appears that there must be as many units as there are physical quantities. However, it is not so. It has been found that if in mechanics we choose arbitrarily units of any three physical quantities we can express the units of all other physical quantities in mechanics in terms of these. Arbitrarily the physical quantities mass, length and time are choosen for this purpose. So any unit of mass, length and time in mechanics is called a fundamental, absolute or base unit. Other units which can be expressed in terms of fundamental units, are called derived units. For example light year or km is a fundamental units as it is a unit of length while s–1, m2 or kg/m are derived units as these are derived from units of time, mass and length respectively. System of units : A complete set of units, both fundamental and derived for all kinds of physical quantities is called system of units. The common systems are given below – (1) CGS system : The system is also called Gaussian system of units. In it length, mass and time have been chosen as the fundamental quantities and corresponding fundamental units are centimetre (cm), gram (g) and second (s) respectively.

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(3) FPS system : In this system foot, pound and second are used respectively for measurements of length, mass and time. In this system force is a derived quantity with unit poundal.

8

(2) MKS system : The system is also called Giorgi system. In this system also length, mass and time have been taken as fundamental quantities, and the corresponding fundamental units are metre, kilogram and second.

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 9 (4) S. I. system : It is known as International system of units, and is infact extended system of units applied to whole physics. There are seven fundamental quantities in this system. These quantities and their units are given in the following table Quantity

Name of Unit

Symbol

Length

metre

m

Mass

kilogram

kg

Time

second

s

Electric Current

ampere

A

Temperature

Kelvin

K

Amount of Substance

mole

mol

Luminous Intensity

candela

cd

Besides the above seven fundamental units two supplementary units are also defined – Radian (rad) for plane angle and Steradian (sr) for solid angle.

Note : 

Apart from fundamental and derived units we also use very frequently practical units.

These may be fundamental or derived units e.g., light year is a practical unit (fundamental) of distance while horse power is a practical unit (derived) of power.  Practical units may or may not belong to a system but can be expressed in any system of units e.g., 1 mile = 1.6 km = 1.6 × 103 m.

1.5 S.I. Prefixes.

Prefix

Symbol

1018

exa

E

1015

peta

P

1012

tera

T

109

giga

G

106

mega

M

103

kilo

k

102

hecto

h

101

deca

da

10–1

deci

d

10–1

centi

c

10–3

milli

m

10–6

micro



Page

Power of 10

9

In physics we have to deal from very small (micro) to very large (macro) magnitudes as one side we talk about the atom while on the other side of universe, e.g., the mass of an electron is 9.1  10–31 kg while that of the sun is 2  1030 kg. To express such large or small magnitudes simultaneously we use the following prefixes :

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 10 10–9

nano

n

10–12

pico

p

10–15

femto

f

10–18

atto

a

1.6 Standards of Length, Mass and Time. (1) Length : Standard metre is defined in terms of wavelength of light and is called atomic standard of length. The metre is the distance containing 1650763.73 wavelength in vacuum of the radiation corresponding to orange red light emitted by an atom of krypton-86. Now a days metre is defined as length of the path travelled by light in vacuum in 1/299,7792, 458 part of a second. (2) Mass : The mass of a cylinder made of platinum-iridium alloy kept at International Bureau of Weights and Measures is defined as 1 kg. On atomic scale, 1 kilogram is equivalent to the mass of 5.0188  1025 atoms of 6C12 (an isotope of carbon). (3) Time : 1 second is defined as the time interval of 9192631770 vibrations of radiation in Cs-133 atom. This radiation corresponds to the transition between two hyperfine level of the ground state of Cs133.

1.7 Practical Units. (1) Length : (i) 1 fermi = 1 fm = 10–15 m (ii) 1 X-ray unit = 1XU = 10–13 m (iii) 1 angstrom = 1Å = 10–10 m = 10–8 cm = 10–7 mm = 0.1 mm (iv) 1 micron = m = 10–6 m (v) 1 astronomical unit = 1 A.U. = 1. 49  1011 m  1.5  1011 m  108 km (vi) 1 Light year = 1 ly = 9.46  1015 m (vii) 1 Parsec = 1pc = 3.26 light year (2) Mass : (i) Chandra Shekhar unit : 1 CSU = 1.4 times the mass of sun = 2.8  1030 kg (ii) Metric tonne : 1 Metric tonne = 1000 kg (iii) Quintal : 1 Quintal = 100 kg (iv) Atomic mass unit (amu) : amu = 1.67  10–27 kg mass of proton or neutron is of the order of 1 amu (3) Time : (i) Year : It is the time taken by earth to complete 1 revolution around the sun in its orbit. (ii) Lunar month : It is the time taken by moon to complete 1 revolution around the earth in its orbit.

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(iii) Solar day : It is the time taken by earth to complete one rotation about its axis with respect to sun. Since this time varies from day to day, average solar day is calculated by taking average of the duration of all the days in a year and this is called Average Solar day.

10

1 L.M. = 27.3 days

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 11 1 Solar year = 365.25 average solar day or

average solar day 

1 the part of solar year 365 .25

(iv) Sedrial day : It is the time taken by earth to complete one rotation about its axis with respect to a distant star. 1 Solar year = 366.25 Sedrial day = 365.25 average solar day Thus 1 Sedrial day is less than 1 solar day. (v) Shake : It is an obsolete and practical unit of time. 1 Shake = 10– 8 sec

1.8 Dimensions of a Physical Quantity. When a derived quantity is expressed in terms of fundamental quantities, it is written as a product of different powers of the fundamental quantities. The powers to which fundamental quantities must be raised in order to express the given physical quantity are called its dimensions. To make it more clear, consider the physical quantity force Force = mass × acceleration 

mass  velocity mass  length/time = mass × length × (time)–2 .... (i)  time time

Thus, the dimensions of force are 1 in mass, 1 in length and – 2 in time. Here the physical quantity that is expressed in terms of the base quantities is enclosed in square brackets to indicate that the equation is among the dimensions and not among the magnitudes. Thus equation (i) can be written as [force] = [MLT–2]. Such an expression for a physical quantity in terms of the fundamental quantities is called the dimensional equation. If we consider only the R.H.S. of the equation, the expression is termed as dimensional formula. Thus, dimensional formula for force is, [MLT – 2].

1.9 Important Dimensions of Complete Physics. Mechanics (1)

Velocity or speed (v)

m/s

[M0L1T –1]

(2)

Acceleration (a)

m/s2

[M0LT –2]

(3)

Momentum (P)

kg-m/s

[M1L1T –1]

(4)

Impulse (I)

Newton-sec or kg-m/s

[M1L1T –1]

(5)

Force (F)

Newton

[M1L1T –2]

(6)

Pressure (P)

Pascal

[M1L–1T –2]

(7)

Kinetic energy (EK)

Joule

[M1L2T –2]

(8)

Power (P)

Watt or Joule/s

[M1L2T –3]

(9)

Density (d)

kg/m3

[M1L– 3T 0]

(10)

Angular displacement ()

Radian (rad.)

[M0L0T 0]

(11)

Angular velocity ( )

Radian/sec

[M0L0T – 1]

(12)

Angular acceleration ()

Radian/sec2

[M0L0T – 2]

(13)

Moment of inertia (I)

kg-m2

Unit

Dimension

[M1L2T0]

11

Quantity

Page

S. N.

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 12 S. N.

Quantity

(14)

Torque ()

Newton-meter

[M1L2T –2]

(15)

Angular momentum (L)

Joule-sec

[M1L2T –1]

(16)

Force constant or spring constant (k)

Newton/m

[M1L0T –2]

(17)

Gravitational constant (G)

N-m2/kg2

[M–1L3T – 2]

(18)

Intensity of gravitational field (Eg)

N/kg

[M0L1T – 2]

(19)

Gravitational potential (Vg)

Joule/kg

[M0L2T – 2]

(20)

Surface tension (T)

N/m or Joule/m2

[M1L0T – 2]

(21)

Velocity gradient (Vg)

Second–1

[M0L0T – 1]

(22)

Coefficient of viscosity ()

kg/m-s

[M1L– 1T – 1]

(23)

Stress

N/m2

[M1L– 1T – 2]

(24)

Strain

No unit

(25)

Modulus of elasticity (E)

N/m2

(26)

Poisson Ratio ()

No unit

[M0L0T 0]

(27)

Time period (T)

Second

[M0L0T1]

(28)

Frequency (n)

Hz

Unit

Dimension

[M0L0T 0] [M1L– 1T – 2]

[M0L0T –1]

Heat S. N.

Quantity

Dimension

(1)

Temperature (T)

Kelvin

[M0L0T0 1]

(2)

Heat (Q)

Joule

[ML2T– 2]

(3)

Specific Heat (c)

Joule/kg-K

[M0L2T– 2 –1]

(4)

Thermal capacity

Joule/K

[M1L2T – 2 –1]

(5)

Latent heat (L)

Joule/kg

[M0L2T – 2]

(6)

Gas constant (R)

Joule/mol-K

[M1L2T– 2 – 1]

(7)

Boltzmann constant (k)

Joule/K

[M1L2T– 2 – 1]

(8)

Coefficient of thermal conductivity (K)

Joule/m-s-K

[M1L1T– 3 – 1]

(9)

Stefan's constant ()

Watt/m2-K4

[M1L0T– 3 – 4]

(10)

Wien's constant (b)

Meter-K

[M0L1To1]

(11)

Planck's constant (h)

Joule-s

[M1L2T–1]

(12)

Coefficient of Linear Expansion ()

Kelvin–1

[M0L0T0 –1]

(13)

Mechanical eq. of Heat (J)

Joule/Calorie

[M0L0T0]

(14)

Vander wall’s constant (a)

Newton-m4

[ML5T– 2]

(15)

Vander wall’s constant (b)

m3

[M0L3T0]

Unit

Electricity Unit

Dimension

(1)

Electric charge (q)

Coulomb

[M0L0T1A1]

(2)

Electric current (I)

Ampere

[M0L0T0A1]

(3)

Capacitance (C)

Coulomb/volt or Farad

[M–1L– 2T4A2]

(4)

Electric potential (V)

Joule/coulomb

M1L2T–3A–1

12

Quantity

Page

S. N.

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 13 S. N.

Quantity

(5)

Permittivity of free space (0)

(6)

Dielectric constant (K)

Unitless

[M0L0T0]

(7)

Resistance (R)

Volt/Ampere or ohm

[M1L2T– 3A– 2]

(8)

Resistivity or Specific resistance ()

Ohm-meter

[M1L3T– 3A– 2]

(9)

Coefficient of Self-induction (L)

volt  second or henery or ohm-second ampere

[M1L2T– 2A– 2]

(10)

Magnetic flux ()

Volt-second or weber

[M1L2T–2A–1]

(11)

Unit

Magnetic induction (B)

Coulomb 2 Newton - meter 2

Joule newton ampere  meter ampere  meter 2

volt  second

or Tesla meter 2 Ampere/meter Ampere-meter2 Newton Joule or 2 or ampere ampere 2  meter

Dimension [M–1L–3T4A2]

[M1L0T– 2A– 1]

(12) (13)

Magnetic Intensity (H) Magnetic Dipole Moment (M)

(14)

Permeability of Free Space (0)

(15)

Surface charge density ()

Coulomb metre 2

[M0L–2T1A1]

(16)

Electric dipole moment (p)

Coulomb  meter

[M0L1T1A1]

(17)

Conductance (G) (1/R)

ohm 1

[M–1L–2T3A2]

(18)

Conductivity () (1/) Current density (J) Intensity of electric field (E) Rydberg constant (R)

ohm 1meter 1 Ampere/m2 Volt/meter, Newton/coulomb m–1

[M–1L–3T3A2]

(19) (20) (21)

Ohm  sec ond henery Volt  second or or meter meter ampere  meter

[M0L– 1T0A1] [M0L2T0A1]

[M1L1T–2A–2]

M0L–2T0A1 M1L1T –3A–1 M0L–1T0

1.10 Quantities Having Same Dimensions. S. N.

Dimension

Quantity

(1) (2) (3) (4) (5) (6) (7) (8)

[M0L0T–1] [M1L2T–2] [M1L–1T–2] [M1L1T–1] [M0L1T–2] [M1L1T–2] [M1L2T–1] [M1L0T–2]

(9)

[M0L0T0]

(10)

[M0L2T–2]

Frequency, angular frequency, angular velocity, velocity gradient and decay constant Work, internal energy, potential energy, kinetic energy, torque, moment of force Pressure, stress, Young’s modulus, bulk modulus, modulus of rigidity, energy density Momentum, impulse Acceleration due to gravity, gravitational field intensity Thrust, force, weight, energy gradient Angular momentum and Planck’s constant Surface tension, Surface energy (energy per unit area) Strain, refractive index, relative density, angle, solid angle, distance gradient, relative permittivity (dielectric constant), relative permeability etc. Latent heat and gravitational potential

(11)

[M0L2T–2–1]

Thermal capacity, gas constant, Boltzmann constant and entropy

(12)

[M0L0T1]

(13)

[M0L0T1]

l g , m k , R g , where l = length

LC , RC where L = inductance, R = resistance, C = capacitance

Page

L/R,

13

g = acceleration due to gravity, m = mass, k = spring constant

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 14

(14)

[ML2T–2]

V2 q2 t, VIt, qV , LI 2 , , CV 2 where I = current, t = time, q = charge, R C L = inductance, C = capacitance, R = resistance I 2 Rt,

1.11 Application of Dimensional Analysis. (1) To find the unit of a physical quantity in a given system of units : Writing the definition or formula for the physical quantity we find its dimensions. Now in the dimensional formula replacing M, L and T by the fundamental units of the required system we get the unit of physical quantity. However, sometimes to this unit we further assign a specific name, e.g., Work = Force  Displacement So

[W] = [MLT–2]  [L] = [ML2T–2]

So its units in C.G.S. system will be g cm2/s2 which is called erg while in M.K.S. system will be kg which is called joule.

m2/s2

Sample problems based on unit finding Problem 1.

a   The equation  P  2  (V  b) = constant. The units of a is V   (a) Dyne  cm 5

Solution : (b)

(b) Dyne  cm 4

(c) Dyne / cm 3

[MNR 1995; AFMC 1995]

(d) Dyne / cm 2

 a  According to the principle of dimensional homogenity [P]   2  V   [a]  [P] [V 2 ]  [ML1T 2 ] [L6 ]  [ML 5 T 2 ] or unit of a = gm × cm 5  sec–2= Dyne  cm4

Problem 2.

If x  at  bt 2 , where x is the distance travelled by the body in kilometre while t the time in seconds, then the units of b are (a) km/s

[CBSE 1993]

(b) km-s

(c) km/s2

Solution : (c)

x From the principle of dimensional homogenity [ x ]  [bt 2 ]  [b ]   2  t 

Problem 3.

The unit of absolute permittivity is (a) Farad - meter

Solution : (b)

 Unit of b = km/s2. [EAMCET (Med.) 1995; Pb. PMT 2001]

(b) Farad / meter

From the formula C  4 0 R   0 

(d) km-s2

(c) Farad/meter 2

(d) Farad

C 4R

By substituting the unit of capacitance and radius : unit of  0  Farad/ meter.

Problem 4.

Unit of Stefan's constant is (a) Js 1

(b) Jm 2 s 1 K 4

(c) Jm 2

(d) Js

Q Q Joule  T 4     Unit of   2 = Jm 2 s 1 K 4 At AtT 4 m  sec  K 4

Solution : (b)

Stefan's formula

Problem 5.

The unit of surface tension in SI system is [MP PMT 1984; AFMC 1986; CPMT 1985, 87; CBSE 1993; Karnataka CET (Engg/Med.) 1999; DCE 2000, 01]

From the formula of surface tension T 

(c) Dyne/cm

(d) Newton/m2

F l

By substituting the S.I. units of force and length, we will get the unit of surface tension = Newton/m

14

Solution : (b)

(b) Newton/m

Page

(a) Dyne / cm 2

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 15

Problem 6.

A suitable unit for gravitational constant is (a) kg metre sec

Solution : (c)

As F 

1

[MNR 1988]

(b) Newton metre

Gm 1m 2

 G

r2

1

2

sec (c) Newton metre kg

2

(d) kg metre sec

1

Fr2 m 1m 2

Substituting the unit of above quantities unit of G = Newton metre 2 kg 2 .

Problem 7.

The SI unit of universal gas constant (R) is [MP Board 1988; JIPMER 1993; AFMC 1996; MP PMT 1987, 94; CPMT 1984, 87; UPSEAT 1999]

(a) Watt K 1 mol 1 Solution : (c)

(d) Erg K 1 mol 1

(b) Newton K 1 mol 1 (c) Joule K 1 mol 1

Ideal gas equation PV  nRT  [R] 

[P][V ] [ML 1 T 2 ][L3 ] [ML 2 T 2 ]   [nT ] [mole ] [K ] [mole ]  [K ]

So the unit will be Joule K 1mol 1 .

(2) To find dimensions of physical constant or coefficients : As dimensions of a physical quantity are unique, we write any formula or equation incorporating the given constant and then by substituting the dimensional formulae of all other quantities, we can find the dimensions of the required constant or coefficient. (i) Gravitational constant : According to Newton’s law of gravitation F  G Substituting the dimensions of all physical quantities [G]  (ii) Plank constant : According to Planck E  h or h 

m1m 2 r2

or G 

Fr 2 m 1m 2

[MLT 2 ][L2 ]  [M 1 L3 T  2 ] [M ][M ]

E



Substituting the dimensions of all physical quantities [h] 

[ML 2 T 2 ]  [ML 2 T 1 ] 1 [T ]

(iii) Coefficient of viscosity : According to Poiseuille’s formula Substituting the dimensions of all physical quantities [] 

pr 4 dV pr 4  or   dt 8l 8 l(dV / dt )

[ML 1 T 2 ][L4 ]  [ML 1 T 1 ] [L][ L3 / T ]

Sample problems based on dimension finding X  3YZ 2 find dimension of Y in (MKSA) system, if X and Z are the dimension of capacity and magnetic field respectively

Solution : (d)

Problem 9.

L T

A

[MP PMT 2003] 1

X  3YZ 2  [Y ]  Dimensions of (a) [LT 1 ]

Solution : (d)

4

(b) ML

[X ] 2

[Z ]

1

0 0



2

(c) M

[M 1 L2 T 4 A 2 ] [MT

2

1 2

3 2

4

L T A

4

(d) M

3 2

L T 8 A4

 [M 3 L2 T 8 A 4 ] .

A ]

, where symbols have their usual meaning, are (b) [L1 T ]

We know that velocity of light C 

(c) [L2 T 2 ]

1

0 0



1

0  0

 C2

[AIEEE 2003]

(d) [L2 T 2 ]

15

(a) M

3 2

Page

Problem 8.

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 16  1  2 2 1 2  So    [LT ] = [L T ] .    0 0

Problem 10. If L, C and R denote the inductance, capacitance and resistance respectively, the dimensional formula for C 2 LR is

[UPSEAT 2002]

(a) [ML 2 T 1 I 0 ] Solution : (b)

(b) [M 0 L0 T 3 I 0 ]

(c) [M 1 L2 T 6 I 2 ]

(d) [M 0 L0 T 2 I 0 ]

 R  R   [C 2 LR ] = C 2 L2  = (LC )2   L  L   

and we know that frequency of LC circuits is given by f 

1 2

1

i.e., the dimension of LC is equal

LC

to [T 2 ] L L and   gives the time constant of L  R circuit so the dimension of is equal to [T]. R R

  R  By substituting the above dimensions in the given formula (LC )2    [T 2 ]2 [T 1 ]  [T 3 ] .  L  

Problem 11. A force F is given by F  at  bt 2 , where t is time. What are the dimensions of a and b [BHU 1998; AFMC 2001]

(a) MLT 3 and ML 2 T 4 (b) MLT 3 and MLT 4 (c) MLT 1 and MLT Solution : (b)

0

(d) MLT 4 and MLT 1

2  F   MLT  From the principle of dimensional homogenity [F]  [at]  [a]        [MLT  t   T  2  F   MLT  Similarly [F]  [bt 2 ]  [b]   2      [MLT 2  t   T 

4

3

]

].

Problem 12. The position of a particle at time t is given by the relation x (t)  

v0   t  (1  c ), where v 0 is a constant  

and   0 . The dimensions of v 0 and  are respectively (a) M 0 L1 T 1 and T 1 Solution : (a)

(b) M 0 L1 T 0 and T 1

(c) M 0 L1 T 1 and LT 2 (d) M 0 L1 T 1 and T

1  From the principle of dimensional homogeneity [ t] = dimensionless  [ ]     [T 1 ] t Similarly [ x ] 

v0 ]  [v 0 ]  [ x ][ ]  [L][T 1 ]  [LT 1 ] . [ ]

Problem 13. The dimensions of physical quantity X in the equation Force  (a) M 1 L4 T 2 Solution : (c)

(b) M 2 L2 T 1

X is given by Density

(c) M 2 L2T 2

(d) M 1 L2 T 1

[X] = [Force] × [Density] = [MLT 2 ]  [ML3 ] = [M 2 L2 T 2 ] .

Problem 14. Number of particles is given by n   D

n 2  n1 crossing a unit area perpendicular to X- axis in unit x 2  x1

time, where n 1 and n 2 are number of particles per unit volume for the value of x meant to x 2 and x 1 . Find dimensions of D called as diffusion constant (c) M 0 LT 3

(n) = Number of particle passing from unit area in unit time =

(d) M 0 L2 T 1

No. of particle [ M 0 L0 T 0 ]  = [L2 T 1 ] At [ L2 ] [T ]

16

Solution : (d)

(b) M 0 L2 T 4

Page

(a) M 0 LT 2

[CPMT 1979]

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 17 [n1 ]  [n2 ]  No. of particle in unit volume = [L3 ] Now from the given formula [D] 

[n][x 2  x1 ] [L2 T 1 ][L]   [L2 T 1 ] .  3 [n2  n1 ] [L ]

Problem 15. E, m, l and G denote energy, mass, angular momentum and gravitational constant respectively, then the dimension of

El 2 m 5 G2

(a) Angle Solution : (a)

are (b) Length

(c) Mass

(d) Time

[E] = energy = [ML2T 2 ] , [m] = mass = [M], [l] = Angular momentum = [ML2T 1 ] [G] = Gravitational constant = [M 1 L3 T 2 ] Now substituting dimensions of above quantities in

El 2 m 5G 2

=

[ML2 T 2 ]  [ML2 T 1 ]2 [M 5 ]  [M 1 L3 T  2 ]2

= [M 0 L0 T 0 ]

i.e., the quantity should be angle.

x Problem 16. The equation of a wave is given by Y  A sin    k  where  is the angular velocity and v is the v



linear velocity. The dimension of k is (a) LT Solution : (b)

(c) T 1

(b) T

(d) T 2

 L  x According to principle of dimensional homogeneity [k]    =    [T ] . v    LT 1 

Problem 17. The potential energy of a particle varies with distance x from a fixed origin as U 

A x x2  B

, where A

and B are dimensional constants then dimensional formula for AB is (a) ML7/2T 2 Solution : (b)

(b) ML11 / 2 T 2

(c) M 2 L9 / 2T 2

(d) ML13 / 2T 3

From the dimensional homogeneity [x 2 ]  [B]  [B] = [L2] As well as U  

[ A] [ x 1 / 2 ] [ x ]  [ B] 2

 [ML 2 T  2 ] 

[ A][L1 / 2 ] 2

 [ A]  [ML7 / 2 T 2 ]

[L ]

Now [ AB ]  [ML7 / 2 T 2 ]  [L2 ]  [ML11 / 2T 2 ]

Problem 18. The dimensions of (a) MLT 1 Solution : (c)

1  0 E 2 (  0 = permittivity of free space ; E = electric field ) is 2

(b) ML 2 T 2

Energy density =

(c) ML 1 T 2

(d) ML2T 1

 ML 2 T 2  1 Energy 1  2 0 E2     [ ML T ] 3 2 Volume  L 

Problem 19. You may not know integration. But using dimensional analysis you can check on some results. In the integral (a) 1

dx x   an sin 1   1  the value of n is 2 1/2 (2ax  x ) a  (b) – 1

(c) 0

(d)

1 2

Let x = length  [ X ]  [L] and [dx ]  [L]

By substituting dimension of each quantity in both sides:

[L] [L  L ] 2

2 1/ 2

 [Ln ]  n  0

17

x By principle of dimensional homogeneity    dimensionless  [a]  [x ]  [L] a

Page

Solution : (c)



genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 18

Problem 20. A physical quantity P 

B 2l 2 where B= magnetic induction, l= length and m = mass. The dimension m

of P is (a) MLT 3 Solution : (b)

(b) ML2T 4 I–2

F = BIL  Dimension of [B]  Now dimension of [P] 

(c) M 2 L2 T 4 I

(d) MLT

2 2

I

2

[ F] [MLT ]  = [MT 2 I 1 ] [I][L] [I][L]

B 2 l 2 [MT 2 I 1 ] 2  [L2 ]  [ML2 T 4 I 2 ]  m [M ] 2ct   2x   cos  , which of the following statements is      

Problem 21. The equation of the stationary wave is y= 2a sin  wrong

Solution : (d)

(a) The unit of ct is same as that of 

(b) The unit of x is same as that of 

(c) The unit of 2c / is same as that of 2x /t

(d) The unit of c/ is same as that of x / 

Here,

2ct



2x

as well as



 2ct   2x  0 0 0 are dimensionless (angle) i.e.    M L T      

 2c   2x  So (i) unit of c t is same as that of  (ii) unit of x is same as that of  (iii)        t 

and (iv)

x c is unit less. It is not the case with .  

(3) To convert a physical quantity from one system to the other : The measure of a physical quantity is nu = constant If a physical quantity X has dimensional formula [MaLbTc] and if (derived) units of that physical quantity in two systems are [M 1a Lb1 T1c ] and [M 2a Lb2 T2c ] respectively and n1 and n2 be the numerical values in the two systems respectively, then n1 [u1 ]  n 2 [u 2 ]  n1 [M 1a Lb1 T1c ]  n 2 [M 2a Lb2 T2c ] a

b

 M   L  T   n 2  n1  1   1   1   M 2   L 2   T2 

c

where M1, L1 and T1 are fundamental units of mass, length and time in the first (known) system and M2, L2 and T2 are fundamental units of mass, length and time in the second (unknown) system. Thus knowing the values of fundamental units in two systems and numerical value in one system, the numerical value in other system may be evaluated. Example : (1) conversion of Newton into Dyne. The Newton is the S.I. unit of force and has dimensional formula [MLT–2]. So 1 N = 1 kg-m/ sec2 a

b

c

1

 M   L  T   k g   m   sec  By using n 2  n1  1   1   1   1        gm   cm   sec   M 2   L 2   T2  1

2

10 3 gm  10 2 cm   sec  2 5  1       10 gm cm sec       1

1

The value of G in C.G.S. system is 6.67  10–8 C.G.S. units while its dimensional formula is [M–1L3T–2] So

G = 6.67  10–8 cm3/g s2

Page

(2) Conversion of gravitational constant (G) from C.G.S. to M.K.S. system

18

 1 N = 105 Dyne

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 19 a

b

1

c

 M   L  T   gm   cm   sec  By using n 2  n1  1   1   1   6.67  10 8        k g   m   sec   M 2   L 2   T2  3

1

 6 .67  10



8

2

 gm   cm   sec   3   2    10 gm  10 cm   sec  3

2

 6 .67  10 11

G = 6.67  10–11 M.K.S. units

Sample problems based on conversion Problem 22. A physical quantity is measured and its value is found to be nu where n  numerical value and u  unit. Then which of the following relations is true (c) n  u

(b) n  u

(a) n  u 2 Solution : (d)

[RPET 2003]

We know P  nu  constant  n1u1  n2u2 or n 

(d) n 

1 u

1 . u

Problem 23. In C.G.S. system the magnitude of the force is 100 dynes. In another system where the fundamental physical quantities are kilogram, metre and minute, the magnitude of the force is (a) 0.036 Solution : (c)

(b) 0.36

(c) 3.6

(d) 36

n1  100 , M1  g , L1  cm , T1  sec and M 2  kg , L2  meter , T2  m inute , x  1 , y  1 , z  2 x

y

 M   L  T  By substituting these values in the following conversion formula n2  n1  1   1   1   M 2   L2   T2  1

1

 gm   cm   sec  n 2  100        kg   meter   minute 

2

2

 gm   cm 1  sec  2 n2  100  3   2     3.6 10 gm  10 cm   60 sec  1

Problem 24. The temperature of a body on Kelvin scale is found to be X K. When it is measured by a Fahrenheit thermometer, it is found to be X F. Then X is (a) 301.25 Solution : (c)

(b) 574.25

Relation between centigrade and Fahrenheit According to problem

[UPSEAT 200]

(c) 313

(d) 40

K  273 F  32  5 9

X  273 X  32   X  313 . 5 9

Problem 25. Which relation is wrong

[RPMT 1997]

(a) 1 Calorie = 4.18 Joules (c) 1 MeV = 1.6 × Solution : (d)

10–13 Joules

(b) 1Å (d)

=10–10 m

1 Newton =10–5 Dynes

Because 1 Newton = 10 5 Dyne. F L . ; where L= length, A= area of A l cross- section of the wire, L  Change in length of the wire when stretched with a force F. The conversion factor to change it from CGS to MKS system is

Problem 26. To determine the Young's modulus of a wire, the formula is Y 

(c) 0.1

We know that the dimension of young's modulus is [ML 1 T 2 ] C.G.S. unit : gm cm 1 sec 2 and M.K.S. unit : kg. m–1 sec–2 .

(d) 0.01

19

Solution : (c)

(b) 10

Page

(a) 1

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 20 1

 M 1   L1      M 2   L2 

By using the conversion formula: n 2  n1 

1

 T1     T2 

2

1

 gm   cm       kg   meter 

1

 sec   sec   

2

 gm   cm  1  sec  2 1 n   0 .1  Conversion factor 2   3   2    n1  10 gm   10 cm   sec  10 1

Problem 27. Conversion of 1 MW power on a new system having basic units of mass, length and time as 10 kg, 1dm and 1 minute respectively is (a) 2.16  10 12 unit Solution : (a)

(b) 1.26  10 12 unit

(c) 2.16  1010 unit

(d) 2  10 14 unit

[P]  [ML 2 T 3 ] x

Using

the

relation

y

z

M  L  T  n 2  n1  1   1   1   1  10 6  M 2   L 2   T2 

1

2

 1 kg   1m   1 s         10 kg   1 dm   1 min 

3

[As

1 MW  10 6 W ] 2

 1kg   10 dm   1 sec   10 6       10 kg   1 dm   60 sec 

3

 2.16  10 12 unit

Problem 28. In two systems of relations among velocity, acceleration and force are respectively v 2  a2  a1 and F2 

F1



2 v1 , 

. If  and  are constants then relations among mass, length and time in two

systems are

 2  3 T1 M1 , L2  2 L1 , T2    

(b) M 2 

(c) M 2 

3 2  M , L  L , T  T1 1 2 3 2 1 2   

(d) M 2 

2 2  [L 2 T21 ]  [L1 T11 ]  

F1



M 1 , L2 

3  L1 , T2  T1 2 3 

2  3 M , L  L , T  T1 1 2 1 2 2 2 3

......(i)

a 2  a1  [L 2 T22 ]  [L1 T12 ] and F2 

 

2

......(ii)

 [M 2 L 2 T2 2 ]  [M 1 L1 T1 2 ] 

1



......(iii)

Dividing equation (iii) by equation (ii) we get M 2 

M1 M  2 12 ( )  B

Squaring equation (i) and dividing by equation (ii) we get L 2  L1 Dividing equation (i) by equation (ii) we get T2  T1

3 3

 2

Problem 29. If the present units of length, time and mass (m, s, kg) are changed to 100m, 100s, and

1 kg then 10

(a) The new unit of velocity is increased 10 times (b) The new unit of force is decreased (c) The new unit of energy is increased 10 times times Solution : (b)

Unit of velocity = m/sec ; in new system =

1 times 1000

(d) The new unit of pressure is increased 1000

100 m m  (same) 100 sec sec

20

v 2  v1

1 2

Page

Solution : (b)

(a) M 2 

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 21 Unit of force 

kg  m 2

sec

Unit of energy 

; in new system 

kg  m 2 sec 2

Unit of pressure 

1 kg  m 1 100 m kg   10 100 sec 100 sec 1000 sec 2

; in new system 

kg m  sec 2

 kg  m 2 1 100 m  100 m  kg  10 sec 2 10 100 sec 100 sec

; in new system 

1 1 1 kg kg  m  10 7 10 100 100 sec  100 sec m  sec 2

Problem 30. Suppose we employ a system in which the unit of mass equals 100 kg, the unit of length equals 1 km and the unit of time 100 s and call the unit of energy eluoj (joule written in reverse order), then (a) 1 eluoj = 104 joule Solution : (a)

(b) 1 eluoj = 10-3 joule

(c) 1 eluoj = 10-4 joule

(d) 1 joule = 103 eluoj

[ E]  [ML 2 T 2 ]

1 eluoj  [100 kg ]  [1km ] 2  [100 sec] 2  100 kg  10 6 m 2  10 4 sec 2  10 4 kg m 2  sec 2  10 4 Joule

Problem 31. If 1gm cms–1 = x Ns, then number x is equivalent to (a) 1  10 1 Solution : (d)

(b) 3  10 2

(c) 6  10 4

(d) 1  10 5

gm - cm s 1  10 3 kg  10 2 m  s 1  10 5 kg  m  s 1 = 10–5 Ns

(4) To check the dimensional correctness of a given physical relation : This is based on the ‘principle of homogeneity’. According to this principle the dimensions of each term on both sides of an equation must be the same. If X  A  (BC )2  DEF , then according to principle of homogeneity [X] = [A] = [(BC)2]  [ DEF ] If the dimensions of each term on both sides are same, the equation is dimensionally correct, otherwise not. A dimensionally correct equation may or may not be physically correct. Example : (1) F  mv 2 / r 2 By substituting dimension of the physical quantities in the above relation – [MLT 2 ]  [M ][LT 1 ]2 /[ L]2

i.e.

[MLT 2 ]  [MT 2 ]

As in the above equation dimensions of both sides are not same; this formula is not correct dimensionally, so can never be physically. (2) s  ut  (1 / 2)at 2 By substituting dimension of the physical quantities in the above relation – [L] = [LT–1][T] – [LT–2][T2] i.e.

[L] = [L] – [L]

As in the above equation dimensions of each term on both sides are same, so this equation is dimensionally correct. However, from equations of motion we know that s  ut  (1 / 2)at 2

Sample problems based on formulae checking R3 GM

(b) T  2

GM R3

(c) T  2

GM R2

(d) T  2

R2 GM

Page

(a) T  2

21

Problem 32. From the dimensional consideration, which of the following equation is correct

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 22 Solution : (a)

R R3 R3  2  2 2 g GM gR

T  2

[As GM = gR2]

Now by substituting the dimension of each quantity in both sides.  L  [T ]     LT  2 

1/2

 [T ]

L.H.S. = R.H.S. i.e., the above formula is Correct.

Problem 33. A highly rigid cubical block A of small mass M and side L is fixed rigidly onto another cubical block B of the same dimensions and of low modulus of rigidity  such that the lower face of A completely covers the upper face of B. The lower face of B is rigidly held on a horizontal surface. A small force F is applied perpendicular to one of the side faces of A. After the force is withdrawn block A executes small oscillations. The time period of which is given by

M L

(a) 2 Solution : (d)

L M

(b) 2

(c) 2

ML



(d) 2

M

L

Given m = mass = [M],  = coefficient of rigidity = [ML 1 T 2 ] , L = length = [L] By substituting the dimension of these quantity we can check the accuracy of the given formulae  [M ]   [T ]  2   [ ] [L] 

1/2

M   =  1  2  ML T L  

1/2

= [T].

L.H.S. = R.H.S. i.e., the above formula is Correct.

Problem 34. A small steel ball of radius r is allowed to fall under gravity through a column of a viscous liquid of coefficient of viscosity. After some time the velocity of the ball attains a constant value known as terminal velocity v T . The terminal velocity depends on (i) the mass of the ball. (ii)  (iii) r and (iv) acceleration due to gravity g. which of the following relations is dimensionally correct (a) vT  Solution : (a)

mg r

(b) vT 

r mg

(c) vT  rmg

(d) vT 

mgr



Given v T = terminal velocity = [LT 1 ] , m = Mass = [M], g = Acceleration due to gravity = [LT 2 ] r = Radius = [L],  = Coefficient of viscosity = [ ] By substituting the dimension of each quantity we can check the accuracy of given formula v T   [LT 1 ] 

[M ][LT 2 ] [ML 1 T 1 ][L]

mg r

= [LT 1 ]

L.H.S. = R.H.S. i.e., the above formula is Correct.

Problem 35. A dimensionally consistent relation for the volume V of a liquid of coefficient of viscosity  flowing per second through a tube of radius r and length l and having a pressure difference p across its end, is (a) V 

Given V = Rate of flow =

(b) V 

l 8 pr 4

(c) V 

8 p l

r

4

(d) V 

p  8lr 4

Volume  [L3 T 1 ] , P = Pressure = [ML1 T 2 ] , r = Radius = [L] sec

By substituting the dimension of each quantity we can check the accuracy of the formula V   [L3 T 1 ] 

[ML1T 2 ] [L4 ] [ML1T 1 ] [L]

= [L3 T 1 ]

P r 4 8l

22

 = Coefficient of viscosity = [ML1 T 1 ] , l = Length = [L]

Page

Solution : (a)

pr 4 8l

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 23 L.H.S. = R.H.S. i.e., the above formula is Correct. 1 2

Problem 36. With the usual notations, the following equation S t  u  a(2 t  1) is

Solution : (c)

(a) Only numerically correct

(b) Only dimensionally correct

(c) Both numerically and dimensionally correct

(d) Neither numerically nor dimensionally correct

Given S t = distance travelled by the body in tth sec.= [LT 1 ] , a = Acceleration = [LT 2 ] , v = velocity = [LT 1 ] ,

t = time = [T]

By substituting the dimension of each quantity we can check the accuracy of the formula St  u 

1 a (2 t  1) 2

 [LT 1 ]  [LT 1 ]  [LT 2 ] [T ]  [LT 1 ]  [LT 1 ]  [LT 1 ]

Since the dimension of each terms are equal therefore this equation is dimensionally correct. And after deriving this equation from Kinematics we can also proof that this equation is correct numerically also.

Problem 37. If velocity v, acceleration A and force F are chosen as fundamental quantities, then the dimensional formula of angular momentum in terms of v, A and F would be (a) FA 1 v Solution : (b)

(b) Fv 3 A 2

(c) Fv 2 A 1

(d) F 2 v 2 A 1

Given, v = velocity = [LT 1 ] , A = Acceleration = [LT 2 ] , F = force = [MLT 2 ] By substituting, the dimension of each quantity we can check the accuracy of the formula [Angular momentum] = Fv 3 A 2 [ML 2 T 1 ]  [MLT

2

] [LT 1 ] 3 [LT 2 ] 2

= [ML2T 1 ] L.H.S. = R.H.S. i.e., the above formula is Correct.

Problem 38. The largest mass (m) that can be moved by a flowing river depends on velocity (v), density (  ) of river water and acceleration due to gravity (g). The correct relation is (a) m  Solution : (d)

 2v 4 g2

(b) m 

v 6 g2

(c) m 

v 4 g3

(d) m 

v 6 g3

Given, m = mass = [M], v = velocity = [LT 1 ] ,  = density = [ML 3 ] , g = acceleration due to gravity = [LT–2] By substituting, the dimension of each quantity we can check the accuracy of the formula m K

 [M ] 

 v6 g3

[ML 3 ][LT 1 ]6 [LT  2 ] 3

= [M] L.H.S. = R.H.S. i.e., the above formula is Correct.

Let time period of a simple pendulum is a function of mass of the bob (m), effective length (l), acceleration due to gravity (g) then assuming the function to be product of power function of m, l and g

Page

Example : (i) Time period of a simple pendulum.

23

(5) As a research tool to derive new relations : If one knows the dependency of a physical quantity on other quantities and if the dependency is of the product type, then using the method of dimensional analysis, relation between the quantities can be derived.

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 24 i.e., T  Km x l y g z ; where K = dimensionless constant If the above relation is dimensionally correct then by substituting the dimensions of quantities – [T] = [M]x [L]y [LT–2]z or [M0L0T1] = [MxLy+zT–2z] Equating the exponents of similar quantities x = 0, y = 1/2 and z = – 1/2 So the required physical relation becomes T  K

l g

The value of dimensionless constant is found (2 ) through experiments so T  2

l g

(ii) Stoke’s law : When a small sphere moves at low speed through a fluid, the viscous force F, opposing the motion, is found experimentally to depend on the radius r, the velocity of the sphere v and the viscosity  of the fluid. So

F = f (, r, v)

If the function is product of power functions of , r and v,

F  K x r y v z ; where K is dimensionless

constant. If the above relation is dimensionally correct [MLT 2 ]  [ML 1 T 1 ] x [L]y [LT 1 ]z or

[MLT 2 ]  [M x L x  y  z T  x  z ]

Equating the exponents of similar quantities x = 1; – x + y + z = 1 and – x – z = – 2 Solving these for x, y and z, we get x = y = z = 1 So eqn (i) becomes F = Krv On experimental grounds, K = 6; so F = 6rv This is the famous Stoke’s law.

Sample problem based on formulae derivation Problem 39. If the velocity of light (c), gravitational constant (G) and Planck's constant (h) are chosen as fundamental units, then the dimensions of mass in new system is (a) c 1 / 2 G1 / 2 h1 / 2 Solution : (c)

(b) c1 / 2 G1 / 2 h 1 / 2

(c) c1 / 2 G 1 / 2 h1 / 2

(d) c 1 / 2 G1 / 2 h1 / 2

Let m  c x G y h z or m  K c x G y h z By substituting the dimension of each quantity in both sides [M 1 L0 T 0 ]  K[LT 1 ] x [M 1 L3 T 2 ]y [ML 2 T 1 ]z  [M y  z Lx 3 y  2 z T  x 2 y z ]

By equating the power of M, L and T in both sides : y  z  1 , x  3y  2z  0 , x  2y  z  0 By solving above three equations x  1 / 2 , y  1 / 2 and z  1 / 2 .  m  c 1 / 2 G 1 / 2 h1 / 2

Problem 40. If the time period (T) of vibration of a liquid drop depends on surface tension (S), radius (r) of the drop and density ( ) of the liquid, then the expression of T is

(c) T  K r 3 / S 1 / 2

Let T  S x r y  z or T = K S x r y  z By substituting the dimension of each quantity in both sides [M 0 L0 T 1 ]  K [MT 2 ] x [L]y [ML 3 ] z  [M x  z Ly 3 z T 2 x ]

(d) None of these

24

Solution : (a)

(b) T  K  1 / 2 r 3 / S

Page

(a) T  K r 3 / S

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 25 x  z  0 , y  3 z  0 , 2 x  1

By equating the power of M, L and T in both sides

By solving above three equations  x  1 / 2 , y  3 / 2 , z  1 / 2

r 3

So the time period can be given as, T  K S 1 / 2 r 3 / 2  1 / 2  K

S

.

Problem 41. If P represents radiation pressure, C represents speed of light and Q represents radiation energy striking a unit area per second, then non-zero integers x, y and z such that P x Q y C z is dimensionless, are [AFMC 1991; CBSE 1992; CPMT 1981, 92; MP PMT 1992]

(a) x  1, y  1, z  1 Solution : (b)

(b) x  1, y  1, z  1

(c) x  1, y  1, z  1

(d) x  1, y  1, z  1

[P x Q y C z ]  M 0 L0 T 0

By substituting the dimension of each quantity in the given expression [ML 1 T 2 ] x [MT 3 ]y [LT 1 ] z  [M x  y L x  z T 2 x 3 y  z ]  M 0 L0 T 0

by equating the power of M, L and T in both sides: x  y  0 , x  z  0 and 2 x  3y  z  0 by solving we get x  1, y  1, z  1 .

Problem 42. The volume V of water passing through a point of a uniform tube during t seconds is related to the crosssectional area A of the tube and velocity u of water by the relation V  A u  t  , which one of the following will be true (a)      Solution : (b)

(b)     

(c)     

(d)     

Writing dimensions of both sides [L3 ]  [L2 ] [LT 1 ]  [ T ]  [L3 T 0 ]  [L2   T    ] By comparing powers of both sides 2    3 and     0 Which give    and  

1 (3   ) i.e.      . 2

Problem 43. If velocity (V), force (F) and energy (E) are taken as fundamental units, then dimensional formula for mass will be (a) V 2 F0 E Solution : (d)

(b) V 0 FE2

(c) VF 2 E0

(d) V 2 F0 E

Let M  V a Fb Ec Putting dimensions of each quantities in both side [M ]  [LT 1 ]a [MLT

2 b

] [ML 2 T 2 ]c

Equating powers of dimensions. We have b  c  1, a  b  2c  0 and a  2b  2c  0 Solving these equations, a  2, b = 0 and c = 1 So M  [V 2 F 0 E]

Problem 44. Given that the amplitude A of scattered light is : (i) Directly proportional to the amplitude (A0) of incident light. (ii) Directly proportional to the volume (V) of the scattering particle (iii) Inversely proportional to the distance (r) from the scattered particle (iv) Depend upon the wavelength (  ) of the scattered light. then: (a) A 



(b) A 

1



2

(c) A 

1



3

(d) A 

1

4

KA 0 Vx r By substituting the dimension of each quantity in both sides  [ L] 

[L] .[ L3 ][Lx ] [L]

25

Let A 

Page

Solution : (b)

1

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 26  [L]  [L3  x ] ;  3  x  1 or x  2

 A  2

1.12 Limitations of Dimensional Analysis. Although dimensional analysis is very useful it cannot lead us too far as, (1) If dimensions are given, physical quantity may not be unique as many physical quantities have same dimensions. For example if the dimensional formula of a physical quantity is [ML 2 T 2 ] it may be work or energy or torque. (2) Numerical constant having no dimensions [K] such as (1/2), 1 or 2 etc. cannot be deduced by the methods of dimensions. (3) The method of dimensions can not be used to derive relations other than product of power functions. For example, s  u t  (1 / 2) a t 2

or

y  a sin  t

cannot be derived by using this theory (try if you can). However, the dimensional correctness of these can be checked. (4) The method of dimensions cannot be applied to derive formula if in mechanics a physical quantity depends on more than 3 physical quantities as then there will be less number (= 3) of equations than the unknowns (>3). However still we can check correctness of the given equation dimensionally. For example T  2 1 mgl can not be derived by theory of dimensions but its dimensional correctness can be checked.

(5) Even if a physical quantity depends on 3 physical quantities, out of which two have same dimensions, the formula cannot be derived by theory of dimensions, e.g., formula for the frequency of a tuning fork f  (d / L2 ) v cannot be derived by theory of dimensions but can be checked.

1.13 Significant Figures. Significant figures in the measured value of a physical quantity tell the number of digits in which we have confidence. Larger the number of significant figures obtained in a measurement, greater is the accuracy of the measurement. The reverse is also true. The following rules are observed in counting the number of significant figures in a given measured quantity. (1) All non-zero digits are significant. Example :

42.3 has three significant figures. 243.4 has four significant figures. 24.123 has five significant figures.

(2) A zero becomes significant figure if it appears between to non-zero digits. Example :

5.03 has three significant figures. 5.604 has four significant figures. 4.004 has four significant figures.

(3) Leading zeros or the zeros placed to the left of the number are never significant. 0.045 has two significant figures. 0.006 has one significant figures. (4) Trailing zeros or the zeros placed to the right of the number are significant.

26

0.543 has three significant figures.

Page

Example :

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 27 Example :

4.330 has four significant figures. 433.00 has five significant figures. 343.000 has six significant figures.

(5) In exponential notation, the numerical portion gives the number of significant figures. 1.32  10–2 has three significant figures.

Example :

1.32  104 has three significant figures.

1.14 Rounding Off. While rounding off measurements, we use the following rules by convention: (1) If the digit to be dropped is less than 5, then the preceding digit is left unchanged. Example : x  7.82 is rounded off to 7.8, again x  3.94 is rounded off to 3.9. (2) If the digit to be dropped is more than 5, then the preceding digit is raised by one. Example : x = 6.87 is rounded off to 6.9, again x = 12.78 is rounded off to 12.8. (3) If the digit to be dropped is 5 followed by digits other than zero, then the preceding digit is raised by one. Example : x = 16.351 is rounded off to 16.4, again x = 6.758 is rounded off to 6.8. (4) If digit to be dropped is 5 or 5 followed by zeros, then preceding digit is left unchanged, if it is even. Example : x = 3.250 becomes 3.2 on rounding off, again x = 12.650 becomes 12.6 on rounding off. (5) If digit to be dropped is 5 or 5 followed by zeros, then the preceding digit is raised by one, if it is odd. Example : x = 3.750 is rounded off to 3.8, again x = 16.150 is rounded off to 16.2.

1.15 Significant Figures in Calculation. In most of the experiments, the observations of various measurements are to be combined mathematically, i.e., added, subtracted, multiplied or divided as to achieve the final result. Since, all the observations in measurements do not have the same precision, it is natural that the final result cannot be more precise than the least precise measurement. The following two rules should be followed to obtain the proper number of significant figures in any calculation. (1) The result of an addition or subtraction in the number having different precisions should be reported to the same number of decimal places as are present in the number having the least number of decimal places. The rule is illustrated by the following examples : (i)

33.3

 (has only one decimal place)

3.11 + 0.313 36.723

 (answer should be reported to one decimal place)

Answer = 36.7 (ii)

3.1421 0.241 + 0.09

 (has 2 decimal places)

3.4731

 (answer should be reported to 2 decimal places)

62.831

 (has 3 decimal places)

– 24.5492

Page

(iii)

27

Answer = 3.47

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 28  (answer should be reported to 3 decimal places after rounding off)

38.2818 Answer = 38.282

(2) The answer to a multiplication or division is rounded off to the same number of significant figures as is possessed by the least precise term used in the calculation. The rule is illustrated by the following examples : (i)

142.06  0.23

 (two significant figures)

32.6738

 (answer should have two significant figures)

Answer = 33 (ii)

51.028  1.31

 (three significant figures)

66.84668 Answer = 66.8

0 .90  0 .2112676 4 .26

(iii) Answer = 0.21

1.16 Order of Magnitude. In scientific notation the numbers are expressed as, Number  M  10 x . Where M is a number lies between 1 and 10 and x is integer. Order of magnitude of quantity is the power of 10 required to represent the quantity. For determining this power, the value of the quantity has to be rounded off. While rounding off, we ignore the last digit which is less than 5. If the last digit is 5 or more than five, the preceding digit is increased by one. For example, (1) Speed of light in vacuum  3  10 8 ms 1  10 8 m / s (ignoring 3 < 5) (2) Mass of electron  9 .1  10 31 kg  10 30 kg (as 9.1 > 5).

Sample problems based on significant figures Problem 45. Each side a cube is measured to be 7.203 m. The volume of the cube up to appropriate significant figures is (a) 373.714 Solution : (c)

Volume  a

3

(b) 373.71  (7 .023 )

3

 373 .715 m

(c) 373.7

(d) 373

3

In significant figures volume of cube will be 373 .7 m 3 because its side has four significant figures.

Problem 46. The number of significant figures in 0.007 m 2 is (a) 1

(b) 2

(c) 3

(d) 4

Solution : (a)

Problem 47. The length, breadth and thickness of a block are measured as 125.5 cm, 5.0 cm and 0.32 cm respectively. Which one of the following measurements is most accurate (a) Length Solution : (a)

(b) Breadth

(c) Thickness

(d) Height

Relative error in measurement of length is minimum, so this measurement is most accurate.

(a) 2.340 kg Solution : (c)

(b) 2.3145 kg.

Total mass  2.3  0.00215  0.01239  2.31 kg

(c) 2.3 kg

(d) 2.31 kg

Page

of the box to the correct number of significant figures is

28

Problem 48. The mass of a box is 2.3 kg. Two marbles of masses 2.15 g and 12.39 g are added to it. The total mass

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 29 Total mass in appropriate significant figures be 2.3 kg.

Problem 49. The length of a rectangular sheet is 1.5 cm and breadth is 1.203 cm. The area of the face of rectangular sheet to the correct no. of significant figures is : (a) 1.8045 cm 2 Solution : (d)

(b) 1.804 cm 2

(c) 1.805 cm 2

(d) 1.8 cm 2

Area  1 .5  1 .203  1 .8045 cm 2  1 . 8 cm 2 (Upto correct number of significant figure).

Problem 50. Each side of a cube is measured to be 5.402 cm. The total surface area and the volume of the cube in appropriate significant figures are :

Solution : (b)

(a) 175.1 cm 2 , 157 cm 2

(b) 175.1 cm 2 , 157.6 cm 3

(c) 175 cm 2 , 157 cm 2

(d) 175.08 cm 2 , 157.639 cm 3

Total surface area = 6  (5 .402)2  175 .09 cm 2  175 .1 cm 2 (Upto correct number of significant figure) Total volume  (5 .402)3  175 .64 cm 3  175.6 cm 3 (Upto correct number of significant figure).

Problem 51. Taking into account the significant figures, what is the value of 9.99 m + 0.0099 m (a) 10.00 m Solution : (a)

(b) 10 m

(c) 9.9999 m

(d) 10.0 m

9.99 m  0.0099 m  9.999 m  10 .00 m (In proper significant figures).

Problem 52. The value of the multiplication 3.124  4.576 correct to three significant figures is (a) 14.295 Solution : (b)

(b) 14.3

(c) 14.295424

(d) 14.305

3.124  4.576  14.295 =14.3 (Correct to three significant figures).

Problem 53. The number of the significant figures in 11.118  10 6 V is (a) 3 Solution : (c)

(b) 4

(c) 5

(d) 6

The number of significant figure is 5 as 10 6 does not affect this number.

Problem 54. If the value of resistance is 10.845 ohms and the value of current is 3.23 amperes, the potential difference is 35.02935 volts. Its value in significant number would be (a) 35 V Solution : (b)

(b) 35.0 V

(c) 35.03 V

(d) 35.025 V

Value of current (3.23 A) has minimum significant figure (3) so the value of potential difference V( IR) have only 3 significant figure. Hence its value be 35.0 V.

1.17 Errors of Measurement. The measuring process is essentially a process of comparison. Inspite of our best efforts, the measured value of a quantity is always somewhat different from its actual value, or true value. This difference in the true value of a quantity is called error of measurement. (1) Absolute error : Absolute error in the measurement of a physical quantity is the magnitude of the difference between the true value and the measured value of the quantity. Let a physical quantity be measured n times. Let the measured value be a1, a2, a3, ….. an. The a  a 2  .... an arithmetic mean of these value is am  1 n Usually, am is taken as the true value of the quantity, if the same is unknown otherwise. By definition, absolute errors in the measured values of the quantity are a1  am  a1

an  am  an

Page

………….

29

a 2  am  a 2

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 30 The absolute errors may be positive in certain cases and negative in certain other cases. (2) Mean absolute error : It is the arithmetic mean of the magnitudes of absolute errors in all the measurements of the quantity. It is represented by a. Thus

a 

| a1 |  | a 2 | ..... | an | n

Hence the final result of measurement may be written as a  am  a This implies that any measurement of the quantity is likely to lie between (a m  a) and (a m  a). (3) Relative error or Fractional error : The relative error or fractional error of measurement is defined as the ratio of mean absolute error to the mean value of the quantity measured. Thus Relative error or Fractional error 

mean absolute error a  mean value am

(4) Percentage error : When the relative/fractional error is expressed in percentage, we call it percentage error. Thus Percentage error 

a  100 % am

1.18 Propagation of Errors. (1) Error in sum of the quantities : Suppose x = a + b Let a = absolute error in measurement of a b = absolute error in measurement of b x = absolute error in calculation of x i.e. sum of a and b. The maximum absolute error in x is x  (a  b) Percentage error in the value of x 

(a  b)  100 % ab

(2) Error in difference of the quantities : Suppose x = a – b Let a = absolute error in measurement of a, b = absolute error in measurement of b x = absolute error in calculation of x i.e. difference of a and b. The maximum absolute error in x is x  (a  b) Percentage error in the value of x 

(a  b)  100 % ab

(3) Error in product of quantities : Suppose x = a  b Let a = absolute error in measurement of a, b = absolute error in measurement of b x = absolute error in calculation of x i.e. product of a and b. The maximum fractional error in x is

x  a b      x b   a

Let a = absolute error in measurement of a,

a b

Page

(4) Error in division of quantities : Suppose x 

30

Percentage error in the value of x = (Percentage error in value of a) + (Percentage error in value of b)

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 31 b = absolute error in measurement of b x = absolute error in calculation of x i.e. division of a and b. The maximum fractional error in x is

x  a b      x b   a

Percentage error in the value of x = (Percentage error in value of a) + (Percentage error in value of b) (5) Error in quantity raised to some power : Suppose x 

an bm

Let a = absolute error in measurement of a, b = absolute error in measurement of b x = absolute error in calculation of x The maximum fractional error in x is

x b   a   n m  x b   a

Percentage error in the value of x = n (Percentage error in value of a) + m (Percentage error in value of b)

Note : 

The quantity which have maximum power must be measured carefully because it's

contribution to error is maximum.

Sample problems based on errors of measurement Problem 55. A physical parameter a can be determined by measuring the parameters b, c, d and e using the relation a = b  c  / d  e  . If the maximum errors in the measurement of b, c, d and e are b1 %,

c 1 %, d1 % and e 1 %, then the maximum error in the value of a determined by the experiment is [CPMT 198

Solution : (d)

(a) ( b1  c1  d1  e1 )%

(b) ( b1  c1  d1  e1 )%

(c) ( b1  c1  d1  e1 )%

(d) ( b1  c1  d1  e1 )%

a  b  c  /d  e 

So maximum error in a is given by

b c d e  a   100  .  100   .  100   .  100   .  100  b c d e  a  max  b1  c1  d1  e1 %

Problem 56. The pressure on a square plate is measured by measuring the force on the plate and the length of the sides of the plate. If the maximum error in the measurement of force and length are respectively 4% and 2%, The maximum error in the measurement of pressure is (a) 1% Solution : (d)

P

(b) 2%

(c) 6%

(d) 8%

F F , so maximum error in pressure (P )  A l2

F l  P   100    100  2  100 = 4% + 2 × 2% = 8%  P F l   max

(a) 5.0  11% Solution : (a)

(b) 5.0  1%

Weight in air  (5.00  0.05) N

(c) 5.0  6%

(d) 1.25  5%

Page

weight in air is (5.00 0.05 ) Newton and weight in water is (4.00  0.05) Newton. Then the relative density along with the maximum permissible percentage error is

31

Problem 57. The relative density of material of a body is found by weighing it first in air and then in water. If the

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 32 Weight in water  (4.00  0.05) N Loss of weight in water  (1.00  0.1) N Now relative density 

weight in air weight loss in water

i.e. R . D 

5.00  0.05 1.00  0.1

Now relative density with max permissible error 

5 .00  0 .05 0 .1      100  5.0  (1  10)% 1 .00  5 .00 1 .00 

 5.0  11%

Problem 58. The resistance R =

V where V= 100  5 volts and i = 10  0.2 amperes. What is the total error in R i

(a) 5% Solution : (b)

R

(b) 7%

(c) 5.2%

(d)

5 % 2

5 0 .2 V I  R   100   100  (5  2)% = 7%   100    100   100  100 10 V I  R  max

V I

Problem 59. The period of oscillation of a simple pendulum in the experiment is recorded as 2.63 s, 2.56 s, 2.42 s, 2.71 s and 2.80 s respectively. The average absolute error is (a) 0.1 s Solution : (b)

(b) 0.11 s

Average value 

(c) 0.01 s

(d) 1.0 s

2 . 63  2 .56  2 . 42  2 . 71  2 .80  2.62 sec 5

Now | T1 |  2.63  2.62  0.01

| T2 |  2.62  2.56  0.06 | T3 |  2.62  2.42  0.20 | T4 |  2.71  2.62  0.09 | T5 |  2.80  2.62  0.18 Mean absolute error T 

| T1 |  | T2 |  | T3 |  | T4 |  | T5 | 0 . 54   0 . 108  0 . 11 sec 5 5

Problem 60. The length of a cylinder is measured with a meter rod having least count 0.1 cm. Its diameter is measured with venier calipers having least count 0.01 cm. Given that length is 5.0 cm. and radius is 2.0 cm. The percentage error in the calculated value of the volume will be (a) 1% Solution : (c)

(b) 2%

(c) 3%

(d) 4%

Volume of cylinder V  r 2 l Percentage error in volume

V 2 r l  100   100   100 V r l

0 .01 0.1    2   100   100   (1  2)% = 3% 2 .0 5 .0  

Problem 61. In an experiment, the following observation's were recorded : L = 2.820 m, M = 3.00 kg, l = 0.087 cm, Diameter D = 0.041 cm Taking g = 9.81 m / s 2 using the formula , Y=

4 Mg

D2l

, the maximum permissible error

in Y is

Y 

4 MgL

D l 2

so maximum permissible error in Y =

(c) 6.50%

(d) 8.42%

 M g L 2D l  Y   100  100       Y g L D l   M

32

Solution : (c)

(b) 4.56%

Page

(a) 7.96%

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 33 1 1 1 1   1     2    100  0.065  100  6.5% 41 87   300 9.81 9820

Problem 62. According to Joule's law of heating, heat produced H  I 2 Rt, where I is current, R is resistance and t is time. If the errors in the measurement of I, R and t are 3%, 4% and 6% respectively then error in the measurement of H is (a)  17% Solution : (b)

(b)  16%

(c)  19%

(d)  25%

H  I2R t



H  2I R t   100       100  (2  3  4  6)%  16% H R t   I

Problem 63. If there is a positive error of 50% in the measurement of velocity of a body, then the error in the measurement of kinetic energy is (a) 25% Solution : (c)

(b) 50%

Kinetic energy E 



(d) 125%

1 mv 2 2

E  m 2v   100      100 E v   m

Here m  0 and 

(c) 100%

v  100  50 % v

E  100  2  50  100 % E

Problem 64. A physical quantity P is given by P=

A

3

1 B2

3 4 C D2

. The quantity which brings in the maximum percentage

error in P is (a) A Solution : (c)

(b) B

(c) C

(d) D

Quantity C has maximum power. So it brings maximum error in P.

Problems based on units and dimensions Number of base SI units is

2.

(b) 7

The unit of Planck's constant is (a) Joule

3.

The dimension of

(d) Joule- s [MP PET 2003]

(b) Volt

(c) Mho

(d) Newton

R are L

[MP PET 2003]

(b) T

(c) T 1

(d) T 2

Dimensions of potential energy are

MLT 1

(a) 1

ML T 6.

(c) Joule/m

The unit of reactance is

(a) T 2 5.

(d) 5 [RPMT 1999; MP PET 2003]

(b) Joule/s

(a) Ohm 4.

(c) 3

1

The dimensions of electric potential are

[MP PET 2003]

(b) ML2 T 2

(c)

ML 1 T 2

(d)

33

(a) 4

[MP PET 2003]

Page

1.

[UPSEAT 2003]

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 34 (a) [ML 2 T 2 Q 1 ] 7.

(b) [MLT 2 Q 1 ]

(c)

1 / 2

2

(a) [ML T 

1

]

(b) [ML T 2

(c) [ML T 

]

1

2

(c) Power of a convex lens

]

(d) None of these [AIEEE 2002]

(b) Force and stress

(c) Angular velocity and frequency

(d) Energy and strain

The dimensions of pressure is equal to

[AIEEE 2002]

(b) Energy per unit volume

(c) Force

(d) Energy

Identify the pair whose dimensions are equal

[AIEEE 2002]

(b) Stress and energy

(c) Force and stress

(d) Force and work

A physical quantity x depends on quantities y and z as follows: x  Ay  B tan Cz, where A, B and C are constants. Which of (a) x and B

(b) C and z

1

[AMU (Eng.) 2001]

(c) y and B / A

(d) x and A

ML 3 T 1 Q 2 is dimension of (a) Resistivity

15.

1

[MP PET 2002]

(b) Planck's constant

the following do not have the same dimensions

14.

1

(d) [ML T 

]

(a) Force and strain

(a) Torque and work 13.

[MP PET 2002] 2

0

Which of the two have same dimensions

(a) Force per unit volume 12.

2

Which of the following quantities is dimensionless (a) Gravitational constant

11.

(d) Stress and Young's modulus

The dimensional formula for Boltzmann's constant is 2

10.

[AIEEE 2003]

(b) Torque and work

(c) Momentum and Planck's constant

9.

(d) ML2T 2Q

The physical quantities not having same dimensions are (a) Speed and ( 0  0 )

8.

ML2T 1Q

[RPET 2000]

(b) Conductivity

(c) Resistance

(d) None of these

Two quantities A and B have different dimensions. Which mathematical operation given below is physically meaningful [CPMT 1997]

(a) A/B 16.

(b) A + B

(c) A – B

(d) None of these

Let [ 0 ] denotes the dimensional formula of the permittivity of the vacuum and [ 0 ] that of the permeability of the vacuum. If M = mass, L= length, T= time and I= electric current, then (a) [ 0 ]  M 1 L3 T 2 I

17.

(c) [ A 1 ]

 0 LV

(b) Charge

(c) Voltage

[MP PET 1993; MP PMT 2003]

(b) Newton-metre2/Coulomb2 (d)

Coulomb2/Newton-metre2

Dimensional formula of capacitance is 1 2

(a) M L T A 4

2

[CPMT 1978; MP PMT 1979; IIT-JEE 1983]

(b) ML T A 2

4

2

The dimensional formula for impulse is (a) MLT 2

22.

(d) Current

The unit of permittivity of free space  0 is (c) Coulomb2/(Newton-metre)2

21.

(d) None of these

; here  0 is the permittivity of free space, L is length, V is potential difference and t is time. The t dimensions of X are same as that of

The quantity X 

(a) Coulomb/Newton-metre 20.

(d) [ 0 ]  ML 2 T 1 I [Roorkee 1994]

(b) [A] 2

(a) Resistance 19.

(c) [ 0 ]  MLT 2 I 2

The dimension of quantity (L / RCV ) is (a) [A]

18.

(b) [ 0 ]  M 1 L3 T 4 I 2

(b) MLT 1

(c)

MLT

4

A

2

(d) M 1 L2 T 4 A 2

[EAMCET 1981; CBSE PMT 1991; CPMT 1978; AFMC 1998; BCECE 2003]

(c)

ML2 T 1

(d) M 2 LT 1

The dimensions of universal gravitational constant are [MP PMT 1984, 87, 97, 2000; CBSE PMT 1988, 92, 2004; MP PET 1984, 96, 99; MNR 1992; DPMT 1984; CPMT 1978, 84, 89, 90, 92, 96; AFMC 1999; NCERT 1975; DPET 1993; AIIMS 2002; RPET 2001;

23.

(b) M 1 L3 T 2

(c)

ML 1 T 2

How many wavelength of Kr 86 are there in one metre (a) 1553164.13

(b) 1650763.73

(d) ML2 T 2

[MNR 1985; UPSEAT 2000]

(c) 652189.63

(d) 2348123.73

Page

(a) M 2 L2 T 2

34

Pb. PMT 2002; UPSEAT 1999; BCECE 2003]

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 35 24.

Light year is a unit of

[MP PMT 1989; AFMC 1991; CPMT 1991]

(a) Time 25.

(b) mass

(c) Distance

(d) Energy

L, C and R represent physical quantities inductance, capacitance and resistance respectively. The combination which has the dimensions of frequency is [IIT-JEE 1984] (a) 1/RC and R/L

(b) 1 / RC and

R/L

(c) 1 / LC

(d) C / L

z

26.

  k e In the relation P  , P is pressure, z is distance , k is Boltzmann constant and  is temperature. The dimensional  formula of  will be

[IIT-JEE (Screening) 2004]

(a) [M 0 L2T 0 ] 27.

28.

(b) [M 1 L2T 1 ]

(c) [M 1 L0 T 0 ]

(d) [M 0 L2T 1 ]

If the acceleration due to gravity be taken as the unit of acceleration and the velocity generated in a falling body in one second as the unit of velocity then (a) The new unit of length is g metre

(b) The new unit of length is 1 metre

(c) The new unit of length is g 2 metre

(d) The new unit of time is

1 second g

The famous Stefan's law of radiation states that the rate of emission of thermal radiation per unit by a black body is proportional to area and fourth power of its absolute temperature that is Q  AT 4 where A  area, T  temperature and

 is a universal constant. In the 'energy- length- time temperature' (E-L-T-K) system the dimension of  is (a)

29.

E 2 T 2 L2 K 2

(b) E 1 T 2 L2 K 1

(c)

ET 1 L3 K 4

(d) ET 1 L2 K 4

2.

The resistive force acting on a body moving with a velocity V through a fluid at rest is given by F  CDV 2 A where, C D = coefficient of drag, A = area of cross-section perpendicular to the direction of motion. The dimensions of C D are (a) ML3T– 2

30.

(b) M-1L– 1T2 (b) [MA– 1 T– 1]

33.

(c) [ML2A– 2 T]

(d) [M2 L–3AT2]

P F where p is the 2l m number of segments of the vibrating string and m is a constant of the string. What are the dimensions of m

The frequency n of vibrations of uniform string of length l and stretched with a force F is given by n 

(a) ML–1 T– 1 32.

(d) M0L0T0

The dimensions of (angular momentum)/(magnetic moment) are : (a) [M3 LT– 2A2]

31.

(c) M-1 L–1 T–2

(b) ML– 3 T0

(c) ML– 2 T0

(d) ML–1 T0

Choose the wrong statement(s) (a) A dimensionally correct equation may be correct

(b) A dimensionally correct equation may be incorrect

(c) A dimensionally incorrect equation may be incorrect

(d) A dimensionally incorrect equation may be incorrect

A certain body of mass M moves under the action of a conservative force with potential energy V given by V 

Kr x  a 2' 2

where x is the displacement and a is the amplitude. The units of K are (a) Watt 34.

(c) Joule-metre

(d) None of these.

The Richardson equation is given by I  AT 2 e  B / kT . The dimensional formula for AB2 is same as that for (a) IT2

35.

(b) Joule

(b) kT

(c) IK2

(d) IK2/T

If the units of force, energy and velocity are 10 N, 100 J and 5 ms–1, the units of length, mass and time will be (a) 10m, 5kg, 1s

(b) 10m, 4kg, 2s

(c) 10m, 4kg, 0.5s

(d) 20m, 5kg, 2s.

Problems based on error of measurement l where l is about 100 cm and is known to 1mm accuracy. g

The period is about 2s. The time of 100 oscillations is measured by a stop watch of least count 0.1 s. The percentage error in g is (a) 0.1%

(b) 1%

(c) 0.2%

(d) 0.8%

35

The period of oscillation of a simple pendulum is given by T  2

Page

36.

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 36 37.

The percentage errors in the measurement of mass and speed are 2% and 3% respectively. How much will be the maximum error in the estimation of the kinetic energy obtained by measuring mass and speed [NCERT 1990; Orissa JEE 1990] (a) 11%

38.

(b) 8%

(c) 5%

(d) 1%

While measuring the acceleration due to gravity by a simple pendulum, a student makes a positive error of 1% in the length of the pendulum and a negative error of 3% in the value of time period. His percentage error in the measurement of g by the





relation g  4 2 l / T 2 will be (a) 2% 39.

(b) 4%

(b)

(b) 3 (b) 3%

(d) 6 (d) 7%

(c) (2.00  0.05) s

(d) (2.00  0.10) s

A body travels uniformly a distance of (13.8 0.2) m in a time (4.0  0.3) s. The velocity of the body within error limits is (b) (3.45  0.3) ms-1

(c) (3.45  0.4) ms-1

(d) (3.45  0.5) ms-1

(c) 8.95%

(d) 9.85%

The percentage error in the above problem is (a) 7%

45.

(c) 4 (c) 5%

(b) (2.00 +0.025) s

(a) (3.45  0.2) ms-1 44.

(d)

The mean time period of second's pendulum is 2.00s and mean absolute error on the time period is 0.05s. To express maximum estimate of error, the time period should be written as (a) (2.00  0.01) s

43.

1 x 2

(c) 2x

Error in the measurement of radius of a sphere is 1%. The error in the calculated value of its volume is (a) 1%

42.

1 x 4

What is the number of significant figures in 0.310×103 (a) 2

41.

(d) 10%

The random error in the arithmetic mean of 100 observations is x; then random error in the arithmetic mean of 400 observations would be (a) 4x

40.

(c) 7%

(b) 5.95%

The unit of percentage error is (a) Same as that of physical quantity (b) Different from that of physical quantity (c) Percentage error is unit less (d) Errors have got their own units which are different from that of physical quantity measured The decimal equivalent of 1/20 upto three significant figures is (a) 0.0500

(b) 38.3937 (b) Percentage error

0 .1  100 5 .3

(b) 3 

(c) Both

(d) None of these

0 .1  100 5 .3

(c)

0 .1  100 3 . 53

(d) 3 

0 .1  100 5 .3

A thin copper wire of length l metre increases in length by 2% when heated through 10ºC. What is the percentage increase in area when a square copper sheet of length l metre is heated through 10ºC (a) 4%

51.

(d) 38.39

The radius of a sphere is (5.3  0.1) cm. The percentage error in its volume is (a)

50.

(c) 38.394

Accuracy of measurement is determined by (a) Absolute error

49.

(d) 5.0 × 10-2

If 97.52 is divided by 2.54, the correct result in terms of significant figures is (a) 38.4

48.

(c) 0.0050

(b) 8%

(c) 16%

(d) None of the above.

In the context of accuracy of measurement and significant figures in expressing results of experiment, which of the following is/are correct (1) Out of the two measurements 50.14 cm and 0.00025 ampere, the first one has greater accuracy (2) If one travels 478 km by rail and 397 m. by road, the total distance travelled is 478 km. (a) Only (1) is correct correct. 1.

2.

(b) Only (2) is correct 3.

4.

5.

(c) Both are correct 6.

7.

8.

(d) None 9.

10.

of

them

is

36

47.

(b) 0.05000

Page

46.

genius PHYSICS by Pradeep Kshetrapal

Units, Dimensions and Measurement 37 b

d

a

c

b

a

c

a

d

c

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

b

a

d

a

a

c

c

d

d

a

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

b

b

b

c

a

a

a

d

d

b

31.

32.

33.

34.

35.

36.

37.

38.

39.

40.

d

c

c

c

b

c

b

c

d

b

41.

42.

43.

44.

45.

46.

47.

48.

49.

50.

b

c

b

c

c

a

a

b

b

a

51.

Page

37

4

genius PHYSICS by Pradeep Kshetrapal Motion In One Dimension 1

2.1 Position. Any object is situated at point O and three observers from three different places are looking for same object, then all three observers will have different observations about the position of point O and no one will be wrong. Because they are observing the object from their different positions. Observer ‘A’ says : Point O is 3 m away in west direction. Observer ‘B’ says : Point O is 4 m away in south direction. Observer ‘C’ says : Point O is 5 m away in east direction. Therefore position of any point is completely expressed by two factors: Its distance from the observer and its direction with respect to observer.

N

W

O

E

S

B C

4m

5m

3m

A

That is why position is characterised by a vector known as position vector. 

Let point P is in a xy plane and its coordinates are (x, y). Then position vector (r ) of point will be xˆi  yˆj and if the point P is in a space and its coordinates are (x, y, z) then position vector can be expressed as  r  xˆi  yˆj  z kˆ .

2.2 Rest and Motion. If a body does not change its position as time passes with respect to frame of reference, it is said to be at rest. And if a body changes its position as time passes with respect to frame of reference, it is said to be in motion. Frame of Reference : It is a system to which a set of coordinates are attached and with reference to which observer describes any event. A passenger standing on platform observes that tree on a platform is at rest. But when the same passenger is passing away in a train through station, observes that tree is in motion. In both conditions observer is right. But observations are different because in first situation observer stands on a platform, which is reference frame at rest and in second situation observer moving in train, which is reference frame in motion. So rest and motion are relative terms. It depends upon the frame of references. Tree is at rest

Tree is in motion

Platform (Frame of reference)

Moving train (Frame of reference)

2 Motion In One Dimension

genius

2.3 Types of Motion. One dimensional

Two dimensional

Three dimensional

Motion of a body in a straight line is called one dimensional motion.

Motion of body in a plane is called two dimensional motion.

Motion of body in a space is called three dimensional motion.

When only one coordinate of the position of a body changes with time then it is said to be moving one dimensionally.

When two coordinates of the position of a body changes with time then it is said to be moving two dimensionally.

When all three coordinates of the position of a body changes with time then it is said to be moving three dimensionally.

e.g.. Motion of car on a straight road.

e.g. Motion of car on a circular turn.

e.g.. Motion of flying kite.

Motion of freely falling body.

Motion of billiards ball.

Motion of flying insect.

2.4 Particle or Point Mass. The smallest part of matter with zero dimension which can be described by its mass and position is defined as a particle. If the size of a body is negligible in comparison to its range of motion then that body is known as a particle. A body (Group of particles) to be known as a particle depends upon types of motion. For example in a planetary motion around the sun the different planets can be presumed to be the particles. In above consideration when we treat body as particle, all parts of the body undergo same displacement and have same velocity and acceleration.

2.5 Distance and Displacement. (1) Distance : It is the actual path length covered by a moving particle in a given interval of time. (i) If a particle starts from A and reach to C through point B as shown in the figure.

C

Then distance travelled by particle  AB  BC  7 m (ii) Distance is a scalar quantity.

4m

(iii) Dimension : [M0L1T0] A

(iv) Unit : metre (S.I.)

3m

B

(2) Displacement : Displacement is the change in position vector i.e., A vector joining initial to final position. (i) Displacement is a vector quantity (ii) Dimension : [M0L1T0] (iii) Unit : metre (S.I.) (iv) In the above figure the displacement of the particle

AC  AB  BC

 | AC |  ( AB)2  (BC)2  2( AB) (BC) cos 90 o = 5m     (v) If S 1 , S 2 , S 3 ........ S n are the displacements of a body then the total (net) displacement is the vector      sum of the individuals. S  S 1  S 2  S 3  ........  S n (3) Comparison between distance and displacement : (i) The magnitude of displacement is equal to minimum possible distance between two positions.

genius PHYSICS by Pradeep Kshetrapal Motion In One Dimension 3

So distance  |Displacement|. (ii) For a moving particle distance can never be negative or zero while displacement can be. (zero displacement means that body after motion has came back to initial position) i.e., Distance > 0 but Displacement > = or < 0 (iii) For motion between two points displacement is single valued while distance depends on actual path and so can have many values. (iv) For a moving particle distance can never decrease with time while displacement can. Decrease in displacement with time means body is moving towards the initial position. (v) In general magnitude of displacement is not equal to distance. However, it can be so if the motion is along a straight line without change in direction. Y   (vi) If rA and rB are the position vectors of particle initially and finally. B s Then displacement of the particle    rAB  rB  rA

 rAB

 rB

A

 rA

and s is the distance travelled if the particle has gone through the path APB.

X

Sample problems based on distance and displacement Problem 1.

A man goes 10m towards North, then 20m towards east then displacement is [KCET (Med.) 1999; JIPMER 1999; AFMC 2003]

(a) 22.5m Solution : (a)

(b) 25m

20 2  10 2  10 5 = 22.5 m.

A body moves over one fourth of a circular arc in a circle of radius r. The magnitude of distance travelled and displacement will be respectively (a)

Solution : (a)

(d) 30m

If we take east as x  axis and north as y  axis, then displacement  20 ˆi  10 ˆj So, magnitude of displacement 

Problem 2.

(c) 25.5m

r 2

(b)

,r 2

r ,r 4

(c) r,

r

(d) r, r

2

 Let particle start from A, its position vector rO A  rˆi Y

 After one quarter position vector rOB  r ˆj.

B

So displacement  rˆj  rˆi O

Magnitude of displacement  r 2 . and distance = one fourth of circumference 

Problem 3.

A

X

2r r  4 2

The displacement of the point of the wheel initially in contact with the ground, when the wheel roles forward half a revolution will be (radius of the wheel is R) (a)

R

 4 2

(b) R  2  4

(c) 2R

(d) R

4 Motion In One Dimension Solution : (b)

genius

Horizontal distance covered by the wheel in half revolution = R

Pnew

So the displacement of the point which was initially in contact with a ground =

2R

(R)2  (2 R)2

 R  2  4.

Pinitial

R

2.6 Speed and Velocity. (1) Speed : Rate of distance covered with time is called speed. (i) It is a scalar quantity having symbol  . (ii) Dimension : [M0L1T–1] (iii) Unit : metre/second (S.I.), cm/second (C.G.S.) (iv) Types of speed : (a) Uniform speed : When a particle covers equal distances in equal intervals of time, (no matter how small the intervals are) then it is said to be moving with uniform speed. In given illustration motorcyclist travels equal distance (= 5m) in each second. So we can say that particle is moving with uniform speed of 5 m/s.

Distance Time Uniform Speed

5m

5m 1 sec

1 sec

5m/ s

5m/ s

5m 1 sec 5m/s

5m 1 sec 5m/ s

5m 1 sec 5m/ s

5m 1m/s 5m/s

(b) Non-uniform (variable) speed : In non-uniform speed particle covers unequal distances in equal intervals of time. In the given illustration motorcyclist travels 5m in 1st second, 8m in 2nd second, 10m in 3rd second, 4m in 4th second etc. Therefore its speed is different for every time interval of one second. This means particle is moving with variable speed. Distance

5m

8m

Time

1 sec

1 sec

5m/ s

8m/ s

Variable Speed

4m

6m

7m

1 sec

1 sec

1 sec

1 sec

10m/ s

4m/ s

6m/ s

7m/ s

10m

(c) Average speed : The average speed of a particle for a given ‘Interval of time’ is defined as the ratio of distance travelled to the time taken. Average speed 

Distance travelled s ; v av  t Time taken

 Time average speed : When particle moves with different uniform speed  1 ,  2 ,  3 ... etc in different time intervals t1 , t 2 , t 3 , ... etc respectively, its average speed over the total time of journey is given as

genius PHYSICS by Pradeep Kshetrapal Motion In One Dimension 5

v av 

Total distance covered Total time elapsed



 t   2 t 2   3 t 3  ...... d1  d 2  d 3  ...... = 11 t1  t 2  t 3  ...... t1  t2  t3  ......

Special case : When particle moves with speed v1 upto half time of its total motion and in rest time it is v  v2 moving with speed v2 then v av  1 2  Distance averaged speed : When a particle describes different distances d 1 , d 2 , d 3 , ...... with different time intervals t1 , t 2 , t 3 , ...... with speeds v 1 , v 2 , v 3 ...... respectively then the speed of particle averaged over the total distance can be given as

 av 

d  d 2  d 3  ...... Total distance covered d  d 2  d 3  ......  1  1 d1 d 2 d 3 Total time elapsed t1  t2  t3  ......    ......

1

then

3

 When particle moves the first half of a distance at a speed of v1 and second half of the distance at speed v2 v av 

then

2

2v 1 v 2 v1  v 2

 When particle covers one-third distance at speed v1, next one third at speed v2 and last one third at speed v3, v av 

3 v1v 2 v 3 v1v 2  v 2 v 3  v 3 v1

(d) Instantaneous speed : It is the speed of a particle at particular instant. When we say “speed”, it usually means instantaneous speed. The instantaneous speed is average speed for infinitesimally small time interval (i.e., t  0 ). Thus Instantaneous speed v  lim

t 0

s ds  t dt

(2) Velocity : Rate of change of position i.e. rate of displacement with time is called velocity. (i) It is a scalar quantity having symbol v . (ii) Dimension : [M0L1T–1] (iii) Unit : metre/second (S.I.), cm/second (C.G.S.) (iv) Types (a) Uniform velocity : A particle is said to have uniform velocity, if magnitudes as well as direction of its velocity remains same and this is possible only when the particles moves in same straight line without reversing its direction. (b) Non-uniform velocity : A particle is said to have non-uniform velocity, if either of magnitude or direction of velocity changes (or both changes). (c) Average velocity : It is defined as the ratio of displacement to time taken by the body   Displacement r ; v av  Averagevelocity  Time taken t

6 Motion In One Dimension

genius (d) Instantaneous velocity : Instantaneous velocity is defined as rate of change of position vector of particles with time at a certain instant of time.     r dr Instantaneous velocity v  lim  t 0 t dt (v) Comparison between instantaneous speed and instantaneous velocity (a) instantaneous velocity is always tangential to the path followed by the particle.

Y

3

When a stone is thrown from point O then at point of projection the instantaneous velocity of stone is v 1 , at point A the instantaneous

2 A

velocity of stone is v 2 , similarly at point B and C are v 3 and v 4

v1

respectively.

O

4

B C

X

Direction of these velocities can be found out by drawing a tangent on the trajectory at a given point. (b) A particle may have constant instantaneous speed but variable instantaneous velocity. Example : When a particle is performing uniform circular motion then for every instant of its circular motion its speed remains constant but velocity changes at every instant. (c) The magnitude of instantaneous velocity is equal to the instantaneous speed. (d) If a particle is moving with constant velocity then its average velocity and instantaneous velocity are always equal. (e) If displacement is given as a function of time, then time derivative of displacement will give velocity.  Let displacement x  A0  A1 t  A2 t 2   dx d Instantaneous velocity v   ( A0  A1 t  A 2 t 2 ) dt dt  v   A1  2 A 2 t For the given value of t, we can find out the instantaneous velocity.   e.g. for t  0 ,Instantaneous velocity v   A1 and Instantaneous speed | v |  A1 (vi) Comparison between average speed and average velocity (a) Average speed is scalar while average velocity is a vector both having same units ( m/s) and dimensions [LT 1 ] . (b) Average speed or velocity depends on time interval over which it is defined. (c) For a given time interval average velocity is single valued while average speed can have many values depending on path followed.     (d) If after motion body comes back to its initial position then v av  0 (as r  0 ) but v av  0 and finite as (s  0) . (e) For a moving body average speed can never be negative or zero (unless t  ) while average velocity  can be i.e. v av  0 while v a = or < 0.

Sample problems based on speed and velocity

genius PHYSICS by Pradeep Kshetrapal Motion In One Dimension 7

Problem 4.

If a car covers 2/5th of the total distance with v1 speed and 3/5th distance with v2 then average speed is [MP PMT 2003]

(a)

Solution : (d)

1 v1v 2 2

(b)

Average speed =

v1  v 2 2

(c)

2v 1v 2 v1  v 2

T otal distance travelled x  T otal time taken t1  t 2

(d)

(2/5)x

(3/5)x

t1

t2

5v 1 v 2 =  (2 / 5) x (3 / 5)x 2v 2  3v 1  v1 v2

x

Problem 5.

Solution : (a)

5v1v 2 3v1  2v 2

A car accelerated from initial position and then returned at initial point, then

[AIEEE 2002]

(a) Velocity is zero but speed increases

(b) Speed is zero but velocity increases

(c) Both speed and velocity increase

(d) Both speed and velocity decrease

As the net displacement = 0 Hence velocity = 0 ; but speed increases.

Note :  Problem 6.

| Average velocity |  1  | Av. speed |  | Av. velocity| | Average speed |

A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km/h. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km/h. The average speed of the man over the interval of time 0 to 40 min. is equal to (a) 5 km/h

Solution : (d)

(b)

25 km/h 4

Time taken in going to market 

(c)

30 km/h 4

(d)

45 km/h 8

2 .5 1  hr  30 min . 5 2

As we are told to find average speed for the interval 40 min., so remaining time for consideration of motion is 10 min. So distance travelled in remaining 10 min = 7 . 5  Hence, average speed =

Problem 7.

(2 .5  1 .25) km T otal distance 45 =  km / hr . T otal time (40 / 60 ) hr. 8

The relation 3 t  3 x  6 describes the displacement of a particle in one direction where x is in metres and t in sec. The displacement, when velocity is zero, is (a) 24 metres

Solution : (d)

10  1 . 25 km . 60

(b) 12 metres

3 x  (3 t  6)  3 x  (3 t  6) 2  x  3t 2  12t  12

3t 

3x  6

v=

dx d  (3 t 2  12 t  12)  6 t  12 dt dt



(c) 5 metres

If velocity = 0 then, 6t  12  0  t  2sec Hence at t = 2, x = 3(2)2 – 12 (2) + 12 = 0 metres.

(d) Zero

8 Motion In One Dimension

Problem 8.

genius

The motion of a particle is described by the equation x  a  bt instantaneous velocity at time 3 sec will be (a) 36 cm/sec

Solution : (b)

(b) 18 cm/sec v=

x  a  bt 2

[AMU (Med.) 2000]

(c) 16 cm/sec

(d) 32 cm/sec

(As b  3cm )

A train has a speed of 60 km/h for the first one hour and 40 km/h for the next half hour. Its average speed in km/h is [JIPMER 1999] (a) 50

Solution : (b)

where a  15 cm and b  3 cm. Its

dx  0  2bt dt

At t = 3sec, v = 2  3  3 = 18 cm / sec

Problem 9.

2

(b) 53.33

(c) 48

Total distance travelled = 60  1  40   Average speed 

(d) 70

1 1 3  80 km and Total time taken = 1 hr  hr  hr 2 2 2

80  53 .33 km/h 3 2

Problem 10. A person completes half of its his journey with speed  1 and rest half with speed  2 . The average speed of the person is (a)   Solution : (b)

[RPET 1993; MP PMT 2001]

1   2

(b)  

2

21  2 1   2

1  2 1   2

(c)  

(d)   1  2

In this problem total distance is divided into two equal parts. So

av

d d  d1  d 2 2 2 =  d1 d 2 d/2 d/2   1 2 1 2



av 

2 1

1



1

2

=

2 1  2 1   2

Problem 11. A car moving on a straight road covers one third of the distance with 20 km/hr and the rest with 60 km/hr. The average speed is (a) 40 km/hr Solution : (d)

(b) 80 km/hr

(c) 46

Let total distance travelled = x and total time taken t1 + t2   Average speed 

2 km /hr 3

(d) 36 km/hr

x / 3 2x / 3  20 60

x 1  = 36km / hr (1 / 3)x (2 / 3)x 1 2   20 60 60 180

2.7 Acceleration. The time rate of change of velocity of an object is called acceleration of the object. (1) It is a vector quantity. It’s direction is same as that of change in velocity (Not of the velocity) (2) There are three possible ways by which change in velocity may occur When only direction of velocity changes Acceleration velocity

perpendicular

e.g. Uniform circular motion

When only magnitude of velocity changes to

Acceleration parallel parallel to velocity

or

e.g. Motion under gravity

anti-

When both magnitude and direction of velocity changes Acceleration has two components one is perpendicular to velocity and another parallel or anti-parallel to velocity e.g. Projectile motion

genius PHYSICS by Pradeep Kshetrapal Motion In One Dimension 9

(3) Dimension : [M0L1T–2] (4) Unit : metre/second2 (S.I.); cm/second2 (C.G.S.) (5) Types of acceleration : (i) Uniform acceleration : A body is said to have uniform acceleration if magnitude and direction of the acceleration remains constant during particle motion.

Note : 

If a particle is moving with uniform acceleration, this does not necessarily imply that

particle is moving in straight line. e.g. Projectile motion. (ii) Non-uniform acceleration : A body is said to have non-uniform acceleration, if magnitude or direction or both, change during motion.     v v 2  v 1 (iii) Average acceleration : a a    t t   v The direction of average acceleration vector is the direction of the change in velocity vector as a  t    v d v (iv) Instantaneous acceleration = a  lim  t 0 t dt (v) For a moving body there is no relation between the direction of instantaneous velocity and direction of acceleration. 

Y

a



 a

2

1 

a



g



g

g



3

O

X

e.g. (a) In uniform circular motion  = 90º always (b) In a projectile motion  is variable for every point of trajectory. (vi) If a force F acts on a particle of mass m, by Newton’s    dv d 2 x  2 (vii) By definition a  dt dt

2nd

  F law, acceleration a  m

  dx   As v   dt  

i.e., if x is given as a function of time, second time derivative of displacement gives acceleration (viii) If velocity is given as a function of position, then by chain rule a 

dv dv dx d    v. dt dx dt dx

dx   as v  dt   

10 Motion In One Dimension

genius (ix) If a particle is accelerated for a time t1 by acceleration a1 and for time t2 by acceleration a2 then a t  a2 t 2 average acceleration is aa  1 1 t1  t2 a1

(x) If same force is applied on two bodies of different masses m 1 and m 2 separately then it produces accelerations a1 and a 2

F

respectively. Now these bodies are attached together and form a combined system and same force is applied on that system so that a be the acceleration of the combined system, then F  m 1  m 2  a 

So,

a2 F

m1

m2

a F

m1

m2

F F F   a a1 a 2

a a 1 1 1  a 1 2   a1  a 2 a a1 a 2

(xi) Acceleration can be positive, zero or negative. Positive acceleration means velocity increasing with time, zero acceleration means velocity is uniform constant while negative acceleration (retardation) means velocity is decreasing with time. (xii) For motion of a body under gravity, acceleration will be equal to “g”, where g is the acceleration due to gravity. Its normal value is 9 .8 m/s 2 or 980 cm/s 2 or 32 feet/s2 .

Sample problems based on acceleration Problem 12. The displacement of a particle, moving in a straight line, is given by s  2t 2  2t  4 where s is in metres and t in seconds. The acceleration of the particle is (a) 2 Solution : (b)

m/s2

(b) 4

m/s2

[CPMT 2001]

(c) 6

Given s  2t 2  2t  4  velocity (v) 

m/s2

(d) 8

m/s2

ds dv  4 t  2 and acceleration (a)   4 (1)  0  4 m / s 2 dt dt

Problem 13. The position x of a particle varies with time t as x  at 2  bt 3 . The acceleration of the particle will be zero at time t equal to (a) Solution : (c)

a b

[CBSE PMT 1997; BHU 1999; DPMT 2000; KCET (Med.) 2000]

(b)

2a 3b

(c)

Given x  at 2  bt 3  velocity (v) 

a 3b

(d) Zero

dx dv  2 at  3 bt 2 and acceleration (a) =  2 a  6 bt. dt dt

When acceleration = 0  2a  6bt  0  t 

2a a  . 6 b 3b

Problem 14. The displacement of the particle is given by y  a  bt  ct 2  dt 4 . The initial velocity and acceleration are respectively (a) b,4 d Solution : (c)

[CPMT 1999, 2003]

(b) b, 2c

Given y  a  bt  ct 2  dt 4  v =

(c) b, 2c

(d) 2c,  4 d

dy  0  b  2 ct  4 dt 3 dt

Putting t  0, vinitial = b So initial velocity = b Now, acceleration (a) 

dv  0  2 c  12 d t 2 dt

Putting t = 0, ainitial = 2c

Problem 15. The relation between time t and distance x is t   x 2   x , where  and  are constants. The retardation is (v is the velocity)

[NCERT 1982]

genius PHYSICS by Pradeep Kshetrapal Motion In One Dimension 11 (a) 2v 3 Solution : (a)

(b) 2 v 3

differentiating time with respect to distance So, acceleration (a) =

(d) 2  2 v 3

(c) 2v 3 dt dx 1  2x    v   dx dt 2x  

v. 2 dv dv dx dv  . =v   2 .v.v 2  2v 3 dt dx dt dx (2x   ) 2

Problem 16. If displacement of a particle is directly proportional to the square of time. Then particle is moving with [RPET 1999]

(a) Uniform acceleration (c) Uniform velocity Solution : (a)

(b) Variable acceleration (d) Variable acceleration but uniform velocity

Given that x  t 2 or x  Kt 2 (where K= constant) dx d  2 Kt and Acceleration (a)   2K dt dt It is clear that velocity is time dependent and acceleration does not depend on time. So we can say that particle is moving with uniform acceleration but variable velocity.

Velocity (v) 

Problem 17. A particle is moving eastwards with velocity of 5 m/s. In 10 sec the velocity changes to 5 m/s northwards. The average acceleration in this time is

[IIT-JEE 1982]

(a) Zero

1

(b)

2 (c) Solution : (b)

1 2 



m/s 2 toward north-east

(d)

m/s 2 toward north-west

1 m/s 2 toward north-west 2

   2  1

 v1

      212 cos 90  5  5  5 2 2 1

2 2

o

2

2

  5 2

 5 2 1   m/s 2 toward northt 10 2 west (As clear from the figure). Average acceleration 

 



2  5m / s 90o 

1  5 m / s

Problem 18. A body starts from the origin and moves along the x-axis such that velocity at any instant is given by (4 t 3  2t) , where t is in second and velocity is in m/s. What is the acceleration of the particle, when it is 2m from the origin? (a) 28 m/s 2 Solution : (b)

(b) 22 m/s 2

(c) 12 m /s 2

Given that   4 t 3  2t



x   dt , x  t 4  t 2  C , at t  0, x  0  C  0 When particle is 2m away from the origin

2  t 4  t 2  t 4  t 2  2  0  (t 2  2) (t 2  1)  0  t  2 sec a





d d  4 t 3  2t  12 t 2  2  a  12t 2  2 dt dt

for t 

2 sec  a  12 

 2

2

 2  a  22 m /s 2

(d) 10 m /s 2

12 Motion In One Dimension

genius

Problem 19. A body of mass 10 kg is moving with a constant velocity of 10 m/s. When a constant force acts for 4 sec on it, it moves with a velocity 2 m/sec in the opposite direction. The acceleration produced in it is [MP PET 1997] (b)  3 m/s 2

(a) 3 m/s 2 Solution : (b)

(c) 0 .3 m/s 2

(d)  0 .3 m/s 2

Let particle moves towards east and by the application of constant force it moves towards west





1  10 m/s and  2  2 m/s . Acceleration   a

Change in velocity  2  1  T ime t

(2)  (10 ) 12   3 m /s 2 4 4

2.8 Position Time Graph. During motion of the particle its parameters of kinematical analysis (u, v, a, r) changes with time. This can be represented on the graph. Position time graph is plotted by taking time t along x-axis and position of the particle on y-axis. Let AB is a position-time graph for any moving particle

genius PHYSICS by Pradeep Kshetrapal Motion In One Dimension 13

From triangle ABC tan   By comparing (i) and (ii)

BC AD y 2  y 1   AC AC t 2  t1

y

…(i)

Position

Change in position y 2  y 1 As Velocity =  Time taken t 2  t1

….(ii)

Velocity = tan

y2 y1

O

D

B

 A t1

C

t2

x

Time

P

 = 0o so v = 0

i.e., line parallel to time axis represents that the particle is at rest. O

T

P

 = 90o so v = 

O

i.e., line perpendicular to time axis represents that particle is changing its position but time does not changes it means the particle possesses infinite velocity. Practically this is not possible.

T

P

 = constant so v = constant, a = 0

i.e., line with constant slope represents uniform velocity of the particle. O

T

P

 is increasing so v is increasing, a is positive.

O

i.e., line bending towards position axis represents increasing velocity of particle. It means the particle possesses acceleration.

T

P

 is decreasing so v is decreasing, a is negative O

i.e., line bending towards time axis represents decreasing velocity of the particle. It means the particle possesses retardation.

T

P

 constant but > 90o so v will be constant but negative 

O

T

i.e., line with negative slope represent that particle returns towards the point of reference. (negative displacement).

P A O

B

Straight line segments of different slopes represent that velocity of the body changes after certain interval of time.

C T

S

P T

O

This graph shows that at one instant the particle has two positions. Which is not possible.

P

The graph shows that particle coming towards origin initially and after that it is moving away from origin. O

T

14 Motion In One Dimension

genius v = tan

It is clear that slope of position-time graph represents the velocity of the particle. Various position – time graphs and their interpretation

Note : 

A

Distance

If the graph is plotted between distance and time then it is always an increasing curve and it never comes back towards origin because distance never decrease with time. Hence such type of distance time graph is valid up to point A only, after point A it is not valid as shown in the figure.

 For two particles having displacement time graph with slopes 1 and 2  tan 1 possesses velocities v1 and v2 respectively then 1   2 tan  2

O

Time

Sample problems based on position-time graph Problem 20. The position of a particle moving along the x-axis at certain times is given below : t (s) x (m)

0

1

2

3

–2

0

6

16

Which of the following describes the motion correctly (a) Uniform, accelerated

(b)

(c) Non-uniform, accelerated Solution : (a)

Instantaneous velocity v 

Uniform, decelerated

(d) There is not enough data for generalisation

x , By using the data from the table t

0  (2) 60 16  6  2 m /s , v 2   6 m /s and v 3   10 m /s i.e. the speed is increasing at a constant 1 1 1 rate so motion is uniformly accelerated. v1 

Problem 21. Which of the following graph represents uniform motion (a)

(b)

(c) s

s

(d) s

s

t t When distance time graph is a straight line with constant slope than motion is uniform. t

Solution : (a)

t

Problem 22. The displacement-time graph for two particles A and B are straight lines inclined at angles of 30o and 60o with the time axis. The ratio of velocities of vA : vB is (a) 1 : 2 Solution : (d)

[CPMT 1990; MP PET 1999; MP PET 2001]

(b) 1 : 3

v  tan  from displacement graph. So

(c) o

(d) 1 : 3

3 :1

vA tan 30 1/ 3    o vB tan 60 3

1 3 3



1 3

(a)

1 3

m/s

(b) 3 m/s (c)

3 m/s

Time (sec)

Problem 23. From the following displacement time graph find out the velocity of a moving body

30o O Displacement (meter)

genius PHYSICS by Pradeep Kshetrapal Motion In One Dimension 15 (d) Solution : (c)

1 3

In first instant you will apply   tan  and say,   tan 30 o 

1

m/s.

3

But it is wrong because formula   tan  is valid when angle is measured with time axis. Here angle is taken from displacement axis. So angle from time axis  90 o  30 o  60 o . Now   tan 60 o  3

Problem 24. The diagram shows the displacement-time graph for a particle moving in a straight line. The average velocity for the interval t = 0, t = 5 is

x 20

(a) 0

10

(b) 6 ms–1

Average velocity =

2

t

4

– 10

(d) 2 ms–1 Solution : (c)

5

O

(c) – 2 ms–1

Total displaceme nt 20   20   10  = –2 m/s = 5 Total time

Problem 25. Figure shows the displacement time graph of a body. What is the ratio of the speed in the first second and that in the next two seconds Displacement

Y

(a) 1 : 2 (b) 1 : 3 (c) 3 : 1

30 20 10 0

1

(d) 2 : 1 Solution: (d)

2

3 Time

X

Speed in first second = 30 and Speed in next two seconds = 15. So that ratio 2 : 1

2.9 Velocity Time Graph. The graph is plotted by taking time t along x-axis and velocity of the particle on y-axis. Distance and displacement : The area covered between the velocity time graph and time axis gives the displacement and distance travelled by the body for a given time interval. Then Total distance | A1 | | A 2 | | A 3 |

+



= Addition of modulus of different area. i.e. s  |  | dt

1

3 t

Total displacement  A1  A 2  A 3

2



= Addition of different area considering their sign. i.e. r   dt

–

As Acceleration =

v  v1 Change in velocity  2 Time taken t 2  t1

From triangle ABC, tan  

BC AD v 2  v 1   AC AC t 2  t1

Velocity

here A1 and A2 are area of triangle 1 and 2 respectively and A3 is the area of trapezium . y Acceleration : Let AB is a velocity-time graph for any moving particle …(i)

v2

D



v1

….(ii)

B

A O

t1

C

t2 Time

x

16 Motion In One Dimension By comparing (i) and (ii) Acceleration (a) = tan  It is clear that slope of velocity-time graph represents the acceleration of the particle.

genius

genius PHYSICS by Pradeep Kshetrapal Motion In One Dimension 17

Velocity

Various velocity – time graphs and their interpretation  = 0, a = 0, v = constant

Time

Velocity

O

i.e., line parallel to time axis represents that the particle is moving with constant velocity.

 = 90o, a = , v = increasing

Time

Velocity

O

i.e., line perpendicular to time axis represents that the particle is increasing its velocity, but time does not change. It means the particle possesses infinite acceleration. Practically it is not possible.

 =constant, so a = constant and v is increasing uniformly with time i.e., line with constant slope represents uniform acceleration of the particle. Time

Velocity

O

 increasing so acceleration increasing Time

Velocity

O

i.e., line bending towards velocity axis represent the increasing acceleration in the body.

 decreasing so acceleration decreasing i.e. line bending towards time axis represents the decreasing acceleration in the body Time

Velocity

O

Positive constant acceleration because  is constant and < 90o but initial velocity of the particle is negative. Time

Velocity

O

Positive constant acceleration because  is constant and < 90o but initial velocity of particle is positive. Time

Velocity

O

Negative constant acceleration because  is constant and > 90o but initial velocity of the particle is positive. Time

Velocity

O

Negative constant acceleration because  is constant and > 90o but initial velocity of the particle is zero. Time

Velocity

O

O

Time

Negative constant acceleration because  is constant and > 90o but initial velocity of the particle is negative.

18 Motion In One Dimension

genius

Sample problems based on velocity-time graph Problem 26. A ball is thrown vertically upwards. Which of the following plots represents the speed-time graph of the

(d)

Time

Time

Time

Time

Solution : (c)

(c)

Speed

(b)

Speed

Speed

(a)

Speed

ball during its flight if the air resistance is not ignored

In first half of motion the acceleration is uniform & velocity gradually decreases, so slope will be negative but for next half acceleration is positive. So slope will be positive. Thus graph 'C' is correct. Not ignoring air resistance means upward motion will have acceleration (a + g) and the downward motion will have (g  a).

Problem 27. A train moves from one station to another in 2 hours time. Its speed-time graph during this motion is shown in the figure. The maximum acceleration during the journey is

[Kerala (Engg.) 2002]

Speed in km/hours

100 80 D

60 40 C

B

20

L

M N 0.25 0.75 1.00

A

E 2.00

1.5 Time in hours

(a) 140 km h–2 Solution : (b)

(b) 160 km h–2

(c) 100 km h–2

(d) 120 km h–2

Maximum acceleration means maximum slope in speed – time graph. that slope is for line CD. So, a max  slope of CD =

60  20 40   160 km h  2 . 1 . 25  1 . 00 0 . 25

Problem 28. The graph of displacement v/s time is S

t

Its corresponding velocity-time graph will be (a)

(b) V

Solution : (a)

(c) V

t

[DCE 2001]

(d) V

V

t

t

t

We know that the velocity of body is given by the slope of displacement – time graph. So it is clear that initially slope of the graph is positive and after some time it becomes zero (corresponding to the peak of the graph) and then it will be negative.

Problem 29. In the following graph, distance travelled by the body in metres is

[EAMCET 1994]

genius PHYSICS by Pradeep Kshetrapal Motion In One Dimension 19 Y v (m/s )

(a) 200 (b) 250 (c) 300

15 10 5

(d) 400 Solution : (a)

0

Distance = The area under v – t graph S 

X

30 20 Time (s)

10

1 (30  10)  10 = 200 metre 2

Problem 30. For the velocity-time graph shown in figure below the distance covered by the body in last two seconds of its motion is what fraction of the total distance covered by it in all the seven seconds [MP PMT/PET 1998; RPET 2001]

Solution : (b)

1 2

(b)

1 4

(c)

1 3

(d)

2 3

v (m/s )

(a)

5 A 0

So required fraction 

3

D

Q

P 1

Distance covered in total 7 seconds = Area of trapezium ABCD  Distance covered in last 2 second = area of triangle CDQ=

C

B

10

5 Time (s)

7

1 (2  6 )  10 = 40 m 2

1  2  10  10 m 2

10 1  40 4

Problem 31. The velocity time graph of a body moving in a straight line is shown in the figure. The displacement and distance travelled by the body in 6 sec are respectively (a) 8 m, 16 m

4 v (m/s)

(b) 16 m, 8 m (c) 16 m, 16 m

Solution : (a)

B G

2 O –2

(d) 8 m, 8 m

A

2

6 4

C D

H

F

E

I

t (sec)

–4

Area of rectangle ABCO = 4  2 = 8 m Area of rectangle CDEF = 2 × (– 2) = – 4 m Area of rectangle FGHI = 2×2 = 4 m Displacement = sum of area with their sign = 8 + (– 4) + 4 = 8 m Distance = sum of area with out sign = 8 + 4 + 4 = 16 m

Problem 32. A ball is thrown vertically upward which of the following graph represents velocity time graph of the ball

(c) Time

(d) Time

Velocity

Time

[CPMT 1993; AMU (Engg.) 2000]

Velocity

(b) Velocity

(a)

Velocity

during its flight (air resistance is neglected)

Time

20 Motion In One Dimension

Solution : (d)

genius

In the positive region the velocity decreases linearly (during rise) and in negative region velocity increase linearly (during fall) and the direction is opposite to each other during rise and fall, hence fall is shown in the negative region. c

Problem 34. The acceleration-time graph of a body is shown below – a

t

The most probable velocity-time graph of the body is 



(a)

Solution : (c)





(b)

(c)

(d)



da  k (constant)  a  kt (by integration) dt



dv  kt  dv  ktdt  dt



dv  k



tdt  v 

t

t

t

t

From given a  t graph acceleration is increasing at constant rate

kt 2 2

i.e., v is dependent on time parabolically and parabola is symmetric about v-axis. and suddenly acceleration becomes zero. i.e. velocity becomes constant. Hence (c) is most probable graph.

Problem 35. Which of the following velocity time graphs is not possible v

(a)

(b)

O

Solution : (d)

v

v

t

v

(c)

O

t

(d)

O

O

t

t

Particle can not possess two velocities at a single instant so graph (d) is not possible.

Problem 36. For a certain body, the velocity-time graph is shown in the figure. The ratio of applied forces for intervals AB and BC is (a) 

1 2

(b) 

1 2

 D

60o

30o A

B

C

t

genius PHYSICS by Pradeep Kshetrapal Motion In One Dimension 21

Solution : (d)

(c) 

1 3

(d) 

1 3

Ratio of applied force = Ratio of acceleration =

a AB tan 30 1 3   = 1 3 aBC tan 120   3

Problem 37. Velocity-time graphs of two cars which start from rest at the same time, are shown in the figure. Graph Velocit y

shows, that (a) Initial velocity of A is greater than the initial velocity of B

A

(b) Acceleration in A is increasing at lesser rate than in B

B

(c) Acceleration in A is greater than in B

t

O

(d) Acceleration in B is greater than in A Solution : (c)

B

A

Time

At a certain instant t slope of A is greater than B (  A   B ), so acceleration in A is greater than B

Problem 38. Which one of the following graphs represent the velocity of a steel ball which fall from a height on to a

Solution : (a)

Time

(d) Time

Velocity

Time

(c)

Velocity

(b)

Velocity

(a)

Velocity

marble floor? (Here  represents the velocity of the particle and t the time)

Time

Initially when ball falls from a height its velocity is zero and goes on increasing when it comes down. Just after rebound from the earth its velocity decreases in magnitude and its direction gets reversed. This process is repeated untill ball comes to at rest. This interpretation is well explained in graph (a).

Problem 39. The adjoining curve represents the velocity-time graph of a particle, its acceleration values along OA, AB Velocity (m/sec)

and BC in metre/sec2 are respectively (a) 1, 0, – 0.5 (b) 1, 0, 0.5 (c) 1, 1, 0.5 (d) 1, 0.5, 0 Solution : (a)

Acceleration along OA 

v 2  v 1 10  0   1m / s 2 t 10

Acceleration along OB 

0 0 10

Acceleration along BC 

0  10  – 0.5 m/s2 20

10

A

B

5 O

10

20 30 40 Time (sec)

2.10 Equations of Kinematics. These are the various relations between u, v, a, t and s for the moving particle where the notations are used as :

22 Motion In One Dimension

genius

u = Initial velocity of the particle at time t = 0 sec v = Final velocity at time t sec a = Acceleration of the particle s = Distance travelled in time t sec sn = Distance travelled by the body in nth sec (1) When particle moves with zero acceleration (i) It is a unidirectional motion with constant speed. (ii) Magnitude of displacement is always equal to the distance travelled. (iii) v = u,

s=ut

[As a = 0]

(2) When particle moves with constant acceleration (i) Acceleration is said to be constant when both the magnitude and direction of acceleration remain constant. (ii) There will be one dimensional motion if initial velocity and acceleration are parallel or anti -parallel to each other. (iii) Equations of motion in scalar from

  u  at s  ut 

1 2 at 2

 2  u 2  2as u v  s t  2 

sn  u 

a (2n  1) 2

Equation of motion in vector from    v  u  at

  1 s  u t  at 2 2    v.v  u.u  2a.s  1   s  (u  v ) t 2    a s n  u  (2n  1) 2

(3) Important points for uniformly accelerated motion (i) If a body starts from rest and moves with uniform acceleration then distance covered by the body in t sec is proportional to t2 (i.e. s  t 2 ). So we can say that the ratio of distance covered in 1 sec, 2 sec and 3 sec is 12 : 22 : 3 2 or 1 : 4 : 9. (ii) If a body starts from rest and moves with uniform acceleration then distance covered by the body in nth sec is proportional to (2n  1) (i.e. s n  (2n  1) So we can say that the ratio of distance covered in I sec, II sec and III sec is 1 : 3 : 5. (iii) A body moving with a velocity u is stopped by application of brakes after covering a distance s. If the same body moves with velocity nu and same braking force is applied on it then it will come to rest after covering a distance of n2s. As  2  u 2  2as  0  u 2  2as  s 

u2 , s  u 2 [since a is constant] 2a

So we can say that if u becomes n times then s becomes n2 times that of previous value. (iv) A particle moving with uniform acceleration from A to B along a straight line has velocities  1 and  2 at A and B respectively. If C is the mid-point between A and B then velocity of the particle at C is equal to

genius PHYSICS by Pradeep Kshetrapal Motion In One Dimension 23

 12   22



2

Sample problems based on uniform acceleration Problem 40. A body A moves with a uniform acceleration a and zero initial velocity. Another body B, starts from the same point moves in the same direction with a constant velocity v . The two bodies meet after a time t. The value of t is [MP PET 2003] (a) Solution : (a)

2v a

(b)

v a

Let they meet after time 't' . Distance covered by body A = and

v 2a

(c)

(d)

v 2a

1 2 at ; Distance covered by body B = vt 2

1 2 2v at  vt  t  . 2 a

Problem 41. A student is standing at a distance of 50metres from the bus. As soon as the bus starts its motion with an

acceleration of 1ms–2, the student starts running towards the bus with a uniform velocity u . Assuming the motion to be along a straight road, the minimum value of u , so that the students is able to catch the bus is [KCET 2003]

(a) 5 Solution : (c)

ms–1

(b) 8

ms–1

(c) 10

ms–1

(d) 12

ms–1

Let student will catch the bus after t sec. So it will cover distance ut. Similarly distance travelled by the bus will be

ut  50 

50 t 1 2 t2  u  at  50  t 2 2 2

To find the minimum value of u,

1 2 at for the given condition 2

(As a  1 m / s 2 )

du  0 , so we get t = 10 sec dt

then u = 10 m/s.

Problem 42. A car, moving with a speed of 50 km/hr, can be stopped by brakes after at least 6m. If the same car is moving at a speed of 100 km/hr, the minimum stopping distance is (a) 6m Solution : (d)

(b) 12m

(c) 18m

v 2  u 2  2as  0  u 2  2as  s  s2  u2  s1  u 1

2

(d) 24m

u2  s  u 2 (As a = constant) 2a

  100       s 2  4 s 1  4  12  24 m .  50   2

Problem 43. The velocity of a bullet is reduced from 200m/s to 100m/s while travelling through a wooden block of thickness 10cm. The retardation, assuming it to be uniform, will be 4

(a) 10  10 m/s2 Solution : (d)

4

(b) 12  10 m/s2

u  200 m / s , v  100 m / s , s  0.1 m

a

u 2  v 2 (200 )2  (100 )2   15  10 4 m / s 2 2s 2  0.1

4

(c) 13.5  10 m/s2

[AIIMS 2001] 4

(d) 15  10 m/s2

24 Motion In One Dimension

genius

Problem 44. A body A starts from rest with an acceleration a 1 . After 2 seconds, another body B starts from rest with an

Solution : (a)

acceleration a 2 . If they travel equal distances in the 5th second, after the start of A, then the ratio a1 : a 2 is equal to [AIIMS 2001] (a) 5 : 9 (b) 5 : 7 (c) 9 : 5 (d) 9 : 7 a a By using S n  u  2n  1  , Distance travelled by body A in 5th second = 0  1 (2  5  1) 2 2 a Distance travelled by body B in 3rd second is = 0  2 (2  3  1) 2 a a a 5 According to problem : 0  1 (2  5  1) = 0  2 (2  3  1)  9 a1  5 a2  1  2 2 a2 9

Problem 45. The average velocity of a body moving with uniform acceleration travelling a distance of 3.06 m is 0.34 ms–1.

Solution : (b)

If the change in velocity of the body is 0.18ms–1 during this time, its uniform acceleration is[EAMCET (Med.) 2000] (a) 0.01 ms–2 (b) 0.02 ms–2 (c) 0.03 ms–2 (d) 0.04 ms–2 Distance 3 .06 Time    9 sec Average velocity 0 .34 and Acceleration 

Change in velocity 0 . 18   0 . 02 m / s 2 . T ime 9

Problem 46. A particle travels 10m in first 5 sec and 10m in next 3 sec. Assuming constant acceleration what is the Solution : (a)

distance travelled in next 2 sec (a) 8.3 m (b) 9.3 m Let initial (t  0) velocity of particle = u

(c) 10.3 m

for first 5 sec of motion s5  10 metre , so by using s  ut  1 a(5 )2  2u  5a  4 2 for first 8 sec of motion s8  20 metre 10  5 u 

20  8 u 

1 a(8 )2  2u  8a  5 2

By solving (i) and (ii) u 

(d) None of above

1 2 at 2

…. (i)

.... (ii)

7 1 m / s a  m / s2 6 3

Now distance travelled by particle in total 10 sec. s10  u  10 

1 a10 2 2

by substituting the value of u and a we will get s10  28.3 m So the distance in last 2 sec = s10 – s8  28.3  20  8.3 m Problem 47. A body travels for 15 sec starting from rest with constant acceleration. If it travels distances S 1 , S 2 and

S 3 in the first five seconds, second five seconds and next five seconds respectively the relation between [AMU (Engg.) 2000] S 1 , S 2 and S 3 is (a) S 1  S 2  S 3 Solution : (c)

(b) 5 S 1  3 S 2  S 3

(c) S 1 

1 1 S2  S3 3 5

Since the body starts from rest. Therefore u  0 . 1 25 a S 1  a(5 ) 2  2 2 S1  S 2 

1 100 a 100 a a a(10 ) 2   S 1 = 75  S2  2 2 2 2

S1  S 2  S 3 

1 225 a 225 a 125a a(15 ) 2   S 2  S1 =  S3  2 2 2 2

(d) S 1 

1 1 S2  S3 5 3

genius PHYSICS by Pradeep Kshetrapal Motion In One Dimension 25 Thus Clearly S 1 

1 1 S2  S3 3 5

Problem 48. If a body having initial velocity zero is moving with uniform acceleration 8 m / sec 2 , the distance travelled

Solution : (a)

by it in fifth second will be (a) 36 metres (b) 40 metres (c) 100 metres 1 1 S n  u  a(2n  1)  0  (8 ) [2  5  1]  36 metres 2 2

(d) Zero

Problem 49. The engine of a car produces acceleration 4m/sec2 in the car, if this car pulls another car of same mass, what will be the acceleration produced (a) 8 m/s2 Solution : (b)

[RPET 1996]

(b) 2 m/s2

1 m / s2 (d) 2

(c) 4 m/s2

1 if F = constant. Since the force is same and the effective mass of system becomes double m a m , a 2  1  2 m/s 2  2 2m

F = ma a 

a2 m  1 a1 m2

Problem 50. A body starts from rest. What is the ratio of the distance travelled by the body during the 4th and 3rd second. [CBSE PMT 1993]

(a) 7/5 Solution : (a)

(b) 5/7

As S n  (2n  1) ,

(c) 7/3

(d) 3/7

S4 7  S3 5

2.11 Motion of Body Under Gravity (Free Fall). The force of attraction of earth on bodies, is called force of gravity. Acceleration produced in the body by the force of gravity, is called acceleration due to gravity. It is represented by the symbol g. In the absence of air resistance, it is found that all bodies (irrespective of the size, weight or composition) fall with the same acceleration near the surface of the earth. This motion of a body falling towards the earth from a small altitude (h t1 Let u is the initial velocity of body then time of ascent t1 

u ga

and h 

u2 2(g  a)

where g is acceleration due to gravity and a is retardation by air resistance and for upward motion both will work vertically downward. For downward motion a and g will work in opposite direction because a always work in direction opposite to motion and g always work vertically downward. So

h

1 (g  a) t 22 2



u2 1  (g  a) t 22  2(g  a) 2

t2 

u=0

u (g  a)(g  a)

1m

t1 

1

1m

t2 

2 1

1m

t3 

3  2

t4 

4  3

1m

28 Motion In One Dimension

genius

Comparing t1 and t2 we can say that t2 > t1 since (g + a ) > (g – a) (10) A particle is dropped vertically from rest from a height. The time taken by it to fall through successive distance of 1m each will then be in the ratio of the difference in the square roots of the integers i.e. 1 , ( 2  1 ), ( 3  2 ).......( 4  3 ),........ .

Sample problems based on motion under gravity Problem 51. If a body is thrown up with the velocity of 15 m/s then maximum height attained by the body is (g = 10 m/s2) [MP PMT 2003]

(a) 11.25 m Solution : (a)

H max 

(b) 16.2 m

(c) 24.5 m

(d) 7.62 m

u2 (15) 2   11 .25 m 2 g 2  10

Problem 52. A body falls from rest in the gravitational field of the earth. The distance travelled in the fifth second of its motion is (g  10 m / s 2 ) [MP PET 2003] (a) 25m (b) 45m (c) 90m (d) 125m g 10 2  5  1  45 m. hn  2n  1  h5 th  Solution : (b) 2 2 Problem 53. If a ball is thrown vertically upwards with speed u, the distance covered during the last t seconds of its ascent is [CBSE 2003]

1 (a) gt 2 2 Solution : (a)

1 (b) ut  gt 2 2

If ball is thrown with velocity u, then time of flight 

(c) (u  gt )t

u g

(d) ut

h

t sec

u  u  velocity after   t  sec : v  u  g  t  = gt. g g     u    t  sec So, distance in last 't' sec : 0 2  (gt) 2  2(g)h. g  1 2  h  gt . 2 Problem 54. A man throws balls with the same speed vertically upwards one after the other at an interval of 2 seconds. What should be the speed of the throw so tha t more than two balls are in the sky at any time (Given g  9.8 m / s 2 )

Solution : (d)

(a) At least 0.8 m/s (b) Any speed less than 19.6 m/s (c) Only with speed 19.6 m/s (d) More than 19.6 m/s Interval of ball throw = 2 sec. If we want that minimum three (more than two) ball remain in air then time of flight of first ball must be 2U greater than 4 sec. i.e. T  4 sec or  4 sec  u  19 .6 m /s g It is clear that for u  19 .6 First ball will just strike the ground (in sky), second ball will be at highest point (in sky), and third ball will be at point of projection or on ground (not in sky).

Problem 55. A man drops a ball downside from the roof of a tower of height 400 meters. At the same time another ball

Solution : (c)

is thrown upside with a velocity 50 meter/sec. from the surface of the tower, then they will meet at which height from the surface of the tower [CPMT 2003] (a) 100 meters (b) 320 meters (c) 80 meters (d) 240 meters Let both balls meet at point P after time t. A h1 400 m

P h2

genius PHYSICS by Pradeep Kshetrapal Motion In One Dimension 29 The distance travelled by ball A ( h1 ) 

1 2 gt 2

.....(i)

1 2 gt .....(ii) 2 By adding (i) and (ii) h1  h2  ut = 400 (Given h  h1  h2  400 . )

The distance travelled by ball B (h2 )  ut 

 t  400 / 50  8 sec and h1  320 m, h2  80 m

Problem 56. A very large number of balls are thrown vertically upwards in quick succession in such a way that the next ball is thrown when the previous one is at the maximum height. If the maximum height is 5m, the number of ball thrown per minute is (take g  10 ms 2 ) (a) 120 Solution : (c)

[KCET (Med.) 2002]

(b) 80

(c) 60

(d) 40

Maximum height of ball = 5m, So velocity of projection  u  2 gh  2  10  5  10 m / s time interval between two balls (time of ascent) =

u 1  1sec  min. g 60

So no. of ball thrown per min = 60

Problem 57. A particle is thrown vertically upwards. If its velocity at half of the maximum height is 10 m/s, then maximum height attained by it is (Take g  10 m/s2) (a) 8 m Solution : (b)

(b) 10 m

(c) 12 m

(d) 16 m

Let particle thrown with velocity u and its maximum height is H then H 

u2 2g

When particle is at a height H / 2 , then its speed is 10 m / s u2 H From equation v 2  u 2  2 gh , 10 2  u 2  2 g    u 2  2 g  u 2  200 4g 2  Maximum height H 

u2 200 = 10m  2 g 2  10

Problem 58. A stone is shot straight upward with a speed of 20 m/sec from a tower 200 m high. The speed with which

Solution : (b)

it strikes the ground is approximately [AMU (Engg.) 1999] (a) 60 m/sec (b) 65 m/sec (c) 70 m/sec (d) 75 m/sec Speed of stone in a vertically upward direction is 20 m/s. So for vertical downward motion we will consider u  20 m / s

v 2  u 2  2 gh  (20)2  2  10  200  v  65 m / s

Problem 59. A body freely falling from the rest has a velocity ‘v’ after it falls through a height ‘h’. The distance it has to fall down for its velocity to become double, is (a) 2h Solution : (b)

[BHU 1999]

(b) 4h

(c) 6h

Let at point A initial velocity of body is equal to zero For path AB : v 2  0  2 gh

… (i)

For path AC : (2v)2  0  2 gx  4 v 2  2 gx

… (ii)

Solving (i) and (ii) x  4 h

(d) 8h u=0 A h x B

v

C

2v

Problem 60. A body sliding on a smooth inclined plane requires 4 seconds to reach the bottom starting from rest at the top. How much time does it take to cover one-fourth distance starting from rest at the top (a) 1 s (b) 2 s (c) 4 s (d) 16 s

30 Motion In One Dimension Solution : (b)

S 

1 2 at  t  2

t2  t1

s2  s1

genius s  As a  constant 

t 4 s/4 1   t2  1   2 s 2 2 s 2

Problem 61. A stone dropped from a building of height h and it reaches after t seconds on earth. From the same

building if two stones are thrown (one upwards and other downwards) with the same velocity u and they reach the earth surface after t1 and t 2 seconds respectively, then [CPMT 1997; UPSEAT 2002; KCET (Engg./Med.) 20 (b) t 

(a) t  t1  t 2 Solution : (c)

For first case of dropping h 

t1  t 2 2

(c) t 

1 g(t 2  t 12 ) 2

1 2 1 2 gt1  gt 2 2

......(i)

For third case of upward throwing h  ut 2   ut 2 

(d) t  t 12 t 22

1 2 gt . 2

For second case of downward throwing h  ut 1    ut 1 

t1 t 2

1 g(t 2  t 22 ) 2

1 2 1 2 gt 2  gt 2 2

.......(ii)

on solving these two equations : 

t1 t 2  t 12  2  t t2 t  t 22

t1 t 2 .

Problem 62. By which velocity a ball be projected vertically downward so that the distance covered by it in 5th second is twice the distance it covers in its 6th second ( g  10 m / s 2 ) (a) 58 .8 m / s Solution : (c)

(b) 49 m / s

(c) 65 m / s

(d) 19 .6 m / s

1 10 10 g (2n  1)  u  [2  5  1]  2 {u  [2  6  1]} 2 2 2  u  45  2  (u  55)  u  65m / s.

By formula hn  u 

Problem 63. Water drops fall at regular intervals from a tap which is 5 m above the ground. The third drop is leaving

Solution : (b)

the tap at the instant the first drop touches the ground. How far above the ground is the second drop at that instant [CBSE PMT 1995] (a) 2.50 m (b) 3.75 m (c) 4.00 m (d) 1.25 m Let the interval be t then from question 1 1 For first drop g(2 t) 2  5 .....(i) For second drop x  gt 2 .....(ii) 2 2 By solving (i) and (ii) x 

5 5 and hence required height h  5   3 . 75 m . 4 4

Problem 64. A balloon is at a height of 81 m and is ascending upwards with a velocity of 12 m / s. A body of 2 kg weight is dropped from it. If g  10 m / s 2 , the body will reach the surface of the earth in (a) 1.5 s Solution : (c)

(b) 4.025 s

(c) 5.4 s

[MP PMT 1994]

(d) 6.75 s

As the balloon is going up we will take initial velocity of falling body  12m / s, h  81m, g  10 m / s 2 By applying h  ut   t

1 2 1 gt ; 81  12 t  (10 )t 2  5 t 2  12t  81  0 2 2

12  144  1620 12  1764  5.4 sec .  10 10

Problem 65. A particle is dropped under gravity from rest from a height h (g  9.8 m / s 2 ) and it travels a distance 9h/25 in the last second, the height h is

[MNR 1987]

genius PHYSICS by Pradeep Kshetrapal Motion In One Dimension 31 (a) 100 m Solution : (b)

(b) 122.5 m

(c) 145 m

1 gn 2 = h 2

Distance travelled in n sec =

Distance travelled in n th sec 

g 9h (2n  1)  2 25

(d) 167.5 m

.....(i) .....(ii)

Solving (i) and (ii) we get. h  122 .5 m .

Problem 66. A stone thrown upward with a speed u from the top of the tower reaches the ground with a velocity 3u. The height of the tower is (a) 3u 2 / g Solution : (b)

(b) 4 u 2 / g

(c) 6u 2 / g

(d) 9u 2 / g

For vertical downward motion we will consider initial velocity = – u. By applying v 2  u 2  2 gh , (3u) 2  (u) 2  2 gh  h 

4u 2 . g

Problem 67. A stone dropped from the top of the tower touches the ground in 4 sec. The height of the tower is about [MP PET 1986; AFMC 1994; CPMT 1997; BHU 1998; DPMT 1999; RPET 1999]

(a) 80 m Solution : (a)

h

(b) 40 m

(c) 20 m

(d) 160 m

1 2 1 gt   10  4 2  80 m . 2 2

Problem 68. A body is released from a great height and falls freely towards the earth. Another body is released from the same height exactly one second later. The separation between the two bodies, two seconds after the release of the second body is (a) 4.9 m Solution : (d)

(b) 9.8 m

(c) 19.6 m

(d) 24.5 m

The separation between two bodies, two second after the release of second body is given by : s=

1 1 g(t 12  t 22 ) =  9 . 8  (3 2  2 2 )  24 . 5 m. 2 2

2.12 Motion With Variable Acceleration. (i) If acceleration is a function of time a  f (t)

then v  u 



t

0

f (t) dt and s  ut 

   f (t) dt dt

(ii) If acceleration is a function of distance a  f (x )

then v 2  u 2  2



x x0

f (x ) dx

(iii) If acceleration is a function of velocity a = f (v)

then t 



v

u

dv and x  x 0  f (v)



v

u

vdv f (v)

Sample problems based on variable acceleration Problem 69. An electron starting from rest has a velocity that increases linearly with the time that is v  kt , where k  2 m / sec 2 . The distance travelled in the first 3 seconds will be

(a) 9 m

(b) 16 m

(c) 27 m

[NCERT 1982]

(d) 36 m

32 Motion In One Dimension

Solution : (a)

x



t2

genius 3

3

v dt 

t1

t2  2 t dt  2    9 m.  2  0 0



Problem 70. The acceleration of a particle is increasing linearly with time t as bt. The particle starts from the origin with an initial velocity v 0 . The distance travelled by the particle in time t will be (a) v 0 t  Solution : (c)



v2

1 2 bt 3

dv =

v1



t2

(b) v 0 t 

a dt =

t1



t2

(d) v 0 t 

1 2 bt 2

t

 bt 2  v 2  v1    2 

2    v 0  bt  2 0



1 3 bt 6

(bt) dt

2    t1

v 0 dt 

(c) v 0 t 

t1

 bt 2  v 2  v1    2 

 S 

1 3 bt 3

[CBSE PMT 1995]



t

bt 2 1 dt  v 0 t  bt 3 2 6

Problem 71. The motion of a particle is described by the equation u  at . The distance travelled by the particle in the first 4 seconds

[DCE 2000]

(a) 4a Solution : (d)

(b) 12a

u  at  4

ds  at dt 4

t 2   s  at dt  a   = 8a.  2  0 0



(c) 6a

(d) 8 a

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PHYSICS

by Pradeep Kshetrapal Motion in Two Dimension 1

The motion of an object is called two dimensional, if two of the three co-ordinates are required to specify the position of the object in space changes w.r.t time. In such a motion, the object moves in a plane. For example, a billiard ball moving over the billiard table, an insect crawling over the floor of a room, earth revolving around the sun etc. Two special cases of motion in two dimension are 1. Projectile motion 2. Circular motion

PROJECTILE MOTION 3.1 Introduction. A hunter aims his gun and fires a bullet directly towards a monkey sitting on a distant tree. If the monkey remains in his position, he will be safe but at the instant the bullet leaves the barrel of gun, if the monkey drops from the tree, the bullet will hit the monkey because the bullet will not follow the linear path.

The path of motion of a bullet will be parabolic and this motion of bullet is defined as projectile motion. If the force acting on a particle is oblique with initial velocity then the motion of particle is called projectile motion.

3.2 Projectile. A body which is in flight through the atmosphere but is not being propelled by any fuel is called projectile. Example: (i) A bomb released from an aeroplane in level flight (ii) A bullet fired from a gun (iii) An arrow released from bow (iv) A Javelin thrown by an athlete

3.3 Assumptions of Projectile Motion. (1) There is no resistance due to air. (2) The effect due to curvature of earth is negligible. (3) The effect due to rotation of earth is negligible. (4) For all points of the trajectory, the acceleration due to gravity ‘g’ is constant in magnitude and direction.

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PHYSICS

by Pradeep Kshetrapal

2 Motion in Two Dimension

3.4 Principles of Physical Independence of Motions. (1) The motion of a projectile is a two-dimensional motion. So, it can be discussed in two parts. Horizontal motion and vertical motion. These two motions take place independent of each other. This is called the principle of physical independence of motions. (2) The velocity of the particle can be resolved into two mutually perpendicular components. Horizontal component and vertical component. (3) The horizontal component remains unchanged throughout the flight. The force of gravity continuously affects the vertical component. (4) The horizontal motion is a uniform motion and the vertical motion is a uniformly accelerated retarded motion.

3.5 Types of Projectile Motion. (1) Oblique projectile motion plane

(2) Horizontal projectile motion X

Y

(3) Projectile motion on an inclined Y X





X Y

3.6 Oblique Projectile. In projectile motion, horizontal component of velocity (u cos), acceleration (g) and mechanical energy remains constant while, speed, velocity, vertical component of velocity (u sin ), momentum, kinetic energy and potential energy all changes. Velocity, and KE are maximum at the point of projection while minimum (but not zero) at highest point. (1) Equation of trajectory : A projectile thrown with velocity u at an angle  with the horizontal. The velocity u can be resolved into two rectangular components. v cos  component along X–axis and u sin  component along Y–axis. For horizontal motion x = u cos  t  t  For vertical motion

y  (u sin ) t 

1 2 gt 2

 x From equation (i) and (ii) y  u sin    u cos  y  x tan  

x …. (i) u cos 

x u sin 

u

P y



O u cos 

…. (ii)  1  x2   g 2  2  2  u cos 

Y

X

   

1 gx 2 2 u 2 cos 2 

This equation shows that the trajectory of projectile is parabolic because it is similar to equation of parabola y = ax – bx2

Note

:  Equation of oblique projectile also can be written as x  y  x tan  1   R 

(where R = horizontal range =

u 2 sin 2 ) g

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PHYSICS

by Pradeep Kshetrapal Motion in Two Dimension 3

Sample problems based on trajectory Problem 1.

The trajectory of a projectile is represented by y  3 x  gx 2 /2 . The angle of projection is (a) 30o

Solution : (c)

(b) 45o

(c) 60o

(d) None of these

By comparing the coefficient of x in given equation with standard equation y  x tan  

gx 2 2u 2 cos 2 

tan   3   60 

Problem 2.

The path followed by a body projected along y-axis is given as by y  3 x  (1 / 2)x 2 , if g = 10 m/s, then the initial velocity of projectile will be – (x and y are in m) (a) 3 10 m/s

Solution : (b)

(b) 2 10 m/s

(c) 10 3 m/s

(d) 10 2 m/s

By comparing the coefficient of x2 in given equation with standard equation y  x tan  

g 2u cos  2

2



gx 2 2u 2 cos 2 

.

1 2

Substituting  = 60o we get u  2 10 m / sec .

Problem 3.

The equation of projectile is y  16 x  (a) 16 m

Solution : (d)

(b) 8 m

5x 2 . The horizontal range is 4 (c) 3.2 m

(d) 12.8 m

x  Standard equation of projectile motion y  x tan  1   R  Given equation : y  16 x 

x  5x 2  or y  16 x 1  64 / 5  4 

By comparing above equations R 

64 =12.8 m. 5

 (2) Displacement of projectile ( r ) : Let the particle acquires a position P having the coordinates (x, y)  just after time t from the instant of projection. The corresponding position vector of the particle at time t is r as shown in the figure.  r  xˆi  yˆj ….(i) Y

The horizontal distance covered during time t is given as x  v x t  x  u cos  t

….(ii)

The vertical velocity of the particle at time t is given as v y  (v 0 )y  gt,

….(iii)

Now the vertical displacement y is given as y  u sin  t  1 / 2 gt 2

 v

 r

y 



O

x

P (x, y)  vx 

 vy

X vi

….(iv)

Putting the values of x and y from equation (ii) and equation (iv) in equation (i) we obtain the position vector at any time t as  1   r  (u cos  ) t ˆi   (u sin  ) t  gt 2  ˆj 2  

 r

1   (u t cos  )2   (u t sin  )  gt 2  2  

2  ut sin  1 / 2 gt 2 gt sin   gt  and   tan 1 (y / x )  tan 1  r ut 1    u (u t cos )  2u  

2

  2u sin   gt   or   tan 1   2u cos      

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PHYSICS

by Pradeep Kshetrapal

4 Motion in Two Dimension

Note : 

The angle of elevation  of the highest point of the projectile

Y

and the angle of projection  are related to each other as

tan  

1 tan  2

u H

 

O

X

R

Sample problems based on displacement Problem 4.

A body of mass 2 kg has an initial velocity of 3 m/s along OE and it is subjected to a force of 4 Newton’s in OF direction perpendicular to OE. The distance of the body from O after 4 seconds will be (a) 12 m

Solution : (c)

(b) 28 m

(c) 20 m

(d) 48 m

Body moves horizontally with constant initial velocity 3 m/s upto 4 seconds  x  ut  3  4  12 m and in perpendicular direction it moves under the effect of constant force with zero initial velocity upto 4 seconds. 

y  ut 

1 1F 1 4 (a) t 2  0    t 2    4 2  16 m 2 2 m  22

So its distance from O is given by d  

Problem 5.

2

y

u

E d

x

 (12)  (16) 2

2

F

d  20 m

A body starts from the origin with an acceleration of 6 m/s2 along the x-axis and 8 m/s2 along the y-axis. Its distance from the origin after 4 seconds will be [MP PMT 1999] (a) 56 m

Solution : (c)

x y 2

O

(b) 64 m

(c) 80 m

Displacement along X- axis : x  u x t 

1 1 a x t 2   6  (4 ) 2  48 m 2 2

Displacement along Y- axis : y  u y t 

1 1 a y t 2   8  (4 ) 2  64 m 2 2

Total distance from the origin 

(d) 128 m

x 2  y 2  (48)2  (64 )2  80 m

(3) Instantaneous velocity v : In projectile motion, vertical component of velocity changes but horizontal component of velocity remains always constant. Example : When a man jumps over the hurdle leaving behind its skateboard then vertical component of his velocity is changing, but not the horizontal component, which matches with the skateboard velocity. As a result, the skateboard stays underneath him, allowing him to land on it.

Let vi be the instantaneous velocity of projectile at time t direction of this velocity is along the tangent to the trajectory at point P.  v i  v x i  v y ˆj  v i  v x2  v y2  u 2 cos 2   (u sin  gt)2

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PHYSICS

by Pradeep Kshetrapal Motion in Two Dimension 5

v i  u 2  g 2 t 2  2u gt sin Direction of instantaneous velocity tan  

vy vx



u sin   gt or u cos 

 

  tan 1 tan  

gt  sec   u 

(4) Change in velocity : Initial velocity (at projection point) u i  u cos  ˆi  u sin  ˆj Final velocity (at highest point) u f  u cos  ˆi  0 ˆj (i) Change in velocity (Between projection point and highest point) u  u f  u i  u sin  ˆj When body reaches the ground after completing its motion then final velocity u f  u cos  ˆi  u sin  ˆj (ii) Change in velocity (Between complete projectile motion) u  u f  ui  2u sin ˆi

Sample problems based on velocity Problem 6.

In a projectile motion, velocity at maximum height is (a)

u cos  2

(b) u cos 

(c)

u sin  2

(d) None of these

Solution : (b)

In a projectile motion at maximum height body possess only horizontal component of velocity i.e. u cos.

Problem 7.

A body is thrown at angle 30o to the horizontal with the velocity of 30 m/s. After 1 sec, its velocity will be (in m/s) (g = 10 m/s2) (a) 10 7

Solution : (a)

(b) 700 10

(c) 100 7

(d)

40

From the formula of instantaneous velocity v  u 2  g 2 t 2  2 u g t sin 

v  (30)2  (10)2  1 2  2  30  10  1  sin 30 o  10 7 m / s

Problem 8.

A projectile is fired at 30o to the horizontal. The vertical component of its velocity is 80 ms–1. Its time of flight is T. What will be the velocity of the projectile at t = T/2 (a) 80 ms–1

Solution : (b)

(b) 80 3 ms–1

(d) 40 ms–1

At half of the time of flight, the position of the projectile will be at the highest point of the parabola and at that position particle possess horizontal component of velocity only. Given uvertical  u sin  80  u  

Problem 9.

(c) (80/ 3 ) ms–1

80 sin 30 o

 160 m / s

u horizontal  u cos   160 cos 30 o  80 3 m / s.

A particle is projected from point O with velocity u in a direction making an angle  with the horizontal. At any instant its position is at point P at right angles to the initial direction of projection. Its velocity at point P is 90o P

(a) u tan

u cot

(c) u cosec

(d) u sec

Horizontal velocity at point ' O'  u cos 

O

Horizontal velocity at point ' P'  v sin In projectile motion horizontal component of velocity remains constant throughout the motion

 90o

v sin

v

 O

P

90o – 

u

u sin

Solution : (b)

(b)

v

u

u cos

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PHYSICS

by Pradeep Kshetrapal

6 Motion in Two Dimension  v sin  u cos   v  u cot 

Problem 10. A particle P is projected with velocity u1 at an angle of 30o with the horizontal. Another particle Q is

thrown vertically upwards with velocity u2 from a point vertically below the highest point of path of P. The necessary condition for the two particles to collide at the highest point is (a) u1  u 2 (b) u1  2u 2 (c) u 1 

u2 2

u1

Q

P

(d) u1  4u 2 Solution : (b)

u2

30o

Both particle collide at the highest point it means the vertical distance travelled by both the particle will be equal, i.e. the vertical component of velocity of both particle will be equal u1  u2 2

u1 sin 30  u 2 

 u1  2u 2

Problem 11. Two seconds after projection a projectile is travelling in a direction inclined at 30o to the horizontal after one more sec, it is travelling horizontally, the magnitude and direction of its velocity are (a) 2 20 m /sec, 60 o Solution : (b)

(b) 20 3 m /sec, 60 o

(c) 6 40 m /sec, 30 o

(d) 40 6 m /sec, 30 o

Let in 2 sec body reaches upto point A and after one more sec upto point B. Total time of ascent for a body is given 3 sec i.e. t 

 u sin   10  3  30

u sin  3 g

v

…..(i)

A u

Horizontal component of velocity remains always constant

u cos  v cos 30

…..(ii)



For vertical upward motion between point O and A v sin 30

o

 u sin   g  2

v sin 30 o  30  20

B

30o

O

u cos

Using v  u  g t As u sin   30 

 v  20 m / s. Substituting this value in equation (ii)

u cos   20 cos 30 o  10 3 …..(iii)

From equation (i) and (iii) u  20 3 and   60 

Problem 12. A body is projected up a smooth inclined plane (length = 20 2 m ) with velocity u from the point M as shown in the figure. The angle of inclination is 45o and the top is connected to a well of diameter 40 m. If the body just manages to cross the well, what is the value of v (a) 40 ms 1 45o

(b) 40 2 ms 1 (c) 20 ms 1 (d) 20 2 ms Solution : (d)

M 40 m

1

At point N angle of projection of the body will be 45°. Let velocity of projection at this point is v. If the body just manages to cross the well then Range  Diameter of well

v 2 sin 2  40 g

v 2  400



As 

 45 

v N R

v  20 m / s

45o

u M

40 m

genius

PHYSICS

by Pradeep Kshetrapal Motion in Two Dimension 7

But we have to calculate the velocity (u) of the body at point M. For motion along the inclined plane (from M to N) Final velocity (v) = 20 m/s, acceleration (a) = – g sin = – g sin 45o, distance of inclined plane (s) = 20 2 m

(20) 2  u 2  2

g

.20 2

[Using v2 = u2 + 2as]

2

u 2  20 2  400  u  20 2 m / s.

Problem 13. A projectile is fired with velocity u making angle  with the horizontal. What is the change in velocity when it is at the highest point (a) u cos Solution : (c)

(b) u

(c) u sin

(d) (u cos – u)

Since horizontal component of velocity remain always constant therefore only vertical component of velocity changes. Initially vertical component u sin Finally it becomes zero. So change in velocity  u sin

(5) Change in momentum : Simply by the multiplication of mass in the above expression of velocity (Article-4).

p  p f  p i   mu sin ˆj

(i) Change in momentum (Between projection point and highest point)

(ii) Change in momentum (For the complete projectile motion) p  p f  p i   2mu sin ˆj (6) Angular momentum : Angular momentum of projectile at highest point of trajectory about the point of projection is given by

L  mvr

Y

 u 2 sin 2   Here r  H   2 g  

P = mv u

u 2 sin 2  m u 3 cos  sin 2  L  m u cos   2g 2g



r



X

O

Sample problems based on momentum and angular momentum Problem 14. A body of mass 0.5 kg is projected under gravity with a speed of 98 m/s at an angle of 30o with the horizontal. The change in momentum (in magnitude) of the body is (a) 24.5 N–s Solution : (b)

(b) 49.0 N–s

(c) 98.0 N–s

(d) 50.0 N–s

Change in momentum between complete projectile motion = 2mu sin  2  0.5  98  sin 30 = 49 N–s.

Problem 15. A particle of mass 100 g is fired with a velocity 20 m sec–1 making an angle of 30o with the horizontal. When it rises to the highest point of its path then the change in its momentum is (a) Solution : (d)

3 kg m sec 1

(b) 1/2 kg m sec–1

(c)

2 kg m sec 1

(d) 1 kg m sec–1

Horizontal momentum remains always constant  So change in vertical momentum ( p ) = Final vertical momentum – Initial vertical momentum  0  mu sin | P |  0 .1  20  sin 30 o  1 kg m / sec .

Problem 16. Two equal masses (m) are projected at the same angle () from two points separated by their range with equal velocities (v). The momentum at the point of their collision is

genius

PHYSICS

by Pradeep Kshetrapal

8 Motion in Two Dimension (b) 2 mv cos

(a) Zero Solution : (a)

(c) – 2 mv cos

(d) None of these

Both masses will collide at the highest point of their trajectory with equal and opposite momentum. So net momentum of the system will be zero. mv cos

mv cos

v

v





Problem 17. A particle of mass m is projected with velocity v making an angle of 45o with the horizontal. The magnitude of the angular momentum of the particle about the point of projection when the particle is at its maximum height is (where g = acceleration due to gravity) (a) Zero Solution : (b)

L

(b) mv3/ (4 2 g)

m u 3 cos  sin 2  mv 3 = 2g (4 2 g)

(c) mv3/ ( 2 g)

(d) mv2/2g

[As  = 45o]

Problem 18. A body is projected from the ground with some angle to the horizontal. What happens to the angular momentum about the initial position in this motion (a) Decreases

(b) Increases

(c) Remains same

(d) First increases and then decreases

Solution : (b)

Problem 19. In case of a projectile, where is the angular momentum minimum (a) At the starting point (b) At the highest point (c) On return to the ground (d) At some location other than those mentioned above Solution : (a)

(7) Time of flight : The total time taken by the projectile to go up and come down to the same level from which it was projected is called time of flight. For vertical upward motion 0 = u sin  – gt  t = (u sin /g) Now as time taken to go up is equal to the time taken to come down so Time of flight T  2 t 

2u sin  g

(i) Time of flight can also be expressed as : T 

2 .u y g

(where uy is the vertical component of initial velocity).

(ii) For complementary angles of projection  and 90o –  (a) Ratio of time of flight =

T1 T 2u sin  / g  = tan   1  tan  T2 2u sin(90   ) / g T2

(b) Multiplication of time of flight = T1 T2 

2u sin  2u cos  2R  T1 T2  g g g

(iii) If t1 is the time taken by projectile to rise upto point p and t2 is the time taken in falling from point p 2u sin   time of flight to ground level then t1  t 2  g or

u sin 

g(t1  t 2 ) 2

Y t1

P h

O

t2 X

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PHYSICS

by Pradeep Kshetrapal Motion in Two Dimension 9

1 2 gt1 2

and height of the point p is given by h  u sin t1 

(t1  t 2 ) 1 t1  gt12 2 2 g t1 t 2 by solving h  2 (iv) If B and C are at the same level on trajectory and the time difference between these two points is t1, similarly A and D are also at the same level and the time difference between these two positions is t2 then hg

t 22  t12 

8h g

Y

h

B A

t1 t2

C D

O

X

Sample problems based on time of flight Problem 20. For a given velocity, a projectile has the same range R for two angles of projection if t1 and t2 are the times of flight in the two cases then (a) t1t 2  R 2 Solution : (b)

(c) t1t 2 

(b) t1t 2  R

As we know for complementary angles t1 t 2 

1 R

(d) t1t 2 

1 R2

2R  t1 t 2  R . g

Problem 21. A body is thrown with a velocity of 9.8 m/s making an angle of 30o with the horizontal. It will hit the ground after a time (a) 1.5 s Solution : (b)

T 

[JIPMER 2001, 2002; KCET (Engg.) 2001]

(b) 1 s

(c) 3 s

(d) 2 s

2u sin  2  9 .8  sin 30 = 1sec  g 9 .8 o

Problem 22. Two particles are separated at a horizontal distance x as shown in figure. They are projected at the same time as shown in figure with different initial speed. The time after which the horizontal distance between the particles become zero is u

u/ 3 60o

30o A

(a) u /2 x Solution : (b)

B

x

(b) x/u

(c) 2u/x

(d) u/x

Let x 1 and x 2 are the horizontal distances travelled by particle A and B respectively in time t.

x1 

u

. cos 30   t

3

x1  x 2 

u

…..(i)

and

x 2  u cos 60 o  t

……(ii)

. cos 30 o  t  u cos 60 o  t  ut  x  ut  t  x / u

3

Problem 23. A particle is projected from a point O with a velocity u in a direction making an angle  upward with the horizontal. After some time at point P it is moving at right angle with its initial direction of projection. The time of flight from O to P is

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PHYSICS

by Pradeep Kshetrapal

10 Motion in Two Dimension

(a) Solution : (b)

u sin  g

(b)

u cosec  g

(c)

u tan  g

(d)

u sec  g

When body projected with initial velocity u by making angle  with the horizontal. Then after time t, (at point P) it’s direction is perpendicular to u . Magnitude of velocity at point P is given by v  u cot . (from sample problem no. 9) For vertical motion : Initial velocity (at point O)  u sin Final velocity (at point P)  v cos  u cot  cos 

90o u sin

Time of flight (from point O to P) = t Applying first equation of motion v  u  g t

 u cot  cos   u sin   g t  t



O



P

u u cos

v cos

(90 – ) v



u sin   u cot  cos  u cosec  u  sin 2   cos 2   g g sin  g

Problem 24. A ball is projected upwards from the top of tower with a velocity 50 ms –1 making angle 30o with the horizontal. The height of the tower is 70 m. After how many seconds from the instant of throwing will the ball reach the ground (a) 2.33 sec Solution : (c)

(b) 5.33 sec

(c) 6.33 sec

(d) 9.33 sec

Formula for calculation of time to reach the body on the ground from the tower of height ‘h’ (If it is thrown vertically up with velocity u) is given by u sin30o u u 2 gh  t  1  1  2  30o g  u  So we can resolve the given velocity in vertical direction and can apply the above formula.

70 m

Initial vertical component of velocity u sin   50 sin 30  25 m / s. 

t

25  2  9 . 8  70  1  1   = 6.33 sec. 9 . 8  (25 ) 2 

Problem 25. If for a given angle of projection, the horizontal range is doubled, the time of flight becomes (a) 4 times Solution : (c)

R

(b) 2 times

(c)

2 times

(d) 1 / 2 times

u sin 2 2u sin  and T  g g 2

 R  u 2 and T  u (If  and g are constant).

In the given condition to make range double, velocity must be increased upto value. So automatically time of flight will becomes

2 times that of previous

2 times.

Problem 26. A particle is thrown with velocity u at an angle  from the horizontal. Another particle is thrown with the same velocity at an angle  from the vertical. The ratio of times of flight of two particles will be (a) tan 2  : 1 Solution : (c)

(b) cot 2  : 1

(c) tan  : 1

For first particles angle of projection from the horizontal is . So T1 

(d) cot  : 1

2u sin  g

For second particle angle of projection from the vertical is . it mean from the horizontal is (90  ).

 T2 

T 2u sin (90   ) 2u cos  . So ratio of time of flight 1  tan  .  T2 g g

Problem 27. The friction of the air causes vertical retardation equal to one tenth of the acceleration due to gravity (Take g = 10 ms–2). The time of flight will be decreased by (a) 0%

(b) 1%

(c) 9%

(d) 11%

genius

PHYSICS

by Pradeep Kshetrapal Motion in Two Dimension 11

Solution : (c)

T g 2u sin   1  2  T T g1 g 2

g 10  11 g 10

g

Fractional decrease in time of flight 

T1  T2 1  T1 11

Percentage decrease = 9%

(8) Horizontal range : It is the horizontal distance travelled by a body during the time of flight. So by using second equation of motion u 2 sin 2   u cos   ( 2 u sin  / g ) R  u cos  T g u 2 sin 2 R g

Y

u

 O

Horizontal range

(i) Range of projectile can also be expressed as : R = u cos × T = u cos   R

2u x u y g

X

2u sin  2 u cos  u sin  2u x u y   g g g

(where ux and uy are the horizontal and vertical component of initial velocity)

(ii) If angle of projection is changed from  to   = (90 – ) then range remains unchanged.

60o 30o

R' 

Blast

u 2 sin 2 ' u 2 sin[ 2(90 o   )] u 2 sin 2   R g g g

So a projectile has same range at angles of projection  and (90 – ), though time of flight, maximum height and trajectories are different. These angles  and 90o –  are called complementary angles of projection and for complementary angles R R u 2 sin 2 / g 1  1 1 of projection ratio of range 1  2 R 2 u sin [2 (90 o   )] / g R2 (iii) For angle of projection 1 = (45 – ) and 2 = (45 + ), range will be same and equal to u2 cos 2/g.

1 and 2 are also the complementary angles.

Y

(iv) Maximum range : For range to be maximum

d  u 2 sin 2  dR 0  0 d   g d   cos 2 = 0 i.e. 2 = 90o   = 45o and Rmax = (u2/g)

u

H

45o

O

Rmax = 4 H

X

genius

PHYSICS

by Pradeep Kshetrapal

12 Motion in Two Dimension i.e., a projectile will have maximum range when it is projected at an angle of 45 o to the horizontal and the maximum range will be (u2/g). When the range is maximum, the height H reached by the projectile H

u 2 sin 2  u 2 sin 2 45 u 2 R max    2g 2g 4g 4

i.e., if a person can throw a projectile to a maximum distance Rmax, The maximum height to which it will rise is  R m ax   4

  . 

(v) Relation between horizontal range and maximum height : R  R u 2 sin 2 / g  2  4 cot  H u sin 2  / 2 g



u 2 sin 2 u 2 sin 2  and H  g 2g

 R  4H cot

(vi) If in case of projectile motion range R is n times the maximum height H i.e.

R = nH



u 2 sin 2 u 2 sin 2  n  tan   [4 / n] or   tan 1 [4 / n] g 2g

The angle of projection is given by   tan 1 [4 / n]

Note

:  If R = H then   tan 1 (4 ) or   76 o . If R = 4H then   tan 1 (1) or   45 o .

Sample problem based on horizontal range Problem 28. A boy playing on the roof of a 10m high building throws a ball with a speed of 10 m/s at an angle of 30o with the horizontal. How far from the throwing point will the ball be at the height of 10 m from the 1 3 ground (g = 10 m/s2, sin 30o = , cos 30o  ) [AIEEE 2003] 2 2 (a) 8.66 m Solution : (a)

(b) 5.20 m

(c) 4.33 m

(d) 2.60 m

Simply we have to calculate the range of projectile

u

u 2 sin 2 (10 ) 2 sin(2  30 ) R  g 10

30o

R  5 3  8.66 meter

10 m

10 m

Problem 29. Which of the following sets of factors will affect the horizontal distance covered by an athlete in a long– jump event

[AMU (Engg.) 2001]

(a) Speed before he jumps and his weight (b) The direction in which he leaps and the initial speed (c) The force with which he pushes the ground and his speed (d) The direction in which he leaps and the weight Solution : (b)

Because range 

(Velocity of projection ) 2  sin 2(Angle of projection) g

Problem 30. For a projectile, the ratio of maximum height reached to the square of flight time is (g = 10 ms–2) [EAMCET (Med.) 2000]

(a) 5 : 4

(b) 5 : 2

(c) 5 : 1

(d) 10 : 1

genius

PHYSICS

by Pradeep Kshetrapal Motion in Two Dimension 13

Solution : (a)

H 

g 10 5 H u 2 sin 2  2u sin  u 2 sin 2  / 2 g    and T   2  2 2 2 8 8 4 2g g T 4 u sin  / g

Problem 31. A cricketer can throw a ball to a maximum horizontal distance of 100 m. The speed with which he throws the ball is (to the nearest integer) (a) 30 ms–1 Solution : (c)

R max 

(b) 42 ms–1

u2 = 100 g

 u  1000

(c) 32 ms–1

(d) 35 ms–1

(when   45  )

 31.62 m / s.

Problem 32. If two bodies are projected at 30o and 60o respectively, with the same velocity, then [CBSE PMT 2000; JIPMER 2002]

Solution : (a)

(a) Their ranges are same

(b) Their heights are same

(c) Their times of flight are same

(d) All of these

Because these are complementary angles.

Problem 33. Figure shows four paths for a kicked football. Ignoring the effects of air on the flight, rank the paths according to initial horizontal velocity component, highest first

y

1

2

3

4 x

0

(a) 1, 2, 3, 4 Solution : (d)

(b) 2, 3, 4, 1

(c) 3, 4, 1, 2

(d) 4, 3, 2, 1

Range  horizontal component of velocity. Graph 4 shows maximum range, so football possess maximum horizontal velocity in this case.

Problem 34. Four bodies P, Q, R and S are projected with equal velocities having angles of projection 15 o, 30o, 45o and 60o with the horizontal respectively. The body having shortest range is (a) P Solution : (a)

(b) Q

(c) R

[EAMCET (Engg.) 2000]

(d) S

Range of projectile will be minimum for that angle which is farthest from 45°.

Problem 35. A particle covers 50 m distance when projected with an initial speed. On the same surface it will cover a distance, when projected with double the initial speed (a) 100 m Solution : (c)

R

(b) 150 m

u 2 sin 2 g

(c) 200 m

R2  u2  R 1  u 1

 R  u 2 so

[RPMT 2000]

(d) 250 m

2

2  2u      R 2  4 R1 = 4  50 = 200 m  u  

Problem 36. A bullet is fired from a canon with velocity 500 m/s. If the angle of projection is 15o and g = 10 m/s2. Then the range is (a) 25 × Solution : (b)

103

Range (R) 

[CPMT 1997]

m

(b) 12.5 ×

103

m

(c) 50 ×

102

m

(d) 25 ×

102

m

500  sin 2  15   12500 m  12 .5  10 3 m u 2 sin 2  g 10 2

Problem 37. A projectile thrown with a speed v at an angle  has a range R on the surface of earth. For same v and , its range on the surface of moon will be (a) R/6 Solution : (b)

R

u sin 2 g 2

(b) 6 R

 R 1/ g

(c) R/36

(d) 36 R

genius

PHYSICS

by Pradeep Kshetrapal

14 Motion in Two Dimension 1    g Moon  6 g Earth   

R Moon g  Earth = 6 R Earth g Moon R Moon  6 R Earth  6 R

Problem 38. A projectile is thrown into space so as to have maximum horizontal range R. Taking the point of projection as origin, the co-ordinates of the point where the speed of the particle is minimum are

 R (b)  R,   2

(a) (R, R) Solution : (c)

R R (c)  ,  2 4

For maximum horizontal Range   45

 R (d)  R,   4 Y

From R  4H cot  = 4H [As  = 45o, for maximum range.] Speed of the particle will be minimum at the highest point of parabola.

R/4 45o

So the co-ordinate of the highest point will be (R/2, R/4)

X

R/2

Problem 39. The speed of a projectile at the highest point becomes

1

2 times its initial speed. The horizontal range of the projectile will be (a)

Solution : (a)

u2 g

(b)

u2 2g

(c)

Velocity at the highest point is given by u cos  

u

u2 3g

(d)

u2 4g

(given)   = 45o

2 Horizontal range R 

u 2 sin 2 u 2 sin(2  45 o ) u 2   g g g

Problem 40. A large number of bullets are fired in all directions with same speed u. What is the maximum area on the ground on which these bullets will spread (a)  Solution : (b)

u2 g

(b) 

u4

(c)  2

g2

u4

(d)  2

g2

u2 g2

The maximum area will be equal to area of the circle with radius equal to the maximum range of projectile Maximum area r   R m ax  2

2

u2    g 

2

4     u [As r  R max  u 2 / g for  = 45o]  g2 

Problem 41. A projectile is projected with initial velocity (6ˆi  8 ˆj) m / sec . If g = 10 ms–2, then horizontal range is (a) 4.8 metre Solution : (b)



(b) 9.6 metre



(c) 19.2 metre

(d) 14.0 metre

Initial velocity  6ˆi  8 Jˆ m / s (given) Magnitude of velocity of projection u  u x2  u y2  6 2  8 2 = 10 m/s Angle of projection tan  

uy ux



4 4 3 8   sin   and cos   3 5 5 6

u 2 sin 2 u 2 2 sin  cos   Now horizontal range R   g g

(10 ) 2  2  10

4 3  5 5  9.6 meter

Problem 42. A projectile thrown with an initial speed u and angle of projection 15o to the horizontal has a range R. If the same projectile is thrown at an angle of 45o to the horizontal with speed 2u, its range will be

genius

PHYSICS

by Pradeep Kshetrapal Motion in Two Dimension 15

(a) 12 R Solution : (c)

R

(b) 3 R

(c) 8 R

(d) 4 R

u sin 2  R  u 2 sin 2 g 2

R2  u2    R 1  u 1 

2

 sin 2 2   sin 2 1

  2u    R 2  R 1    u  

2

 sin 90 o   sin 30 o 

   8 R1  

Problem 43. The velocity at the maximum height of a projectile is half of its initial velocity of projection u. Its range on the horizontal plane is (a) Solution : (a)

[MP PET 1993]

2

2

2

(b) u / 3 g

3u / 2 g

2

(c) 3u / 2 g

(d) 3u / g

If the velocity of projection is u then at the highest point body posses only u cos u cos  

u (given) 2

   60 

u 2 sin( 2  60 ) 3 u2  g 2 g

Now R 

Problem 44. A projectile is thrown from a point in a horizontal place such that its horizontal and vertical velocity component are 9.8 m/s and 19.6 m/s respectively. Its horizontal range is (a) 4.9 m Solution : (d)

(b) 9.8 m

We know R 

2u x u y g



(c) 19.6 m

(d) 39.2 m

2  9 . 8  19 . 6  39 .2 m 9 .8

Where u x  horizontal component of initial velocity, u y  vertical component of initial velocity.

Problem 45. A particle is projected with a velocity v such that its range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is (where g is acceleration due to gravity) (a) Solution : (a)

4v 2 5g

4g

(b)

(c)

5v 2

v2 g

(d)

4v 2 5g

We know R  4H cot 

2H  4 H cot 

Range 

 cot  

u . 2 . sin  . cos   g 2

As R  2H given 

1 1 2 ; sin   ; cos   2 5 5

2u 2

2

.

1

5 g

5



4u 2 5g

Problem 46. The range R of projectile is same when its maximum heights are h1 and h2. What is the relation between R and h1 and h2

(a) R  h1 h 2 Solution : (d)

[EAMCET (Med.) 2000]

(b) R  2 h1 h 2

(c) R  2 h1 h 2

(d) R  4 h1 h 2

For equal ranges body should be projected with angle  or (90 o   ) from the horizontal. And for these angles : h1 

u 2 cos 2  u 2 sin 2  and h 2  2g 2g

by multiplication of both height : h1 h 2 

u 2 sin 2 cos 2 4g2



1 16

 u 2 sin 2    g 

   

2

 16 h1 h 2  R 2  R  4 h1 h 2

Problem 47. A grasshopper can jump maximum distance 1.6 m. It spends negligible time on the ground. How far can it go in 10 seconds

genius

PHYSICS

by Pradeep Kshetrapal

16 Motion in Two Dimension (a) 5 2 m Solution : (c)

(b) 10 2 m

(c) 20 2 m

(d) 40 2 m

Horizontal distance travelled by grasshopper will be maximum for   45

u2  1 .6 m g

R max 

 u  4 m / s.

Horizontal component of velocity of grasshopper u cos   4 cos 45  2 2 m / s

1.6 m

Total distance covered by it in10 sec. S  u cos   t  2 2  10  20 2 m

Problem 48. A projectile is thrown with an initial velocity of v  aˆi  bˆj, if the range of projectile is double the maximum height reached by it then (a) a = 2b Solution : (c)

(b) b = a

Angle of projection   tan 1

vy vx

 tan 1

(c) b = 2a

b a

From formula R  4 H cot   2H  cot   From equation (i) and (ii)

 tan  

b a

1  tan   2 2

(d) b = 4a …(i) …(ii)

[As R = 2H given]

b = 2a

(9) Maximum height : It is the maximum height from the point of projection, a projectile can reach. So, by using v 2  u 2  2as

Y

0  (u sin  )  2 gH

H 

u

u 2 sin 2  2g

 O

(i) Maximum height can also be expressed as H 

(ii) H max 

u y2

X

(where u y is the vertical component of initial velocity).

2g

u2 2g

Max. height

2

(when sin2 = max = 1 i.e.,  = 90o)

i.e., for maximum height body should be projected vertically upward. So it falls back to the point of projection after reaching the maximum height. (iii) For complementary angles of projection  and 90o –  Ratio of maximum height = 

H1 sin 2  u 2 sin 2  / 2 g   tan 2   2 2 o 2 H 2 u sin (90   )2 g cos 

H1  tan 2  H2

Sample problem based on maximum height Problem 49. A cricketer can throw a ball to a maximum horizontal distance of 100 m. With the same effort, he throws the ball vertically upwards. The maximum height attained by the ball is (a) 100 m Solution : (d)

Rmax 

2

u  100 m g

(b) 80 m (when   45  )

(c) 60 m

[UPSEAT 2002]

(d) 50 m

genius

PHYSICS

by Pradeep Kshetrapal Motion in Two Dimension 17

u 2  100  10  1000

H max 

1000 u2   50 metre . (when   90  ) 2  10 2g

Problem 50. A ball thrown by one player reaches the other in 2 sec. the maximum height attained by the ball above the point of projection will be about (a) 10 m Solution : (c)

T

(b) 7.5 m

2u sin   2 sec g H 

Now

[Pb. PMT 2002]

(given)

(c) 5 m

(d) 2.5 m

u sin   10

u 2 sin 2  (10 ) 2   5 m. 2g 2  10

Problem 51. Two stones are projected with the same magnitude of velocity, but making different angles with

horizontal. The angle of projection of one is /3 and its maximum height is Y, the maximum height attained by the other stone with as /6 angle of projection is (a) Y

Solution : (d)

(b) 2 Y

(c) 3 Y

(d)

Y 3

When two stones are projected with same velocity then for complementary angles  and (90o –  ) Ratio of maximum heights :

H H1  Y  tan 2   tan 2  3  H 2  1  3 3 3 H2

Problem 52. If the initial velocity of a projectile be doubled. Keeping the angle of projection same, the maximum height reached by it will

(a) Remain the same Solution : (c)

H

u 2 sin 2 2g

(b) Be doubled

(c) Be quadrupled

(d) Be halved

[As   constant]

H  u 2

If initial velocity of a projectile be doubled then H will becomes 4 times.

Problem 53. Pankaj and Sudhir are playing with two different balls of masses m and 2m respectively. If Pankaj throws his ball vertically up and Sudhir at an angle , both of them stay in our view for the same period. The height attained by the two balls are in the ratio (a) 2 : 1 Solution : (b)

(c) 1 : cos

(b) 1 : 1

Time of flight for the ball thrown by Pankaj T1 

2u1 g

Height of the ball thrown by Sudhir H 2  H1 u 2 / 2g  2 1 2 =1 H 2 u 2 cos  / 2 g

u 12 2g u 22



u2 90o – 

m

2u1 2u 2 cos   u1  u 2 cos   g g

Height of the ball thrown by Pankaj H 1 



u1

2u 2 sin(90 o   ) 2u 2 cos   g g

Time of flight for the ball thrown by Sudhir T2  According to problem T1  T2 

(d) 1 : sec

2m

Short Trick : Maximum height H  T2

sin (90   )  2g 2

[As u1  u 2 cos  ]

o

u 22

cos  2g 2

H 1  T1  H 2  T 2 H  1 1 H2

  

2

(As T1 = T2)

Problem 54. A boy aims a gun at a bird from a point, at a horizontal distance of 100 m. If the gun can impart a velocity

of 500 ms–1 to the bullet. At what height above the bird must he aim his gun in order to hit it (take g = 10 ms–2)

genius

PHYSICS

by Pradeep Kshetrapal

18 Motion in Two Dimension [CPMT 1996]

(a) 20 cm Solution : (a)

(b) 10 cm

(c) 50 cm

(d) 100 cm

Time taken by bullet to travel a horizontal distance of 100 m is given by t 

100 1  sec 500 5

In this time the bullet also moves downward due to gravity its vertical displacement 2

h

1 2 1 1 g t   10     1 / 5 m = 20 cm 2 2 5

So bullet should be fired aiming 20 cm above the bird to hit it.

Problem 55. The maximum horizontal range of a projectile is 400 m. The maximum height attained by it will be (a) 100 m Solution : (a)

(b) 200 m

(c) 400 m

(d) 800 m

R max  400 m [when   45  ] So from the Relation R  4H cot   400  4 H cot 45  H  100 m.

Problem 56. Two bodies are projected with the same velocity. If one is projected at an angle of 30 o and the other at an angle of 60o to the horizontal, the ratio of the maximum heights reached is

[EAMCET (Med.) 1995; Pb. PMT 2000; AIIMS 2001]

(a) 3 : 1 Solution : (b)

(b) 1 : 3

(c) 1 : 2

(d) 2 : 1

H1 sin 2  1 sin 2 30 o 1    2 2 o H2 3 sin  2 sin 60

Problem 57. If time of flight of a projectile is 10 seconds. Range is 500 m. The maximum height attained by it will be [RPMT 1997; RPET 1998]

(a) 125 m Solution : (a)

(b) 50 m

(c) 100 m

(d) 150 m

2u sin  u 2 sin 2  (50 )2   125 m .  10 sec  u sin   50 so H  2g 2  10 g

T 

Problem 58. A man can throw a stone 80 m. The maximum height to which he can raise the stone is (a) 10 m Solution : (d)

(b) 15 m

(c) 30 m

(d) 40 m

The problem is different from problem no. (54). In that problem for a given angle of projection range was given and we had find maximum height for that angle. But in this problem angle of projection can vary, R max 

u2  80 m [for   45  ] g

But height can be maximum when body projected vertically up H max 

u2 1 u2 u 2 sin 2 90 o    2 g 2  g 2g

  = 40  

m

Problem 59. A ball is thrown at different angles with the same speed u and from the same points and it has same range in both the cases. If y1 and y2 be the heights attained in the two cases, then y1  y 2  (a) Solution : (c)

u2 g

(b)

2u 2 g

(c)

u2 2g

(d)

u2 4g

Same ranges can be obtained for complementary angles i.e.  and 90o – 

y1 

u 2 sin 2  u 2 cos 2  and y 2  2g 2g

 y1  y 2 

u 2 sin 2  u 2 cos 2  u 2   2g 2g 2g

(10) Projectile passing through two different points on same height at time t1 and t2 : If the particle passes two points situated at equal height y at t  t1 and t  t 2 , then (i) Height (y): y  u sin  t1 

1 2 gt1 2

.....(i)

Y

u

t = t1

t = t2

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PHYSICS

by Pradeep Kshetrapal Motion in Two Dimension 19

y  u sin  t 2 

1 2 .....(ii) gt2 2 Comparing equation (i) with equation (ii) and

gt1  t 2  2 Substituting this value in equation (i) u sin 

1  t  t2  y  g 1  t1  gt 12 2  2 

(ii) Time (t1 and t2): y  u sin t 

 y

gt1 t 2 2

1 2 gt 2   2 gy u sin     t 1  1   u sin  g   

2u sin  2y t  t 0 g g 2

  2 gy u sin    t1  1  1   u sin  g   

2   u sin   and t 2    g   

  2 gy   1  1    u sin   

2        

2        

(11) Motion of a projectile as observed from another projectile : Suppose two balls A and B are projected simultaneously from the origin, with initial velocities u1 and u2 at angle 1 and 2, respectively with the horizontal. Y The instantaneous positions of the two balls are given by Ball A :

x1 = (u1 cos1)t

y 1  (u1 sin 1 ) t 

1 2 gt 2

1 2 gt 2 The position of the ball A with respect to ball B is given by Ball B :

x2 = (u2 cos2)t

u1 y1 – y2 u2

y 2  (u 2 sin 2 ) t 

1 2

B X

O

x  x 1  x 2  (u1 cos  1  u 2 cos  2 ) t y  y1  y 2  (u1 sin  1  u 2 sin  2 ) t

Now

y  u 1 sin  1  u 2 sin  2    constant  x  u 1 cos  1  u 2 cos  2 

Thus motion of a projectile relative to another projectile is a straight line. (12) Energy of projectile : When a projectile moves upward its kinetic energy decreases, potential energy increases but the total energy always remain constant. If a body is projected with initial kinetic energy K(=1/2 mu2), with angle of projection  with the horizontal then at the highest point of trajectory

1 1 (i) Kinetic energy  m (u cos )2  mu 2 cos 2  2 2 

K '  K cos  u 2 sin 2  1  mu 2 sin 2  2g 2

u cos K = Kcos2

u K

2

(ii) Potential energy  mgH  mg

Y



2 2   As H  u sin   2g 

X

   

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PHYSICS

by Pradeep Kshetrapal

20 Motion in Two Dimension

(iii) Total energy = Kinetic energy + Potential energy 

1 1 mu 2 cos 2   mu 2 sin 2  2 2

1 mu 2 = Energy at the point of projection. 2 This is in accordance with the law of conservation of energy. =

Sample problems based on energy Problem 60. A projectile is projected with a kinetic energy K. Its range is R. It will have the minimum kinetic energy, Solution : (b)

after covering a horizontal distance equal to [UPSEAT 2002] (a) 0.25 R (b) 0.5 R (c) 0.75 R (d) R Projectile possess minimum kinetic energy at the highest point of the trajectory i.e. at a horizontal distance R / 2.

Problem 61. A projectile is fired at 30o with momentum p. Neglecting friction, the change in kinetic energy when it Solution : (a)

returns to the ground will be (a) Zero (b) 30% (c) 60% (d) 100% According to law of conservation of energy, projectile acquire same kinetic energy when it comes at same level.

Problem 62. A particle is projected making angle 45o with horizontal having kinetic energy K. The kinetic energy at highest point will be (a)

2 Solution : (b)

[CBSE PMT 2000, 01; AIEEE 2002]

K (b) 2

K

(c) 2K

(d) K

Kinetic energy at the highest point K '  K cos 2   K cos 2 45 o  K / 2

Problem 63. Two balls of same mass are projected one vertically upwards and the other at angle 60 o with the vertical.

Solution : (c)

The ratio of their potential energy at the highest point is (a) 3 : 2 (b) 2 : 1 (c) 4 : 1 1 Potential energy at the highest point is given by PE  mu 2 sin 2  2 For first ball   90   (PE)1 

1 mu 2 2

For second ball   (90 o  60 o )  30  from the horizontal  (PE) 2  

PEI (PE)II

(d) 4 : 3

1 1 mu 2 sin 2 30   mu 2 2 8

 4 :1

Problem 64. In the above problem, the kinetic energy at the highest point for the second ball is K. What is the kinetic energy for the first ball (a) 4 K Solution : (d)

(b) 3 K

KE at the highest point KE 

(c) 2 K

(d) Zero

1 mu 2 cos 2  2

For first ball  = 90o  KE = 0

Problem 65. A ball is thrown at an angle  with the horizontal. Its initial kinetic energy is 100 J and it becomes 30 J at the highest point. The angle of projection is (a) 45o Solution : (d)

(b) 30o

(c) cos–1 (3/10)

KE at highest point K '  K cos 2 

30  100 cos 2   cos 2  

 3  3     cos 1   10 10  

(d) cos 1 ( 3 / 10 )

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PHYSICS

by Pradeep Kshetrapal Motion in Two Dimension 21

3.7 Horizontal Projectile. A body be projected horizontally from a certain height ‘y’ vertically above the ground with initial velocity u. If friction is considered to be absent, then there is no other horizontal force which can affect the horizontal motion. The horizontal velocity therefore remains constant and so the object covers equal distance in horizontal direction in equal intervals of time. (1) Trajectory of horizontal projectile : The horizontal displacement x is governed by the equation x = ut  t 

x u

…. (i)

The vertical displacement y is governed by y 

1 2 gt 2

u

O

…. (ii)

X

y x

(since initial vertical velocity is zero) By substituting the value of t in equation (ii) y 

1 gx2 2 u2

P(x, y)

Y

Sample problems based on trajectory Problem 66. An aeroplane is flying at a constant horizontal velocity of 600 km/hr at an elevation of 6 km towards a point directly above the target on the earth’s surface. At an appropriate time, the pilot releases a ball so that it strikes the target at the earth. The ball will appear to be falling (a) On a parabolic path as seen by pilot in the plane (b) Vertically along a straight path as seen by an observer on the ground near the target (c) On a parabolic path as seen by an observer on the ground near the target (d) On a zig-zag path as seen by pilot in the plane

Solution : (c) X

Y

The path of the ball appears parabolic to a observer near the target because it is at rest. But to a Pilot the path appears straight line because the horizontal velocity of aeroplane and the ball are equal, so the relative horizontal displacement is zero.

Problem 67. The barrel of a gun and the target are at the same height. As soon as the gun is fired, the target is also released. In which of the following cases, the bullet will not strike the target

(a) Range of projectile is less than the initial distance between the gun and the target (b) Range of projectile is more than the initial distance between the gun and the target (c) Range of projectile is equal to the initial distance between the gun and target (d) Bullet will always strike the target Solution : (a)

Condition for hitting of bullet with target initial distance between the gun and target  Range of projectile.

Problem 68. A ball rolls off top of a staircase with a horizontal velocity u m/s. If the steps are h metre high and b mere wide, the ball will just hit the edge of nth step if n equals to

genius

PHYSICS

by Pradeep Kshetrapal

22 Motion in Two Dimension

(a) Solution : (c)

hu 2

(b)

gb 2

u28 gb 2

By using equation of trajectory y 

nh 

g(nb ) 2 2u

n 

2

2 hu 2

(c)

gx 2

gb 2

for given condition

2u 2

hb 2

u h

2 hu 2 gb

2u 2 g

(d)

b nh

2

nb

 (2) Displacement of Projectile (r ) : After time t, horizontal displacement x  ut and vertical displacement 1 y  gt 2 . 2

 1 So, the position vector r  ut ˆi  gt 2 ˆj 2 Therefore

 gt  r  ut 1     2u 

2

 gt    2u 

  tan 1 

and

 gy  u   2 

  as t   

  tan 1 

2y g

   

x

(3) Instantaneous velocity : Throughout the motion, the horizontal component of the velocity is vx = u. The vertical component of velocity increases with time and is given by vy = 0 + g t = g t (From v = u + g t)   So, v  v ˆi  v ˆj = v  u ˆi  g t ˆj x

u

O

X



y

 r

P (x,y)

vx



vy

v

Y

y

 gt2

i.e.

v u

Again

 v  uˆi  2 gy ˆj

i.e.

v  u 2  2 gy

2

 gt  u 1    u

2

Direction of instantaneous velocity : tan  

vy vx



 vy    v  x

  t a n1 

 2 gy  tan 1   u 

   

or

 gt   u

  tan 1 

Where  is the angle of instantaneous velocity from the horizontal.

Sample problems based on velocity Problem 69. A body is projected horizontally from the top of a tower with initial velocity 18 ms–1. It hits the ground at angle 45o. What is the vertical component of velocity when it strikes the ground (a) 9 ms–1 Solution : (c)

(b) 9 2 ms–1

When the body strikes the ground

(c) 18 ms–1

(d) 18 u = 18

2 ms–1

genius

PHYSICS

by Pradeep Kshetrapal Motion in Two Dimension 23

tan 45 o 

vy vx



vy 18

=1

v y  18 m / s.

Problem 70. A man standing on the roof of a house of height h throws one particle vertically downwards and another particle horizontally with the same velocity u. The ratio of their velocities when they reach the earth’s surface will be (a) Solution : (c)

2 gh  u 2 : u

(b) 1 : 2

(c) 1 : 1

(d)

For first particle : v 2  u 2  2 gh  v  u 2  2 gh For second particle : v  v x2  v y2  u 2 



2 gh

2

 u 2  2 gh

2 gh

u

u



2 gh  u 2 :

h v

So the ratio of velocities will be 1 : 1.

vx vy

Problem 71. A staircase contains three steps each 10 cm high and 20 cm wide.

u

What should be the minimum horizontal velocity of a ball rolling off the uppermost plane so as to hit directly the lowest plane

v

h b

(a) 0.5 m/s Solution : (c)

(b) 1 m/s

Formula for this condition is given by n   3

2  10  u 2 10  20 2

(c) 2 m/s

(d) 4 m/s

2hu 2 gb 2

 u 2  200 cm / sec  2 m / sec

where h = height of each step, b = width of step, u = horizontal velocity of projection, n = number of step.

(4) Time of flight : If a body is projected horizontally from a height h with velocity u and time taken by the body to reach the ground is T, then

h0 T 

1 gT 2 (for vertical motion) 2

2h g

Sample problems based on time of flight Problem 72. Two bullets are fired simultaneously, horizontally and with different speeds from the same place. Which bullet will hit the ground first (a) The faster one

(b) Depends on their mass

(c) The slower one

(d) Both will reach simultaneously

Solution : (d)

Problem 73. An aeroplane is flying at a height of 1960 m in horizontal direction with a velocity of 360 km/hr. When it is vertically above the point. A on the ground, it drops a bomb. The bomb strikes a point B on the ground, then the time taken by the bomb to reach the ground is (a) 20 2 sec

(b) 20 sec

(c) 10 2 sec

(d) 10 sec

genius

PHYSICS

by Pradeep Kshetrapal

24 Motion in Two Dimension

Solution : (b)

t

2h  g

2  1960 = 20 sec 9.8

(5) Horizontal range : Let R is the horizontal distance travelled by the body

1 R  uT  0 T 2 (for horizontal motion) 2 R u

u h

2h g

Range

Sample problems based on horizontal range Problem 74. A bomb is dropped on an enemy post by an aeroplane flying with a horizontal velocity of 60 km/hr and at a height of 490 m. How far the aeroplane must be from the enemy post at time of dropping the bomb, so that it may directly hit the target. (g = 9.8 m/s2) (a) Solution : (b)

100 m 3

S  ut  u 

(b)

500 m 3

(c)

200 m 3

(d)

400 m 3

500 2h 5 2  490  m  60   3 g 18 9.8

Problem 75. A body is thrown horizontally with velocity

2 gh from the top of a tower of height h. It strikes the level ground through the foot of tower at a distance x from the tower. The value of x is 2h h (a) h (b) (c) 2 h (d) 3 2

Solution : (c)

x u

2h  g

2 gh 

2h g

u 

 x  2h

2 gh

h

x

Problem 76. An aeroplane moving horizontally with a speed of 720 km/h drops a food packet, while flying at a height of 396.9 m. The time taken by a food packet to reach the ground and its horizontal range is (Take g = 9.8 m/sec 2 ) (a) 3 sec and 2000 m Solution : (d)

Time of descent t 

(b) 5 sec and 500 m

2h  g

(c) 8 sec and 1500 m

2  396 .9 9 .8

(d) 9 sec and 1800 m u = 720 km/h

 t  9 sec and horizontal distance S  u  t

 720  5  S    9  1800 m  18 

s

(6) If projectiles A and B are projected horizontally with different initial velocity from same height and third particle C is dropped from same point then (i) All three particles will take equal time to reach the ground. (ii) Their net velocity would be different but all three particle possess same vertical component of velocity. (iii) The trajectory of projectiles A and B will be straight line w.r.t. particle C.

h

C

A

B

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PHYSICS

by Pradeep Kshetrapal Motion in Two Dimension 25

(7) If various particles thrown with same initial velocity but indifferent direction then A u

u A

B

u C

D

E u

u

h A

E

(i) They strike the ground with same speed at different times irrespective of their initial direction of velocities. (ii) Time would be least for particle E which was thrown vertically downward. (iii) Time would be maximum for particle A which was thrown vertically upward.

3.8 Projectile Motion on an Inclined Plane. Let a particle be projected up with a speed u from an inclined plane which makes an angle  with the horizontal velocity of projection makes an angle  with the inclined plane. We have taken reference x-axis in the direction of plane. Y Hence the component of initial velocity parallel and perpendicular  u

to the plane are equal to u cos  and u sin respectively i.e. u||  u cos  and u   u sin  .



The component of g along the plane is g sin  and perpendicular to the plane is g cos as shown in the figure i.e. a||   g sin  and

O t =0

a   g cos  .

Therefore the particle decelerates at a rate of g sin  as it moves from O to P. (1) Time of flight : We know for oblique projectile motion T  or we can say T 

2u sin  g

2u  a

 Time of flight on an inclined plane T 

2u sin  g cos 

(2) Maximum height : We know for oblique projectile motion H  or we can say

H 

P t =T

ax=–g sin 

u 2 2a 

 Maximum height on an inclined plane H 

u 2 sin 2  2 g cos 

(3) Horizontal range : For one dimensional motion s  ut 

1 2 at 2

u 2 sin 2  2g



 g

ay= g cos 

X

genius

PHYSICS

by Pradeep Kshetrapal

26 Motion in Two Dimension

Horizontal range on an inclined plane R  u|| T 

R  u cos T 

1 g sin T 2 2

 2u sin R  u cos   g cos By solving R 

1 a|| T 2 2

 1  2u sin   g sin   2  g cos

  

2

2u 2 sin  cos(   ) g cos 2 

(i) Maximum range occurs when  







4 2 (ii) The maximum range along the inclined plane when the projectile is thrown upwards is given by R max 

u2 g (1  sin  )

(iii) The maximum range along the inclined plane when the projectile is thrown downwards is given by R max 

u2 g (1  sin  )

Sample problem based on inclined projectile Problem 77. For a given velocity of projection from a point on the inclined plane, the maximum range down the plane is three times the maximum range up the incline. Then, the angle of inclination of the inclined plane is (a) 30o (b) 45o (c) 60o (d) 90o

Solution : (a)

Maximum range up the inclined plane (R max )up 

u2 g(1  sin  )

Maximum range down the inclined plane (R max )down  and according to problem :

u2 g(1  sin  )

u2 u2  3 g(1  sin  ) g(1  sin  )

By solving  = 30o Problem 78. A shell is fired from a gun from the bottom of a hill along its slope. The slope of the hill is  = 30o, and the angle of the barrel to the horizontal  = 60o. The initial velocity v of the shell is 21 m/sec. Then distance of point from the gun at which shell will fall (a) 10 m (b) 20 m (c) 30 m (d) 40 m Solution : (c) Here u = 21 m/sec,  = 30o,  =  –  = 60o – 30o = 30o 2u 2 sin  cos(   ) 2  (21) 2  sin 30 o cos 60 o Maximum range R    30 m g cos 2  9 .8  cos 2 30 o

Problem 79. The maximum range of rifle bullet on the horizontal ground is 6 km its maximum range on an inclined of 30o will be (a) 1 km Solution : (c)

(b) 2 km

(c) 4 km u2  6 km (given) Maximum range on horizontal plane R  g Maximum range on a inclined plane R max  Putting  = 30o R m ax 

u2 g (1  sin 30 ) o



u2 g (1  sin  )

2  u 2 3  g

 2    6  4 km .  3 

(d) 6 km

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PHYSICS

by Pradeep Kshetrapal Motion in Two Dimension 27

CIRCULAR MOTION Circular motion is another example of motion in two dimensions. To create circular motion in a body it must be given some initial velocity and a force must then act on the body which is  always directed at right angles to instantaneous velocity. v2  v3

Since this force is always at right angles to the displacement due to the initial velocity therefore no work is done by the force on the particle. Hence, its kinetic energy and thus speed is unaffected. But due to simultaneous action of the force and the velocity the particle follows resultant path, which in this case is a circle. Circular motion can be classified into two types – Uniform circular motion and non-uniform circular motion.

F F

 r 

 v2

 v1 B

 r2 O



 r1

A

B

 r2 O

r  r  r  2r.r cos

F

 v4

(1) Displacement and distance : When particle moves in a circular path describing an angle  during time t (as shown in the figure) from the position  A to the position B, we see that the magnitude of the position vector r   (that is equal to the radius of the circle) remains constant. i.e., r1  r2  r and the direction of the position vector changes from time to time.  (i) Displacement : The change of position vector or the displacement r of the particle from position A to the position B is given by referring the figure.     r  r2  r1    r  r  r2  r1 r  r12  r22  2r1r2 cos 2

F

 v1

3.9 Variables of Circular Motion.

Putting r1  r2  r we obtain



 r



2

 r1

A

  2r 2 1  cos    2r 2  2 sin 2  2 

 r  2r sin



2 (ii) Distance : The distanced covered by the particle during the time t is given as d = length of the arc AB = r  d r   (iii) Ratio of distance and displacement :  cosec( / 2) r 2r sin  / 2 2

Sample problems based on distance and displacement Problem 80. A particle is rotating in a circle of radius r. The distance traversed by it in completing half circle would be

(a) r (b)  r (c) 2r (d) Zero Solution : (b) Distance travelled by particle = Semi-circumference = r. Problem 81. An athlete completes one round of a circular track of radius 10 m in 40 sec. The distance covered by him in 2 min 20 sec is [Kerala PMT 2002] (a) 70 m (b) 140 m (c) 110 m (d) 220 m Total time of motion 140 sec No. of revolution (n)    3 .5 Solution : (d) Time period 40 sec 22  10 = 220 m. Distance covered by an athlete in revolution  n (2 r)  3.5 (2 r)  3 .5  2  7 Problem 82. A wheel covers a distance of 9.5 km in 2000 revolutions. The diameter of the wheel is [RPMT 1999; BHU 2000]

(a) 15 m

(b) 7.5 m

(c) 1.5 m

(d) 7.5 m

genius

PHYSICS

by Pradeep Kshetrapal

28 Motion in Two Dimension

Solution : (c)

Distance  n (2 r)  9.5  10 3  2000  (D)

 D

9.5  10 3  1 .5 m. 2000  

(2) Angular displacement () : The angle turned by a body moving on a circle from some reference line is called angular displacement. (i) Dimension = [M0L0T0] (as  = arc / radius) . (ii) Units = Radian or Degree. It is some times also specified in terms of fraction or multiple of revolution. (iii) 2 rad  360 o  1 Revolution (iv) Angular displacement is a axial vector quantity. Its direction depends upon the sense of rotation of the object and can be given by Right Hand Rule; which states that if the curvature of the fingers of right hand represents the sense of rotation of the object, then the thumb, held perpendicular to the curvature of the fingers, represents the direction of angular displacement vector. S (v) Relation between linear displacement and angular displacement s    r or

O



r

s  r

Sample problem based on angular displacement Problem 83. A flywheel rotates at constant speed of 3000 rpm. The angle described by the shaft in radian in one Solution : (c)

second is (a) 2  (b) 30  (c) 100  Angular speed  3000 rpm  50 rps  50  2 rad / sec  100 rad / sec

(d) 3000 

i.e. angle described by the shaft in one second is 100  rad.

Problem 84. A particle completes 1.5 revolutions in a circular path of radius 2 cm. The angular displacement of the particle will be – (in radian) (a) 6  (b) 3  Solution : (b)

(c) 2 

(d) 

1 revolution mean the angular displacement of 2 rad

1.5 revolution means 1.5  2 = 3  rad .

(3) Angular velocity () : Angular velocity of an object in circular motion is defined as the time rate of change of its angular displacement. (i) Angular velocity  =

angle traced  d   Lt  t 0 t time taken dt

d dt 0 0 (ii) Dimension : [M L T–1] (iii) Units : Radians per second (rad.s–1) or Degree per second. (iv) Angular velocity is an axial vector. 



(v) Relation between angular velocity and linear velocity v    r Its direction is the same as that of . For anticlockwise rotation of the point object on the circular path, the direction of , according to Right hand rule is along the axis of circular path directed upwards. For clockwise rotation of the point object on the circular path, the direction of  is along the axis of circular path directed downwards.

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PHYSICS

by Pradeep Kshetrapal Motion in Two Dimension 29

Note : 

It is important to note that nothing actually moves in the direction of the angular velocity vector

 . The direction of  simply represents that the rotational motion is taking place in a plane perpendicular to it. time.

(vi) For uniform circular motion  remains constant where as for non-uniform motion  varies with respect to

Sample problems based on angular velocity Problem 85. A scooter is going round a circular road of radius 100 m at a speed of 10 m/s. The angular speed of the scooter will be (a) 0.01 rad/s Solution : (b)

 

(b) 0.1 rad/s

(c) 1 rad/s

(d) 10 rad/s

v 10   0 . 1 rad / sec r 100

Problem 86. The ratio of angular velocity of rotation of minute hand of a clock with the angular velocity of rotation of the earth about its own axis is (a) 12 Solution : (c)

(b) 6

 Minute hand 

(c) 24

(d) None of these

 Minute hand 2 rad 2 Rad 2 rad   24 : 1 and  Earth   60 min 24 hr 24  60 min  Earth

Problem 87. A particle P is moving in a circle of radius ‘a’ with a uniform speed v. C is the centre of the circle and AB is

a diameter. When passing through B the angular velocity of P about A and C are in the ratio [NCERT 1982] (a) 1 : 1

Solution : (b)

(b)

1:2

(c) 2 : 1

(d)

4:1

Angular velocity of P about A

A 

v 2a

Angular velocity of P about C

C 

v a



A 1:2 C

A

P

C

B

a

Problem 88. The length of the seconds hand of a watch is 10 mm. What is the change in the angular speed of the watch after 15 seconds

(b) (10 / 2) mms 1

(a) Zero Solution : (a)

(d) 10 2 mms 1

(c) (20 /  ) mms 1

Angular speed of seconds hand of watch is constant and equal to

2 rad   rad / sec . So change in 60 sec 30

angular speed will be zero.

(4) Change in velocity : We want to know the magnitude and direction of the change in velocity of the particle which is performing uniform circular motion as it moves from A to B during time t as shown in figure. The change in velocity vector is given as    v1 v  v 2  v 1   v  v2 2    v1 v  v 2  v 1  v  v12  v 22  2v1 v 2 cos or For uniform circular motion v1  v 2  v

In vector form v    r





So v  2v 2 1  cos   2v sin 2  The direction of v is shown in figure that can be given as 180 o     90 o   / 2 2 Note :  Relation between linear velocity and angular velocity.





B



O



A

  v  v 2  v1

 v 1

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PHYSICS

by Pradeep Kshetrapal

30 Motion in Two Dimension

Sample problems based on velocity Problem 89. If a particle moves in a circle describing equal angles in equal times, its velocity vector [CPMT 1972, 74; JIPMER 1997]

(a) Remains constant

(b) Changes in magnitude

(c) Changes in direction

(d) Changes both in magnitude and direction

Solution : (c) In uniform circular motion velocity vector changes in direction but its magnitude always remains constant. | v 1 | | v 2 | | v 3 | | v 4 | = constant

 v2

 v1

 v3

 v4

Problem 90. A body is whirled in a horizontal circle of radius 20 cm. It has angular velocity of 10 rad/s. What is its linear velocity at any point on circular path (a) 10 m/s Solution : (b)

(b) 2 m/s

[CBSE PMT 1996; JIPMER 2000]

(c) 20 m/s

(d)

2 m/s

v  r   0.2  10  2 m/s

Problem 91. The linear velocity of a point on the equator is nearly (radius of the earth is 6400 km) (a) 800 km/hr Solution : (b)

(b) 1600 km/hr

v  r w  6400 km 

(c) 3200 km/hr

(d) 6400 km/hr

2 rad  1675 km / hr  1600 km / hr 24 hr

Problem 92. A particle moves along a circle with a uniform speed v. After it has made an angle of 60o its speed will be (a) v 2

(b)

v

(c)

v

2 Solution : (d)

(d) v

3

Uniform speed means speed of the particle remains always constant.

Problem 93. A particle is moving along a circular path of radius 2 m and with uniform speed of 5 ms–1. What will be the change in velocity when the particle completes half of the revolution (a) Zero Solution : (b)

(c) 10 2 ms 1

(b) 10 ms–1

 180 o   v  2v sin    2  5 sin  2  2

(d) 10 / 2 ms 1

   

180o

 2  5 sin 90 o  10 m / s

 v2





Problem 94. What is the value of linear velocity, if   3ˆi  4 ˆj  kˆ and r  5ˆi  6 ˆj  6 kˆ (a) 6ˆi  2ˆj  3kˆ

Solution : (b)

(b) 18ˆi  13 ˆj  2kˆ

ˆi    v   r  3

Jˆ 4

kˆ 1

5

6

6

 v1

(c) 4ˆi  13 ˆj  6kˆ

[Pb. PMT 2000]

(d) 6ˆi  2 ˆj  8 kˆ

 v  (24  6) i  (18  5) Jˆ  (18  20 ) kˆ  18 ˆi  13 Jˆ  2kˆ

Problem 95. A particle comes round circle of radius 1 m once. The time taken by it is 10 sec. The average velocity of motion is

(a) 0.2  m/s Solution : (d)

[JIPMER 1999]

(b) 2  m/s

(c) 2 m/s

(d) Zero

In complete revolution total displacement becomes zero. So the average velocity will be zero.

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PHYSICS

by Pradeep Kshetrapal Motion in Two Dimension 31

Problem 96. Two particles of mass M and m are moving in a circle of radii R and r. If their time-periods are same, what will be the ratio of their linear velocities (a) MR : mr Solution : (c)

(b) M : m

r v1  1 1 . v 2 r2 2

[CBSE PMT 2001]

(c) R : r

Time periods are equal i.e.  1   2 

(d) 1 : 1

v1 r R  1  v2 r2 r

(5) Time period (T) : In circular motion, the time period is defined as the time taken by the object to complete one revolution on its circular path. (i) Units : second.

(ii) Dimension : [M0L0T]

(iii) Time period of second’s hand of watch = 60 second. (iv) Time period of minute’s hand of watch = 60 minute (v) Time period of hour’s hand of watch = 12 hour (6) Frequency (n) : In circular motion, the frequency is defined as the number of revolutions completed by the object on its circular path in a unit time. (i) Units : s–1 or hertz (Hz). (ii) Dimension : [M0L0T–1]

Note : 

Relation between time period and frequency : If n is the frequency of revolution of an object in circular motion, then the object completes n revolutions in 1 second. Therefore, the object will complete one revolution in 1/n second. T  1 / n

 Relation between angular velocity, frequency and time period : Consider a point object describing a uniform circular motion with frequency n and time period T. When the object completes one revolution, the angle traced at its axis of circular motion is 2  radians. It means,  2 when time t = T,   2 radians. Hence, angular velocity    ( T = 1/n)  2n t T 2   2n T  If two particles are moving on same circle or different coplanar concentric circles in same direction with different uniform angular speeds A and B respectively, the angular velocity of B relative to A will be

 rel   B   A So the time taken by one to complete one revolution around O with respect to the other (i.e., time in which B complete one revolution around O with respect to the other (i.e., time in which B completes one more or less revolution around O than A) T

2

 rel



T T 2  1 2  2   1 T1  T2

2   as T     

Special case : If  B   A ,  rel  0 and so T = ., particles will maintain their position relative to each other. This is what actually happens in case of geostationary satellite (1 = 2 = constant) (7) Angular acceleration () : Angular acceleration of an object in circular motion is defined as the time rate of change of its angular velocity. (i) If  be the change in angular velocity of the object in time interval t and t + t, while moving on a circular path, then angular acceleration of the object will be d 2  d  2  t 0 t dt dt

  Lt (ii) Units : rad. s–2

(iii) Dimension : [M0L0T–2]



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PHYSICS

by Pradeep Kshetrapal

32 Motion in Two Dimension (iv) Relation between linear acceleration and angular acceleration a    r (v) For uniform circular motion since  is constant so  

d 0 dt

(vi) For non-uniform circular motion   0

Note : 

Relation between linear (tangential) acceleration and angular acceleration a    r

 For uniform circular motion angular acceleration is zero, so tangential acceleration also is equal to zero.  For non-uniform circular motion a  0 (because   0).

Sample problems based on angular acceleration Problem 97. A body is revolving with a uniform speed v in a circle of radius r. The angular acceleration of the body is v (b) Zero r away from the centre

(a)

Solution : (b)

(c)

v2 along the radius and towards the centre r

In uniform circular motion  constant so  

(d)

v2 along the radius and r

d 0 dt

Problem 98. The linear acceleration of a car is 10m/s2. If the wheels of the car have a diameter of 1m, the angular acceleration of the wheels will be (a) 10 rad/sec2 Solution : (b)

(b) 20 rad/sec2

Angular accelerati on 

(c) 1 rad/sec2

(d) 2 rad/sec2

linear accelerati on 10   20 rad / sec 2 radius 0 .5

Problem 99. The angular speed of a motor increases from 600 rpm to 1200 rpm in 10 s. What is the angular acceleration of the motor (a) 600 rad sec 2 Solution : (d)

 

 2  1 t



(b) 60 rad sec 2

(c) 60 rad sec 2

(d) 2 rad sec 2

2 (n 2  n 1 ) 2 (1200  600 ) rad  2 rad / sec 2  t 10  60 sec 2

3.10 Centripetal Acceleration. (1) Acceleration acting on the object undergoing uniform circular motion is called centripetal acceleration. path.

(2) It always acts on the object along the radius towards the centre of the circular (3) Magnitude of centripetal acceleration a 

v 4   2 r  4 n 2 r  2 r r T 2

2

ac

v

(4) Direction of centripetal acceleration : It is always the same as that of  . When  t decreases,  also decreases. Due to which  becomes more and more perpendicular to  . When  t  0,  becomes perpendicular to the velocity vector. As the velocity vector of the particle at an instant acts along

the tangent to the circular path, therefore  and hence the centripetal acceleration vector acts along the radius of the circular path at that point and is directed towards the centre of the circular path.

Sample problems based on centripetal acceleration Problem 100. If a cycle wheel of radius 4 m completes one revolution in two seconds. Then acceleration of the cycle will be

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PHYSICS

by Pradeep Kshetrapal Motion in Two Dimension 33

(a)  2 m / s 2 Solution : (c)

(b) 2 2 m / s 2

(c)

4 2 m / s 2

(d) 8 m /s 2

[Pb. PMT 2001]

Given r  4 m and T  2 seconds.

4 2

 ac 

r

T2

4 2 (2) 2

4  4 2 m / s 2

Problem 101. A stone is tied to one end of a spring 50 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 10 revolutions in 20 s, what is the magnitude of acceleration of the stone (a) 493 Solution : (a)

cm/sec2

Time period 

a c 

4 2

(c) 860

cm/sec2

[Pb. PMT 2000]

(d) 990 cm/sec2

Total time 20   2 sec No. of revolution 10

.r 

T2

(b) 720

cm/sec2

4 2 (2) 2

 (1 / 2) m / s 2  4 .93 m / s 2  493 cm / s 2

Problem 102. A particle moves with a constant speed v along a circular path of radius r and completes the circle in time T. What is the acceleration of the particle (a) mg Solution : (b)

ac 

(b)

2v T

[Orissa JEE 2002]

(c)

r 2 T

(d)

v 2 T

v2  2  2 v   2 r  v = v   T r  T 

Problem 103. If the speed of revolution of a particle on the circumference of a circle and the speed gained in falling through a distance equal to half the radius are equal, then the centripetal acceleration will be (a) Solution : (d)

g 2

(b)

g 4

(c)

g 3

(d) g

Speed gain by body falling through a distance h is equal to v  2 gh  v

gr 

2g

r 2

[As h 

r given] 2

v2 g r

Problem 104. Two cars going round curve with speeds one at 90 km/h and other at 15 km/h. Each car experiences same acceleration. The radii of curves are in the ratio of

Solution : (d)

(a) 4 : 1

(b) 2 : 1

Centripetal acceleration =

v 12 v 22  (given) r1 r2

v r  1   1 r2  v 2

2

[EAMCET (Med.) 1998]

(c) 16 : 1

(d) 36 : 1

 36  90       1  15   2

Problem 105. A wheel of radius 0.20m is accelerated from rest with an angular acceleration of 1 rad / s 2 . After a rotation of 90 o the radial acceleration of a particle on its rim will be (a)  m / s 2 Solution : (d)

(b) 0 .5  m / s 2

(d) 0 .2  m / s 2

(c) 2 . 0 m / s 2

From the equation of motion Angular speed acquired by the wheel,  22   12  2  0  2  1  Now radial acceleration  2 r    0.2  0.2 m / s 2

 2

  22  

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PHYSICS

by Pradeep Kshetrapal

34 Motion in Two Dimension

3.11 Centripetal Force. According to Newton's first law of motion, whenever a body moves in a straight line with uniform velocity, no force is required to maintain this velocity. But when a body moves along a circular path with uniform speed, its direction changes continuously i.e. velocity keeps on changing on account of a change in direction. According to Newton's second law of motion, a change in the direction of motion of the body can take place only if some external force acts on the body. Due to inertia, at every point of the circular path; the body tends to move along the tangent to the circular path at that point (in figure). Since every body has directional inertia, a velocity v cannot change by itself and as such we have to apply a force. But this force should v be such that it changes the direction of velocity and not its magnitude. This is possible only if the force acts perpendicular to the direction of velocity. Because the F velocity is along the tangent, this force must be along the radius (because the F F radius of a circle at any point is perpendicular to the tangent at that point). F Further, as this force is to move the body in a circular path, it must acts towards the centre. This centre-seeking force is called the centripetal force. v Hence, centripetal force is that force which is required to move a body in a circular path with uniform speed. The force acts on the body along the radius and towards centre. (1) Formulae for centripetal force : F 

v

mv 2 m 4 2 r  m  2 r  m 4 2 n 2 r  r T2

(2) Centripetal force in different situation Situation

Centripetal Force

A particle tied to a string and whirled in a horizontal circle

Tension in the string

Vehicle taking a turn on a level road

Frictional force exerted by the road on the tyres

A vehicle on a speed breaker

Weight of the body or a component of weight

Revolution of earth around the sun

Gravitational force exerted by the sun

Electron revolving around the nucleus in an atom

Coulomb attraction exerted by the protons in the nucleus

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PHYSICS

by Pradeep Kshetrapal Motion in Two Dimension 35

A charged particle describing a circular path in a magnetic field

Magnetic force exerted by the agent that sets up the magnetic field

3.12 Centrifugal Force. It is an imaginary force due to incorporated effects of inertia. When a body is rotating in a circular path and the centripetal force vanishes, the body would leave the circular path. To an observer A who is not sharing the motion along the circular path, the body appears to fly off tangential at the point of release. To another observer B, who is sharing the motion along the circular path (i.e., the observer B is also rotating with the body with the same velocity), the body appears to be stationary before it is released. When the body is released, it appears to B, as if it has been thrown off along the radius away from the centre by some force. In reality no force is actually seen to act on the body. In absence of any real force the body tends to continue its motion in a straight line due to its inertia. The observer A easily relates this events to be due to inertia but since the inertia of both the observer B and the body is same, the observer B can not relate the above happening to inertia. When the centripetal force ceases to act on the body, the body leaves its circular path and continues to moves in its straight-line motion but to observer B it appears that a real force has actually acted on the body and is responsible for throwing the body radially out-words. This imaginary force is given a name to explain the effects on inertia to the observer who is sharing the circular motion of the body. This inertial force is called centrifugal force. Thus centrifugal force is a fictitious force which has significance only in a rotating frame of reference.

Sample problems based on centripetal and centrifugal force Problem 106.

A ball of mass 0.1 kg is whirled in a horizontal circle of radius 1 m by means of a string at an initial speed of 10 r.p.m. Keeping the radius constant, the tension in the string is reduced to one quarter of its initial value. The new speed is (a) 5 r.p.m.

Solution : (a)

(b) 10 r.p.m.

(c) 20 r.p.m.

(d) 14 r.p.m.

Tension in the string T  m  r  m 4 n r 2

T  n 2 or n  T 

n2  n1

T2  T1

2

2

[As m and r are constant] n 10 T /4  5 rpm  n2  1  2 2 T

Problem 107. A cylindrical vessel partially filled with water is rotated about its vertical central axis. It’s surface will [RPET 2000]

(a) Rise equally Solution : (b)

(b) Rise from the sides (c) Rise from the middle (d) Lowered equally

Due to the centrifugal force.

Problem 108. A proton of mass 1.6 × 10 –27 kg goes round in a circular orbit of radius 0.10 m under a centripetal force of 4 × 10–13 N. then the frequency of revolution of the proton is about (a) 0.08 × (c) 8 × Solution : (a)

108

108

cycles per sec

(b) 4 ×

cycles per sec

108

(d) 12 ×

[Kerala PMT 2002]

cycles per sec

108

cycles per sec

F  4  10 13 N ; m  1 .6  10 27 kg ; r  0.1 m

Centripetal force F  m 4 2 n 2 r  n 

F 4m  2r

 8  10 6 cycles / sec  0 .08  10 8 cycle / sec .

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PHYSICS

by Pradeep Kshetrapal

36 Motion in Two Dimension

Problem 109. Three identical particles are joined together by a thread as shown in figure. All the three particles are moving in a horizontal plane. If the velocity of the outermost particle is v0, then the ratio of tensions in the three sections of the string is

A

O l

(a) 3 : 5 : 7 Solution : (d)

B l

C l

(b) 3 : 4 : 5

(c) 7 : 11 : 6

(d) 3 : 5 : 6

Let the angular speed of the thread is  For particle ‘C’  T3  m  2 3l O

For particle ‘B’ T2  T3  m  2 2l  T2  m  2 5 l

T1

A

B

T2

T3

C

For particle ‘C’ T1  T2  m  2 l  T1  m  2 6 l  T3 : T2 : T1  3 : 5 : 6

Problem 110. A stone of mass 1 kg tied to the end of a string of length 1 m, is whirled in a horizontal circle with a uniform angular velocity of 2 rad/s. The tension of the string is (in N) (a) 2 Solution : (c)

(b)

1 3

(c) 4

[KCET 1998]

(d)

1 4

T  m  2 r  1  (2) 2  (1)  4 Newton

Problem 111. A cord can bear a maximum force of 100 N without breaking. A body of mass 1 kg tied to one end of a cord of length 1 m is revolved in a horizontal plane. What is the maximum linear speed of the body so that the cord does not break (a) 10 m/s Solution : (a)

(b) 20 m/s

(c) 25 m/s

(d) 30 m/s

mv2 and for critical condition this tension will be r 100  1  v max  10 m / s  100  v m2 ax  1

Tension in cord appears due to centrifugal force T 



equal to breaking force (100 N)

2 m v max r

Problem 112. A mass is supported on a frictionless horizontal surface. It is attached to a string and rotates about a fixed

centre at an angular velocity 0. If the length of the string and angular velocity are doubled, the tension in the string which was initially T0 is now [AIIMS 1985] (a) T0

Solution : (d)

(b) T0/2

T  m 2l 

T2   2  T1   1

  

2

 l2   l1

(c) 4T0

(d) 8T0

 T  2   2l    2       T2  8T0 T0     l   2

Problem 113. A stone is rotated steadily in a horizontal circle with a period T by a string of length l. If the tension in the string is kept constant and l increases by 1%, what is the percentage change in T (a) 1% Solution : (b)

Tension 

(b) 0.5%

m 4 l 2

T2

(c) 2%

 l  T 2 or T  l

Percentage change in Time period  

(d) 0.25%

[Tension and mass are constant]

1 (percentage change in length) 2

[If % change is very small]

1 (1 %)  0.5% 2

Problem 114. If mass speed and radius of rotation of a body moving in a circular path are all increased by 50%, the necessary force required to maintain the body moving in the circular path will have to be increased by (a) 225%

(b) 125%

(c) 150%

(d) 100%

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PHYSICS

by Pradeep Kshetrapal Motion in Two Dimension 37

Solution : (b)

Centripetal force F 

mv2 r

m  v  m   v   2 2  If m, v and r are increased by 50% then let new force F'  r  r   2  Percentage increase in force

2



9 9 m v2  F 4 4 r

F'  F F 500  100   100 %  %  125 % F F 4

Problem 115. Two masses m and M are connected by a light string that passes through a smooth hole O at the centre of a table. Mass m lies on the table and M hangs vertically. m is moved round in a horizontal circle with O as the centre. If l is the length of the string from O to m then the frequency with which m should revolve so that M remains stationary is l

m

O M

(a) Solution : (a)

1 2

Mg m l

(b)

1

Mg ml



(c)

1 2

ml Mg

(d)

1



ml Mg

‘m’ Mass performs uniform circular motion on the table. Let n is the frequency of revolution then centrifugal force  m 4 2 n 2 l For equilibrium this force will be equal to weight Mg

l

m 4 2 n 2 l  Mg n 

1 2

Mg ml

m

M

Problem 116. A particle of mass M moves with constant speed along a circular path of radius r under the action of a force F. Its speed is (a)

Solution : (a)

[MP PMT 2002]

rF m

Centripetal force F 

(b)

F r

m v2 v  r

(c)

Fmr

(d)

F mr

rF m

Problem 117. In an atom for the electron to revolve around the nucleus, the necessary centripetal force is obtained from the following force exerted by the nucleus on the electron (a) Nuclear force

(b) Gravitational force (c) Magnetic force

(d) Electrostatic force

Solution : (d)

Problem 118. A motor cycle driver doubles its velocity when he is having a turn. The force exerted outwardly will be [AFMC 2002]

(a) Double Solution : (c)

F

m v2  F  v 2 or r

(b) Half F2  v 2    F1  v 1 

(c) 4 times 2

(d)

1 times 4

2

 2v      4  F2  4 F1  v 

Problem 119. A bottle of soda water is grasped by the neck and swing briskly in a vertical circle. Near which portion of the bottle do the bubbles collect (a) Near the bottom

(b) In the middle of the bottle

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PHYSICS

by Pradeep Kshetrapal

38 Motion in Two Dimension (c) Near the neck Solution : (c)

(d) Uniformly distributed in the bottle

Due to the lightness of the gas bubble they feel less centrifugal force so they get collect near the neck of the bottle. They collect near the centre of circular motion i.e. near the neck of the bottle.

Problem 120. A body is performing circular motion. An observer O1 is sitting at the centre of the circle and another observer O 2 is sitting on the body. The centrifugal force is experienced by the observer (a) O1 only Solution : (b)

(b) O 2 only

(c) Both by O1 and O 2

(d) None of these

Centrifugal force is a pseudo force, which is experienced only by that observer who is attached with the body performing circular motion.

3.13 Work done by Centripetal Force. The work done by centripetal force is always zero as it is perpendicular to velocity and hence instantaneous displacement. Work done = Increment in kinetic energy of revolving body Work done = 0 Also

S 90o

W = F . S = F  S cos

F

90o

= FS cos =0 Example : (i) When an electron revolve around the nucleus in hydrogen atom in a particular orbit, it neither absorb nor emit any energy means its energy remains constant. (ii) When a satellite established once in a orbit around the earth and it starts revolving with particular speed, then no fuel is required for its circular motion.

Sample problem based on work done Problem 121. A particle does uniform circular motion in a horizontal plane. The radius of the circle is 20 cm. The centripetal force acting on the particle is 10 N. It’s kinetic energy is (a) 0.1 Joule Solution : (d)

(b) 0.2 Joule

(c) 2.0 Joule

(d) 1.0 Joule

mv2  10 N (given)  m v 2  10  r  10  0.2  2 r Kinetic energy

1 1 m v 2  (2)  1 Joule . 2 2

Problem 122. A body of mass 100 g is rotating in a circular path of radius r with constant velocity. The work done in one complete revolution is (a) 100r Joule Solution : (d)

[AFMC 1998]

(b) (r/100) Joule

(c) (100/r) Joule

(d) Zero

Because in uniform circular motion work done by the centripetal force is always zero.

Problem 123. A particle of mass m is describing a circular path of radius r with uniform speed. If L is the angular momentum of the particle about the axis of the circle, the kinetic energy of the particle is given by [CPMT 1995] (a) L2 / mr 2 Solution : (b)

(b) L2 / 2mr 2

Rotational kinetic energy E 

L2 L2  2I 2m r 2

(c) 2 L2 / mr 2

(d) mr 2 L

(As for a particle I  m r 2 )

3.14 Skidding of Vehicle on a Level Road. When a vehicle turns on a circular path it requires centripetal force. If friction provides this centripetal force then vehicle can move in circular path safely if Friction force  Required centripetal force mg

m2r

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PHYSICS

by Pradeep Kshetrapal Motion in Two Dimension 39

 mg 

mv 2 r

v safe  rg



This is the maximum speed by which vehicle can turn in a circular path of radius r, where coefficient of friction between the road and tyre is .

Sample problem based on skidding of vehicle on a level road Problem 124. Find the maximum velocity for overturn for a car moved on a circular track of radius 100 m . The coefficient of friction between the road and tyre is 0.2 (a) 0.14 m / s Solution : (d)

v m ax 

(b) 140 m / s

(c) 1.4 km / s

(d) 14 m / s

 r g  0 .2  100  10  10 2  14 m / s

Problem 125. When the road is dry and the coefficient of friction is  , the maximum speed of a car in a circular path is 10 m / s . If the road becomes wet and    (a) 5 m / s Solution : (d)

v

 

 2

, what is the maximum speed permitted

(b) 10 m / s

2  1

v2  v1

(c) 10 2 m / s

(d) 5 2 m / s

/2 1 1 10  5 2m/s   v2  v1  v 2   2 2 2

Problem 126. The coefficient of friction between the tyres and the road is 0.25. The maximum speed with which a car can be driven round a curve of radius 40 m with skidding is (assume g = 10 ms–2) (a) 40 ms 1 Solution : (d)

v m ax 

(b) 20 ms 1

(c) 15 ms 1

(d) 10 ms 1

 r g  0.25  40  10  10 m / s

3.15 Skidding of Object on a Rotating Platform. On a rotating platform, to avoid the skidding of an object (mass m) placed at a distance r from axis of rotation, the centripetal force should be provided by force of friction. Centripetal force = Force of friction m2r = mg 

 max  (g / r) ,

Hence maximum angular velocity of rotation of the platform is

(g / r) , so that object will not skid on it.

3.16 Bending of a Cyclist. A cyclist provides himself the necessary centripetal force by leaning inward on a horizontal track, while going round a curve. Consider a cyclist of weight mg taking a turn of radius r with velocity v. In order to provide the necessary centripetal force, the cyclist leans through angle  inwards as shown in figure. The cyclist is under the action of the following forces : The weight mg acting vertically downward at the centre of gravity of cycle and the cyclist. The reaction R of the ground on cyclist. It will act along a line-making angle  with the vertical. The vertical component R cos of the normal reaction R will balance the weight of the cyclist, while the horizontal component R sin  will provide the necessary centripetal force to the cyclist. R sin  

mv 2 r

…..(i)

R R cos

mv2/r

 R sin

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PHYSICS

by Pradeep Kshetrapal

40 Motion in Two Dimension and R cos  = mg Dividing equation (i) by (ii), we have

…..(ii)

m v2 r R sin   R cos  mg

tan  

or

v2 rg

…..…(iii)

v2  Therefore, the cyclist should bend through an angle   tan 1    rg    It follows that the angle through which cyclist should bend will be greater, if (i) The radius of the curve is small i.e. the curve is sharper (ii) The velocity of the cyclist is large.

Note : 

For the same reasons, an ice skater or an aeroplane has to bend inwards, while taking a turn.

Sample problem based on bending of cyclist Problem 127. A boy on a cycle pedals around a circle of 20 metres radius at a speed of 20 metres/sec. The combined mass of the boy and the cycle is 90kg . The angle that the cycle makes with the vertical so that it may not fall is (g  9 .8 m / sec 2 ) (a) 60.25 o Solution : (b)

[MP PMT 1995]

(b) 63.90 o

(c) 26.12 o

(d) 30.00 o

r  20 m, v  20 m / s, m  90 kg , g  9.8 m / s 2 (given) v2 

 20  20 

1   tan 1    tan 1    tan (2)  63.90 r g 20  10    

Problem 128. If a cyclist moving with a speed of 4.9 m / s on a level road can take a sharp circular turn of radius 4m , then coefficient of friction between the cycle tyres and road is (a) 0.41 (b) 0.51 (c) 0.71 Solution : (d)

v  4.9 m / s, r  4 m and g  9 .8 m / s



2

(d) 0.61

(given)

v2 4 .9  4 .9   0.61 rg 4  9 .8

Problem 129. A cyclist taking turn bends inwards while a car passenger taking same turn is thrown outwards. The reason is

[NCERT 1972]

(a) Car is heavier than cycle (b) Car has four wheels while cycle has only two (c) Difference in the speed of the two (d) Cyclist has to counteract the centrifugal force while in the case of car only the passenger is thrown by this force Solution : (d)

3.17 Banking of a Road. For getting a centripetal force cyclist bend towards the centre of circular path but it is not possible in case of four wheelers. Therefore, outer bed of the road is raised so that a vehicle moving on it gets automatically inclined towards the centre.

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PHYSICS

by Pradeep Kshetrapal Motion in Two Dimension 41

In the figure (A) shown reaction R is resolved into two components, the component R cos balances weight of vehicle

R cos   mg



……(i)

and the horizontal component R sin  provides necessary centripetal force as it is directed towards centre of desired circle mv 2 ...…(ii) r Dividing (ii) by (i), we have R sin  

Thus

v2 rg

tan  

 2r

v  or tan   g rg

R cos 

R



l

h

R sin 

...... (iii)





 x

mg

Fig. (A)

Fig. (B)

[As v = r ]

...... (iv)

If l = width of the road, h = height of the outer edge from the ground level then from the figure (B) h h .......(v) [since  is very small] tan    x l tan  

From equation (iii), (iv) and (v)

Note : 

v 2  2 r v h    g rg l rg

If friction is also present between the tyres and road then

 Maximum safe speed on a banked frictional road v 

  tan  v2  rg 1   tan 

rg(  tan  ) 1   tan 

Sample problems based on banking of a road Problem 130. For traffic moving at 60 km / hr along a circular track of radius 0.1 km , the correct angle of banking is [MNR 1993]

 (50 / 3)  (b) tan 1    100  9 . 8  2

2

(a) Solution : (b)

(60) 0 .1

v  60 km / hr 

50 m / s , r  0.1km  100 m, 3

Angle of banking tan  

 100  9 . 8  (c) tan 1  2   (50 / 3) 

(d) tan 1 60  0.1  9.8

g  9 .8 m / s 2 (given)

2    v2  v2   tan 1  50 / 3   or   tan 1   rg rg  100  9 . 8 

Problem 131. A vehicle is moving with a velocity v on a curved road of width b and radius of curvature R . For counteracting the centrifugal force on the vehicle, the difference in elevation required in between the outer and inner edges of the road is (a) Solution : (a)

v 2b Rg

(b)

For Banking of road tan   

v 2b v2 h v 2l    h rg l rg Rg

rb Rg

(c)

vb 2 Rg

h v2 and tan   l rg

[As l  b and r  R given]

(d)

vb R2g

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PHYSICS

by Pradeep Kshetrapal

42 Motion in Two Dimension

Problem 132. The radius of curvature of a road at a certain turn is 50m . The width of the road is 10m and its outer edge is 1.5m higher than the inner edge. The safe speed for such an inclination will be (a) 6.5 m / s Solution : (b)

(b) 8.6 m / s

(c) 8 m / s

(d) 10 m / s

h  1 .5 m, r  50 m, l  10 m, g  10 m / s 2 (given)

v2 h  rg l

v

hrg  l

1 .5  50  10  8.6 m / s 10

Problem 133. Keeping the banking angle same to increase the maximum speed with which a car can travel on a curved road by 10%, the radius of curvature of road has to be changed from 20m to (a) 16m Solution : (c)

tan  

(b) 18 m

v2 rg

(c) 24 .25m

[EAMCET 1991]

(d) 30.5 m

 r  v 2 (if  is constant) 2

r2  v 2   1 . 1v      = 1.21  r2 = 1.21  r1 = 1.21  20 = 24.2 m r1  v 1   v  2

Problem 134. The slope of the smooth banked horizontal road is p . If the radius of the curve be r , the maximum velocity with which a car can negotiate the curve is given by (a) prg Solution : (b)

tan  

(b)

prg

v v p v rg rg

prg

2

(c) p / rg

(d)

p / rg

2

3.18 Overturning of Vehicle. When a car moves in a circular path with speed more than maximum speed then it overturns and it’s inner wheel leaves the ground first Weight of the car = mg Speed of the car = v Radius of the circular path = r R2 G R1 Distance between the centre of wheels of the car = 2a h Height of the centre of gravity (G) of the car from the road level = h 2a F Reaction on the inner wheel of the car by the ground = R1 mg Reaction on the outer wheel of the car by the ground = R2 When a car move in a circular path, horizontal force F provides the required centripetal force mv 2 .......(i) R For rotational equilibrium, by taking the moment of forces R1, R2 and F about G

i.e., F 

Fh  R1 a  R 2 a

As there is no vertical motion so R1 + R2 = mg By solving (i), (ii) and (iii)

and

.......(ii) .......(iii)

R1 

1  v 2h  M g   2  ra 

.......(iv)

R2 

1  v 2h  M g   2  ra 

.......(v)

It is clear from equation (iv) that if v increases value of R1 decreases and for R1 = 0

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PHYSICS

by Pradeep Kshetrapal Motion in Two Dimension 43

v 2h g ra

or

v

gra h

i.e. the maximum speed of a car without overturning on a flat road is given by v 

gra h

Sample problems based on overturning of vehicle Problem 135. The distance between two rails is 1.5m . The centre of gravity of the train at a height of 2m from the ground. The maximum speed of the train on a circular path of radius 120 m can be (a) 10 .5 m / s Solution : (c)

(b) 42 m / s

(c) 21 m / s

(d) 84 m / s

Height of centre of gravity from the ground h = 2m, Acceleration due to gravity g = 10 m / s 2 , Distance between two rails 2a = 1.5m, Radius of circular path r = 120 m (given)

v max 

gra h

 v max 

10  120  0.75  21 .2 m / s 2

Problem 136. A car sometimes overturns while taking a turn. When it overturns, it is

[AFMC 1988]

(a) The inner wheel which leaves the ground first (b) The outer wheel which leaves the ground first (c) Both the wheels leave the ground simultaneously (d) Either wheel leaves the ground first Solution : (a)

Problem 137. A car is moving on a circular path and takes a turn. If R 1 and R 2 be the reactions on the inner and outer wheels respectively, then (a) R1  R 2 Solution : (b)

(b) R1  R 2

Reaction on inner wheel R 1 

M 2

(c) R1  R 2

(d) R1  R 2

 v 2h M  v 2h  g   and Reaction on outer wheel R 2  g   2  r a  r a  

 R1  R 2 .

Problem 138. A train A runs from east to west and another train B of the same mass runs from west to east at the same speed along the equator. A presses the track with a force F1 and B presses the track with a force F2 (a) F1  F2 (b) F1  F2 (c) F1  F2 (d) The information is insufficient to find the relation between F1 and F2 Solution : (a)

We know that earth revolves about its own axis from west to east. Let its angular speed is  e and the angular speed of the train is  t For train A : Net angular speed = (  e   t ) because the sense of rotation of train is opposite to that of earth So reaction of track R1  F1  m g  m ( e   t ) 2 R For train B : Net angular speed = ( e   t ) because the sense of rotation of train is same as that of earth So reaction of track R 2  F2  m g  m ( e   t ) 2 R So it is clear that F1  F2

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PHYSICS

by Pradeep Kshetrapal

44 Motion in Two Dimension

3.19 Motion of Charged Particle in Magnetic Field. When a charged particle having mass m , charge q enters perpendicularly in a magnetic field B, with velocity v then it describes a circular path of radius r. Because magnetic force (qvB) works in the perpendicular direction of v and it provides required centripetal force Magnetic force = Centripetal force mv 2 qvB = r

 radius of the circular path r 

mv qB























































F



























v

  

q 

 

B



3.20 Reaction of Road on Car. (1) When car moves on a concave bridge then R

mv 2 Centripetal force = R  mg cos   r

and reaction R  mg cos  

mv 2 r



v

 mg cos

Concave bridge

mg

(2) When car moves on a convex bridge Centripetal force = mg cos   R 

mv r

R

2

v 

mg cos

mg



mv 2 and reaction R  mg cos   r

Convex bridge

Sample problem based on reaction of road Problem 139. The road way bridge over a canal is in the form of an arc of a circle of radius 20m . What is the minimum speed with which a car can cross the bridge without leaving contact with the ground at the highest point (g  9 . 8 m / s 2 ) (a) 7 m / s Solution : (b)

(b) 14 m / s

(c) 289 m / s

At the highest point of the bridge for critical condition mg   v m ax 

(d) 5 m / s

mv2 mv2 0  mg r r

gr  9 .8  20  196  14 m / s

Problem 140. A car moves at a constant speed on a road as shown in the figure. The normal force exerted by the road on the car is N A and N B when it is at the points A and B respectively (a) N A  N B A

(b) N A  N B (c) N A  N B (d) All possibilities are there Solution : (c)

From the formula N  mg 

mv 2 r

Nr

B

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PHYSICS

by Pradeep Kshetrapal Motion in Two Dimension 45

As rA < rB

 NA  NB

Problem 141. A car while travelling at a speed of 72 km / hr . Passes through a curved portion of road in the form of an arc of a radius 10m . If the mass of the car is 500 kg the reaction of the car at the lowest point P is (a) 25 kN

72 km/hr

(b) 50 kN (c) 75 kN

P

(d) None of these Solution : (a)

v  72

km  20 m /s, r  10 m, h

m  500 kg (given)

Reaction at lowest point R  mg 

m v2 r

 500  10 

500  (20)2  25000 N  25 KN 10

3.21 Non-Uniform Circular Motion. If the speed of the particle in a horizontal circular motion changes with respect to time, then its motion is said to be non-uniform circular motion. Consider a particle describing a circular path of radius r with centre at O. Let at an instant the particle be at P and  be its linear velocity and  be its angular velocity.    Then, …..(i)    r Differentiating both sides of w.r.t. time t we have 





d  d    dr   r   dt dt dt



…..(ii)

dv  Here,  a, (Resultant acceleration) dt 

d    (Angular acceleration) dt

     a    r     a

 at

  ac



.….(iii)

dr    (Linear velocity) dt

Thus the resultant acceleration of the particle at P has two component accelerations (1) Tangential acceleration : a t    r It acts along the tangent to the circular path at P in the plane of circular path.

at a

 O

ac

P

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PHYSICS

by Pradeep Kshetrapal

46 Motion in Two Dimension 



According to right hand rule since  and r are perpendicular to each other, therefore, the magnitude of tangential acceleration is given by | a t | |   r |   r sin 90 o   r.

(2) Centripetal (Radial) acceleration : a c    v It is also called centripetal acceleration of the particle at P. It acts along the radius of the particle at P. According to right hand rule since  and  are perpendicular to each other, therefore, the magnitude of centripetal acceleration is given by    | a c | |   |    sin 90 o =     ( r)   2 r   2 / r (3) Tangential and centripetal acceleration in different motions Centripetal acceleration

Tangential acceleration

Net acceleration

ac = 0

at = 0

a=0

Uniform translatory motion

ac = 0

at  0

a = at

Accelerated translatory motion

ac  0

at = 0

a = ac

Uniform circular motion

ac  0

at  0

Note : 

a  ac2  a t2

Type of motion

Non-uniform circular motion

Here at governs the magnitude of v while a c its direction of motion.

(4) Force : In non-uniform circular motion the particle simultaneously possesses two forces Centripetal force : Fc  ma c 

mv 2  mr  2 r

Tangential force : Ft  ma t Net force : Fnet  ma = m ac2  at2

Note : 

  In non-uniform circular motion work done by centripetal force will be zero since Fc  v

 In non uniform circular motion work done by tangential of force will not be zero since Ft  0  Rate of work done by net force in non-uniform circular = rate of work done by tangential force   dW i.e. P  Ft .v dt

Sample problems based on non-uniform circular motion Problem 142. The kinetic energy k of a particle moving along a circle of radius R depends on the distance covered. It is given as K.E. = as2 where a is a constant. The force acting on the particle is [MNR 1992; JIPMER 2001, 2002] (a) 2 a Solution : (b)

s2 R

1/2

 s2  (b) 2as1  2   R  

(c) 2as

(d) 2a

R2 s

In non-uniform circular motion two forces will work on a particle Fc and Ft So the net force FNet 

Fc2  Ft2

Centripetal force Fc 

2as2 mv 2  R R

….(i) ….(ii)

[As kinetic energy

1 mv 2  as 2 given] 2

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PHYSICS

by Pradeep Kshetrapal Motion in Two Dimension 47

Again from :

1 2as 2 2a mv 2  as 2  v 2   vs 2 m m

Tangential acceleration a t 

at  v

2a 2a s m m

Ft

dv dv ds d  2a   .  at  s  .v dt ds dt ds  m 

Fc

2 as 2a  m m

and Ft  ma t  2as

….(iii)

Now substituting value of Fc and Ft in equation (i)

 FNet

 2as 2    R 

2

2      2as2  2as 1  s   R 2   

1/2

Problem 143. A particle of mass m is moving in a circular path of constant radius r such that its centripetal acceleration a c is varying with time t as ac  k 2 rt 2 , where k is a constant. The power delivered to the particle by the forces acting on it is (a) 2mk 2r 2 t Solution : (b)

ac  k 2 r t 2 

(b) mk 2r 2 t

(c)

v2  k 2r t 2  v 2  k 2 r 2 t 2 r

Tangential acceleration at 

mk 4 r 2 t 5 3

(d) Zero

 v krt

dv kr dt

As centripetal force does not work in circular motion. So power delivered by tangential force P  Ftv  m at v =m(kr) krt  mk 2r 2t

Problem 144.

A simple pendulum is oscillating without damping. When the displacement of the bob is less than  maximum, its acceleration vector a is correctly shown in

 a

(a)

(b)

(c)

(d)

 a

Solution : (c)

ac  centripetal accelerati on ,

 a

at  tangential acceleration ,

 aN

aN  net acceleration  Resultant of ac and at

aN =

a c2  a t2

 a

 ac

 at

Problem 145. The speed of a particle moving in a circle of radius 0.1m is v  1.0t where t is time in second. The resultant acceleration of the particle at t  5 s will be (a) 10 m / s 2 Solution : (c)

v  1.0 t  a t  and ac 

(b) 100 m / s 2

(c) 250 m / s 2

dv  1m / s2 dt

v 2 (5)2   250 m / s 2 r 0 .1

[At t  5 sec, v  5 m / s ]

 a N  ac2  at2  (250)2  1 2  a N  250 m / s 2 (approx.)

(d) 500 m / s 2

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PHYSICS

by Pradeep Kshetrapal

48 Motion in Two Dimension

Problem 146. A particle moving along the circular path with a speed v and its speed increases by ‘g’ in one second. If the radius of the circular path be r, then the net acceleration of the particle is 1

2

(a)

v g r

(b)

v

2

r2

 g2

2

Solution : (c)

at  g (given) and ac 

1

v4 2 (c)  2  g 2   r 

v2  v2 and a N  at2  ac2     g 2  r  r 

v2 2 (d)   g    r v4 r2

 g2

Problem 147. A car is moving with speed 30 m/sec on a circular path of radius 500 m. Its speed is increasing at the rate of 2m/sec2. What is the acceleration of the car (a) 2 Solution : (b)

m/s2

at = 2m/s2 and a c 

(b) 2.7

m/s2

[Roorkee 1982; RPET 1996; MH CET 2002; MP PMT 2003]

(c) 1.8 m/s2

(d) 9.8 m/s2

v2 30  30   1.8 m /s 2  a  a t2  ac2  2 2  (1.8) 2  2.7m/s 2 . r 500

Problem 148. For a particle in circular motion the centripetal acceleration is

[CPMT 1998]

(a) Less than its tangential acceleration

(b) Equal to its tangential acceleration

(c) More than its tangential acceleration acceleration

(d) May

be

more

or

less

than

its

tangential

Solution : (d)

Problem 149. A particle is moving along a circular path of radius 3 meter in such a way that the distance travelled measured along the circumference is given by S  (a) 1.3 m/s2 Solution : (b)

s

(b) 13 m/s2

t2 t3 . The acceleration of particle when t  2 sec is  2 3 (c) 3 m/s2 (d) 10 m/s2

ds dv d t2 t3  v  t  t 2 and a t   (t  t 2 )  1  2t  dt dt dt 2 3

At t = 2 sec, v = 6 m/s and at  5 m / s 2 , a c 

v2 36   12 m /s 2 r 3

a N  ac2  a t2  (12)2  (5)2  13 m / s 2 .

3.22 Equations of Circular Motion. For accelerated motion

For retarded motion

 2  1   t

 2  1   t

1   1 t   t 2 2

1   1 t   t 2 2

1 = Initial angular velocity of particle

 22   12  2 

 22   12  2 

 = Angular acceleration of particle

 n  1 

 2

(2n  1)

 n  1 

 2

(2n  1)

Where

2 = Final angular velocity of particle  = Angle covered by the particle in time t n = Angle covered by the particle in nth second

Sample problems based on equation of circular motion Problem 150. The angular velocity of a particle is given by   1 .5 t  3 t 2  2 , the time when its angular acceleration ceases to be zero will be (a) 25 sec

(b) 0.25 sec

(c) 12 sec

(d) 1.2 sec

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PHYSICS

by Pradeep Kshetrapal Motion in Two Dimension 49

Solution : (b)

  1 .5 t  3 t 2  2 and  

d  1 .5  6 t dt

 0  1.5  6 t  t 

1 .5  0 .25 sec 6

Problem 151. A wheel is subjected to uniform angular acceleration about its axis. Initially its angular velocity is zero. In

the first 2 sec , it rotates through an angle  1 . In the next 2 sec , it rotates through an additional angle  2 . The ratio of 1 /  2 is [AIIMS 1982] (a) 1

Solution : (c)

(b) 2

(c) 3

From equation of motion    1 t 

1  0 

1  (2) 2 = 2 2

(d) 5

1  t2 2

[As 1  0, t  2 sec,    1 ]

…..(i)

For second condition

1   2  0 

1  (4 ) 2 2

[As 1  0, t  2  2  4 sec,    1   2 ]

1   2  8

….(ii)

From (i) and (ii)  1  2,  2  6 

2 3 1

Problem 152. If the equation for the displacement of a particle moving on a circular path is given by ( )  2t 3  0.5 , where  is in radians and t in seconds, then the angular velocity of the particle after 2 sec from its start is [AIIMS 1998]

Solution : (c)

(a) 8 rad / sec

(b) 12 rad / sec

  2t 3  0.5 and  

d  6t 2 dt

(c) 24 rad / sec

(d) 36 rad / sec

at t = 2 sec,   6 (2) 2  24 rad / sec

Problem 153. A grinding wheel attained a velocity of 20 rad/sec in 5 sec starting from rest. Find the number of revolutions made by the wheel (a) Solution : (c)

 rev/ sec 25

(b)

1 rev/sec 

(c)

25



rev/sec

(d) None of these

1  0,  2  20 rad / sec , t  5 sec 

 2  1 t



20  0  4 rad / sec 2 5

From the equation    1 t 

1 1  t 2  0  (4 ) . (5 ) 2  50 rad 2 2

2 rad means 1 revolution.  50 Radian means

50 25 or rev. 2 

Problem 154. A grind stone starts from rest and has a constant angular acceleration of 4.0 rad/sec2. The angular displacement and angular velocity, after 4 sec. will respectively be (a) 32 rad, 16 rad/sec Solution : (a)

(b) 16 rad, 32 rad/sec (c) 64 rad, 32 rad/sec

(d) 32 rad, 64 rad/sec

1  0,   4 rad / sec , t  4 sec 2

Angular displacement    1 t 

1 1  t 2  0  4 (4 ) 2  32 rad. 2 2

 Final angular  2  1   t  0  4  4  16 rad / sec

Problem 155. An electric fan is rotating at a speed of 600 rev/minute. When the power supply is stopped, it stops after 60 revolutions. The time taken to stop is

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PHYSICS

by Pradeep Kshetrapal

50 Motion in Two Dimension (a) 12 s Solution : (a)

(b) 30 s

(c) 45 s

(d) 60 s

1  600 rev / min  10 rev / sec ,  2  0 and   60 rev From the equation  22   12  2  0  (10 ) 2  2  60   

100 5  120 6

Again  2  1   t  0  1   t

t

1 10  6   12 sec .  5

3.23 Motion in Vertical Circle. This is an example of non-uniform circular motion. In this motion body is under the influence of gravity of earth. When body moves from lowest point to highest point. Its speed decrease and becomes minimum at highest point. Total mechanical energy of the body remains conserved and KE converts into PE and vice versa. (1) Velocity at any point on vertical loop : If u is the initial velocity imparted to body at lowest point then. Velocity of body at height h is given by C

v  u  2 gh  u  2 gl(1  cos ) 2

2

[As h = l – l cos = l (1 – cos)]

where l in the length of the string

l

D

O

B v



P

h

u

A (2) Tension at any point on vertical loop : Tension at general point P, According to Newton’s second law of motion. C Net force towards centre = centripetal force

T  mg cos  

mv 2 l

or

m T  [u 2  gl(2  3 cos )] l

T  mg cos  

mv 2 l

D

l O

[As v  u 2  2 gl(1  cos ) ]

(3) Velocity and tension in a vertical loop at different positions Position

Angle

Velocity

Tension

A

0o

u

mu 2  mg l

B

90o

u 2  2 gl

mu 2  2mg l

C

180o

u 2  4 gl

mu 2  5mg l

D

270o

u 2  2 gl

mu 2  2mg l

T A  TB  TC and TB = TD T A  TB  3mg , T A  TC  6 mg

and

TB  TC  3mg

(4) Various conditions for vertical motion :

B

T

P

 A

It is clear from the table that :



mg

mg cos + mv2/r

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PHYSICS

by Pradeep Kshetrapal Motion in Two Dimension 51

Velocity at lowest point

Condition

u A  5 gl

Tension in the string will not be zero at any of the point and body will continue the circular motion.

uA 

Tension at highest point C will be zero and body will just complete the circle.

5 gl ,

Particle will not follow circular motion. Tension in string become zero somewhere between points B and C whereas velocity remain positive. Particle leaves circular path and follow parabolic trajectory.

2 gl  u A  5 gl ,

2 gl

Both velocity and tension in the string becomes zero between A and B and particle will oscillate along semi-circular path.

u A  2 gl

velocity of particle becomes zero between A and B but tension will not be zero and the particle will oscillate about the point A.

uA 

Note : 

K.E. of a body moving in horizontal circle is same throughout the path but the K.E. of the body

moving in vertical circle is different at different places.  If body of mass m is tied to a string of length l and is projected with a horizontal velocity u then : Height at which the velocity vanishes is h 

u2 2g

Height at which the tension vanishes is h 

u 2  gl 3g

(5) Critical condition for vertical looping : If the tension at C is zero, then body will just complete revolution in the vertical circle. This state of body is known as critical state. The speed of body in critical state is called as critical speed. From the above table

TC =

mu 2  5 mg  0  u  5 gl l

It means to complete the vertical circle the body must be projected with minimum velocity of 5 gl at the lowest point. (6) Various quantities for a critical condition in a vertical loop at different positions : Quantity

Point A

Point B

5 gl

3 gl

gl

3 gl

gl (3  2 cos  )

5g l

3g l

g l

3g l

g (3  2 cos  ) l

Tension in String (T)

6 mg

3 mg

0

3 mg

3mg (1  cos  )

Kinetic Energy (KE)

5 mgl 2

3 mgl 2

1 mgl 2

3 mgl 2

mgl (3  2 cos  ) 2

0

mgl

2 mgl

mgl

mgl (1  cos  )

5 mgl 2

5 mgl 2

5 mgl 2

5 mgl 2

5 m gl 2

Linear velocity (v) Angular velocity ( )

Potential Energy (PE) Total Energy (TE)

Point C

Point D

Point P

(7) Motion of a block on frictionless hemisphere : A small block of mass m slides down from the top of a frictionless hemisphere of radius r. The component of the force of gravity (mg cos) provides required centripetal force but at point B it's circular motion ceases and the block lose contact with the surface of the sphere.

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PHYSICS

by Pradeep Kshetrapal

52 Motion in Two Dimension mv 2 r For point A and B, by law of conservation of energy

For point B, by equating the forces, mg cos  

.....(i) A B

Total energy at point A = Total energy at point B K.E.(A) + P.E.(A) = K.E.(B) + P.E.(B)



h



r

mg

1 0 + mgr = mv 2  mgh  v  2 g(r  h) .....(ii) 2 and from the given figure h  r cos

(r – h)

.....(iii)

By substituting the value of v and h from eqn (ii) and (iii) in eqn (i) h m mg    r r





2 g (r  h)

h  2 (r  h)  h 



2

2 r 3

i.e. the block lose contact at the height of

2 r from the ground. 3

and angle from the vertical can be given by cos 

h 2  r 3

   cos 1

2 . 3

Sample problems based on vertical looping Problem 156. A small block is shot into each of the four tracks as shown below. Each of the tracks rises to the same

height. The speed with which the block enters the track is the same in all cases. At the highest point of the track, the normal reaction is maximum in (a)

(b)

(d)

v

v

Solution : (a)

(c)

v

v

Normal reaction at the highest point of the path R 

mv 2  mg r

For maximum R, value of the radius of curvature (r ) should be minimum and it is minimum in first condition.

Problem 157. A stone tied to string is rotated in a vertical circle. The minimum speed with which the string has to be rotated [EAMCET (Engg.) 1998; CBSE PMT 1999]

(a) Decreases with increasing mass of the stone

(b) Is independent of the mass of the stone

(c) Decreases with increasing in length of the string of the string Solution : (b)

(d)

Is independent of the length

v  5 g r for lowest point of vertical loop.

v  m 0 i.e. it does not depends on the mass of the body.

Problem 158. A mass m is revolving in a vertical circle at the end of a string of length 20 cms. By how much does the tension of the string at the lowest point exceed the tension at the topmost point (a) 2 mg Solution : (c)

(b) 4 mg

TLowest point  THighest point  6 mg (Always)

(c) 6 mg

(d) 8 mg

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PHYSICS

by Pradeep Kshetrapal Motion in Two Dimension 53

Problem 159. In a simple pendulum, the breaking strength of the string is double the weight of the bob. The bob is

released from rest when the string is horizontal. The string breaks when it makes an angle  with the vertical (a)   cos 1 (1 / 3)

Solution : (c)

(c)   cos 1 (2 / 3)

(b)   60 o

(d)   0 o

Let the string breaks at point B. Tension  mg cos  

m v B2  Breaking strength r

m v B2  2 mg ….(i) r If the bob is released from rest (from point A) then velocity acquired by it at point B  mg cos  

vB  vB 

2 gh

A

2 g r cos 

....(ii)

[As h= r cos  ]

By substituting this value in equation (i) mg cos  



m (2 g r cos  )  2 mg r

or 3mg cos   2 mg  cos  

T

h = r cos

r

B mg cos

 C

2 2    cos 1   3 3

mv2/r

mg

Problem 160. A toy car rolls down the inclined plane as shown in the fig. It goes around the loop at the bottom. What is the relation between H and h

h H

(a) Solution : (d)

H 2 h

(b)

r

H 3 h

(c)

H 4 h

(d)

H 5 h

When car rolls down the inclined plane from height H, then velocity acquired by it at the lowest point v  2g H

….(i)

and for looping of loop, velocity at the lowest point should be v  5 g r

….(ii)

From eqn (i) and (ii) v 

2 gH 

From the figure H  h  2r  r 

5 gr  H 

5r 2

….(iii)

H h 2

Substituting the value of r in equation (iii) we get H 

H 5 H  h 5    h 2 2 

Problem 161. The mass of the bob of a simple pendulum of length L is m . If the bob is left from its horizontal position then the speed of the bob and the tension in the thread in the lowest position of the bob will be respectively. (a) Solution : (a)

2 gL and 3mg

(b) 3mg and

2 gL

(c) 2mg and

2 gl

By the conservation of energy Potential energy at point A = Kinetic energy at point B

A

(d) 2 gl and 3mg l

T B mg mv2/L

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PHYSICS

by Pradeep Kshetrapal

54 Motion in Two Dimension mg l 

1 m v 2  v  2 gl 2 m mv 2  T  mg  (2 gl)  T  3 mg l l

and tension  mg 

Problem 162. A stone of mass m is tied to a string and is moved in a vertical circle of radius r making n revolutions per minute. The total tension in the string when the stone is at its lowest point is (a) m {g  ( n r) / 900 } (b) m (g   nr ) 2 2

Solution : (a)

(c) m (g   nr)

2

[Kerala (Engg.) 2001]

(d) m (g  n 2r 2 )

Tension at lowest point T  mg  m w 2 r  mg  m 4 2 n 2 r If n is revolution per minute then T  mg  m 4 2

  2n 2r  n2 m 2n 2r  m g  r  mg   900  3600 900 

Problem 163. A particle is kept at rest at the top of a sphere of diameter 42m. When disturbed slightly, it slides down. At what height h from the bottom, the particle will leave the sphere (a) 14 m (b) 28 m (c) 35 m Solution : (c)

[IMS-BHU 2003]

(d) 7 m

Let the particle leave the sphere at height ‘h’ from the bottom We know for given condition x  and h  r  x  r 

2 r 3

v



5 5 2 r  r   21  35 m 3 3 3

x

[As r= 21 m] r

h

Problem 164. A bucket tied at the end of a 1.6 m long string is whirled in a vertical circle with constant speed. What should be the minimum speed so that the water from the bucket does not spill, when the bucket is at the highest position (Take g = 10 m/sec2) (a) 4 m / sec Solution : (a)

v

(b) 6.25 m / sec

(c) 16m / sec

(d) None of these

g r  10  1 . 6  16  4 m / s

Problem 165. The ratio of velocities at points A, B and C in vertical circular motion is A

O

B

C

(a) 1 : 9 : 25 Solution : (d)

v A : v B : vc 

(b) 1 : 2 : 3

(c) 1 : 3 : 5

(d) 1 : 3 : 5

g r : 3 g r : 5g r  1 : 3 : 5

Problem 166. The minimum speed for a particle at the lowest point of a vertical circle of radius R, to describe the circle is ‘v’. If the radius of the circle is reduced to one-fourth its value, the corresponding minimum speed will be

[EAMCET (Engg.) 1999]

(a) Solution : (b)

v 4

v  5g r

(b) v 

r So

v 2

v2  v1

(c) 2v r2  r1

(d) 4v

r/4 1   v2  v / 2 r 2

Problem 167. A body slides down a frictionless track which ends in a circular loop of diameter D, then the minimum height h of the body in term of D so that it may just complete the loop, is

[AIIMS 2000]

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PHYSICS

by Pradeep Kshetrapal Motion in Two Dimension 55

(a) h  Solution : (b)

5D 2

(b) h 

5D 4

5 5  D 5D r     2 4 2 2

We know h 

(c) h 

3D 4

(d) h 

D 4

[For critical condition of vertical looping]

Problem 168. A can filled with water is revolved in a vertical circle of radius 4m and the water just does not fall down. The time period of revolution will be (a) 1 sec Solution : (d)

[CPMT 1985; RPET 1999]

(b) 10 sec

At highest point mg  m  2 r  g 

(c) 8 sec

4 2 T2

r  10 

4 2 4 T2

(d) 4 sec

 T 2  16  T  4 sec

Problem 169. A particle is moving in a vertical circle. The tensions in the string when passing through two positions at angles 30o and 60o from vertical (lowest position) are T1 and T2 respectively, Then (a) T1  T2 Solution : (b)

(b) T1  T2

(c) T1  T2

(d) T1  T2

mv 2 r As  increases T decreases T  mg cos  



So T1  T2

T

r

mg

 mg cos + mv2/r

Problem 170. A mass of 2kg is tied to the end of a string of length 1m. It is, then, whirled in a vertical circle with a constant speed of 5 ms 1 . Given that g  10 ms 2 . At which of the following locations of tension in the string will be 70 N

Solution : (b)

(a) At the top

(b) At the bottom

(c) When the string is horizontal

(d) At none of the above locations

Centrifugal force F 

mv 2 2  (5)2   50 Newton r 1

Weight  mg  2  10  20 Newton T

Tension = 70 N (sum of above two forces) i.e. the mass is at the bottom of the vertical circular path

mg 

mv

2

r

Problem 171. With what angular velocity should a 20 m long cord be rotated such that tension in it, while reaching the highest point, is zero (a) 0.5 rad/sec Solution : (d)



g  r

(b) 0.2 rad/sec

(c) 7.5 rad/sec

(d) 0.7 rad/sec

10  0 .5  0.7 rad / sec 20

Problem 172. A body of mass of 100 g is attached to a 1m long string and it is revolving in a vertical circle. When the string makes an angle of 60 o with the vertical then its speed is 2 m / s . The tension in the string at

  60 o will be (a) 89 N Solution : (b)

T  mg cos  

(b) 0.89 N

(c) 8 .9 N

(d) 0.089 N

0 .1  (2) 2 mv 2  0.1  9.8  cos 60   0.49  0.4  0.89 Newton r 1

Problem 173. A body of mass 2kg is moving in a vertical circle of radius 2m . The work done when it moves from the lowest point to the highest point is

genius

PHYSICS

by Pradeep Kshetrapal

56 Motion in Two Dimension (a) 80 J Solution : (a)

(b) 40 J

(c) 20 J

(d) 0

work done = change in potential energy  2 mgr  2  2  10  2  80 J

Problem 174. A body of mass m is tied to one end of a string of length l and revolves vertically in a circular path. At the lowest point of circle, what must be the K.E. of the body so as to complete the circle (a) 5 mgl Solution : (c)

(b) 4 mgl

(c) 2 .5 mgl

Minimum velocity at lowest point to complete vertical loop  5 gl So minimum kinetic energy 

1 1 5 m (v 2 )  m ( 5 gl ) 2  mgl  2 .5 mgl 2 2 2

(d) 2 mgl

[RPMT 1996]

genius

PHYSICS

by Pradeep Kshetrapal Motion in Two Dimension 57

3.24 Conical Pendulum. This is the example of uniform circular motion in horizontal plane. A bob of mass m attached to a light and in-extensible string rotates in a horizontal circle of radius r with constant angular speed  about the vertical. The string makes angle  with vertical and appears tracing the surface of a cone. So this arrangement is called conical pendulum. S The force acting on the bob are tension and weight of the bob. From the figure T sin  

mv 2 r

T cos   mg

and

….(ii)

v2  (1) Tension in the string : T  mg 1     rg    T 

mg  cos 

mgl l r 2



….(i)

T sin

2

(2) Angle of string from the vertical : tan  

T cos

 P

O

mv2/r

mg S

[As cos  

2

T

l r ] l 2

h  l

2

 l

h

v2 rg

O

r

P

(3) Linear velocity of the bob : v  gr tan  (4) Angular velocity of the bob :  

g tan   r

(5) Time period of revolution : TP  2

g  h

g l cos 

l cos h l2  r2 r  2  2  2 g g g g tan 

Sample problems based on conical pendulum Problem 175. A point mass m is suspended from a light thread of length l, fixed at O, is whirled in a horizontal circle at constant speed as shown. From your point of view, stationary with respect to the mass, the forces on the mass are O l m T

(a)

(b)

T

T

(c)

F

Solution : (c)

F

W Centrifugal force W (F) works radially outward,

Weight (w) works downward

T

(d)

W

F

F

W

Tension (T) work along the string and towards the point of suspension

Problem 176. A string of length L is fixed at one end and carries a mass M at the other end. The string makes 2/

revolutions per second around the vertical axis through the fixed end as shown in the figure, then tension in the string is [BHU 2002] S



L T M

genius

PHYSICS

by Pradeep Kshetrapal

58 Motion in Two Dimension

(a) ML Solution : (d)

(b) 2 ML

T sin   M R

..... (i)

T sin   M L sin 

..... (ii)

2

2

(c) 4 ML

(d) 16 ML



T

From (i) and (ii)



2

2

2 T  M L  M 4 n L  M 4    L  16 ML    2 2

2

L

m2R

R

Problem 177. A string of length 1m is fixed at one end and a mass of 100 gm is attached at the other end. The string makes 2 /  rev/sec around a vertical axis through the fixed point. The angle of inclination of the string with the vertical is ( g  10 m / sec 2 ) (a) tan 1 Solution : (d)

5 8

(b) tan 1

8 5

(c) cos 1

8 5

(d) cos 1

5 8

For the critical condition, in equilibrium T sin   m  2 r and T cos   mg

 tan  



 r 2

T

T cos



g

T sin

4 2 (2 /  ) 2 . 1 8 4 2 n 2 r    5 g 10

m2R mg

Sample problems (Miscellaneous) Problem 178. If the frequency of the rotating platform is f and the distance of a boy from the centre is r, which is the area swept out per second by line connecting the boy to the centre (a) rf Solution : (c)

(c) r 2 f

(b) 2rf

(d) 2r 2 f

Area swept by line in complete revolution  r 2 If frequency of rotating platform is f per second, then Area swept will be  r 2 f per second.

Problem 179. Figure below shows a body of mass M moving with uniform speed v along a circle of radius R. What is the change in speed in going from P1 to P2

P2

v

P3

P1 R P4

(a) Zero Solution : (a)

(b)

2v

(c) v / 2

(d) 2 v

In uniform circular motion speed remain constant.  change in speed is zero.

Problem 180. In the above problem, what is change in velocity in going from P1 to P2 (a) Zero Solution : (b)

(b)

2v

(c) v / 2

(d) 2 v

2v  90   2v Change in velocity  2v sin( / 2)  2v sin    2v sin 45   2  2

Problem 181. In the above problem, what is the change in angular velocity in going from P1 to P2 (a) Zero

(b)

2v/R

(c) v / 2 R

(d) 2 v / R

genius

PHYSICS

by Pradeep Kshetrapal Motion in Two Dimension 59

Solution : (a)

Angular velocity remains constant, so change in angular velocity = Zero.

Problem 182. A particle of mass m is fixed to one end of a light spring of force constant k and unstretched length l . The system is rotated about the other end of the spring with an angular velocity  , in gravity free space. The increase in length of the spring will be (a) (b)

m  2l k



m  2l

l

m

k  m 2 m  2l

(c)

k  m 2 (d) None of these Solution : (b)

In the given condition elastic force will provides the required centripetal force k x  m  2r



k x  m  (l  x )  k x  m  l  m  x  x (k  m  )  m  l 2

2

2

2

2

r = (l+x)

m

m l 2

x 

k m2

Problem 183. A uniform rod of mass m and length l rotates in a horizontal plane with an angular velocity  about a vertical axis passing through one end. The tension in the rod at a distance x from the axis is (a) Solution : (d)

1 m 2x 2

(b)

1 x2 m 2 2 l

(c)

1 x  m  2 l 1   2 l 

(d)

1 m 2 2 [l  x 2 ] 2 l

Let rod AB performs uniform circular motion about point A. We have to calculate the tension in the rod at a distance x from the axis of rotation. Let mass of the small segment at a distance x is dm

m2 m  So dT  dm  2 x    dx .  2 x  [x d x] l  l  l

Integrating both sides



dT 

x

 T 

m 2l

2

l

2

 x2

m l

2 l



dx l

x dx  T 

x

m 2  x 2    l  2  x

B

A x



Problem 184. A long horizontal rod has a bead which can slide along its length, and initially placed at a distance L from one end A of the rod. The rod is set in angular motion about A with constant angular acceleration  . If the coefficient of friction between the rod and the bead is  , and gravity is neglected, then the time after which the bead starts slipping is (a) Solution : (a)

 

(b)

 

(c)

1



(d) Infinitesimal

Let the bead starts slipping after time t For critical condition 2 Frictional force provides the centripetal force m  L   R   m  at mL 

 m (t)2L = mL  t  

(As  = t)

A L

Problem 185. A smooth table is placed horizontally and an ideal spring of spring constant k  1000 N / m and unextended length of 0.5m has one end fixed to its centre. The other end is attached to a mass of 5 kg which is moving in a circle with constant speed 20m / s . Then the tension in the spring and the extension of this spring beyond its normal length are

genius

PHYSICS

by Pradeep Kshetrapal

60 Motion in Two Dimension (a) 500 N , 0.5 m Solution : (a)

(b) 600 N , 0.6 m

(c) 700 N , 0.7 m

(d) 800 N , 0.8 m

k  1000 , m  5 kg, l  0.5 m, v  20 m / s (given)

mv2 mv2 5 (20)2  1000 x   x  0 .5 m  r lx 0.5  x 1 and Tension in the spring = kx  1000   500 N 2 Problem 186. A particle describes a horizontal circle at the mouth of a funnel type vessel as shown in figure. The surface of the funnel is frictionless. The velocity v of the particle in terms of r and  will be Restoring force = kx 

(a) v  rg / tan  (b) v 

rg tan 

(c) v 

rg cot 

r

 h

(d) v  rg / cot 

Solution : (c)

For uniform circular motion of a particle

mv2  R cos  ….(i) r

and mg  R sin 

R

….(ii)

R sin



R cos

Dividing (i) by (ii)

mv



2

r

r

2

v  cot   v  rg cot  rg



R

k

P

m L

Problem 187. Figure shows a smooth track, a part of which is a circle of radius R. A block of mass m is pushed against a spring constant k fixed at the left end and is then released. Find the initial compression of the spring so that the block presses the track with a force mg when it reaches the point P [see. Fig], where the radius of the track is horizontal

(a)

mgR 3k

(b)

3 gR mk

(c)

3mgR k

(d)

3mg Solution kR

mv P2  mg  v P  R From this we can easily calculate the required velocity at the lowest point of circular track. v p2  v L2  2 gR (by using formula : v 2  u 2  2 gh ) (c) For the given condition, centrifugal force at P should be equal to mg i.e.

v L  v P2  2 gR  Rg  2 gR  3 gR It means the block should possess kinetic energy  And by the law of conservation of energy

1 1 mv L2  m  3 gR 2 2

1 2 1 kx  3 m  g R  x  2 2

3m g R . k

Rg

:

genius PHYSICS

by Pradeep Kshetrapal

Newton’s Laws of Motion 1

1

R sin



R R cos

4.1 Point Mass. (1) An object can be considered as a point object if during motion in a given time, it covers distance much greater than its own size. (2) Object with zero dimension considered as a point mass. (3) Point mass is a mathematical concept to simplify the problems.

4.2 Inertia. (1) Inherent property of all the bodies by virtue of which they cannot change their state of rest or uniform motion along a straight line by their own is called inertia. (2) Inertia is not a physical quantity, it is only a property of the body which depends on mass of the body. (3) Inertia has no units and no dimensions (4) Two bodies of equal mass, one in motion and another is at rest, possess same inertia because it is a factor of mass only and does not depend upon the velocity.

4.3 Linear Momentum. (1) Linear momentum of a body is the quantity of motion contained in the body. (2) It is measured in terms of the force required to stop the body in unit time. (3) It is measured as the product of the mass of the body and its velocity i.e., Momentum = mass × velocity. If a body of mass m is moving with velocity v then its linear momentum p is given by p  m v (4) It is a vector quantity and it’s direction is the same as the direction of velocity of the body. (5) Units : kg-m/sec [S.I.], g-cm/sec [C.G.S.] (6) Dimension : [MLT

1

]

(7) If two objects of different masses have same momentum, the lighter body possesses greater velocity.

v p = constant

p  m 1v1  m 2 v 2 = constant



v1 m 2  v 2 m1

i.e. v 

(8) For a given body p  v p

m= constant

v

1 m

[As p is constant]

m

(9) For different bodies at same velocities p  m p

v = constant

m

2

2 Newton’s Laws of Motion

2 4.4 Newton’s First Law.

A body continue to be in its state of rest or of uniform motion along a straight line, unless it is acted upon by some external force to change the state. (1) If no net force acts on a body, then the velocity of the body cannot change i.e. the body cannot accelerate. (2) Newton’s first law defines inertia and is rightly called the law of inertia. Inertia are of three types : Inertia of rest, Inertia of motion, Inertia of direction (3) Inertia of rest : It is the inability of a body to change by itself, its state of rest. This means a body at rest remains at rest and cannot start moving by its own. Example : (i) A person who is standing freely in bus, thrown backward, when bus starts suddenly. When a bus suddenly starts, the force responsible for bringing bus in motion is also transmitted to lower part of body, so this part of the body comes in motion along with the bus. While the upper half of body (say above the waist) receives no force to overcome inertia of rest and so it stays in its original position. Thus there is a relative displacement between the two parts of the body and it appears as if the upper part of the body has been thrown backward.

Note : 

If the motion of the bus is slow, the inertia of motion will be transmitted to the body of the person uniformly and so the entire body of the person will come in motion with the bus and the person will not experience any jerk.

(ii) When a horse starts suddenly, the rider tends to fall backward on account of inertia of rest of upper part of the body as explained above. (iii) A bullet fired on a window pane makes a clean hole through it while a stone breaks the whole window because the bullet has a speed much greater than the stone. So its time of contact with glass is small. So in case of bullet the motion is transmitted only to a small portion of the glass in that small time. Hence a clear hole is created in the glass window, while in case of ball, the time and the area of contact is large. During this time the motion is transmitted to the entire window, thus creating the cracks in the entire window. Cracks by the ball Hole by the bullet (iv) In the arrangement shown in the figure : (a) If the string B is pulled with a sudden jerk then it will experience tension while due to inertia of rest of mass M this force will not be transmitted to the string A and so the string B will break. A (b) If the string B is pulled steadily the force applied to it will be transmitted from string B to A through the mass M and as tension in A will be greater than in B by Mg M (weight of mass M) the string A will break. (v) If we place a coin on smooth piece of card board covering a glass and strike the B card board piece suddenly with a finger. The cardboard slips away and the coin falls into the glass due to inertia of rest. (vi) The dust particles in a durree falls off when it is beaten with a stick. This is because the beating sets the durree in motion whereas the dust particles tend to remain at rest and hence separate. (4) Inertia of motion : It is the inability of a body to change itself its state of uniform motion i.e., a body in uniform motion can neither accelerate nor retard by its own. Example : (i) When a bus or train stops suddenly, a passenger sitting inside tends to fall forward. This is because the lower part of his body comes to rest with the bus or train but the upper part tends to continue its motion due to inertia of motion.

genius PHYSICS

by Pradeep Kshetrapal

Newton’s Laws of Motion 3

3

(ii) A person jumping out of a moving train may fall forward. (iii) An athlete runs a certain distance before taking a long jump. This is because velocity acquired by running is added to velocity of the athlete at the time of jump. Hence he can jump over a longer distance. (5) Inertia of direction : It is the inability of a body to change by itself direction of motion. Example : (i) When a stone tied to one end of a string is whirled and the string breaks suddenly, the stone flies off along the tangent to the circle. This is because the pull in the string was forcing the stone to move in a circle. As soon as the string breaks, the pull vanishes. The stone in a bid to move along the straight line flies off tangentially. (ii) The rotating wheel of any vehicle throw out mud, if any, tangentially, due to directional inertia. (iii) When a car goes round a curve suddenly, the person sitting inside is thrown outwards. Sample problem based on Newton’s first law

Problem 1.

When a bus suddenly takes a turn, the passengers are thrown outwards because of [AFMC 1999; CPMT 2000, 2001]

(a) Inertia of motion

(b) Acceleration of motion

(c) Speed of motion

(d) Both (b) and (c)

Solution : (a)

Problem 2.

A person sitting in an open car moving at constant velocity throws a ball vertically up into air. The ball fall [EAMCET (Med.) 1995]

Solution : (d)

(a) Outside the car

(b) In the car ahead of the person

(c) In the car to the side of the person

(d) Exactly in the hand which threw it up

Because the horizontal component of velocity are same for both car and ball so they cover equal horizontal distances in given time interval.

4.5 Newton’s Second Law. (1) The rate of change of linear momentum of a body is directly proportional to the external force applied on the body and this change takes place always in the direction of the applied force.    (2) If a body of mass m, moves with velocity v then its linear momentum can be given by p  m v and if 

force F is applied on a body, then   dp dp F FK dt dt  d p or F dt   d   dv or F  (m v )  m  ma dt dt    F  ma

(K = 1 in C.G.S. and S.I. units)

 dv (As a   acceleration produced in the body) dt

Force = mass  acceleration

Sample problem based on Newton’s second law Problem 3.

A train is moving with velocity 20 m/sec. on this, dust is falling at the rate of 50 kg/min. The extra force required to move this train with constant velocity will be [RPET 1999] (a) 16.66 N

(b) 1000 N

(c) 166.6 N

(d) 1200 N

dm 50  20   16 .66 N 60 dt

Solution : (a)

Force F  v

Problem 4.

A force of 10 Newton acts on a body of mass 20 kg for 10 seconds. Change in its momentum is [MP PET 2002]

4

4 Newton’s Laws of Motion

4 (a)

5 kg m/s

(b) 100 kg m/s

(c) 200 kg m/s

(d) 1000 kg m/s

F u=0

Body remains at rest. Here force is trying to change the state of rest.

v=0

Solu tion : (b)

Pro

F

Body starts moving. Here force changes the state of rest. u=0

ble m 5.

v>0

(a) F v>u

u0

u

F

v Fl) In this condition the two bodies will move in the same direction (i.e. of applied force) but with different acceleration. Here force of kinetic friction  k mg will oppose the motion of A while will cause the motion of B. F  Fk  m a A F  Fk m (F   k mg )  m

Free body diagram of A

i.e. a A  aA

Note

:

Fk  M a B

maA

i.e. A Fk

F



aB  aB 

Fk M

 k mg M

As both the bodies are moving in the same direction.

Free body diagram of B MaB FK B

genius PHYSICS

by Pradeep Kshetrapal

Newton’s Laws of Motion 47

47

Acceleration of body A relative to B will be a  a A  a B  So, A will fall from B after time t 

2L  a

MF   k mg (m  M ) mM

2 m ML MF   k mg (m  M )

(iv) If there is friction between B and floor (where Fl    (M  m ) g = limiting friction between B and floor, Fk = kinetic friction between A and B) B will move only if Fk  Fl and then Fk  Fl  M a B However if B does not move then static friction will work (not limiting friction) between body B and the floor i.e. friction force = applied force (= Fk) not Fl .

MaB FK B Fl

(2) A force F is applied to the lower body, then following four situations are possible (i) When there is no friction (a) B will move with acceleration (F/M) while A will remain at rest (relative to ground) as there is no pulling force on A. F a B    and a A  0 M

(b) As relative to B, A will move backwards with acceleration (F/M) and so will fall from it in time t.  t

2L  a

2 ML F

(ii) If friction is present between A and B only and F < Fl (where F = Pseudo force on body A and Fl = limiting friction between body A and B) F (a) Both the body will move together with common acceleration a  M m mF (b) Pseudo force on the body A, F   ma  and Fl   s mg mM mF (c) F   Fl    s mg  F   s (m  M ) g mM F So both bodies will move together with acceleration a A  a B  if F   s [m  M ] g mM (iii) If friction is present between A and B only and F > Fl (where Fl = s (m + M)g = limiting friction between body B and surface) Both the body will move with different acceleration. Here force of kinetic friction  k mg will oppose the motion of B while will cause the motion of A. ma A  k mg

F  Fk  Ma B

Free body diagram of A

i.e. a A   k g

A

maA Fk

Note

:

[ F   k mg ] i.e. a B  M

As both the bodies are moving in the same direction

Acceleration of body A relative to B will be

Free body diagram of B MaB FK B

F

48

48 Newton’s Laws of Motion

48  F   k g(m  M )  a  a A  a B    M  

Negative sign implies that relative to B, A will move backwards and will fall it after time t

2L  a

2 ML F   k g(m  M )

(iv) If there is friction between B and floor : The system will move only if F  Fl' then replacing F by F  Fl . The entire case (iii) will be valid.

However if F  Fl the system will not move and friction between B and floor will be F while between A and B is zero.

Sample problems based on body resting on another Problem 19. A 4 kg block A is placed on the top of a 8 kg block B which rests on a smooth table. A just slips on B when a force of 12 N is applied on A. Then the maximum horizontal force on B to make both A and B move together, is (a) 12 N Solution : (c)

(b) 24 N

(c) 36 N

(d) 48 N

Maximum friction i.e. limiting friction between A and B, Fl = 12 N. If F is the maximum value of force applied on lower body such that both body move together It means Pseudo force on upper body is just equal to limiting friction

 F   4  F'  Fl  m    F  12  F  36 N . m  M  4 8 

Problem 20. A body A of mass 1 kg rests on a smooth surface. Another body B of mass 0.2 kg is placed over A as shown. The coefficient of static friction between A and B is 0.15. B will begin to slide on A if A is pulled with a force greater than

Solution : (a)

(a) 1.764 N

(b) 0.1764 N

B

(c) 0.3 N

(d) It will not slide for any F

A

B will begin to slide on A if Pseudo force is more than limiting friction

 F   F  F'  Fl  m    s R  m    0 .15 mg F  1.764 N m  M   m  M 

Problem 21. A block A of mass 2 kg rests on another block B of mass 8 kg which rests on a horizontal floor. The coefficient of friction between A and B is 0.2, while that between B and floor is 0.5. When a horizontal force of 25 N is applied on the block B, the force of friction between A and B is [IIT-JEE 1993] (a) Zero Solution : (a)

(b) 3.9 N

(c) 5.0 N

(d) 49 N

Limiting friction between the block B and the surface

FBS  BS .R  0.5 m  M  g  0.5(2  8)10  50 N but the applied force is 25 N so the lower block will not move i.e. there is no pseudo force on upper block A. Hence there will be no force of friction between A and B.

5.14 Motion of an Insect in the Rough Bowl.

A B

2kg 8kg

Surface

25 N

genius PHYSICS

by Pradeep Kshetrapal

Newton’s Laws of Motion 49

49

The insect crawl up the bowl up to a certain height h only till the component of its weight along the bowl is balanced by limiting frictional force. Let m = mass of the insect, r = radius of the bowl,  = coefficient of friction for limiting condition at point A O r R  mg cos  ......(i) and Fl  mg sin  ......(ii)  Fl

Dividing (ii) by (i)

tan   

So

Fl  R

r2  y 2  y

As Fl

 R 

 1 h  r  y  r 1   1  2 

y

A mg sin

mg cos

or y 

R

r

h

mg

1 2

  1  ,  h  r 1    1  2  

   

Problem 22. An insect crawls up a hemispherical surface very slowly (see the figure). The coefficient of friction between the insect and the surface is 1/3. If the line joining the centre of the hemispherical surface to the insect makes an angle  with the vertical, the maximum possible value of  is given by [IIT-JEE (Screening) 2001] (a) cot   3 (b) tan   3  (c) sec   3 (d) cosec   3 Solution : (a)

From the above expression, for the equilibrium R  mg cos  and F  mg sin  . Substituting these value in F  R we get tan    or cot  

1



 3.

5.15 Minimum Mass Hung From the String to Just Start the Motion. (1) When a mass m1 placed on a rough horizontal plane : Another mass m 2 hung from the string connected by pulley, the tension (T) produced in string will try to start the motion of mass m 1 . At limiting condition T  Fl 

m 2 g  R



m 2 g   m1 g

R F1

T

m1

T

m1g

 m 2  m 1 this is the minimum value of m 2 to start the motion.

Note

:  In the above condition Coefficient of friction  

m2

m2 m1

(2) When a mass m1 placed on a rough inclined plane : Another mass m 2 hung from the string connected by pulley, the tension (T) produced in string will try to start the motion of mass m 1 . At limiting condition For m 2 T  m 2 g For m 1

...... (i) R

T  m 1 g sin   F  T  m 1 g sin   R

 T  m 1 g sin   m 1 g cos 

T

m1

......(ii)

From equation (i) and (ii) m 2  m 1 [sin    cos  ]

T

m1g sin + F

 m1g

m2 m1g cos

m2g

50

50 Newton’s Laws of Motion

50 this is the minimum value of m 2 to start the motion

Note

:  In the above condition Coefficient of friction 

 m2  tan    m 1 cos  



Sample problems based on hung mass Problem 23. Two blocks of mass M1 and M 2 are connected with a string passing over a pulley as shown in the figure. The block M1 lies on a horizontal surface. The coefficient of friction between the block M1 and horizontal surface is . The system accelerates. What additional mass m should be placed on the block M1 so that the system does not accelerate M 2  M1 (a)

m



(b)

M2



M1

 M1

(c) M 2 

M1

M2



(d) (M 2  M1 ) Solution : (b)

By comparing the given condition with general expression M M M2  m  M1  2  m  2  M 1  m  M1  

Problem 24. The coefficient of kinetic friction is 0.03 in the diagram where mass m 2  20 kg and m1  4 kg . The acceleration of the block shall be (g  10 ms 2 )

T

20 kg m2

(a) 1.8 ms 2

T

(b) 0.8 ms 2

m1

(c) 1.4 ms 2 (d) 0.4 ms 2 Solution : (c)

m2a F

Let the acceleration of the system is a From the F.B.D. of m 2

T

m2

T  F  m 2 a  T  m 2 g  m 2a  T  0.03  20  10  20a  T  6  20a From the FBD of m1

.....(i)

m1

m1 g  T  m1a  4  10  T  4 a  40  T  4 a

....(ii)

Solving (i) and (ii) a  1.4 m / s . 2

5.16 Maximum Length of Hung Chain .

T

F m1g

m1 a

4 kg

genius PHYSICS

by Pradeep Kshetrapal

Newton’s Laws of Motion 51

51

A uniform chain of length l is placed on the table in such a manner that its l' part is hanging over the edge of table with out sliding. Since the chain have uniform linear density therefore the ratio of mass or ratio of length for any part of the chain will be equal. ( l – l m mass hanging from the table We know   2  [From article 5.15] ) m1 mass lying on the table l

 For this expression we can rewrite above expression in the following manner length hanging from the table [As chain have uniform linear density]  length lying on the table



l l  l

by solving l  

l (  1)

Problem 25. A heavy uniform chain lies on a horizontal table top. If the coefficient of friction between the chain and the table surface is 0.25, then the maximum fraction of the length of the chain that can hang over one edge of the table is [CBSE PMT 1990] (a) 20% (b) 25% (c) 35% (d) 15% Solution : (a)

From the expression  l' 

    0 .25   l   l'   l   1  0 .25  1   

[As  = 0.25]

0 . 25 l l = 20% of the length of the chain. 5 1 . 25

5.17 Coefficient of Friction Between Body and Wedge . A body slides on a smooth wedge of angle  and its time of descent is t. S



S Smooth wedge



Rough wedge

If the same wedge made rough then time taken by it to come down becomes n times more (i.e. nt) The length of path in both the cases are same. For smooth wedge 1 S  u t  at 2 2 1 .....(i) S  (g sin ) t 2 2 [As u  0 and a  g sin  ] From equation (i) and (ii)

For rough wedge 1 S  u t  at 2 2 1 S  g (sin   cos )(nt)2 .....(ii) 2 [As u  0 and a  g (sin    cos  )]

1 1 (g sin ) t 2 = g (sin   cos ) (nt) 2 2 2



sin   (sin    cos  ) n 2

52

52 Newton’s Laws of Motion

52 

 

  tan  1 

1  n 2 

Problem 26. A body takes just twice the time as long to slide down a plane inclined at 30 o to the horizontal as if the plane were frictionless. The coefficient of friction between the body and the plane is (a) Solution : (a)

3 4

(b)



  tan  1  

(c)

3

4 3

(d)

[JIPMER 1999]

3 4

1  1  3  .  tan 30 1  2   2  4 n  2  

5.18 Stopping of Block Due to Friction. (1) On horizontal road (i) Distance travelled before coming to rest : A block of mass m is moving initially with velocity u on a rough surface and due to friction it comes to rest after covering a distance S. Retarding force F  ma  R

ma   mg

 

a  g .

From

v  u  2aS  0  u  2  g S 2

S

2

2

[As v  0, a  g]

v=0

u

2



S 

u 2 g

or

S 

P2 2 m 2 g

(ii) Time taken to come to rest From equation v  u  a t  0  u   g t

[As momentum P = mu] [As v  0, a   g]

u g (iii) Force of friction acting on the body We know, F = ma (v  u) So, Fm t mu [As v = 0] F t  u  F   mg  As t  g   t 

(2) On inclined road : When block starts with velocity u its kinetic energy will be converted into potential energy and some part of it goes against friction and after travelling distance S it comes to rest i.e. v = 0. And we know that retardation a  g [sin    cos  ] By substituting the value of v and a in the following equation

v=0

v  u  2a S 2



2

0  u 2  2 g [sin    cos  ] S

S 

u2 2 g (sin    cos  )

S u



Sample problems based on motion of body on rough surface

genius PHYSICS

by Pradeep Kshetrapal

Newton’s Laws of Motion 53

53

Problem 27. A marble block of mass 2 kg lying on ice when given a velocity of 6 m/s is stopped by friction in 10s. Then the coefficient of friction is (a) 0.01 (b) 0.02 Solution : (d)

v  u  at  u  g t  0

 

(c) 0.03

(d) 0.06

u 6   0 .06 . gt 10  10

Problem 28. A 2 kg mass starts from rest on an inclined smooth surface with inclination 30 o and length 2 m. How much will it travel before coming to rest on a surface with coefficient of friction 0.25 (a) 4 m (b) 6 m (c) 8 m (d) 2 m Solution : (a)

v

v 2  u 2  2aS  0  2  g sin 30  2 Let it travel distance ‘S’ before coming to rest S

20

v2 20   4 m. 2 g 2  0 .25  10

5.19 Stopping of Two Blocks Due to Friction. When two masses compressed towards each other and suddenly released then energy acquired by each block will be dissipated against friction and finally block comes to rest i.e., F × S = E [Where F = Friction, S = Distance covered by block, E = Initial kinetic energy of the block]  FS 

P2 2m

[Where P = momentum of block] m1

P2  mg  S  2m

 S 

[As F =  mg]

A

B

m1

m2

m2 S2

S1

P2 2 m 2 g

In a given condition P and  are same for both the blocks. m  S 1 So S  2  1   2  S 2  m1  m

2

5.20 Velocity at the Bottom of Rough Wedge. A body of mass m which is placed at the top of the wedge (of height h) starts moving downward on a rough inclined plane. Loss of energy due to friction = FL (Work against friction) PE at point A = mgh u=0 KE at point B =

1 mv 2 2

L

m h

By the law of conservation of energy 1 i.e. mv 2  mgh  FL 2 v

A

m B v

2 (mgh  FL) m

5.21 Sticking of a Block With Accelerated Cart. When a cart moves with some acceleration toward right then a pseudo force (ma) acts on block toward left. This force (ma) is action force by a block on cart. Now block will remain static w.r.t. block. If friction force R  mg 

ma  mg [As R  ma ]

a F

F

M CART

ma

m mg

R

m

54

54 Newton’s Laws of Motion

54 g



a



a m in 

 g

 This is the minimum acceleration of the cart so that block does not fall. and the minimum force to hold the block together Fmin  (M  m ) a min

OR

Fm in  (M  m )

g



5.22 Sticking of a Person With the Wall of Rotor. A person with a mass m stands in contact against the wall of a cylindrical drum (rotor). The coefficient of friction between the wall and the clothing is . If Rotor starts rotating about its axis, then person thrown away from the centre due to centrifugal force at a particular speed w, the person stuck to the wall even the floor is removed, because friction force balances its weight in this condition. From the figure. Friction force (F) = weight of person (mg) F  R = mg R FC  [Here, Fc= centrifugal force]  Fc  mg 

m  m2 in r  mg



 m in 

mg

g r

Sample problems (Miscellaneous) Problem 29. A motorcycle is travelling on a curved track of radius 500m if the coefficient of friction between road and tyres is 0.5. The speed avoiding skidding will be

[MH

CET

(Med.)

2001]

(a) 50 m/s Solution : (a)

v

(b) 75 m/s

(c) 25 m/s

(d) 35 m/s

rg  0 . 5  500  10  50 m /s.

Problem 30. A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of

Solution : (a)

friction between the block and the wall is 0.2. The weight of the block is (a) 2 N (b) 20 N 10 N R (c) 50 N (d) 100 N For equilibrium Weight (W) = Force of friction (F) W  R  0.2  10 = 2N

[AIEEE 2003] F 10 N W

Problem 31. A body of mass 2 kg is kept by pressing to a vertical wall by a force of 100 N. The friction between wall Solution : (b)

and body is 0.3. Then the frictional force is equal to [Orissa JEE 2003] (a) 6 N (b) 20 N (c) 600 N (d) 700 N For the given condition Static friction = Applied force = Weight of body = 2 × 10 = 20 N.

Problem 32. A fireman of mass 60kg slides down a pole. He is pressing the pole with a force of 600 N. The coefficient of friction between the hands and the pole is 0.5, with what acceleration will the fireman slide down (g = 10 m/s 2 ) [Pb. PMT 2002] (a) 1 m/s2 (b) 2.5 m/s2 (c) 10 m/s2 (d) 5 m/s2

genius PHYSICS

by Pradeep Kshetrapal

Newton’s Laws of Motion 55

55

Solution : (d)

Friction = R = 0.5 × 600 = 300 N, Weight = 600 N W  F 600  300  ma = W – F  a  m 60

F R

600 N W

 a  5 m /s 2

Problem 33. The system shown in the figure is in equilibrium. The maximum value of W, so that the maximum value of static frictional force on 100 kg. body is 450 N, will be (a) 100 N (b) 250 N (c) 450 N (d) 1000 N Solution : (c)

For vertical equilibrium T1 sin 45 o  W For horizontal equilibrium T2  T1 cos 45 o 

W

 T1 

W o

45o

100 kg

W sin 45 o

cos 45 o = W

T1 sin 45o T1 F

T2

sin 45 and for the critical condition T2 = F  W = T2 = F = 450 N

45o T1 cos 45o W

 Basic level 1.

When a body is moving on a surface, the force of friction is called (a) Static friction

2.

[Kerala (Engg.) 2001]

(c) Sand

(d) Graphite

(b) 89 kg wt

(c) 49 kg wt

(d) 5 kg wt

A vehicle of mass m is moving on a rough horizontal road with momentum P. If the coefficient of friction between the tyres and the road be , then the stopping distance is [CBSE PMT 2001] (a)

5.

(b) Ball bearings

(d) Rolling friction

A block of mass 10 kg is placed on an inclined plane. When the angle of inclination is 30o, the block just begins to slide down the plane. The force of static friction is (a) 10 kg wt

4.

(c) Limiting friction

Which one of the following is not used to reduce friction (a) Oil

3.

(b) Dynamic friction

[MP PET 2002]

P 2 m g

(b)

P2 2 m g

(c)

P 2 m 2g

(d)

P2 2 m 2 g

A box is lying on an inclined plane what is the coefficient of static friction if the box starts sliding when an angle of inclination is 60o [KCET (Engg./Med.) 2000]

(a) 1.173

6.

(b) 1.732

(c) 2.732

A brick of mass 2 kg begins to slide down on a plane inclined at an angle of

45o

(d) 1.677 with the horizontal. The force of friction will be [CPMT 2000]

56

56 Newton’s Laws of Motion

56 (a) 19.6 sin 45o

7.

8.

(b) 19.6 cos 45o

(c) 9.8 sin 45o

(d) 9.8 cos 45o

To avoid slipping while walking on ice, one should take smaller steps because of the

[BHU 1999]

(a) Friction of ice is large

(b) Larger normal reaction

(c) Friction of ice is small

(d) Smaller normal reaction

Two bodies having the same mass, 2 kg each have different surface areas 50 m 2 and 100 m 2 in contact with a horizontal plane. If the coefficient of friction is 0.2, the forces of friction that come into play when they are in motion will be in the ratio [EAMCET (Med.) 1999]

(a) 1 : 1

9.

(b) 1 : 2

11.

12.

13.

(d) 1 : 4

Starting from rest, a body slides down a 45 o inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is [CBSE PMT 1990] (a) 0.33

10.

(c) 2 : 1

(b) 0.25

(c) 0.75

(d) 0.80

Brakes of very small contact area are not used although friction is independent of area, because friction (a) Resists motion

(b) Causes wear and tear

(c) Depends upon the nature of materials

(d) Operating in this case is sliding friction

The angle between frictional force and the instantaneous velocity of the body moving over a rough surface is (a) Zero

(b) /2

(c) 

(d) Equal to the angle of friction

What happens to the coefficient of friction, when the normal reaction is halved (a) Halved

(b) Doubled

(c) No change

(d) Depends on the nature of the surface

What can be inferred regarding the limiting frictional force in the following four figures R

R

R

R A

mg

mg

(a) FA  FB  FC  FD

14.

18.

(c)

FA  FB  FC  FD

(d)

FA  FB  FC  FD

(b) 0.89

(c) 0.49

(d) 0.1

(b) 4.9 m / sec 2

(c) 3.92 m / sec 2

(d) 1 m / sec 2

A particle is projected along a line of greatest slope up a rough plane inclined at an angle of 45 o with the horizontal. If the 1 coefficient of friction is , then the retardation is 2 (a)

17.

FA  FB  FC  FD

mg

mg

A 60 kg body is pushed with just enough force to start it moving across a floor and the same force continues to act afterwards. The coefficients of static and sliding friction are 0.5 and 0.4 respectively. The acceleration of the body is (a) 6 m / sec 2

16.

(b)

D

A force of 98 Newton is required to drag a body of mass 100 kg on ice. The coefficient of friction will be (a) 0.98

15.

C

B

g 2

(b)

g 2 2

(c)

g  1 1   2  2

(d)

g  1 1   2  2

A block moves down a smooth inclined plane of inclination . Its velocity on reaching the bottom is v. If it slides down a rough inclined plane of same inclination its velocity on reaching the bottom is v/n, where n is a number greater than 0. The coefficient of friction  is given by

1  (a)   tan  1  2   n 

1  (b)   cot  1  2   n 

1

1 2  (c)   tan  1  2   n 

1

(d)

1 2    cot  1  2   n 

genius PHYSICS

by Pradeep Kshetrapal

Newton’s Laws of Motion 57

57

19.

Consider a car moving along a straight horizontal road with a speed of 72 km/hr. If the coefficient of static friction between the tyres and the road is 0.5, the shortest distance in which the car can be stopped is (g  10 m / s 2 ) (a) 30 m

20.

(b) 40 m

(c) 72 m

(d) 20 m

All the surfaces shown in the figure are rough. The direction of friction on B due to A is (a) Zero (b) To the left

BA

F

(c) Upwards (d) Downwards

21.

A body of mass M just starts sliding down an inclined plane (rough) with inclination , such that tan = 1/3. The force acting on the body down the plane in this position is (a)

Mg

(b)

Mg 3

2 Mg 3

(c)

(d)

Mg 10

 Advance level 22.

Consider the following statements Assertion (A) : It is difficult to move a cycle along the road with its brakes on. Reason (R) : Sliding friction is greater than rolling friction. Of these statements

[AIIMS 2002]

(a) Both A and R are true and the R is a correct explanation of the A (b) Both A and R are true but the R is not a correct explanation of the A (c) A is true but the R is false (d) Both A and R are false (e) A is false but the R is true

23.

A body is sliding down an inclined plane having coefficient of friction 0.5. If the normal reaction is twice that of the resultant downward force along the incline, the angle between the inclined plane and the horizontal is [EAMCET (Engg.) 2000] (a) 15o

24.

(d) 60o

(b) 0.7  9 .8  3 N

(c) 9 . 8  3 N

(d) 0.7  9.8 N

A body of weight W is lying at rest on a rough horizontal surface. If the angle of friction is , then the minimum force required to move the body along the surface will be (a) W tan

26.

(c) 45o

A block of mass 2 kg rests on a rough inclined plane making an angle of 30 o with the horizontal. The coefficient of static friction between the block and the plane is 0.7. The frictional force on the block is [IIT-JEE 1980] (a) 9.8 N

25.

(b) 30o

(b) W cos

(c) W sin

(d) W cos

A block of mass M is placed on a rough horizontal surface as shown in the figure. A force F = Mg acts on the block. It is inclined to the vertical at an angle . The coefficient of friction is . The block can be pushed along the surface only when (a) tan    F = Mg

(b) cot    (c)

 M

tan  / 2  

(d) cot  / 2  

27.

A plane is inclined at an angle  with the horizontal. A body of mass m rests on it. If the coefficient of friction is , then the minimum force that has to be applied parallel to the inclined plane to make the body just move up the inclined plane is (a) mg sin 

(b)  mg cos

(c)  mg cos – mg sin

(d)  mg cos + mg sin

58

58 Newton’s Laws of Motion

58 28.

A block of mass m is placed on another block of mass M which itself is lying on a horizontal surface. The coefficient of friction between the two blocks is  1 and that between the block of mass M and horizontal surface is  2 . What maximum horizontal force can be applied to the lower block so that the two blocks move without separation (a) (M + m) (2  1 )g

m

(b) (M – m) (2  1 )g

M

(c) (M – m) (2  1 )g (d) (M + m) (2  1 )g

29.

A block of mass M1 is placed on a slab of mass M2. The slab lies on a frictionless horizontal surface. The coefficient of static friction between the block and slab is 1 and that of dynamic friction is 2. A force F acts on the block M1. Take g = 10 ms–2. If M1 = 10 kg, M2 = 30 kg, 1 = 0.5, 2 = 0.15 and F = 40 N, what will be the acceleration with which the slab will move (a) 5 ms 2 (b) 2 ms 2 (c) 1 ms 2 (d) Zero

30.

In the above problem if F = 100 N, what will be the acceleration with which the slab will move (a) 5 ms 2

31.

(b) 2 ms 2

(c) 1 ms 2

(d) None of these

A block X of mass 4 kg is lying on another block Y of mass 8 kg. As shown in the figure. When the force acting on X is 12N, block X is on the verge of slipping on Y. The force F in Newton necessary to make both X and Y move simultaneously will be (a) 36 (b) 3.6 (c) 0.36 (d) 3.6

32.

Two masses 10 kg and 5 kg are connected by a string passing over a pulley as shown. If the coefficient of friction be 0.15, then the minimum weight that may be placed on 10 kg to stop motion is (a) 18.7 kg (b) 23.3 kg (c) 32.5 kg (d) 44.3 kg

33.

Two blocks of mass M1 and M2 are connected with a string which passes over a smooth pulley. The mass M1 is placed on a rough inclined plane as shown in the figure. The coefficient of friction between the block and the inclined plane is . What should be the maximum mass M2 so that block M1 slides downwards (a) M 2  M1(sin    cos  ) (b) M 2  M1 (sin    cos  ) (c) M 2  M1 /(sin    cos  ) (d) M 2  M1 /(sin    cos  )

34.

A car starts from rest to cover a distance s. the coefficient of friction between the road and the tyres is . The minimum time in which the car can cover the distance is proportional to (a) 

35.

(b)



(c)

1 

(d)

1



An engine of mass 50,000 kg pulls a coach of mass 40,000 kg. If there is a resistance of 1 N per 100 kg acting on both the engine and the coach, and if the driving force of the engine be 4,500 N, then the acceleration of the engine is

genius PHYSICS

by Pradeep Kshetrapal

Newton’s Laws of Motion 59

59

(a) 0.08 m / s 2

36.

(c) 0.04 m / s 2

(d) None of these

(c) 500 N

(d) 1000 N

In the above question, then tension in the coupling is (a) 2,000 N

37.

(b) Zero (b) 1,500 N

An aeroplane requires for take off a speed of 72 km/h. The run of the ground is 100m.The mass of the plane is 10 4 kg and the coefficient of friction between the plane and the ground is 0.2. The plane accelerates uniformly during take off. What is the acceleration of the plane (a) 1 m/s2

38.

(d) 4 m/s2

(b) tan  = 2 tan 

(c) tan  = 3 tan 

(d) tan  = 3 tan 

A body is on a rough horizontal plane. A force is applied to the body direct towards the plane at an angle  with the vertical. If  is the angle of friction then for the body to move along the plane (a)   

40.

(c) 3 m/s2

The force required to just move a body up an inclined plane is double the force required to just prevent it from sliding down. If  is angle of friction and  is the angle which incline makes with the horizontal then (a) tan  = tan 

39.

(b) 2 m/s2

(b)   

(c)   

(d)  can take up any value

In the arrangement shown W1  200 N , W2  100 N ,   0.25 for all surfaces in contact. The block W1 just slides under the block W2 (a) A pull of 50 N is to be applied on W1 (b) A pull of 90 N is to be applied on W1 (c) Tension in the string AB is 10 2 N (d) Tension in the string AB is 20 2 N

41.

A board of mass m is placed on the floor and a man of mass M is standing on the board as shown. The coefficient of friction between the board and the floor is . The maximum force that the can exert on he rope so that the board does not slip on the floor is (a) F  (M  m)g (b) F  mg (c)

F

(d) F 

42.

Mg  1

 (M  m )g  1

A body slides over an inclined plane forming an angle of 45° with the horizontal. The distance x travelled by the body in time t is described by the equation x  kt 2 , where k  1.732. The coefficient of friction between the body and the plane has a value (a)   0.5

43.

(b)   1

(c)

  0.25

(d)   0.75

Two blocks A and B of masses m and M respectively are placed on each other and their combination rests on a fixed horizontal surface C. A light string passing over the smooth light pulley is used to connect A and B as shown. The coefficient of sliding friction between all surfaces in contact is . If A is dragged with a force F then for both A and B to move with a uniform speed we have (a) F  (M  m)g (b) F  mg (c)

F  (3 M  m)g

(d) F  (3m  M)g

44.

A force of 100 N is applied on a block of mass 3 kg as shown in figure. The coefficient of friction between the surface of the block is 1/4. The friction force acting on the block is (a) 15 N downwards

(b) 25 N upwards

(c)20 N downwards (d)20 N upwards F = 100 N

60

60 Newton’s Laws of Motion

60

Answer Sheet (Practice problems) 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

b

c

d

d

b

a

c

a

c

b

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

c

c

a

d

d

c

a

b

c

d

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

a

c

a

a

c

d

d

c

d

a

31.

32.

33.

34.

35.

36.

37.

38.

39.

40.

b

b

d

c

a

b

c

a

b, d

d

41.

42.

43.

44

a

d

c

b

Work, Energy Power, and Collision 1

6.1 Introduction. The terms 'work', 'energy' and 'power' are frequently used in everyday language. A farmer clearing weeds in his field is said to be working hard. A woman carrying water from a well to her house is said to be working. In a drought affected region she may be required to carry it over large distances. If she can do so, she is said to have a large stamina or energy. Energy is thus the capacity to do work. The term power is usually associated with speed. In karate, a powerful punch is one delivered at great speed. In physics we shall define these terms very precisely. We shall find that there is at best a loose correlation between the physical definitions and the physiological pictures these terms generate in our minds. Work is said to be done when a force applied on the body displaces the body through a certain distance in the direction of force.

6.2 Work Done by a Constant Force. Let a constant force F be applied on the body such that it makes an angle  with the horizontal and body is displaced through a distance s F sin

By resolving force F into two components :

F

(i) F cos in the direction of displacement of the body.

 F cos

(ii) F sin in the perpendicular direction of displacement of the body. Since body is being displaced in the direction of F cos , therefore work

s

done by the force in displacing the body through a distance s is given by W  (F cos  ) s  Fs cos 

or

W  F.s

Thus work done by a force is equal to the scalar or dot product of the force and the displacement of the body. If a number of force F 1 , F 2 , F 3 ...... F n are acting on a body and it shifts from position vector r 1 to position vector r 2 then W  (F1  F 2  F 3  .... F n ).( r 2  r 1 )

6.3 Nature of Work Done. Positive work

Negative work

Positive work means that force (or its component) is parallel to displacement

Negative work means that force (or its component) is opposite to displacement i.e.

Collection by Pradeep Kshetrapal for students at genius academy, Maxwell classes Korba, Gupta classes Kusmunda

2 Work, Energy, Power and Collision Direction of motion F

0 o    90 o



F

Direction of motion

o  90 o    180

s

The positive work signifies that the external force favours the motion of the body. Example: (i) When a person lifts a body from the ground, the work done by the (upward) lifting force is positive

s

The negative work signifies that the external force opposes the motion of the body. Example: (i) When a person lifts a body from the ground, the work done by the (downward) force of gravity is negative.

F man

Fg

s

s

(ii) When a lawn roller is pulled by applying a force along the handle at an acute angle, work done by the applied force is positive.

(ii) When a body is made to slide over a rough surface, the work done by the frictional force is negative.

F

s

F

s

(iii) When a spring is stretched, work done by the external (stretching) force is positive.

(iii) When a positive charge is moved towards another positive charge. The work done by electrostatic force between them is negative.

s F

+

F

s

+

Maximum work : Wmax  F s

Minimum work : Wmin   F s

When cos  maximum  1 i.e.   0 o

When cos   minimum  1 i.e   180 o

It means force does maximum work when angle between force and displacement is zero.

It means force does minimum [maximum negative] work when angle between force and displacement is 180 o.

Zero work Under three condition, work done becomes zero W  Fs cos  0 (1) If the force is perpendicular to the displacement [ F  s ] Example: (i)

When a coolie travels on a horizontal platform with a load on his head, work done against gravity by the coolie is zero.(Work done against friction is +ve)

(ii) When a body moves in a circle the work done by the centripetal force is always zero. (iii) In case of motion of a charged particle in a magnetic field as force [F  q(v  B)] is always perpendicular to motion, work done by this force is always zero. Collection by Pradeep Kshetrapal for students at genius academy, Maxwell classes Korba, Gupta classes Kusmunda

s

Fg

Work, Energy Power, and Collision 3 (2) If there is no displacement [s = 0] Example: (i) When a person tries to displace a wall or heavy stone by applying a force then it does not move, the work done is zero.

F s

0

(ii) A weight lifter does work in lifting the weight off the ground but does not work in holding it up. (3) If there is no force acting on the body [F = 0] Example: Motion of an isolated body in free space.

Sample Problems based on work done by constant force Problem 1.

A body of mass 5 kg is placed at the origin, and can move only on the x-axis. A force of 10 N is acting on it in a direction making an angle of 60 o with the x-axis and displaces it along the xaxis by 4 metres. The work done by the force is (a) 2.5 J

Solution : (d)

Problem 2.

(b) 7.25 J (c) 40 J  o Work done  F.s  F s cos   10  4  cos 60  20 J

A force F  (5ˆi  3ˆj) N is applied over a particle which displaces it from its origin to the point r  (2ˆi  1ˆj) metres. The work done on the particle is (a) –7 J

Solution : (c)

Problem 3.

(d) 20 J

(b) +13 J (c) +7 J  Work done  F.r  (5 i  3 ˆj).(2ˆi  ˆj)  10  3  7 J

(d) +11 J

A horizontal force of 5 N is required to maintain a velocity of 2 m/s for a block of 10 kg mass sliding over a rough surface. The work done by this force in one minute is (a) 600 J

(b) 60 J

(c) 6 J

(d) 6000 J

Solution : (a)

Work done = Force  displacement = F  s = F  v  t = 5  2  60 = 600 J.

Problem 4.

A box of mass 1 kg is pulled on a horizontal plane of length 1 m by a force of 8 N then it is raised vertically to a height of 2m, the net work done is (a) 28 J

Solution : (a)

(b) 8 J

(c) 18 J

(d) None of above

Work done to displace it horizontally = F  s = 8  1 = 8 J Work done to raise it vertically F  s = mgh = 1  10  2 = 20 J  Net work done = 8 +20 = 28 J

Problem 5.

A 10 kg satellite completes one revolution around the earth at a height of 100 km in 108 minutes. The work done by the gravitational force of earth will be (a) 108  100  10 J

Solution : (d)

(b)

108  10 J 100

(c)

100  10 J 108

(d) Zero

Work done by centripetal force in circular motion is always equal to zero.

6.4 Work Done by a Variable Force. When the magnitude and direction of a force varies with position, the work done by such a force for an infinitesimal displacement is given by dW  F. d s The total work done in going from A to B as shown in the figure is Collection by Pradeep Kshetrapal for students at genius academy, Maxwell classes Korba, Gupta classes Kusmunda

4 Work, Energy, Power and Collision W 



B

A

F. d s 



B

(F cos  )ds

A

B

In terms of rectangular component F  Fx ˆi  Fy ˆj  Fz kˆ

ds

d s  dx ˆi  dyˆj  dz kˆ



A

B

W 

 (F ˆi  F ˆj  F kˆ ).(dx ˆi  dyˆj  dz kˆ )

or W 



x

A

xB

xA

y

Fx dx 

F

z



yB

yA

Fy dy 



zB

zA

Fz dz

Sample Problems based on work done by variable force

Problem 6.

 A position dependent force F  (7  2 x  3 x 2 ) N acts on a small abject of mass 2 kg to displace

it from x  0 to x  5m . The work done in joule is (a) 70 J Solution : (d)

(b) 270 J

Work done 



x2

F dx 

x1

Problem 7.

5

 (7  2 x  3 x

[CBSE PMT 1994]

(c) 35 J 2

0

(d) 135 J

)dx  [7 x  x 2  x 3 ]50  35  25  125  135 J

A particle moves under the effect of a force F = Cx from x = 0 to x = x1. The work done in the process is [CPMT 1982]

(a) Cx 12 Solution : (b)

(b)

Work done 



x2

x1

Problem 8.

F dx 



x1

0

1 Cx 12 2

x2  Cx dx  C    2 

(c) Cx 1 x1

 0

(d) Zero

1 C x 12 2

The vessels A and B of equal volume and weight are immersed in water to a depth h. The vessel A has an opening at the bottom through which water can enter. If the work done in immersing A and B are W A and WB respectively, then (a) WA  WB

(b) WA  WB

(c) WA  WB

(d) WA    WB

Solution : (b)

When the vessels are immersed in water, work has to be done against up-thrust force but due to opening at the bottom in vessel A, up-thrust force goes on decreasing. So work done will be less in this case.

Problem 9.

Work done in time t on a body of mass m which is accelerated from rest to a speed v in time t1 as a function of time t is given by (a)

Solution : (d)

1 v 2 m t 2 t1

(b) m

v 2 t t1

2

(c)

1 1 v 1  Work done = F.s = ma. a t 2   m a 2 t 2  m  2 2  t1 2 

1  mv  2   t 2  t1  2

 2  t 

(d)

  v given   As accelerati on (a)  t1  

6.5 Dimension and Units of Work. Dimension : As work = Force  displacement 

1 v2 2 m t 2 t12

[W] = [Force]  [Displacement]  [MLT 2 ]  [L]  [ML 2 T 2 ]

Collection by Pradeep Kshetrapal for students at genius academy, Maxwell classes Korba, Gupta classes Kusmunda

Work, Energy Power, and Collision 5

Units : The units of work are of two types Absolute units

Gravitational units

Joule [S.I.]: Work done is said to be one Joule, when 1 Newton force displaces the body through 1 meter in its own direction.

kg-m [S.I.]: 1 Kg-m of work is done when a force of 1kg-wt. displaces the body through 1m in its own direction.

From W = F.s

From W = F s

1 Joule = 1 Newton  1 metre

1 kg-m = 1 kg-wt  1 metre = 9.81 N  1 metre = 9.81 Joule

Erg [C.G.S.] : Work done is said to be one erg when 1 dyne force displaces the body through 1 cm in its own direction.

gm-cm [C.G.S.] : 1 gm-cm of work is done when a force of 1gm-wt displaces the body through 1cm in its own direction.

From W = F s

From W = F s 1 gm-cm = 1gm-wt  1cm. = 981 dyne 

1 Erg  1Dyne  1cm

1cm

Relation between Joule and erg 1 Joule = 1 N  1 m = 105 dyne  102 cm

= 981 erg

= 107 dyne  cm = 107 Erg

6.6 Work Done Calculation by Force Displacement Graph. Let a body, whose initial position is x i , is acted upon by a variable force (whose magnitude is changing continuously) and consequently the body acquires its final position x f . Let F be the average value of variable force within the interval dx from position x to (x + dx) i.e. for small displacement dx. The work done will be the area of the shaded strip of width dx. The work done on the body in displacing it from position x i to x f will be equal to the sum of areas of all the such strips Force dW  F dx

W 

W 



xf



xf

xi

xi

dW 



xf

xi

 F

F dx

(Area of strip of width dx )

O

xi x

W  Area under curve Between x i and x f

xf Displacement

dx

i.e. Area under force displacement curve with proper algebraic sign represents work done by the force.

Sample problems based on force displacement graph

Problem 10. A 10 kg mass moves along x-axis. Its acceleration as a function of its position is shown in the

(a) 8  10 2 J

(b) 16  10 2 J

(c) 4  10 4 J

(d) 1 .6  10 3 J

a (cm/sec2)

figure. What is the total work done on the mass by the force as the mass moves from x  0 to [AMU (Med.) 2000] x  8 cm 20 15 10 5

Solution : (a) Work done on the mass = mass  covered area between the graph and displacement axis on a-t graph. 1 0 2 4 6 8 x (cm) = 10  (8  10  2 )  20  10  2 = 8  10 2 J. 2 Problem 11. The relationship between force and position is shown in the figure given (in one dimensional case). The work done by the force in displacing a body from x  1 cm to x  5 cm is

Collection by Pradeep Kshetrapal for students at genius academy, Maxwell classes Korba, Gupta classes Kusmunda

6 Work, Energy, Power and Collision (a) 20 ergs

20 10 0 10 20

Force (dyne)

(b) 60 ergs (c) 70 ergs (d) 700 ergs Solution : (a) erg.

1

2

3

4

5 6 x (cm)

Work done = Covered area on force-displacement graph = 1  10 + 1  20 – 1  20 + 1  10 = 20

Problem 12. The graph between the resistive force F acting on a body and the distance covered by the body is

shown in the figure. The mass of the body is 25 kg and initial velocity is 2 m/s. When the distance covered by the body is 5 m , its kinetic energy would be 50 J 40 J 20 J 10 J

20

F (Newton)

(a) (b) (c) (d)

10

0 1 2 3 4 x (m) 1 1 mu 2   25  (2) 2  50 J 2 2 Final kinetic energy = Initial energy – work done against resistive force (Area between graph and 1  50   4  20  50  40  10 J. displacement axis) 2

Solution : (d)

Initial kinetic energy of the body 

6.7 Work Done in Conservative and Non-Conservative Field . (1) In conservative field work done by the force (line integral of the force i.e.

 F.d l )

is

independent of the path followed between any two points. W A B  W A B  W AB Path I

or

Path II

I

A

Path III

 F.d l   F.d l   F.d l Path I

Path II

B

II III

Path III

(2) In conservative field work done by the force (line integral of the force i.e.

 F.d l ) over a closed

path/loop is zero. W A  B  WB  A  0 or

 F.dl  0

A

B

Conservative force : The forces of these type of fields are known as conservative forces. Example : Electrostatic forces, gravitational forces, elastic forces, magnetic forces etc and all the central forces are conservative in nature. If a body of man m lifted to height h from the ground level by different path as shown in the figure B

B

II

I

Work done through different paths WI  F. s  mg  h  mgh

 A

A

B

B

II I

l A

I V A

h  mgh sin  mgh 1  0  mgh 2  0  mgh 3  0  mgh 4  mg (h1  h2  h3  h4 )  mgh

WII  F. s  mg sin  l  mg sin  WIII

h



WIV  F. d s  mgh Collection by Pradeep Kshetrapal for students at genius academy, Maxwell classes Korba, Gupta classes Kusmunda

Work, Energy Power, and Collision 7

It is clear that WI  WII  WIII  WIV  mgh . Further if the body is brought back to its initial position A, similar amount of work (energy) is released from the system it means WAB  mgh and WBA  mgh . Hence the net work done against gravity over a round strip is zero. WNet  WAB  WBA  mgh  (mgh )  0 i.e. the gravitational force is conservative in nature. Non-conservative forces : A force is said to be non-conservative if work done by or against the force in moving a body from one position to another, depends on the path followed between these two positions and for complete cycle this work done can never be a zero. Example: Frictional force, Viscous force, Airdrag etc. If a body is moved from position A to another position B on a rough table, work done against frictional force shall depends on the length of the path between A and B and not only on the position A and B. W AB  mgs Further if the body is brought back to its initial position A, work has to be done against the frictional force, which always opposes the motion. Hence the net work done against the friction over a round trip is not zero. R s WBA  mgs . F WNet  WAB  WBA  mgs  mgs  2 mgs  0. i.e. the friction is a non-conservative force.

Sample problems based on work done in conservative and non-conservative field Problem 13. If W1 , W2 and W3 represent the work done in moving a particle from A to B along three different paths 1, 2 and 3 respectively (as shown) in the gravitational field of a point mass m, find the correct relation B

Solution : (b)

(a) W1  W2  W3

(b) W1  W2  W3

(c) W1  W2  W3

(d) W2  W1  W3

1

m 2 3

A

As gravitational field is conservative in nature. So work done in moving a particle from A to B does not depends upon the path followed by the body. It always remains same.

Problem 14. A particle of mass 0.01 kg travels along a curve with velocity given by 4ˆi  16kˆ ms-1. After some time, its velocity becomes 8ˆi  20 ˆj ms 1 due to the action of a conservative force. The work done on particle during this interval of time is (a) 0.32 J Solution : (d)

(b) 6.9 J

(c) 9.6 J

(d) 0.96 J

v1  4 2  16 2  272 and v 2  8 2  20 2  464

Work done = Increase in kinetic energy 

1 1 m [v 22  v 12 ]   0 . 01[464  272 ]  0 . 96 J . 2 2

6.8 Work Depends on Frame of Reference. With change of frame of reference (inertial) force does not change while displacement may change. So the work done by a force will be different in different frames. Examples : (1) If a porter with a suitcase on his head moves up a staircase, work done by the upward lifting force relative to him will be zero Collection by Pradeep Kshetrapal for students at genius academy, Maxwell classes Korba, Gupta classes Kusmunda h

8 Work, Energy, Power and Collision (as displacement relative to him is zero) while relative to a person on the ground will be mgh. (2) If a person is pushing a box inside a moving train, the work done in the frame of train will F. s while in the frame of earth will be F. (s  s 0 ) where s 0 is the displacement of the train relative

to the ground.

6.9 Energy. The energy of a body is defined as its capacity for doing work. (1) Since energy of a body is the total quantity of work done therefore it is a scalar quantity. (2) Dimension: [ML 2 T 2 ] it is same as that of work or torque. (3) Units : Joule [S.I.], erg [C.G.S.] Practical units : electron volt (eV), Kilowatt hour (KWh), Calories (Cal) Relation between different units: 1 Joule = 10 7 erg 1 eV = 1.6  10 19 Joule 1 KWh = 3.6  10 6 Joule 1 Calorie = 4 .18 Joule (4) Mass energy equivalence : Einstein’s special theory of relativity shows that material particle itself is a form of energy. The relation between the mass of a particle m and its equivalent energy is given as where c = velocity of light in vacuum. E  mc 2 27 kg then E  931 MeV  1 .5  10 10 Joule . If m  1 amu  1 .67  10 If m  1kg then E  9  10 16 Joule Examples : (i) Annihilation of matter when an electron (e  ) and a positron (e  ) combine with each other, they annihilate or destroy each other. The masses of electron and positron are converted into energy. This energy is released in the form of  -rays. e  e   Each  photon has energy = 0.51 MeV. Here two  photons are emitted instead of one  photon to conserve the linear momentum.

(ii) Pair production : This process is the reverse of annihilation of matter. In this case, a photon ( ) having energy equal to 1.02 MeV interacts with a nucleus and give rise to electron (e  ) and positron (e  ) . This energy is converted into matter.  (Photon)

e– + e+

(iii) Nuclear bomb : When the nucleus is split up due to mass defect (The difference in the mass of nucleons and the nucleus) energy is released in the form of  -radiations and heat. (5) Various forms of energy (i) Mechanical energy (Kinetic and Potential) (ii) Chemical energy (iii) Electrical energy (iv) Magnetic energy (v) Nuclear energy (vi) Sound energy (vii) Light energy (viii) Heat energy

Collection by Pradeep Kshetrapal for students at genius academy, Maxwell classes Korba, Gupta classes Kusmunda

Work, Energy Power, and Collision 9

(6) Transformation of energy : Conversion of energy from one form to another is possible through various devices and processes. Mechanical  electrical

N

Light  Electrical

Chemical  electrical

Dynamo

Cathode +

Anode

S

–

Photoelectric cell Chemical  heat

Primary cell

Sounds  Electrical

Heat  electrical Fe Hot

Coal Burning

Microphone

Heat  Mechanical

Engine

Thermo-couple

Electrical  Mechanical

Motor Electrical  Sound

G

Cu

Electrical  Heat

Heater Electrical  Chemical

Anode +

Speaker

Cold

Electrical  Light

Cathod – e

Voltameter

Bulb Electrolyte

Sample problems based on energy Problem 15. A particle of mass ‘m’ and charge ‘q’ is accelerated through a potential difference of ‘V’ volt. Its energy is

[UPSEAT 2001]

q q (a) qV (b) mqV (c)  V (d) mV m  Solution : (a) Energy of charged particle = charge  potential difference = qV Problem 16. An ice cream has a marked value of 700 kcal. How many kilowatt hour of energy will it deliver to the body as it is digested (a) 0.81 kWh (b) 0.90 kWh (c) 1.11 kWh (d) 0.71 kWh 3 700  10  4 .2  0 .81 kWh Solution : (a) 700 k cal = 700  10 3  4.2 J  [As 3 .6  10 6 J  1 kWh ] 3 .6  10 6 Problem 17. A metallic wire of length L metres extends by l metres when stretched by suspending a weight Mg to it. The mechanical energy stored in the wire is Collection by Pradeep Kshetrapal for students at genius academy, Maxwell classes Korba, Gupta classes Kusmunda

10 Work, Energy, Power and Collision (a) 2 Mgl Solution : (c)

(b) Mgl

Elastic potential energy stored in wire U 

(c)

Mgl 2

(d)

Mgl 4

1 Mgl Fx  . 2 2

6.10 Kinetic Energy.

The energy possessed by a body by virtue of its motion is called kinetic energy. Examples : (i) Flowing water possesses kinetic energy which is used to run the water mills. (ii) Moving vehicle possesses kinetic energy. (iii) Moving air (i.e. wind) possesses kinetic energy which is used to run wind mills. (iv) The hammer possesses kinetic energy which is used to drive the nails in wood.

(v) A bullet fired from the gun has kinetic energy and due to this energy the bullet penetrates into a target.

(1) Expression for kinetic energy : Let v u= m = mass of the body, u = Initial velocity of the body (= 0) 0 F F = Force acting on the body, a = Acceleration of the body s = Distance travelled by the body, v = Final velocity of the body s From v 2  u 2  2as 2 v  v 2  0  2as  s  2a Since the displacement of the body is in the direction of the applied force, then work done by the force is v2 W  F  s  ma  2a 1  W  mv 2 2 1 This work done appears as the kinetic energy of the body KE  W  mv 2 2 (2) Calculus method : Let a body is initially at rest and force F is applied on the body to displace it through ds along its own direction then small work done dW  F.d s  F ds  [As F = ma] dW  m a ds  

dv    As a  dt   

dv ds dt ds dW  mdv . dt dW  m

ds    v …….(i)  As dt   Therefore work done on the body in order to increase its velocity from zero to v is given by



dW  m v dv

v

W

v v 2  1 mv dv  m v dv  m    mv 2 0 0 2  2 0



v



This work done appears as the kinetic energy of the body KE 

1 mv 2 . 2

1 m (v . v ) 2 As m and v .v are always positive, kinetic energy is always positive scalar i.e. kinetic energy can never be negative. In vector form KE 

Collection by Pradeep Kshetrapal for students at genius academy, Maxwell classes Korba, Gupta classes Kusmunda

Work, Energy Power, and Collision 11

(3) Kinetic energy depends on frame of reference : The kinetic energy of a person of 1 mass m, sitting in a train moving with speed v, is zero in the frame of train but mv 2 in the frame of 2 the earth. 1 (4) Kinetic energy according to relativity : As we know E  mv 2 . 2 But this formula is valid only for (v F electromagnetic > F gravitational . (7) The ratio of gravitational force to electrostatic force between two electrons is of the order of 10 43 . (8) It is a conservative force i.e. work done by it is path independent or work done in moving a particle round a closed path under the action of gravitational force is zero. (9) It is an action reaction pair i.e. the force with which one body (say earth) attracts the second body (say moon) is equal to the force with which moon attracts the earth. This is in accordance with Newton's third law of motion.

Note : 

The law of gravitation is stated for two point masses, therefore for any two arbitrary

finite size bodies, as shown in the figure, It can not be applied as there is not unique value for the separation.

m2

m1

m1

m2

r r=?

But if the two bodies are uniform spheres then the separation r may be taken as the distance between their centres because a sphere of uniform mass behave as a point mass for any point lying outside it.

Sample problems based on Newton’s law of gravitation Problem 1.

The gravitational force between two objects does not depend on (a) Sum of the masses

(b) Product of the masses

(c) Gravitational constant

(d) Distance between the masses

[RPET 2003]

genius PHYSICS Gravitation 3

Gravitatio nal constant  product of the masses . (Distance between the masses) 2

Solution : (a)

F

Problem 2.

Mass M is divided into two parts xM and (1 – x)M. For a given separation, the value of x for which the gravitational attraction between the two pieces becomes maximum is [EAMCET 2001] (a)

Solution : (a)

1 2

(b)

Gravitational force F 

Problem 3.

(c) 1

(d) 2

Gm 1m 2 GxM (1  x )M GM 2   2 x (1  x ) r2 r2 r

For maximum value of force



3 5

 dF d  GM 2 x 0  (1  x )  0  2 dx dx  r 

d (x  x 2 )  0  1  2 x  0  x  1 / 2 dx

The mass of the moon is about 1.2% of the mass of the earth. Compared to the gravitational force the earth exerts on the moon, the gravitational force the moon exerts on earth [SCRA 1998] (a) Is the same

(b) Is smaller

(c) Is greater

(d) Varies with its phase

Solution : (a)

Earth and moon both exerts same force on each other.

Problem 4.

Three identical point masses, each of mass 1kg lie in the x-y plane at points (0, 0), (0, 0.2m) and (0.2m, 0). The net gravitational force on the mass at the origin is

Solution : (a)

(a) 1.67  10 9 (ˆj  ˆj)N

(b) 3.34  10 10 (ˆi  ˆj)N

(c) 1.67  10 9 (ˆi  ˆj)N

(d) 3.34  10 10 (ˆi  ˆj)N

Let particle A lies at origin, particle B and C on y and x -axis respectively

FAC 

Y

G m A m B ˆ 6 .67  10 11  1  1 ˆ i  1 .67  10 9 ˆi N i  2 rAB (0 . 2) 2

B (0, 0.2) FAB

Similarly FAB  1 . 67  10 9 ˆj N (0, 0) A FAC

 Net force on particle A F  F AC  F AB  1 . 67  10 9 (ˆi  ˆj) N

Problem 5.

X

Four particles of masses m, 2m, 3m and 4m are kept in sequence at the corners of a square of side a. The magnitude of gravitational force acting on a particle of mass m placed at the centre of the square will be (a)

Solution : (c)

(0.2, 0) C

24 m 2 G

(b)

a2

6m 2G

(c)

a2

4 2 Gm 2

(d) Zero

a2

If two particles of mass m are placed x distance apart then force of attraction

Gmm  F (Let) x2

Now according to problem particle of mass m is placed at the centre (P) of square. Then it will experience four forces

FPA  force at point P due to particle A  Similarly FPB 

G 2mm x2

 2 F , FPC 

Hence the net force on P

G3mm x2

D 4m

Gmm F x2

 3 F and FPD 

3m C

FPD P m

G 4 mm x2

F net  F PA  F PB  F PC  F PD  2 2 F

 4F

FPC

FPA A m

FPB

2m B

genius PHYSICS 4 Gravitation  F net  2 2

Gmm x



2

4 2 Gm 2 a2

Gm 2

2 2

(a / 2 )

[x 

2

a

 half of the diagonal of the square]

2

.

8.4 Acceleration Due to Gravity. The force of attraction exerted by the earth on a body is called gravitational pull or gravity. We know that when force acts on a body, it produces acceleration. Therefore, a body under the effect of gravitational pull must accelerate. The acceleration produced in the motion of a body under the effect of gravity is called acceleration due to gravity, it is denoted by g. Consider a body of mass m is lying on the surface of earth then gravitational force on the body is given by F 

GMm R2

…..(i)

Where M = mass of the earth and R = radius of the earth. If g is the acceleration due to gravity, then the force on the body due to earth is given by Force = mass  acceleration or

F = mg

m

…..(ii) mg

GMm From (i) and (ii) we have mg  R2 

g

GM R2



g

G 4 3   R   2 R 3 



g

4  GR 3

R

Earth

…..(iii)

4 [As mass (M) = volume ( R 3 ) × density ()] 3 …..(iv)

Important points GM 4   GR it is clear that its value depends upon the mass radius and 3 R2 density of planet and it is independent of mass, shape and density of the body placed on the surface of the planet. i.e. a given planet (reference body) produces same acceleration in a light as well as heavy body. (i) From the expression g 

(ii) The greater the value of (M / R 2 ) or R, greater will be value of g for that planet. (iii) Acceleration due to gravity is a vector quantity and its direction is always towards the centre of the planet. (iv) Dimension [g] = [LT–2] (v) it’s average value is taken to be 9.8 m/s2 or 981 cm/sec2 or 32 feet/sec2, on the surface of the earth at mean sea level.

genius PHYSICS Gravitation 5

(vi) The value of acceleration due to gravity vary due to the following factors : (a) Shape of the earth, (b) Height above the earth surface, (c) Depth below the earth surface and (d) Axial rotation of the earth.

Sample problems based on acceleration due to gravity Problem 6.

Acceleration due to gravity on moon is 1/6 of the acceleration due to gravity on earth. If the ratio of   5 densities of earth (m ) and moon (  e ) is  e   then radius of moon Rm in terms of Re will be [MP PMT 20  m  3 (a)

Solution : (a)

5 Re 18

1 Re 6

(b)

Acceleration due to gravity g 

(c)

3 Re 18

1

(d)

2 3

4 g  R GR  g   R or m  m . m 3 ge  e Re

[As

Re

gm  1 5 and e   ge 6 m 3

(given)] 

Problem 7.

Rm  gm   Re  ge

  e    m

 1 5     6 3

 Rm 

5 Re 18

A spherical planet far out in space has a mass M 0 and diameter D0 . A particle of mass m falling freely near the surface of this planet will experience an acceleration due to gravity which is equal to [MP PMT 1987; DPMT 2002]

(a) GM 0 / Solution : (c)

(b) 4 mGM 0 /

D02

We know g 

GM R

2



GM (D / 2)

2



(c) 4 GM 0 /

D02

4 GM D2

If mass of the planet  M 0 and diameter of the planet  D0 . Then g 

Problem 8.

(d) GmM 0 / D02

D 02

4 GM 0

.

D02

The moon's radius is 1/4 that of the earth and its mass is 1/80 times that of the earth. If g represents the acceleration due to gravity on the surface of the earth, that on the surface of the moon is [MP PMT 1997; RPET 2000; MP PET 2000, 2001]

g (a) 4

Solution : (b)

g (b) 5

Acceleration due to gravity g  gmoon  gearth 

GM R2

g (c) 6



gmoon M R2  1  4   moon . 2earth     gearth M earth Rmoon  80  1 

(d)

g 8

2

16 g  . 80 5

Problem 9.

If the radius of the earth were to shrink by 1% its mass remaining the same, the acceleration due to gravity on the earth's surface would [IIT-JEE 1981; CPMT 1981; MP PMT 1996, 97; Roorkee 1992; MP PET 1999] (a) Decrease by 2% (b) Remain unchanged (c) Increase by 2% (d) Increase by 1%

Solution : (c)

We know g 

1 [As R decreases, g increases] R2

So % change in g  2 (% change in R )  2  1%  2%  acceleration due to gravity increases by 2%.

Problem 10. Mass of moon is 7.34  10 22 kg . If the acceleration due to gravity on the moon is 1.4m/s2, the radius of the moon is (G  6.667  10 11 Nm 2 / kg 2 ) (a) 0.56  10 4 m Solution : (b)

We know g 

GM R2

(b) 1.87  10 6 m  R

GM  g

[AFMC 1998]

(c) 1.92  10 6 m

(d) 1.01  10 8 m

6 .67  10 11  7 .34  10 22  1.87  10 6 m . 1 .4

Problem 11. A planet has mass 1/10 of that of earth, while radius is 1/3 that of earth. If a person can throw a stone

on earth surface to a height of 90m, then he will be able to throw the stone on that planet to a height [RPMT 199 (a) 90m (b) 40m (c) 100m (d) 45m

genius PHYSICS 6 Gravitation Solution : (c)

Acceleration due to gravity g 

GM R2



g planet g earth

M planet  R earth   M earth  R planet

2

2    1   3   9  10  1  10 

If a stone is thrown with velocity u from the surface of the planet then maximum height H  H planet H earth



gearth g planet

 H planet 

u2 2g

10 10  H earth   90 = 100 metre. 9 9

Problem 12. The radii of two planets are respectively R1 and R 2 and their densities are respectively 1 and  2 . The ratio of the accelerations due to gravity at their surfaces is (a) g 1 : g 2 

1 R12

:

2

(b) g1 : g2  R1 R2 : 1 2

R 22

(c) g1 : g 2  R1  2 : R 2 1 Solution : (d)

(d) g1 : g 2  R1 1 : R 2  2

Acceleration due to gravity g 

4 GR 3

 g1 : g 2  R1 1 : R 2  2 .

8.5 Variation in g Due to Shape of Earth. Earth is elliptical in shape. It is flattened at the poles and bulged out at the equator. The equatorial GM radius is about 21 km longer than polar radius, from g  2 R At equator

GM ge  2 Re

At poles

GM gp  2 Rp

From (i) and (ii)

R p2 ge  2 gp Re

gp

......(i)

Rp Re

.....(ii)

ge

Since R equator  R pole  g pole  g equator and g p  g e  0 .018 ms 2 Therefore the weight of body increases as it is taken from equator to the pole.

Sample problems based on variation in g due to shape of the earth Problem 13. Where will it be profitable to purchase 1 kg sugar (by spring balance) (a) At poles Solution : (b)

(b) At equator

(c) At 45° latitude

[RPET 1996]

(d) At 40°latitude

At equator the value of g is minimum so it is profitable to purchase sugar at this position.

Problem 14. Force of gravity is least at

[CPMT 1992]

(a) The equator (c) A point in between equator and any pole

(b) The poles (d) None of these

Solution : (a) g

8.6 Variation in g With Height. Acceleration due to gravity at the surface of the earth

g

GM R2

.....(i)

Acceleration due to gravity at height h from the surface of the earth g' 

GM (R  h) 2

.....(ii)

h r

g R O

genius PHYSICS Gravitation 7

 R  g'  g  R h

From (i) and (ii) =g

2

.....(iii)

R2 r2

.....(iv)

[As r = R + h]

Important points (i) As we go above the surface of the earth, the value of g decreases because g  

1 . r2

(ii) If r   then g   0 , i.e., at infinite distance from the earth, the value of g becomes zero. (iii) If h  R i.e., height is negligible in comparison to the radius then from equation (iii) we get 2

h  R   g   g   g1   R R h 

2

2h    g 1   R 

[As h  R ]

(iv) If h  R then decrease in the value of g with height : Absolute decrease g  g  g   Fractional decrease

2hg R

g g  g  2h   g g R

Percentage decrease

g 2h  100 %   100 % g R

Sample problems based on variation in g with height Problem 15. The acceleration of a body due to the attraction of the earth (radius R) at a distance 2R from the surface of the earth is (g = acceleration due to gravity at the surface of the earth) (a)

g 9

(b)

g  R  1  R       g R h R  2 R 9   2

Solution : (a)

g 3

(c)

2

 g 

g 4

(d) g

g . 9

Problem 16. The height of the point vertically above the earth's surface, at which acceleration due to gravity becomes 1% of its value at the surface is (Radius of the earth = R) (a) 8R Solution : (b)

(b) 9R

(c) 10 R

 R  Acceleration due to gravity at height h is given by g  g  R h 

g  R   g  100  R  h

2



(d) 20R 2

R 1   h  9R . R  h 10

Problem 17. At surface of earth weight of a person is 72 N then his weight at height R/2 from surface of earth is (R = radius of earth) (a) 28N

[CBSE PMT 2000; AIIMS 2000]

(b) 16N

(c) 32N

(d) 72N

genius PHYSICS 8 Gravitation   R R   W  W  W R  R  h R 2  2

Solution : (c)

Weight of the body at height R,

2

 2    W  2   4 W  4  72  32 N .  9 9 3  

Problem 18. If the distance between centres of earth and moon is D and the mass of earth is 81 times the mass of moon, then at what distance from centre of earth the gravitational force will be zero (a) D/2 Solution : (d)

(b) 2D/3

(c) 4D/3

(d) 9D/10

If P is the point where net gravitational force is zero then FPA  FPB

Gm 1m Gm 2m  2 x (d  x )2

d P

m1

FPA

A

By solving x 

B

d–x

m1  m 2

For the given problem d  D , m1  earth, m 2  moon and m1  81m 2  m 2 

So x 

x

m1 d

m2

FPB

m1 D m1  m 2

m1 D



m1 

m1 81

D



1

1 9



m1 81

9D 10

8.7 Variation in g With Depth. Acceleration due to gravity at the surface of the earth

g

GM 4   GR R2 3

…..(i)

g d

Acceleration due to gravity at depth d from the surface of the earth

R r

P

g

O

4 g    G(R  d ) 3

…..(ii)

d  From (i) and (ii) g   g 1   R 

Important points (i) The value of g decreases on going below the surface of the earth. From equation (ii) we get g   (R  d ) . So it is clear that if d increase, the value of g decreases. (ii) At the centre of earth d  R  g   0 , i.e., the acceleration due to gravity at the centre of earth becomes zero. (iii) Decrease in the value of g with depth Absolute decrease g  g  g   Fractional decrease

dg R

g g  g  d   g g R

genius PHYSICS Gravitation 9

Percentage decrease

g d  100 %   100 % g R

(iv) The rate of decrease of gravity outside the earth ( if h  R ) is double to that of inside the earth.

Sample problems based on variation in g with depth Problem 19. Weight of a body of mass m decreases by 1% when it is raised to height h above the earth's surface. If the body is taken to a depth h in a mine, change in its weight is (a) 2% decrease Solution : (b)

(b) 0.5% decrease

[KCET 2003; MP PMT 2003]

(c) 1% increase

(d) 0.5% increase

g 2h  100  100 %   1% g R

Percentage change in g when the body is raised to height h , Percentage change in g when the body is taken into depth d,

g d h  100 %   100 %   100 % [As g R R

d h]  Percentage decrease in weight 

1  2h  1  100   1 %   0 .5 % .  2 R  2

Problem 20. The depth at which the effective value of acceleration due to gravity is

g is (R = radius of the earth) 4 [MP PET 2003]

(a) R Solution : (b)

(b)

3R 4

(c)

R 2

(d)

R 4

3R d g d   g  g  1     g 1    d  4 R 4 R    

Problem 21. Assuming earth to be a sphere of a uniform density, what is the value of gravitational acceleration in a mine 100 km below the earth's surface (Given R = 6400km) (a) 9.66 m / s 2 Solution : (a)

Acceleration  9 .8 

(b) 7.64 m / s 2 due

to

gravity

at

depth

(c) 5.06 m / s 2

(d) 3.10 m / s 2

d 100  1     g   g 1    g 1    9 .8 1  64  R   6400   

d,

63  9 . 66 m / s 2 . 64

Problem 22. The depth d at which the value of acceleration due to gravity becomes

1 times the value at the n

surface, is [R = radius of the earth] (a)

Solution : (b)

R n

n 1 (b) R    n 

(c)

R n

2

d g d d 1   n 1 g  g  1     g  1     1   d   R R n R R n      n 

 n  (d) R   n 1

genius PHYSICS 10 Gravitation

8.8 Variation in g Due to Rotation of Earth. As the earth rotates, a body placed on its surface moves along the circular path and hence experiences centrifugal force, due to it, the apparent weight of the body decreases. 

Since the magnitude of centrifugal force varies with the latitude of the place, therefore the apparent weight of the body varies with latitude due to variation in the magnitude of centrifugal force on the body.

P

r

mg



Fc

mg

If the body of mass m lying at point P, whose latitude is , then due to rotation of earth its apparent weight can be given by m g   mg  Fc or

m g   (mg )2  (Fc )2  2mg Fc cos(180  )



m g   (mg )2  (m  2 R cos )2  2mg m  2 R cos  ( cos ) [As Fc  m  2 r  m  2 R cos  ]

By solving we get g   g   2 R cos 2 

Note : 

The latitude at a point on the surface of the earth is defined as the angle, which the

line joining that point to the centre of earth makes with equatorial plane. It is denoted by  .  For the poles   90 o and for equator   0 o

Important points (i) Substituting   90 o in the above expression we get g pole  g   2 R cos 2 90 o 

g pole  g

…..(i)

i.e., there is no effect of rotational motion of the earth on the value of g at the poles. (ii) Substituting   0 o in the above expression we get g eqator  g   2 R cos 2 0 o 

g equator  g   2 R

…..(ii)

i.e., the effect of rotation of earth on the value of g at the equator is maximum. From equation (i) and (ii) g pole  g equator  R  2  0 .034 m / s 2 (iii) When a body of mass m is moved from the equator to the poles, its weight increases by an amount m (g p  g e )  m  2 R

(iv) Weightlessness due to rotation of earth : As we know that apparent weight of the body decreases due to rotation of earth. If  is the angular velocity of rotation of earth for which a body at the equator will become weightless g   g   2 R cos 2 



0  g   2 R cos 2 0 o



g 2R

[As   0 o for equator] 



g R

genius PHYSICS Gravitation 11

or time period of rotation of earth T 

2



Substituting the value of R  6400  10 3 m



1 rad  1.25  10 3 800 sec

Note :  This time is about

R g

 2

and

g  10 m / s 2 we get

and

T  5026 .5 sec  1.40 hr .

1 times the present time period of earth. Therefore if earth starts 17

rotating 17 times faster then all objects on equator will become weightless.  If earth stops rotation about its own axis then at the equator the value of g increases by  2 R and consequently the weight of body lying there increases by m  2 R .  After considering the effect of rotation and elliptical shape of the earth, acceleration due to gravity at the poles and equator are related as g p  g e  0 .034  0 .018 m / s 2

 g p  g e  0 .052 m / s 2

Sample problems based on variation in g due to rotation of the earth Problem 23. The angular velocity of the earth with which it has to rotate so that acceleration due to gravity on 60° latitude becomes zero is (Radius of earth = 6400 km. At the poles g = 10 ms–2) (a) 2.5×10–3rad/sec Solution : (a)

(b) 5.0×10–1rad/sec (c) 10  10 1 rad / sec

(d) 7.8  10 2 rad / sec

Effective acceleration due to gravity due to rotation of earth g  g   2 R cos 2   0  g   2 R cos 2 60 o    

 2R 4

g  

4g g 2 rad 2  R R 800 sec

[As g  0 and   60 o ]

1 rad  2 . 5  10  3 . 400 sec

Problem 24. If earth stands still what will be its effect on man's weight (a) Increases Solution : (a)

(b) Decreases

(c) Remains same

(d) None of these

When earth stops suddenly, centrifugal force on the man becomes zero so its effective weight increases.

Problem 25. If the angular speed of earth is increased so much that the objects start flying from the equator, then the length of the day will be nearly (a) 1.5 hours Solution : (a)

(b) 8 hours

Time period for the given condition T  2

(c) 18 hours

(d) 24 hours

R  1 .40 hr  1 .5 hr nearly. g

8.9 Mass and Density of Earth. Newton’s law of gravitation can be used to estimate the mass and density of the earth. As we know g  

M

GM gR 2 , so we have M  G R2

9 .8  (6 .4  10 6 ) 2  5 .98  10 24 kg  10 25 kg 6 .67  10 11

genius PHYSICS 12 Gravitation and as we know g  



4 3g  GR , so we have   3 4GR

3  9.8  5478 .4 kg / m 3 11 6 4  3.14  6.67  10  6.4  10

8.10 Inertial and Gravitational Masses. (1) Inertial mass : It is the mass of the material body, which measures its inertia. If an external force F acts on a body of mass mi, then according to Newton’s second law of motion F  m i a or m i 

F a

Hence inertial mass of a body may be measured as the ratio of the magnitude of the external force applied on it to the magnitude of acceleration produced in its motion.

Important points (i) It is the measure of ability of the body to oppose the production of acceleration in its motion by an external force. (ii) Gravity has no effect on inertial mass of the body. (iii) It is proportional to the quantity of matter contained in the body. (iv) It is independent of size, shape and state of body. (v) It does not depend on the temperature of body. (vi) It is conserved when two bodies combine physically or chemically. (vii) When a body moves with velocity v , its inertial mass is given by m

m0 v2 1 2 c

, where m0 = rest mass of body, c = velocity of light in vacuum,

(2) Gravitational Mass : It is the mass of the material body, which determines the gravitational pull acting upon it. If M is the mass of the earth and R is the radius, then gravitational pull on a body of mass m g is given by

F

GMm g R

2

or m g 

F F  2 (GM / R ) E

Here m g is the gravitational mass of the body, if E  1 then m g  F Thus the gravitational mass of a body is defined as the gravitational pull experienced by the body in a gravitational field of unit intensity, (3) Comparison between inertial and gravitational mass (i) Both are measured in the same units. (ii) Both are scalars (iii) Both do not depends on the shape and state of the body (iv) Inertial mass is measured by applying Newton’s second law of motion where as gravitational mass is measured by applying Newton’s law of gravitation. (v) Spring balance measure gravitational mass and inertial balance measure inertial mass.

genius PHYSICS Gravitation 13

(4) Comparison between mass and weight of the body Mass (m)

Weight (W)

It is a quantity of matter contained in a body.

It is the attractive force exerted by earth on any body.

Its value does not change with g

Its value changes with g.

Its value can never be zero for any material particle.

At infinity and at the centre of earth its value is zero.

Its unit is kilogram and its dimension is [M].

Its unit is Newton or kg-wt and dimension are [ MLT

2

]

It is determined by a physical balance.

It is determined by a spring balance.

It is a scalar quantity.

It is a vector quantity.

Sample problems based on inertial and gravitational mass Problem 26. Gravitational mass is proportional to gravitational (a) Field

(b) Force

[AIIMS 1998]

(c) Intensity

(d) All of these

Solution : (d)

Problem 27. The ratio of the inertial mass to gravitational mass is equal to (a) 1/2

(b) 1

[CPMT 1978]

(c) 2

(d) No fixed number

Solution : (b)

8.11 Gravitational Field. The space surrounding a material body in which gravitational force of attraction can be experienced is called its gravitational field. Gravitational field intensity : The intensity of the gravitational field of a material body at any point in its field is defined as the force experienced by a unit mass (test mass) placed at that point, provided the unit mass (test mass) itself does not produce any change in the field of the body. So if a test mass m at a point in a gravitational field experiences a force F then I 

F m

Important points (i) It is a vector quantity and is always directed towards the centre of gravity of body whose gravitational field is considered. (ii) Units : Newton/kg or m/s2 (iii) Dimension : [M0LT–2] (iv) If the field is produced by a point mass M and the test mass m is at a distance r from it then by GMm Newton’s law of gravitation F  Test mass r2

F GMm / r 2 then intensity of gravitational field I   m m 

I

r

m

M Source point

GM r2

(v) As the distance (r) of test mass from the point mass (M ) , increases, intensity of gravitational field decreases

genius PHYSICS 14 Gravitation

I

GM 1 ;  I 2 2 r r

M2 I1

M1

(vi) Intensity of gravitational field I  0 , when r   .

I2

I3

(vii) Intensity at a given point (P) due to the combined effect of different point masses can be calculated by vector sum of different intensities

M3

Inet  I1  I 2  I 3  ........

(viii) Point of zero intensity : If two bodies A and B of different masses m 1 and m 2 are d distance apart. Let P be the point of zero intensity i.e., the intensity at this point is equal and apposite due to two bodies A and B and if any test mass placed at this point it will not experience any force. For point P

I1  I 2  0 

By solving

x

 Gm 1

m1 d m1  m 2

x

2



Gm 2 (d  x ) 2

and (d  x ) 

d A m1

0

P

x I1

B m2

I2

m2 d m1  m 2

(ix) Gravitational field line is a line, straight or curved such that a unit mass placed in the field of another mass would always move along this line. Field lines for an isolated mass m are radially inwards. (x) As I 

d–x

m

GM GM and also g  2  I  g 2 r R

Thus the intensity of gravitational field at a point in the field is equal to acceleration of test mass placed at that point.

Sample problems based on gravitational field M where M is the mass of Earth, find the distance of the point where 81 gravitational field due to Earth and Moon cancel each other, from the Moon. Given that distance between Earth and Moon is 60 R. Where R is the radius of Earth

Problem 28. Knowing that mass of Moon is

(a) 2 R Solution : (c)

Point of zero intensity x 

(b) 4 R

(c) 6 R

(d) 8 R

m1 d m1  m 2

d

mass of the earth m1  M , Mass of the moon m2 

M 81

m1

and distance between earth & moon d  60 R Point of zero intensity from the Earth x 

M  60 R M 

So distance from the moon  60 R  54 R  6 R .

M 81

P

Earth



9  60 R  54 R 10

x

Moon

m2

genius PHYSICS Gravitation 15

Problem 29. The gravitational potential in a region is given by V  (3 x  4 y  12 z) J /kg. The modulus of the gravitational field at (x  1, y  0, z  3) is (a) 20 N kg 1 Solution : (b)

(b) 13 N kg 1



 V ˆ V ˆ V ˆ  I   i j k    3ˆi  4 ˆj  12kˆ  x  y  z  

(c) 12 N kg 1



(d) 5 N kg 1

[As V  (3 x  4 y  12 z) (given)]

It is uniform field Hence its value is same every where | I |  3 2  4 2  12 2  13 Nkg 1 .

Problem 30. The magnitudes of the gravitational field at distance r1 and r2 from the centre of a uniform sphere of radius R and mass M are F1 and F2 respectively. Then (a)

F1 r  1 if r1  R and r2  R F2 r2

(b)

F1 r2  22 if r1  R and r2  R F2 r1

(c)

F1 r  1 if r1  R and r2  R F2 r2

(d)

F1 r2  12 if r1  R and r2  R F2 r2

Solution : (a, b) We know that gravitational force  Intensity 

1 r



2

when r  R

[As I 

GM r2

]

F1 r2  22 if r1  R and r2  R F2 r1

and gravitational force  Intensity  r I

when r  R

[As

4  Gr ] 3

F1 r  1 if r1  R and r2  R . F2 r2



Problem 31. Infinite bodies, each of mass 3kg are situated at distances 1m, 2m, 4m, 8m....... respectively on x-axis. The resultant intensity of gravitational field at the origin will be (a) G Solution : (d)

(b) 2G

(c) 3G

Intensity at the origin I  I1  I 2  I 3  I 4  .......



GM GM GM GM  2  2  2  .......... r12 r2 r3 r4

(d) 4G 3kg

O

3kg

3kg

3kg

1m 2m

1 1 1 1   GM  2  2  2  2  .......  2 4 8 1 

4m 8m

1 1 1    GM 1     .........  4 16 64     1  GM  1  1  4   GM 

     

4 4  G  3   4G 3 3

[As sum of G.P. 

a ] 1r

[As M  3kg given]

Problem 32. Two concentric shells of mass M1 and M 2 are having radii r1 and r2 . Which of the following is the correct expression for the gravitational field on a mass m.

genius PHYSICS 16 Gravitation (a) I  (b) I 

G(M 1  M 2 ) r2 G(M 1  M 2 ) r2

(c) I  G (d) I  Solution : (d)

M2 r2

GM 1 r2

for r  r1 M2

r2 M1

for r  r2

r1

r

for r1  r  r2

m

for r1  r  r2

Gravitational field on a mass m due to outer shell (radius r2 ) will be zero because the mass is placed inside this shell. But the inner shell (radius r1 ) behaves like point mass placed at the centre so I 

GM 1 r2

for r1  r  r2

Problem 33. A spherical shell is cut into two pieces along a chord as shown in the figure. P is a point on the plane of the chord. The gravitational field at P due to the upper part is I1 and that due to the lower part is I2 . What is the relation between them (a) I1  I2 P

(b) I1  I2 (c) I1  I2 (d) No definite relation Solution : (c)

Intensity at P due to upper part  I1 and Intensity at P due to lower part  I 2 Net Intensity at p due to spherical shell I1  I 2  0  I1   I 2

Problem 34. A uniform ring of mass m is lying at a distance 1.73 a from the centre of a sphere of mass M just over the sphere where a is the small radius of the ring as well as that of the sphere. Then gravitational force exerted is (a)

(c)

GMm 8a 2 3

GMm a

Solution : (d)

(b)

(1.73 a)2

(d) 1.73

2

m

GMm

1.73 a

GMm

a

8a 2

Intensity due to uniform circular ring at a point on its axis I 

 Force on sphere F 

GMmr (a  r ) 2

2 3/2



a

GMm 3 a (a  ( 3 a) ) 2

2 3/2



M

Gmr (a  r 2 ) 3 / 2 2

GMm 3 a 2 3/2

(4 a )



3 GMm 8a 2

[As r 

3a ]

8.13 Gravitational Potential. At a point in a gravitational field potential V is defined as negative of work done per unit mass in shifting a test mass from some reference point (usually at infinity) to the given point i.e.,

genius PHYSICS Gravitation 17

V  

I

F.d r W   I .d r  m m





[As

F  I] m

dV dr

i.e., negative gradient of potential gives intensity of field or potential is a scalar function of position whose space derivative gives intensity. Negative sign indicates that the direction of intensity is in the direction where the potential decreases.

Important points (i) It is a scalar quantity because it is defined as work done per unit mass. (ii) Unit : Joule/kg or m2/sec2 (iii) Dimension : [M0L2T–2]  GM  V   I dr     2  dr  r 



(iv) If the field is produced by a point mass then 

V 



GM c r

[As I  

GM ] r2

[Here c = constant of integration]

Assuming reference point at  and potential to be zero there we get

0

GM  c c  0 

GM r

Gravitational potential V  



(v) Gravitational potential difference : It is defined as the work done to move a unit mass from one point to the other in the gravitational field. The gravitational potential VA VB difference in bringing unit test mass m from point A to point B under the m m M gravitational influence of source mass M is A B W  V  VB  V A  A  B  GM m

1 1    rB rA

rA

  

rB

(vi) Potential due to large numbers of particle is given by scalar addition of all the potentials. V  V1  V2  V3  ..........



GM GM GM   ........ r1 r2 r3

 G

in

 i 1

Mi ri

M M1 r5 M 5

r1 P r4

r2 2

r3

M

M

4

3

(vii) Point of zero potential : It is that point in the gravitational field, if the

A m1

B

unit mass is shifted from infinity to that point then net work done will be equal to zero. Let m1 and m2 are two masses placed at d distance apart and P is the point of zero potential in between the two masses. Net potential for point P  VA  VB  0

m2

P

x

d–x d

genius PHYSICS 18 Gravitation 



Gm 1 Gm 2  0 x dx

By solving x 

m1d m1  m 2

Sample problems based on gravitational potential Problem 35. In some region, the gravitational field is zero. The gravitational potential in this region (a) Must be variable Solution : (b)

As I  

(b) Must be constant (c) Cannot be zero

[BVP 2003]

(d) Must be zero

dV , if I  0 then V = constant. dx

Problem 36. The gravitational field due to a mass distribution is E  K / x 3 in the x - direction (K is a constant). Taking the gravitational potential to be zero at infinity, its value at a distance x is (a) K / x Solution : (d)



V   E dx  

(c) K / x 2

(b) K / 2 x

x

K 3

dx 

K 2x 2

(d) K / 2 x 2

.

Problem 37. The intensity of gravitational field at a point situated at a distance of 8000 km from the centre of the earth is 6 N / kg . The gravitational potential at that point is – (in Joule / kg) (a) 8  10 6 Solution : (c)

(b) 2.4  10 3

(c) 4.8  10 7

I

Gravitational intensity at point P, V 

(d) 6.4  10 14

GM r2

and

gravitational

potential

GM r

 V  I  r  6 N / kg  8000 km  4.8  10 7

Joule . kg

Problem 38. The gravitational potential due to the earth at infinite distance from it is zero. Let the gravitational potential at a point P be 5 J / kg . Suppose, we arbitrarily assume the gravitational potential at infinity to be + 10 J / kg , then the gravitational potential at P will be (a) 5 J / kg Solution : (b)

(c) 15 J / kg

(b) 5 J / kg

(d) 15 J / kg

Potential increases by 10 J / kg every where so it will be 10  5  5 J / kg at P

Problem 39. An infinite number of point masses each equal to m are placed at x =1. x = 2, x = 4, x = 8 ……… What is the total gravitational potential at x = 0 (a)  Gm (b) 2Gm Solution : (b)

 Gm Gm Gm     .........  Net potential at origin V    r2 r3  r1 

  1 1 1 1 1   Gm       Gm  1  1 1 2 4 8   2 

(d)  8 Gm

(c)  4 Gm

O

m

m

m

m

1m

    2Gm   

2m 4m 8m

Problem 40. Two bodies of masses m and M are placed a distance d apart. The gravitational potential at the position where the gravitational field due to them is zero is V, then G Gm GM (a) V   (m  M ) (b) V   (c) V   d d d G V   ( m  M )2 d Solution : (d)

If P is the point of zero intensity, then x 

M M  m

.d

and

(d)

dx 

m M  m

d

genius PHYSICS Gravitation 19 Now potential at point P, V  V1  V2  

GM GM  x dx

Substituting the value of x and d – x we get V  

G d

m

M

. 2

8.15 Gravitational Potential Energy. The gravitational potential energy of a body at a point is defined as the amount of work done in bringing the body from infinity to that point against the gravitational force. r

GMm 1  W  dx  GMm   2  x  x 



r

M

r

m

GMm W  r This work done is stored inside the body as its gravitational potential energy GMm U   r

Important points (i) Potential energy is a scalar quantity. (ii) Unit : Joule (iii) Dimension : [ML2T–2] (iv) Gravitational potential energy is always negative in the gravitational field because the force is always attractive in nature. (v) As the distance r increases, the gravitational potential energy becomes less negative i.e., it increases. (vi) If r   then it becomes zero (maximum) (vii) In case of discrete distribution of masses  Gm 1 m 2 Gm 2 m 3    ........  Gravitational potential energy U   u i    r23  r12 

(viii) If the body of mass m is moved from a point at a distance r1 to a point at distance r2 (r1  r2 ) then change in potential energy U 



r2

r1

1 1 1 1  GMm dx  GMm    or U  GMm    2 x  r2 r1   r1 r2 

As r1 is greater than r2 , the change in potential energy of the body will be negative. It means that if a body is brought closer to earth it's potential energy decreases. (ix) Relation between gravitational potential energy and potential U   

GMm   GM   m  r  r 

U  mV

(x) Gravitational potential energy at the centre of earth relative to infinity.

3 GMm  3 GM  U centre  m Vcentre  m     2 R  2 R  (xi) Gravitational potential energy of a body at height h from the earth surface is given by Uh  

GMm gR 2 m mgR   h Rh Rh 1 R

genius PHYSICS 20 Gravitation

8.16 Work Done Against Gravity. If the body of mess m is moved from the surface of earth to a point at distance h above the surface of earth, then change in potential energy or work done against gravity will be 1 1  W  U  GMm     r1 r2 



1  1 W  GMm   R R  h  



W

[As r1  R and r2  R  h ]

GMmh mgh  h h  R 2 1   1  R R 

[As

GM  g] R2

Important points (i) When the distance h is not negligible and is comparable to radius of the earth, then we will use above formula.  n  (ii) If h  nR then W  mgR   n 1

(iii) If h  R then W 

1 mgR 2

(iv) If h is very small as compared to radius of the earth then term h / R can be neglected

W

From

h    As R  0   

mgh  mgh 1h/R

Sample problems based on potential Energy Problem 41. Energy required to move a body of mass m from an orbit of radius 2R to 3 R is (a)

GMm 12R

Solution : (d)

2

(b)

GMm 3R

2

(c)

GMm 8R

(d)

GMm 6R

 GMm   GMm  GMm GMm  Work done = Change in potential energy  U 2  U1        3R 2R r2   r1   GMm  . 6R

Problem 42. A body of mass m kg. starts falling from a point 2R above the earth’s surface. Its kinetic energy when it has fallen to a point ‘R’ above the earth’s surface [R-Radius of earth, M-Mass of earth, G-Gravitational constant] [MP PMT 2002]

(a) Solution : (b)

1 GMm 2 R

(b)

1 GMm 6 R

(c)

2 GMm 3 R

(d)

1 GMm 3 R

When body starts falling toward earth’s surface its potential energy decreases so kinetic energy increases. Increase in kinetic energy = Decrease in potential energy Final kinetic energy – Initial kinetic energy = Initial potential energy – Final potential energy  GMm   GMm      Final kinetic energy – 0    r1   r2  

genius PHYSICS Gravitation 21 

Final

kinetic

  GMm    R  h1

energy

  GMm        Rh  2   

GMm GMm 1 GMm  GMm   GMm    .       3R 2R 6 R  R  2R   R  R 

Problem 43. A body of mass m is taken from earth surface to the height h equal to radius of earth, the increase in potential energy will be

[CPMT 1971, 97; IIT-JEE 1983; CBSE PMT 1991; Haryana CEE 1996; CEET Bihar 1995; MNR 1998; RPET 2000]

(a) mgR Solution : (b)

Work done 

(b)

1 mgR 2

(c) 2mgR

(d)

1 mgR 4

1 mgh mgR  mgR . , If h  R then work done  2 1h/R 1R/R

Problem 44. If mass of earth is M, radius is R and gravitational constant is G, then work done to take 1 kg mass from earth surface to infinity will be (a) Solution : (b)

GM 2R

(b)

GM R

(c)

2 GM R

(d)

GMm  GMm  Work done = U final  U initial  U  U R  0      R R  

GM 2R

[As m  1kg ]

Problem 45. Three particles each of mass 100 gm are brought from a very large distance to the vertices of an equilateral triangle whose side is 20 cm in length. The work done will be (a) 0 .33  10 11 Joule Solution : (d)

(b)  0 .33  10 11 Joule (c) 1 .00  10 11 Joule

(d)  1.00  10 11 Joule

Potential energy of three particles system

U

m1

Gm 1 m 2 Gm 2 m 3 Gm 1 m 3   r12 r23 r13

r

r

Given m 1  m 2  m 3  100 gm and r12  r23  r13  2cm   6 . 67  10 11  (10 1 )  (10 1 )  11 Joule .  U 3   1 .00  10 2 20  10  

m2

r

m3

Problem 46. A boy can jump to a height h on ground level. What should be the radius of a sphere of density d such that on jumping on it, he escapes out of the gravitational field of the sphere  4  Gd  (a)    3 gh 

Solution : (c)

1/2

 4 gh  (b)    3 Gd 

1/2

 3 gh  (c)    4 Gd 

1/2

 3 Gd  (d)    4  gh 

1/2

When a boy jumps from a ground level up to height h then its velocity of jumping v  …..(i) and for the given condition this will become equal to escape velocity vescape =

2GM  R

2 gh

2G  4  3   R .d  R 3 

…..(ii) Equating (i) and (ii)

2 gh  R

8  3 gh  Gd  R    3  4  Gd 

1/2

.

8.17 Escape Velocity. The minimum velocity with which a body must be projected up so as to enable it to just overcome the gravitational pull, is known as escape velocity. The work done to displace a body from the surface of earth (r = R) to infinity ( r   ) is

W



GMm dx  GMm R x2



1 1    R   

genius PHYSICS 22 Gravitation 

W

GMm R

This work required to project the body so as to escape the gravitational pull is performed on the body by providing an equal amount of kinetic energy to it at the surface of the earth. If v e is the required escape velocity, then kinetic energy which should be given to the body is

1 GMm mv e2  2 R



ve  2 

or



ve 

2GM R



ve 

2 gR

4  GR  R  3

ve  R

8 G 3

1 mv e2 2

[As GM  gR 2 ] [As g 

4  GR ] 3

Important points (i) Escape velocity is independent of the mass and direction of projection of the body. (ii) Escape velocity depends on the reference body. Greater the value of (M / R) or (gR ) for a planet, greater will be escape velocity. (iii) For the earth as g  9 .8 m / s 2 and R  6400 km  v e  2  9 . 8  6 . 4  10 6  11 . 2km / sec

 (iv) A planet will have atmosphere if the velocity of molecule in its atmosphere v rms  

3 RT   is lesser M 

than escape velocity. This is why earth has atmosphere (as at earth v rms  v e ) while moon has no atmosphere (as at moon v rms  v e ) (v) If body projected with velocity lesser than escape velocity ( v  v e ) it will reach a certain maximum height and then may either move in an orbit around the planet or may fall down back to the planet. (vi) Maximum height attained by body : Let a projection velocity of body (mass m ) is v , so that is attains a maximum height h . At maximum height, the velocity of particle is zero, so kinetic energy is zero. By the law of conservation of energy Total energy at surface = Total energy at height h .





GMm 1 GMm  mv 2   0 R 2 Rh



v2 1  GMh 1  GM     R(R  h) 2 R R  h



2GM R  h R  1 2 h v R h



h

 v2  R R  2 R  2   2 GM  ve ve  v 2    1  2  1  v R  v2

[As v e 

2GM 2GM   v e2 ] R R

(vii) If a body is project with velocity greater than escape velocity (v  v e ) then by conservation of energy.

genius PHYSICS Gravitation 23

Total energy at surface = Total energy at infinite

1 GMm 1 mv 2   m(v )2  0 2 R 2 2GM  v ' 2  v 2  v e2 R

i.e.,

(v )2  v 2 



v  v 2  ve2

2GM  v e2 ] R

[As

i.e, the body will move in interplanetary or inter stellar space with velocity

v 2  v e2 .

(viii) Energy to be given to a stationary object on the surface of earth so that its total energy becomes zero, is called escape energy. Total energy at the surface of the earth  KE  PE  0  Escape energy 



GMm R

GMm R

(ix) If the escape velocity of a body is equal to the velocity of light then from such bodies nothing can escape, not even light. Such bodies are called black holes. The radius of a black hole is given as

R

2GM C2

[As C 

2 GM , where C is the velocity of light] R

Sample problems based on escape velocity Problem 47. For a satellite escape velocity is 11 km / s . If the satellite is launched at an angle of 60 o with the vertical, then escape velocity will be (a) 11 km / s

(b) 11 3 km / s

(c)

11

km / s

(d) 33 km / s

3 Solution : (a)

Escape velocity does not depend upon the angle of projection.

Problem 48. The escape velocity from the earth is about 11 km / s . The escape velocity from a planet having twice the radius and the same mean density as the earth, is [MP PMT 1987; UPSEAT 1999; AIIMS 2001; MP PET 2001, 2003] (a) 22 km/s (b) 11 km/s (c) 5.5 km/s (d) 15.5 km/s Solution : (a)

2Gm 8  v e  R if  = constant. Since the planet having double radius in  GR 2 R 3 comparison to earth therefore the escape velocity becomes twice i.e. 22 km/ s . ve 

Problem 49. A projectile is projected with velocity kv e in vertically upward direction from the ground into the space. ( v e is escape velocity and k < 1). If air resistance is considered to be negligible then the

maximum height from the centre of earth to which it can go, will be (R = radius of earth) [Roorkee 1999; RPET 1 R R R R (a) 2 (b) 2 (c) (d) 2 k 1 k 1 k 1 1k Solution : (c)

From the law of conservation of energy Difference in potential energy between ground and maximum height = Kinetic energy at the point of projection 1 1 mgh 1  m(kv e ) 2  mk 2 v e2  m k 2 ( 2 g R ) 2 2 2 1h/R 2

By solving height from the surface of earth h 

Rk 2 1k2

[As v e 

2 gR ]

genius PHYSICS 24 Gravitation So height from the centre of earth r  R  h  R 

R Rk 2 .  2 1k 1k2

Problem 50. If the radius of earth reduces by 4% and density remains same then escape velocity will [MP PET 1991; MP PMT 1995]

(a) Reduce by 2% Solution : (c)

(b) Increase by 2%

(c) Reduce by 4%

(d) Increase by 4%

Escape velocity v e  R  and if density remains constant v e  R So if the radius reduces by 4% then escape velocity also reduces by 4%.

Problem 51. A rocket of mass M is launched vertically from the surface of the earth with an initial speed V. Assuming the radius of the earth to be R and negligible air resistance, the maximum height attained by the rocket above the surface of the earth is

R  gR  (b) R  (c)  1 2  gR   2V   1    2V 2  Kinetic energy given to rocket at the surface of earth

R  2 gR  (d) R  2  1   2 gR   V   2  1  V  = Change in potential energy of the rocket in reaching from ground to highest point

1 mgh  mv 2  2 1  h/R

v2 g   1 1 2  h R

(a)

Solution : (c)



v2R 1 2 gR  v 2 1 1 2g 1 2g 1   h     2   2  h h R v h v R 2 gR  v 2 v2R  h

R  2 gR   2  1  v 

Problem 52. A body of mass m is situated at a distance 4 R e above the earth’s surface, where R e is the radius of earth. How much minimum energy be given to the body so that it may escape mgR e mgR e (a) mgRe (b) 2mgRe (c) (d) 5 16 Solution : (c)

Potential energy of the body at a distance 4 R e from the surface of earth

U

mgR e mgR e mgR e   1  4 5 1  h / Re

[As h  4 R e (given)]

So minimum energy required to escape the body will be

mgR e . 5

8.18 Kepler’s Laws of Planetary Motion. Planets are large natural bodies rotating around a star in definite orbits. The planetary system of the star sun called solar system consists of nine planets, viz., Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune and Pluto. Out of these planets Mercury is the smallest, closest to the sun and so hottest. Jupiter is largest and has maximum moons (12). Venus is closest to Earth and brightest. Kepler after a life time study work out three empirical laws which govern the motion of these planets and are known as Kepler’s laws of planetary motion. These are, (1) The law of Orbits : Every planet moves around the sun in an elliptical orbit with sun at one of the foci. (2) The law of Area : The line joining the sun to the planet sweeps out equal areas in equal interval of time. i.e. areal velocity is constant. According to this law planet will move slowly when it is farthest from sun and more rapidly when it is nearest to sun. It is similar to law of conservation of angular momentum.

Sun r

dA

v dt

genius PHYSICS Gravitation 25

Areal velocity 

dA 1 r(vdt) 1   rv dt 2 dt 2

dA L  dt 2m



[As L  mvr ; rv 

L ] m

(3) The law of periods : The square of period of revolution (T ) of any planet around sun is directly proportional to the cube of the semi-major axis of the orbit. r r  T 2  a 3 or T 2   1 2   2 

3

E

Proof : From the figure AB  AF  FB

2a  r1  r2  a 

r1  r2 2

Perigee

where a = semi-major axis

Sun F

A

r1

r1  Shortest distance of planet from sun (perigee).

Apogee

r2 a

C

B

D

r2  Largest distance of planet from sun (apogee).

Important data Planet

Semi-major axis

Period

T2/a3

a (1010 meter)

T(year)

(10–34 year2/meter3)

Mercury

5.79

0.241

2.99

Venus

10.8

0.615

3.00

Earth

15.0

1.00

2.96

Mars

22.8

1.88

2.98

Jupiter

77.8

11.9

3.01

Saturn

143

29.5

2.98

Uranus

287

84.0

2.98

Neptune

450

165

2.99

Pluto

590

248

2.99

Note:: 

Solar System

Kepler's laws are valid for satellites also.

8.19 Velocity of a Planet in Terms of Eccentricity. Applying the law of conservation of angular momentum at perigee and apogee mv p rp  mv a ra



vp va



ra a  c 1  e   rp a  c 1  e

[As rp  a  c,

ra  a  c and eccentricity e 

Applying the conservation of mechanical energy at perigee and apogee

1 1 1 GMm 1 GMm 2 2 2 2 mv p   mv a   v p  v a  2GM    2 rp 2 ra  rp ra  

va

2

 ra 2  rp 2   ra  rp     2 GM   2  ra rp   rp 

[As v p 

v a ra ] rp

c ] a

genius PHYSICS 26 Gravitation 2 GM  rp  2 GM 2    va  ra  rp  ra  a

va 



2

 a  c  2 GM   a ac

1  e    1  e 

Thus the speeds of planet at apogee and perigee are 2 GM  1  e   , a 1  e 

va 

vp 

2 GM  1  e    a 1  e 

Note::  The gravitational force is a central force so torque on planet relative to sun is always zero, hence angular momentum of a planet or satellite is always constant irrespective of shape of orbit.

8.20 Some Properties of the Planet. Mercur y

Venus

Earth

Mars

Jupiter

Saturn

Uranus

Neptune

Pluto

Mean distacne from sun, 106 km

57.9

108

150

228

778

1430

2870

4500

5900

Period of revolution, year

0.241

0.615

1.00

1.88

11.9

29.5

84.0

165

248

Orbital speed, km/s

47.9

35.0

29.8

24.1

13.1

9.64

6.81

5.43

4.74

Equatiorial diameter, km

4880

12100

12800

6790

143000

120000

51800

49500

2300

0.0558

0.815

1.000

0.107

318

95.1

14.5

17.2

0.002

Mass (Earth =1) Density (Water =1)

5.60

5.20

5.52

3.95

1.31

0.704

1.21

1.67

2.03

Surface value of g, m/s2

3.78

8.60

9.78

3.72

22.9

9.05

7.77

11.0

0.5

Escape velocity, km/s

4.3

10.3

11.2

5.0

59.5

35.6

21.2

23.6

1.1

0

0

1

2

16+ring

18+rings

17+rings

8+rings

1

Known satellites

Sample problems based on Kepler’s law Problem 53. The distance of a planet from the sun is 5 times the distance between the earth and the sun. The Time period of the planet is (a) 5 Solution : (a)

3/2

years

[UPSEAT 2003]

(b) 5

2/3

years

(c) 5

1/3

(d) 5

years

1/2

years

3/2 Tearth  5 (3 / 2)  1 year  5 3 / 2 years . According to Kepler’s law T  R 3 / 2  T planet  (5 )

Problem 54. In planetary motion the areal velocity of position vector of a planet depends on angular velocity ( ) and the distance of the planet from sun (r). If so the correct relation for areal velocity is dA dA dA dA r 2 r   r2  r (a) (b) (c) (d) dt dt dt dt Solution : (c)

dA L 1 mvr   r 2  dt 2m 2m 2 

[As Angular momentum L  mvr and v = r ]

dA  r 2 . dt

Problem 55. The planet is revolving around the sun as shown in elliptical path. The correct option is

[UPSEAT 2002]

B

(a) The time taken in travelling DAB is less than that for BCD (b) The time taken in travelling DAB is greater than that for BCD

S C

A

(c) The time taken in travelling CDA is less than that for ABC (d) The time taken in travelling CDA is greater than that for ABC Solution : (a)

D

When the planet passes nearer to sun then it moves fast and vice-versa. Hence the time taken in travelling DAB is less than that for BCD.

genius PHYSICS Gravitation 27

Problem 56. The distance of Neptune and Saturn from sun are nearly 10 13 and 1012 meters respectively. Assuming

that they move in circular orbits, their periodic times will be in the ratio [NCERT 1975; CBSE PMT 1994; MP PET 2 (a) Solution : (c)

(b) 100

10

Kepler’s third law T 2  R 3 

TNep tune TSaturn

(c) 10 10  R Nep tune       R Saturn 

3/2

 10 13    12   10 

(d) 1 / 10 3/2

 10 10 .

Problem 57. The maximum and minimum distance of a comet from the sun are 8  10 12 m and 1.6  10 12 m. If its Solution : (a)

velocity when nearest to the sun is 60 m/s, what will be its velocity in m/s when it is farthest (a) 12 (b) 60 (c) 112 (d) 6 According to conservation of angular momentum mv min rmax  mv max rmin  constant  vmin  vmax 

 1 .6  10 12 rmin  60   12 rmax  8  10

   12m / s  

Problem 58. A satellite A of mass m is at a distance of r from the centre of the earth. Another satellite B of mass 2m is at distance of 2r from the earth’s centre. Their time periods are in the ratio of Solution : (d)

(a) 1 : 2 (b) 1 : 16 (c) 1 : 32 (d) 1 : 2 2 Time period does not depend upon the mass of satellite, it only depends upon the orbital radius. According to Kepler's law

T1  r1    T 2  r2 

3/2

 r     2r 

3/2



1

.

2 2

Problem 59. A planet moves around the sun. At a given point P, it is closed from the sun at a distance d 1 and has a speed v 1 . At another point Q. when it is farthest from the sun at a distance d 2 , its speed will be [MP PMT 1987] (a) Solution : (c)

d 12 v 1

(b)

d 22

d 2 v1 d1

(c)

d 1 v1

(d)

d2

According to law of conservation of angular momentum mv1d1  mv 2 d 2

 v2 

d 22 v 1 d 12

d1v1 . d2

8.21 Orbital Velocity of Satellite. Satellites are natural or artificial bodies describing orbit around a planet under its gravitational attraction. Moon is a natural satellite while INSAT-1B is an artificial satellite of earth. Condition for establishment of artificial satellite is that the centre of orbit of satellite must coincide with centre of earth or satellite must move around great circle of earth. Orbital velocity of a satellite is the velocity required to put the satellite into its orbit around the earth. For revolution of satellite around the earth, the gravitational pull provides the required centripetal force.

mv 2 GMm  r r2 

v v

Satellite h r

GM r 2

gR g R Rh Rh

R Earth

v

[As GM  gR 2 and r  R  h ]

Important points (i) Orbital velocity is independent of the mass of the orbiting body and is always along the tangent of the orbit i.e., satellites of deferent masses have same orbital velocity, if they are in the same orbit. (ii) Orbital velocity depends on the mass of central body and radius of orbit.

genius PHYSICS 28 Gravitation (iii) For a given planet, greater the radius of orbit, lesser will be the orbital velocity of the satellite

v  1 / r .

(iv) Orbital velocity of the satellite when it revolves very close to the surface of the planet GM GM GM v [As h  0 and GM  gR 2 ] v   gR r Rh R For the earth v  9.8  6.4  10 6  7.9 k m / s  8 km / sec v

(v) Close to the surface of planet 

v

ve

GM R

[As v e 

i.e., v escape  2 v orbital

2

It means that if the speed of a satellite orbiting close to the earth is made 41%) then it will escape from the gravitational field.

2 times (or increased by

(vi) If the gravitational force of attraction of the sun on the planet varies as F  velocity varies as v 

1 rn  1

2GM ] R

1 then the orbital rn

.

Sample problems based on orbital velocity Problem 60. Two satellites A and B go round a planet P in circular orbits having radii 4R and R respectively. If the speed of the satellite A is 3V, the speed of the satellite B will be (a) 12 V (b) 6 V (c) 3/2 V Solution : (b)

GM r

Orbital velocity of satellite v 

 v

1



r

vB  vA

(d) 3/2 V rA rB



vB  3V

4R  v B  6V . R

Problem 61. A satellite is moving around the earth with speed v in a circular orbit of radius r. If the orbit radius is

Solution : (b)

decreased by 1%, its speed will (a) Increase by 1% (b) Increase by 0.5% (c) Decrease by 1% (d) Decrease by 0.5% 1 Gm Orbital velocity v   v [If r decreases then v increases] r r Percentage change in v 

1 1 (Percentage change in r)  (1%)  0.5%  orbital velocity increases by 2 2

0.5%.

Problem 62. If the gravitational force between two objects were proportional to 1/R; where R is separation between them, then a particle in circular orbit under such a force would have its orbital speed v proportional to

[CBSE PMT 1994; JIPMER 2001, 02]

Solution : (b)

(a) 1 / R 2 1 If F  n then v  R

(b) R 0

1 R n 1

(c) R 1 1  R0 . here n  1 v  1 1 R

;

(d) 1 / R

Problem 63. The distance between centre of the earth and moon is 384000 km. If the mass of the earth is 6  10 24 kg and G  6 .67  10 11 Nm 2 / kg 2 . The speed of the moon is nearly (a) 1 km / sec Solution : (a)

Orbital velocity v 

(b) 4 km / sec GM  r

6 . 67  10

11

(c) 8 km / sec  6  10

24

38400  10 3

8.22 Time Period of Satellite. It is the time taken by satellite to go once around the earth. 

T

Circumfere nce of the orbit orbital velocity

[MH CET 2002]

(d) 11.2 km / sec

v  1.02 km / sec  1km / sec (Approx.)

genius PHYSICS Gravitation 29

2r r  2r v GM



T



T  2



T  2

r3  2 GM

R  h 3 g R2

[As v  r3 gR 2

 2

GM ] r

[As GM  gR 2 ] R h 1   g R

3/2

[As r  R  h ]

Important points r3 it is clear that time period is independent of the mass of orbiting body and GM depends on the mass of central body and radius of the orbit

(i) From T  2

(ii)

T  2



T2 

r3 GM

4 2 3 r i.e., T 2  r 3 GM

This is in accordance with Kepler’s third law of planetary motion r becomes a (semi major axis) if the orbit is elliptic. (iii) Time period of nearby satellite, From

T  2

r3 R3 R  2  2 2 GM gR g

[As h  0 and GM  gR 2 ]

For earth R  6400 km and g  9 .8 m / s 2 T  84 .6 minute  1.4 hr

(iv) Time period of nearby satellite in terms of density of planet can be given as r3 R3 2 R 3  T  2  2   1/2 GM GM  4  3 G. 3 R     1/2

3 G

(v) If the gravitational force of attraction of the sun on the planet varies as F 

T r

1 then the time period varies as rn

n 1 2

(vi) If there is a satellite in the equatorial plane rotating in the direction of earth’s rotation from west to east, then for an observer, on the earth, angular velocity of satellite will be ( S   E ) . The time interval between the two consecutive appearances overhead will be T

T T 2  S E  s   E TE  TS

2    As T     

If  S   E , T   i.e. satellite will appear stationary relative to earth. Such satellites are called geostationary satellites.

Sample problems based on time period

genius PHYSICS 30 Gravitation

Problem 64. A satellite is launched into a circular orbit of radius ‘R’ around earth while a second satellite is launched into an orbit of radius 1.02 R. The percentage difference in the time periods of the two satellites is [EAMCET 2003] (a) 0.7 Solution : (d)

(b) 1.0

(c) 1.5

(d) 3

Orbital radius of second satellite is 2% more than first satellite So from T  (r) 3 / 2 , Percentage increase in time period  

3 (Percentage increase in orbital radius) 2

3 (2%) = 3%. 2

Problem 65. Periodic time of a satellite revolving above Earth’s surface at a height equal to R, where R the radius of Earth, is [g is acceleration due to gravity at Earth’s surface]

Solution : (b)

(a) 2

2R g

T  2

(R  h) 3 GM

R g

(b) 4 2

 2

(R  R ) 3 gR

2

R g

(c) 2

(d) 8

8R R  4 2 g g

 2

R g

[As h  R (given)].

Problem 66. An earth satellite S has an orbit radius which is 4 times that of a communication satellite C. The period of revolution of S is

[MP PMT 1994; DCE 1999]

(a) 4 days Solution : (b)

(b) 8 days

(c) 16 days

(d) 32 days

Orbital radius of satellite rs  4 rc (given) From Kepler’s law T  r 3 / 2 

T s  rs    Tc  s c 

3/2

 (4 ) 3 / 2  Ts  8 Tc  8  1 day = 8 days.

Problem 67. One project after deviation from its path, starts moving round the earth in a circular path at radius equal to nine times the radius at earth R, its time period will be

Solution : (b)

(a) 2

R g

T  2

r3 (9 R)3  2 (9 ) 3 / 2  2 2 2 gR gR

(b) 27  2

R g

(c) 

R g

R R  27  2 g g

(d) 8  2

R g

[As r  9 R (given)].

Problem 68. A satellite A of mass m is revolving round the earth at a height ‘r’ from the centre. Another satellite B of mass 2m is revolving at a height 2r. The ratio of their time periods will be (a) 1 : 2 Solution : (d)

(b) 1 : 16

(c) 1 : 32

Time period depends only upon the orbital radius

T1  r1    T 2  r2 

8.23 Height of Satellite. As we know, time period of satellite T  2

By squaring and rearranging both sides



 T 2g R2 h   2  4

  

1/3

R

r3  2 GM

(R  h) 3 gR 2

g R 2T 2 3  R  h  4 2

[CBSE PMT 1993]

(d) 1 : 2 2 3/2

 r     2r 

3/2



1 2 2

.

genius PHYSICS Gravitation 31

By knowing the value of time period we can calculate the height of satellite the surface of the earth.

Sample problems based on height Problem 69. Given radius of earth ‘R’ and length of a day ‘T’ the height of a geostationary satellite is [G – Gravitational constant, M – Mass of earth]

 4 2 GM (a)   T2  Solution : (c)

   

1/3

 4GM  (b)    R2 

 T 2 gR 2 From the expression h   2  4

   

1/3

R

[MP PMT 2002]

1/3

R

 GMT 2 (c)   4 2 

 GMT 2  h  4 2 

   

   

1/3

 GMT 2 (d)   4 2 

R

   

1/3

R

1/3

R

[As gR 2  GM ]

Problem 70. A satellite is revolving round the earth in circular orbit at some height above surface of earth. It takes 5.26  10 3 seconds to complete a revolution while its centripetal acceleration is 9 .92 m / s 2 . Height of satellite above surface of earth is (Radius of earth 6 .37  10 6 m ) (a) 70 km Solution : (c)

(b) 120 km

Centripetal acceleration (ac )  From equation (i) and (ii) r 

(c) 170 km

[MP PET 1993]

(d) 220 km

2r v2 and T  v r

ac T 2 4 2

 Rh 

9.32  (5.26  10 3 ) 2 4  2

h  6.53  10 6  R  6.53  10 6  6.37  10 6  160  10 3 m  160km  170km .

8.24 Geostationary Satellite. The satellite which appears stationary relative to earth is called geostationary or geosynchronous satellite, communication satellite. A geostationary satellite always stays over the same place above the earth such a satellite is never at rest. Such a satellite appears stationary due to its zero relative velocity w.r.t. that place on earth. The orbit of a geostationary satellite is known as the parking orbit.

Important points (i) It should revolve in an orbit concentric and coplanar with the equatorial plane. (ii) It sense of rotation should be same as that of earth about its own axis i.e., in anti-clockwise direction (from west to east). (iii) Its period of revolution around the earth should be same as that of earth about its own axis.  T  24 hr  86400 sec

(iv) Height of geostationary satellite As

T  2

r3 (R  h)3  2  24 hr GM GM

Substituting the value of G and M we get R  h  r  42000 km  7 R  height of geostationary satellite from the surface of earh h  6 R  36000 km

genius PHYSICS 32 Gravitation (v) Orbital velocity of geo stationary satellite can be calculated by v 

GM r

Substituting the value of G and M we get v  3 .08 km / sec

8.25 Angular Momentum of Satellite. Angular momentum of satellite L  mvr GM r r



Lm



L  m 2 GMr

[As v 

GM ] r

i.e., Angular momentum of satellite depend on both the mass of orbiting and central body as well as the radius of orbit.

Important points (i) In case of satellite motion, force is central so torque = 0 and hence angular momentum of satellite is conserved i.e., L  constant (ii) In case of satellite motion as areal velocity

dA 1 (r)(vdt) 1   rv dt 2 dt 2

Sun r

dA L  dt 2m



dA

v dt

[As L  mvr ]

But as L  constant,  areal velocity (dA/dt) = constant which is Kepler’s II law i.e., Kepler’s II law or constancy of areal velocity is a consequence of conservation of angular momentum.

Sample problems based on angular momentum Problem 71. The orbital angular momentum of a satellite revolving at a distance r from the centre is L. If the distance is increased to 16r, then new angular momentum will be L (a) 16 L (b) 64 L (c) 4 Solution : (d)

(d) 4 L

Angular momentum L  m 2 GMr  L  r L2  L1

r2  r1

16 r 4 r

L 2  4 L1  4 L

Problem 72. Angular momentum of a planet of mass m orbiting around sun is J, areal velocity of its radius vector will be (a)

1 mJ 2

(b)

J 2m

(c)

m 2J

(d)

1 2mJ

Solution : (b)

8.26 Energy of Satellite. When a satellite revolves around a planet in its orbit, it possesses both potential energy (due to its position against gravitational pull of earth) and kinetic energy (due to orbital motion).

genius PHYSICS Gravitation 33

(1) Potential energy : U  mV  (2) Kinetic energy : K 

 GM 2   2  As V  r , L  m GMr   

 GMm  L2  r mr 2

1 GMm L2 mv 2   2 2r 2 mr 2

(3) Total energy : E  U  K 

  As v  

GM   r 

 GMm GMm  GMm  L2    r 2r 2r 2mr 2

Important points (i) Kinetic energy, potential energy or total energy of a satellite depends on the mass of the satellite and the central body and also on the radius of the orbit. (ii) From the above expressions we can say that Kinetic energy (K) = – (Total energy) Potential energy (U) = 2 (Total energy) Potential energy (K) = – 2 (Kinetic energy) (iii) Energy graph for a satellite

(iv) Energy distribution in elliptical orbit

+ Satellite

r

Energy

K

Perigee K.E. = max P.E. = min

O E

U

Focus

a Semi major axis

Apogee K.E. = min P.E. = max

r rmin

rmax

(B)

– (A)

(v) If the orbit of a satellite is elliptic then (a) Total energy (E) 

 GMm  constant ; where a is semi-major axis . 2a

(b) Kinetic energy (K) will be maximum when the satellite is closest to the central body (at perigee) and maximum when it is farthest from the central body (at apogee) (c) Potential energy (U) will be minimum when kinetic energy = maximum i.e., the satellite is closest to the central body (at perigee) and maximum when kinetic energy = minimum i.e., the satellite is farthest from the central body (at apogee). (vi) Binding Energy : Total energy of a satellite in its orbit is negative. Negative energy means that the satellite is bound to the central body by an attractive force and energy must be supplied to remove it from the orbit to infinity. The energy required to remove the satellite from its orbit to infinity is called Binding Energy of the system, i.e., Binding Energy (B.E.)   E 

GMm 2r

8.Change in the orbit of Satelite When the satellite is transferred to a higher orbit (r2  r1 ) then variation in different quantities can be shown by the following table Quantities Orbital velocity

Variation Decreases

Relation with r

v

1 r

Time period

Increases

T  r3 / 2

genius PHYSICS 34 Gravitation Linear momentum

Decreases

P

1 r

Angular momentum

Increases

Kinetic energy

Decreases

Potential energy

Increases

Total energy

Increases

Binding energy

Decreases

Note:: 

L r K

1 r

U

1 r

E

1 r

BE 

1 r

Work done in changing the orbit W  E 2  E1  GMm   GMm      W    2r2   2r1   W

GMm 2

r2 r1

M Earth

1 1      r1 r2 

Sample problems based on Energy Problem 73. Potential energy of a satellite having mass ‘m’ and rotating at a height of 6.4  10 6 m from the earth centre is [AIIMS 2000; CBSE PMT 2001; BHU 2001]

(a) 0.5 mgRe Solution : (a)

(b)  mgRe

Potential energy    Potential energy  

(c) 2 mgRe

GMm GMm GMm   r Re  h 2Re

(d) 4 mgRe

[As h  R e (given)]

gR e2 m  0.5mgR e 2Re

2 [As GM  gR ]

Problem 74. In a satellite if the time of revolution is T, then kinetic energy is proportional to (a) Solution : (d)

1 T

(b)

1 T

2

(c)

1

(d) T 2 / 3

T3

Time period T  r 3 / 2  r  T 2 / 3 and Kinetic energy 

1 1  2/3  T r T

2 / 3

.

Problem 75. Two satellites are moving around the earth in circular orbits at height R and 3R respectively, R being the radius of the earth, the ratio of their kinetic energies is (a) 2 Solution : (a)

(b) 4

(c) 8

(d) 16

r1  R  h1  R  R  2 R and r2  R  h2  R  3 R  4 R Kinetic energy 

(KE )1 r 1 4R 2   2   . r (KE ) 2 r1 2R 1

8.28 Weightlessness. The weight of a body is the force with which it is attracted towards the centre of earth. When a body is stationary with respect to the earth, its weight equals the gravity. This weight of the body is known as its static or true weight.

genius PHYSICS Gravitation 35

We become conscious of our weight, only when our weight (which is gravity) is opposed by some other object. Actually, the secret of measuring the weight of a body with a weighing machine lies in the fact that as we place the body on the machine, the weighing machine opposes the weight of the body. The reaction of the weighing machine to the body gives the measure of the weight of the body. The state of weightlessness can be observed in the following situations. (1) When objects fall freely under gravity : For example, a lift falling freely, or an airship showing a feat in which it falls freely for a few seconds during its flight, are in state of weightlessness. (2) When a satellite revolves in its orbit around the earth : Weightlessness poses many serious problems to the astronauts. It becomes quite difficult for them to control their movements. Everything in the satellite has to be kept tied down. Creation of artificial gravity is the answer to this problem. (3) When bodies are at null points in outer space : On a body projected up, the pull of the earth goes on decreasing, but at the same time the gravitational pull of the moon on the body goes on increasing. At one particular position, the two gravitational pulls may be equal and opposite and the net pull on the body becomes zero. This is zero gravity region or the null point and the body in question is said to appear weightless.

8.29 Weightlessness in a Satellite. A satellite, which does not produce its own gravity moves around the earth in a circular orbit under the GM action of gravity. The acceleration of satellite is 2 towards the centre of earth. r If a body of mass m placed on a surface inside a satellite moving around the earth. Then force on the body are Satellit (i) The gravitational pull of earth 

GMm r2

r R

(ii) The reaction by the surface  R By Newton’s law

h

Earth

m

e ma

GMm r

2

v

GmM R m a r2 GmM  GM  R m  2  2 r  r 



R=0

Thus the surface does not exert any force on the body and hence its apparent weight is zero. A body needs no support to stay at rest in the satellite and hence all position are equally comfortable. Such a state is called weightlessness.

Important points (i) One will find it difficult to control his movement, without weight he will tend to float freely. To get from one spot to the other he will have to push himself away from the walls or some other fixed objects. (ii) As everything is in free fall, so objects are at rest relative to each other, i.e., if a table is withdrawn from below an object, the object will remain where it was without any support. (iii) If a glass of water is tilted and glass is pulled out, the liquid in the shape of container will float and will not flow because of surface tension. (iv) If one tries to strike a match, the head will light but the stick will not burn. This is because in this situation convection currents will not be set up which supply oxygen for combustion (v) If one tries to perform simple pendulum experiment, the pendulum will not oscillate. It is because there will not be any restoring torque and so T  2 (L / g )   . [As g   0 ]

genius PHYSICS 36 Gravitation (vi) Condition of weightlessness can be experienced only when the mass of satellite is negligible so that it does not produce it own gravity. e.g. Moon is a satellite of earth but due to its own weight it applies gravitational force of attraction on the body placed on its surface and hence weight of the body will not be equal to zero at the surface of the moon.

Sample problems based on weightlessness in satellite Problem 76. The time period of a simple pendulum on a freely moving artificial satellite is (a) Zero Solution : (d)

T  2

(b) 2 sec l l  2  g 0

(c) 3 sec

(d) Infinite

[As g   0 in the satellite]

Problem 77. The weight of an astronaut, in an artificial satellite revolving around the earth, is (a) Zero (c) More than that on the earth

(b) Equal to that on the earth (d) Less than that on the earth

Solution : (a)

Problem 78. A ball is dropped from a spacecraft revolving around the earth at a height of 120 km. What will happen to the ball (a) It will continue to move with velocity v along the original orbit of spacecraft (b) If will move with the same speed tangentially to the spacecraft (c) It will fall down to the earth gradually (d) It will go very far in the space Solution : (a)

Because ball possess same initial tangential speed as that of space craft.. So it also feels the condition of weightlessness.

Sample problems (Miscellaneous) Problem 79. Two particles of equal mass go round a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is (a) v  Solution : (c)

1 2R

1 Gm

(b) v 

Gm 2R

[CBSE PMT 1995]

(c) v 

1 2

Gm R

Both the particles moves diametrically opposite position along the circular path of radius R and the gravitational force provides required centripetal force mv 2 Gmm 1  v  2 R 2 (2 R)

Gm R

(d) v 

m

4 Gm R

2R m

Problem 80. Two types of balances, the beam balance and the spring balance are commonly used for measuring weight in shops. If we are on the moon, we can continue to use (a) Only the beam type balance without any change (b) Only the spring balance without any change (c) Both the balances without any change (d) Neither of the two balances without making any change Solution : (a)

Because in beam type balance effect of less gravitation force works on both the Pans. So it is neutralizes but in spring balance weight of the body decreases so apparent weight varies with actual weight.

Problem 81. During a journey from earth to the moon and back, the greatest energy required from the space-ship rockets is to overcome (a) The earth’s gravity at take off (b) The moon’s gravity at lunar landing (c) The moon’s gravity at lunar take off (d) The point where the pull of the earth and moon are equal but opposite Solution : (a)

[CBSE PMT 1991]

genius PHYSICS Gravitation 37 1 of its present value, the length of the day will be approximately n

Problem 82. If the radius of earth contracts 24 h n

(a) Solution : (b)

(b)

24 n2

(d) 24 n 2 h

(c) 24 n h

h

Conservation of angular momentum L  I  

2 2 MR 2   constant  T  R 2 5 T

[If M remains

same] R T  2   2 T1  R 1

2

 R /n      R  

2

1



n

2

 T2 

24 hr n2

[As T1  24 hr ].

Problem 83. A body released from a height h takes time t to reach earth’s surface. The time taken by the same body released from the same height to reach the moon’s surface is (a) t Solution : (c)

(b) 6t

(c)

6t t 2h  moon  t earth g

If body falls from height h then time of descent t 

(d)

t 6

g earth  6  t moon  6 t . g moon

Problem 84. A satellite is revolving round the earth with orbital speed v 0 . If it stops suddenly, the speed with which it will strike the surface of earth would be ( v e  escape velocity of a particle on earth’s surface) (a) Solution : (d)

v e2 v0

(b) v 0

v e2  v 02

(c)

(d)

v e2  2v 02

Applying conservation of mechanical energy between A and B point GMm GMm GMm 1  GMm  1 mv 2    mv 2    ; 2 R r r 2 R  

 v2 

B

2 Gm 2 Gm   v e2  2v 02  v  ve2  2v02 R r

[As escape velocity v e 

u=0

v

A r

2Gm , orbital velocity v 0  R

Gm ] r

Problem 85. The escape velocity for a planet is v e . A tunnel is dug along a diameter of the planet and a small body is dropped into it at the surface. When the body reaches the centre of the planet, its speed will be v v (a) v e (b) e (c) e (d) Zero 2 2 Solution : (b)

Gravitational potential at the surface of the earth V s   Gravitational potential at the centre of earth Vc   By the conservation of energy

GM R

3 GM 2R

1 mv 2  m Vs  Vc  2

v2  2

v2 GM  3  GM  gR  e   1  R R 2 2 

 v

ve

[As v e 

2 gR ]

2

Problem 86. A small body of superdense material, whose mass is twice the mass of the earth but whose size is very

small compared to the size of the earth, starts from rest at a height H  R above the earth’s surface, and reaches the earth’s surface in time t. Then t is equal to (a)

Solution : (c)

2H / g

(b)

H/g

(c)

2H / 3g

As the masses of the body and the earth are comparable, they will move towards their centre of mass, which remains stationary.

(d) M

4 H / 3g

2H 3

H C.M. 3

P Earth H

2M

Body

genius PHYSICS 38 Gravitation Hence the body of mass 2m move through distance and time to reach the earth surface 

2h  g

H . 3

2H / 3  g

2H 3g

Problems based on Newton’s law of gravitation

 Basic level 1.

The force of gravitation is (a) Repulsive

2.

(d) Non - conservative

(b) Becomes four times

(c) Is reduced to half

[CPMT 1973; AMU (Med.) 2000]

(d) Is reduced to a quarter

(b) 1/2

(c) 1/4

(d) 1/5

Three particles each of mass m are placed at the three corners of an equilateral triangle. The centre of the triangle is at a distance x from either corner. If a mass M be placed at the centre, what will be the net gravitational force on it (a) Zero

5.

(c) Conservative

A mass M is split into two parts, m and M – m, which are then separated by a certain distance. What ratio of m/M maximizes the gravitational force between the two parts [AMU 2000] (a) 1/3

4.

(b) Electrostatic

If the distance between two masses is doubled, the gravitational attraction between them (a) Is doubled

3.

[AIIMS 2002]

(b) 3GMm / x 2

(c)

2GMm / x 2

(d)

GMm / x 2

Two identical spheres are placed in contact with each other. The force of gravitation between the sph eres will be proportional to (R = radius of each sphere) (a) R

(b) R2

(c) R4

(d) None of these

 Advance level 6.

A solid sphere of uniform density and radius R applies a gravitational force of attraction equal to F1 on a particle placed at A, distance 2 R from the centre of the sphere. A spherical cavity of radius R/2 is now made in the sphere as shown in the figure. The sphere with cavity now applies a gravitational force F2 on the same particle placed at A. The ratio F2 / F1 will be (a) 1/2 (b) 3 (c) 7

R

R

A

(d) 7/9

7.

Three uniform spheres of mass M and radius R each are kept in such a way that each touches the other two. The magnitude of the gravitational force on any of the spheres due to the other two is (a)

8.

3 GM 2 4 R2

(b)

3 GM 2 2 R2

(c)

3 GM 2 R2

(d)

3 GM 2 2 R2

A mass of 10kg is balanced on a sensitive physical balance. A 1000 kg mass is placed below 10 kg mass at a distance of 1m. How much additional mass will be required for balancing the physical balance (a) 66  10 15 kg

(b) 6.7  10 8 kg

(c) 66  10 12 kg

(d)

Problems based on acceleration due to gravity

 Basic level

6.7  10 6 kg

genius PHYSICS Gravitation 39 9.

If R is the radius of the earth and g the acceleration due to gravity on the earth's surface, the mean density of the earth is [CPMT 1990; CBSE 1995; BHU 1998; MH CET (Med.) 1999; Kerala PMT 2002]

(a) 4G / 3 gR

10.

(b) 3R / 4 gG

(b) m / 8

(d)

Rg / 12G

(c) m / 4

(d) m

The diameters of two planets are in the ratio 4 : 1 and their mean densities in the ratio 1: 2. The acceleration due to gravity on the planets will be in ratio [ISM Dhanbad 1994] (a) 1 : 2

12.

3 g / 4RG

A mass 'm' is taken to a planet whose mass is equal to half that of earth and radius is four times that of earth. The mass of the body on this planet will be [RPMT 1989, 97] (a) m / 2

11.

(c)

(b) 2 : 3

(c) 2 : 1

(d) 4 : 1

The acceleration due to gravity on the moon is only one sixth that of earth. If the earth and moon are assumed to have the same density, the ratio of the radii of moon and earth will be (a)

1 6

(b)

1 (6)1 / 3

(c)

1 36

(d)

1 (6 )2 . / 3

 Advance level 13.

14.

Let g be the acceleration due to gravity at earth's surface and K be the rotational kinetic energy of the earth. Suppose the earth's radius decreases by 2% keeping all other quantities same, then [BHU 1994; JIPMER 2000] (a) g decreases by 2% and K decreases by 4%

(b) g decreases by 4% and K increases by 2%

(c) g increases by 4% and K decreases by 4%

(d) g decreases by 4% and K increase by 4%

Clock A based on spring oscillations and a clock B based on oscillations of simple pendulum are synchronised on earth. Both are taken to mars whose mass is 0.1 times the mass of earth and radius is half that of earth. Which of the following statement is correct (a) Both will show same time (b) Time measured in clock A will be greater than that in clock B (c) Time measured in clock B will be greater than that in clock A (d) Clock A will stop and clock B will show time as it shows on earth

Problems based on variation in g with height

 Basic level 15.

A body weight W Newton at the surface of the earth. Its weight at a height equal to half the radius of the earth will be [UPSEAT 2002]

(a)

16.

W 2

(b)

2W 3

(c)

4W 9

(d)

8W 27

The value of g on the earth's surface is 980 cm/sec2. Its value at a height of 64 km from the earth's surface is (Radius of the earth R = 6400 Kilometers) (a) 960 .40 cm / sec

17.

2

(b) 984 .90 cm / sec

[MP PMT 1995] 2

(c) 982 .45 cm / sec

2

(d)

977 .55 cm / sec

(d)

R 2 hg

2

The decrease in the value of g at height h from earth's surface is (a)

2h R

(b)

2h g R

(c)

h g R

 Advance level 18.

A simple pendulum has a time period T1 when on earth's surface and T2 when taken to a height R above the earth's surface, where R is the radius of earth. The value of T2 / T1 is [IIT-JEE (Screening) 2001] (a) 1

19.

20.

(b)

2

(c) 4

(d) 2

A pendulum clock is set to give correct time at the sea level. This clock is moved to hill station at an altitude of 2500m above the sea level. In order to keep correct time of the hill station, the length of the pendulum [SCRA 1994] (a) Has to be reduced

(b) Has to be increased

(c) Needs no adjustment

(d) Needs no adjustment but its mass has to be increased

Which of the following correctly indicates the approximate effective values of g on various parts of a journey to the moon (values are in metres/sec2) Before take off from earth

One minute after lift – off

In earth orbit

on the moon

genius PHYSICS 40 Gravitation

21.

(a)

9.80

9.80

0

1.6

(b)

9.80

0.98

0

1.6

(c)

9.80

0.00

0

9.8×6

(d)

9.80

7.00

0

1.6

Two blocks of masses m each are hung from a balance. The scale pan A is at height H 1 whereas scale pan B is at height H 2 . The error in weighing when H 1  H 2 and R being the radius of earth is

 1  2H 1 (a) mg  R 

  

H H (b) 2mg  1  2 R  R

  

(c)

H  H 2mg  2  1  R R  

(d)

2mg

A

m m

H1

B H2

H 2 H1 H1  H 2

Problems based on variation in g with depth

 Basic level 22.

If the value of 'g' acceleration due to gravity, at earth surface is 10 m / s 2 , its value in m / s 2 at the centre of the earth, which is assumed to be a sphere of radius 'R' metre and uniform mass density is [AIIMS 2002] (a) 5

23.

(b) 10/R

(d) Zero

The loss in weight of a body taken from earth's surface to a height h is 1%. The change in weight taken into a mine of depth h will be (a) 1% loss

24.

(c) 10/2R

(b) 1% gain

(c) 0.5% gain

(d) 0.5% loss

The weight of body at earth's surface is W. At a depth half way to the centre of the earth, it will be (assuming uniform density in earth) (a) W

(b) W/2

(c) W/4

(d) W/8

 Advance level 25.

A particle would take a time t to move down a straight tunnel from the surface of earth (supposed to be a homogeneous t sphere) to its centre. If gravity were to remain constant this time would be t'. The ratio of will be t' (a)



(b)

2 2

26.

 2

(c)

2 3



(d)

3

Suppose a vertical tunnel is dug along the diameter of earth assumed to be a sphere of uniform mass having density  . If a body of mass m is thrown in this tunnel, its acceleration at a distance y from the centre is given by

27.

(a)

4 G ym 3

(c)

4 y 3

(b)

3 Gy 4

(d)

4 G y 3

m y

A tunnel is dug along the diameter of the earth. If a particle of mass m is situated in the tunnel at a distance x from the centre of earth then gravitational force acting on it, will be (a)

GM e m x R e3

(b)

GM e m R e2

(c)

GM e m x2

(d)

GM e m (R e  x ) 2

Problems based on variation in g due to shape of the earth

 Basic level 28.

The acceleration due to gravity at pole and equator can be related as

[DPMT 2002]

genius PHYSICS Gravitation 41 (a) g p  ge

29.

(b) g p  ge  g

g p  ge  g

(d)

g p  ge

Weight of a body is maximum at (a) Moon

30.

(c)

[AFMC 2001]

(b) Poles of earth

(c) Equator of earth

(d) Centre of earth

The value of 'g' at a particular point is 9.8m/s2. Suppose the earth suddenly shrinks uniformly to half its present size without losing any mass. The value of 'g' at the same point (assuming that the distance of the point from the centre of earth does not shrink) will now be [NCERT 1984; DPMT 1999] (a) 4.9m / sec 2

(b) 3.1m / sec 2

(c) 9.8 m / sec 2

(d)

19 .6m / sec 2

 Advance level 31.

The acceleration due to gravity increases by 0.5% when we go from the equator to the poles. What will be the time period of the pendulum at the equator which beats seconds at the poles (a) 1.950 s

(b) 1.995 s

(c) 2.050 s

(d) 2.005 s

Problems based on gravitational field

 Basic level 32.

There are two bodies of masses 100 kg and 10000 kg separated by a distance 1m. At what distance from the smaller body, the intensity of gravitational field will be zero [BHU 1997] (a)

33.

1 m 9

(b)

1 m 10

I

(a)

(d)

10 m 11

I

(b)

I

(c)

(d)

r =a r =a r r =a r r r =a The curve depicting the dependence of intensity of gravitational field on the distance r from the centre of the earth is I

(a) O

I

I

(b)

r=R

r=R

r

r

I

(c) O

r

35.

1 m 11

Which one of the following graphs represents correctly the variation of the gravitational field (F) with the distance (r) from the centre of a spherical shell of mass M and radius a I

34.

(c)

(d) O

r=R

O

r

r=R

r

A thin spherical shell of mass M and radius R has a small hole. A particle of mass m is released at the mouth of the hole. Then (a) The particle will execute simple harmonic motion inside the shell

m

(b) The particle will oscillate inside the small, but the oscillations are not simple harmonic R

(c) The particle will not oscillate, but the speed of the particle will go on increasing

M

(d) None of these

 Advance level 36.

A solid sphere of uniform density and radius 4 units is located with its centre at the origin O of coordinates. Two spheres of equal radii 1 unit with their centres at A(– 2, 0, 0) and B(2, 0, 0) respectively are taken out of the solid leaving behind spherical cavities as shown in figure (a) The gravitational force due to this object at the origin is zero (b) The gravitational force at the point B (2, 0, 0) is zero (c) The gravitational potential is the same at all points of the circle y 2  z 2  36

[IIT-JEE 1993]

y A

B O

z

x

genius PHYSICS 42 Gravitation (d) The gravitational potential is the same at all points on the circle y 2  z 2  4

37.

(a)

(c)

38.

Gm along x axis l 2 Gm l2

(b)

along x axis

(d)

Gm along y axis l 2 Gm l2

(c)

Gm 1 r22

G(m 1  m 2 )m r12

,

(b)

G(m 1  m 2 )m Gm 2 , ,0 r12 r22

O

B

x

(d)

Gm 2 r22

, 0,

Gm 1

B r2 C

r12

r3

G(m 1  m 2 )m Gm 1 , 2 ,0 r12 r2

m2 m 1

A r 1

A sphere of mass M and radius R2 has a concentric cavity of radius R1 as shown in figure. The force F exerted by the sphere on a particle of mass m located at a distance r from the centre of sphere varies as (0  r  )

F

R2

F

r

(a)

F

r

(b)

F

r

(c)

r

(d)

A spherical hole is made in a solid sphere of radius R. The mass of the sphere before hollowing was M. The gravitational field at the centre of the hole due to the remaining mass is (a) Zero (c)

41.

A

along y axis

R1

40.

l

Two concentric shells of different masses m 1 and m 2 are having a sliding particle of mass m. The forces on the particle at position A, B and C are (a) 0,

39.

y

Gravitational field at the centre of a semicircle formed by a thin wire AB of mass m and length l is

(b)

GM

(d)

2R 2

GM 8R 2

R

GM R2

A point P lies on the axis of a ring of mass M and radius a, at a distance a from its centre C. A small particle starts from P and reaches C under gravitational attraction only. Its speed at C will be

2GM a

(a)

(b)

2 GM a

  1  1    2  

2GM ( 2  1) a

(c)

(d) Zero

Problems based on gravitational potential

 Basic level 42.

If V is the gravitational potential on the surface of the earth, then what is its value at the centre of the earth (a) 2V

43.

(b) 3V

(c)

3 V 2

(d)

2 V 3

The diagram showing the variation of gravitational potential of earth with distance from the centre of earth is V

O

V

V r=R r

O

r=R r

O

V r=R r

O

r=R r

genius PHYSICS Gravitation 43 (a)

44.

(b)

(c)

By which curve will the variation of gravitational potential of a hollow sphere of radius R with distance be depicted V

V

(a)

45.

(d)

(b)

O

r=R

r=R

O

r

V

V

(c) r

(d)

r=R

O

r=R

O

r

r

Two concentric shells have mass M and m and their radii are R and r respectively, where R  r . What is the gravitational potential at their common centre (a) 

GM R

(b) 

Gm r

M m  G   r R

(c)

M m (d)  G    r R

 Advance level 46.

A person brings a mass of 1 kg from infinity to a point A. Initially the mass was at rest but it moves with a speed of 2 m/s as it reaches A. The work done by the person on the mass is – 3 J. The potential of A is (a) – 3 J/kg

47.

(b) – 2 J/kg

(c) – 5 J/kg

(d) – 7 J/kg

A thin rod of length L is bent to form a semicircle. The mass of the rod is M. What will be the gravitational potential at the centre of the circle (a) 

GM L

(b) 

GM 2L



(c)

GM

(d) 

2L

GM L

Problems based on escape velocity

 Basic level 48.

The escape velocity of a planet having mass 6 times and radius 2 times as that of earth is (a)

49.

(b) 3 Ve

3 Ve

(d) m 1

(c) m 0

(b) 2 times

2 times

(c)

[RPMT 2000]

(d) 4 times

3 times

The orbital velocity of a satellite at a height h above the surface of earth is v. The value of escape velocity from the same location is given by [J&K CET 2000] (a)

52.

(b) m

[CPMT 1978; RPMT 1999; AIEEE 2002]

How many times is escape velocity (v e ) , of orbital velocity (v 0 ) for a satellite revolving near earth (a)

51.

(d) 2Ve

2 Ve

The escape velocity of a particle of mass m varies as (a) m 2

50.

(c)

[CPMT 1999; MP PET 2003]

How

(b) v

2v

much

energy

will

be

v

(c) necessary

for

making

(d)

2

a

body

of

500

kg

v 2

escape

from

the

earth

[g  9.8 m / s , radius of earth  6.4  10 m] 2

6

[MP PET 1999]

(a) About 9 .8  10 6 J

53.

(b) About 6.4  10 8 J

(c) About 3.1  10 10 J

(d) About 27 .4  10 12 J

The escape velocity of a body on the surface of the earth is 11 .2 km / s . If the earth’s mass increases to twice its present value and the radius of the earth becomes half, the escape velocity would become [CBSE PMT 1997] (a) 5.6 km / s

(b) 11 .2 km / s (remain unchanged)

genius PHYSICS 44 Gravitation (c)

54.

A rocket is launched with velocity 10 km/s. If radius of earth is R, then maximum height attained by it will be (a) 2R

55.

(d) 44 .8 km / s

22 .4 km / s (b) 3R

(c) 4R

[RPET 1997]

(d) 5R

A missile is launched with a velocity less then the escape velocity. The sum of its kinetic and potential energy is

[MP PET 1995]

(a) Positive (b) Negative (c) Zero (d) May be positive or negative depending upon its initial velocity

56.

v e and v p denotes the escape velocity from the earth and another planet having twice the radius and the same mean density as the earth. Then (a) v e  v p

57.

[NCERT 1974; MP PMT 1994]

(b) ve  v p / 2

(c) ve  2v p

(d) ve  v p / 4

The magnitude of the potential energy per unit mass of the object at the surface of earth is E. Then the escape velocity of the object is (a)

2E

(b) 4 E 2

(c)

E

(d)

E/2

 Advance level 58.

A ball of mass m is fired vertically upwards from the surface of the earth with velocity nv e , where v e is the escape velocity and n < 1. Neglecting air resistance, to what height will the ball rise? (Take radius of the earth as R ) (a) R / n 2

59.

(b) R /(1  n 2 )

(c)

Rn 2 /(1  n 2 )

(d) Rn 2

The masses and radii of the earth and moon are M 1 , R1 and M 2 , R 2 respectively. Their centres are distance d apart. The minimum velocity with which a particle of mass m should be projected from a point midway between their centres so that it escape to infinity is [MP PET 1997]

(a) 2

60.

G (M 1  M 2 ) d

(b) 2

2G (M 1  M 2 ) d

(c)

2

Gm (M 1  M 2 ) d

(d) 2

Gm (M 1  M 2 ) d (R 1  R 2 )

A body is projected with a velocity 2ve, where ve is the escape velocity. Its velocity when it escapes the gravitational field of the earth is (a)

7v e

(b)

5v e

(c)

3v e

(d) v e

Problems based on energy

 Basic level 61.

Escape velocity of a body of 1 kg mass on a planet is 100 m/sec. Gravitational potential energy of the body at the planet is [MP PMT 2002]

(a) – 5000 J

62.

63.

(b) – 1000 J (c) – 2400 J (d) 5000 J R A body of mass m rises to a height h  from the earth’s surface where R is earth’s radius. If g is acceleration due to gravity 5 at the earth’s surface, the increase in potential energy is [CPMT 1989] 4 5 6 mgh mgh mgh (a) mgh (b) (c) (d) 5 6 7 The work done is bringing three particles each of mass 10 g from large distances to the vertices of an equilateral triangle of side 10 cm. (a) 1  10 13 J

64.

(b) 2  10 13 J

(c)

4  10 11 J

(d) 1  10 11 J

The potential energy due to gravitational field of earth will be maximum at (a) Infinite distance

(b) The poles of earth

(c) The centre of earth

(d) The equator of earth

 Advance level 65.

The radius and mass of earth are increased by 0.5%. Which of the following statement is false at the surface of the earth [Roorkee 2000]

genius PHYSICS Gravitation 45

66.

(a) g will increase

(b) g will decrease

(c) Escape velocity will remain unchanged

(d) Potential energy will remain unchanged

Two identical thin rings each of radius R are coaxially placed at a distance R. If the rings have a uniform mass distribution and each has mass m 1 and m 2 respectively, then the work done in moving a mass m from centre of one ring to that of the other is (a) Zero

(b)

Gm (m 1  m 2 )( 2  1)

(c)

2R

Gm 2 (m 1  m 2 ) R

(d)

Gm 1 m 2 ( 2  1) m2R

Problems based on orbital velocity of satellite

 Basic level 67.

The orbital velocity of an artificial satellite in a circular orbit just above the earth’s surface is v. For a satellite orbiting at an altitude of half of the earth’s radius, the orbital velocity is [Kerala (Engg.) 2001] (a)

68.

3 V 2

(b)

(b) (a / b) v

(d)

2 V 3

(c) ( a / b ) v

(d) ( b / a ) v

For the moon to cease to remain the earth’s satellite its orbital velocity has to increase by a factor of (a) 2

70.

2 V 3

(c)

The speed of a satellite is v while revolving in an elliptical orbit and is at nearest distance ‘a’ from earth. The speed of satellite at farthest distance ‘b’ will be [RPMT 1995] (a) (b / a) v

69.

3 V 2

(b)

(c) 1 / 2

2

(d)

[MP PET 1994]

3

Two artificial satellites A and B are at a distances rA and rB above the earth’s surface. If the radius of earth is R, then the ratio of their speeds will be 1/2

r  R  (a)  B   rA  R 

r  R  (b)  B   rA  R 

2

r (c)  B  rA

   

2

r (d)  B  rA

   

1/2

 Advance level 71.

When a satellite going round earth in a circular orbit of radius r and speed v, losses some of its energy. Then r and v change as [EAMCET (Med.) 2000]

(a) r and v both will increase

(b)

(c) r will decrease and v will increase

72.

r and v both will decrease

(d) r will increase and v will decrease

A satellite is revolving around a planet of mass M in an elliptical orbit of semi-major axis a. The orbital velocity of the satellite at a distance r from the focus will be   2 1  (a) GM     r a  

1/2

  1 2  (b) GM     r a  

1/2

(c)

 1  2 GM  2  2 r a  

  

1/2

 2  1 (d) GM  2  2 r a  

  

1/2

Problems based on time period of satellite

 Basic level 73.

A geo-stationary satellite is orbiting the earth at a height of 6 R above the surface of earth, R being the radius of earth. The time period of another satellite at a height of 2.5 R from the surface of earth is [UPSEAT 2002; AMU (Med.) 2002] (a) 10 hr

74.

(c) 6 hr

(d) 6 2 hr

Time period of revolution of a satellite around a planet of radius R is T. Period of revolution around another planet. Whose radius is 3R but having same density is [CPMT 1981] (a) T

75.

(b) (6 / 2 ) hr

(b) 3T

(c) 9T

(d) 3 3 T

A satellite is orbiting around the earth with a period T. If the earth suddenly shrinks to half its radius without change in mass, the period of revolution of the satellite will be (a) T / 2

(b) T / 2

(c) T

(d) 2T

genius PHYSICS 46 Gravitation 76.

A satellite is orbiting around the earth in the equitorial plane rotating from west to east as the earth does. If  e be the angular speed of the earth and  s be that of satellite, then the satellite will repeatedly appear at the some location after a time t = (a)

77.

2 s  c

(b)

2  s  c

(c)

 s  c

(d)

 s  c

Suppose the gravitational force varies inversely as the nth power of distance. Then, the time period of a planet in circular orbit of radius R around the sun will be proportional to (a) R n

(b) R

n 1 2

(c)

R

n 1 2

(d) R n

 Advance level 78.

A geostationary satellite orbits around the earth in a circular orbit of radius 36000 km. Then, the time period of a satellite orbiting a few hundred kilometres above the earth’s surface (R Earth  6400 km ) will approximately be [IIT-JEE (Screening) 2002] (a) 1/2 h

79.

(b) 1 h

(c) 2 h

(d) 4 h

If the distance between the earth and the sun becomes half its present value, the number of days in a year would have been [IIT-JEE 1996; RPET 1996]

(a) 64.5

80.

(b) 129

(c) 182.5

(d) 730

A satellite is launched into a circular orbit of radius R around the earth. A second satellite is launched into an orbit of radius (1.01) R. The period of the second satellite is larger than that of the first one by approximately [IIT-JEE 1995] (a) 0.5%

(b) 1.0%

(c) 1.5%

(d) 3.0%

81.

A satellite moves eastwards very near the surface of the earth in the equatorial plane of the earth with speed v 0 . Another satellite moves at the same height with the same speed in the equatorial plane but westwards. If R = radius of the earth about its own axis, then the difference in the two time period as observed on the earth will be approximately equal to 4Rv 0 4Rv 0 4Rv 0 2Rv 0 (a) (b) (c) (d) R 2 4  v 02 R 2 2  v 02 R 2 2  v 02 R 2 2  v 02

82.

A “double star” is a composite system of two stars rotating about their centre of mass under their mutual gravitational attraction. Let us consider such a “double star” which has two stars of masses m and 2m at separation l. If T is the time period of rotation about their centre of mass then, (a) T  2

83.

l3 mG

(b) T  2

l3 2mG

(c) T  2

l3 3mG

(d) T  2

l3 4 mG

A space probe projected from the earth moves round the moon in a circular orbit at a distance equal to its radius R R moon  where R = radius of the earth. Its rocket launcher moves in circular orbit around the earth at a distance equal to R 4 from its surface. The ratio of the times taken for one revolution by the probe and the rocket launcher is M   , where M  mass of the earth   M moon  80   (a)

3 :2

(b)

5 :2

(c) 1 : 1

(d) 2 : 3

Problems based on height of satellite

 Basic level 84.

The distance of a geo-stationary satellite from the centre of the earth (Radius R = 6400 km) is nearest to (a) 5 R

85.

(b) 7 R

(c) 10 R

[AFMC 2001]

(d) 18 R

An artificial satellite is moving in a circular orbit around the earth with a speed equal to half the escape speed from the earth. If R is the radius of the earth then the height of the satellite above the surface of the earth is (a)

R 2

(b)

2R 3

(c)

R

(d) 2R

 Advance level 86.

If the angular velocity of a planet about its own axis is halved, the distance of geostationary satellite of this planet from the centre of the planet will become (a) (2)1 / 3 times

(b) (2)3 / 2 times

(c) (2) 2 / 3 times

Problems based on energy of satellite

(d) 4 times

genius PHYSICS Gravitation 47

 Basic level 87.

A satellite moves around the earth in a circular orbit with speed v. If m is the mass of the satellite, its total energy is [CBSE PMT 1991] (a) 

88.

(b)

1 mv 2 2

(c)

3 mv 2 2

(d)

1 mv 2 4

The minimum energy required to launch a satellite of mass m from the surface of earth of radius R in a circular orbit at an altitude 2 R is (mass of earth is M) (a)

89.

1 mv 2 2

5 GmM 6R

(b)

2GmM 3R

(c)

The masses of moon and earth are 7.36  10 22 kg

GmM 2R

(d)

GmM 3R

and 5 .98  10 24 kg respectively and their mean separation is

3.82  10 5 km . The energy required to break the earth-moon system is

(a) 12 .4  10 32 J

90.

(b) 3.84  10 28 J

(d) 2.96  10 20 J

A body placed at a distance R0 form the centre of earth, starts moving from rest. The velocity of the body on reaching at the earth’s surface will be ( R e  radius of earth and M e  mass of earth)

 1 1   (a) GM e    R R 0   e

91.

(c) 5.36  10 24 J

 1 1   (b) 2GM e    R R 0   e

(c) GM e

1 1  Re R0

 1 1   2GM e    R R 0   e

(d)

If total energy of an earth satellite is zero, it means that (a) The satellite is bound to earth (b) The satellite may no longer be bound to earth’s field (c) The satellite moves away from the orbit along a parabolic path (d) The satellite escapes in a hyperbolic path

 Advance level 92.

By what percent the energy of a satellite has to be increased to shift it from an orbit of radius r to (a) 66.7%

93.

(b) 33.3%

(c) 15%

3 r 2

(d) 20.3%

A mass m is raised from the surface of the earth to a point distant R(  1) from the centre of the earth and then put into a circular orbit to make it an artificial satellite. The total work done to complete this job is (a) mgR (2   1)

(b) mgR (2   1)

(c) mgR (  1)

(d) mgR

2  1 2

Problems based on angular momentum of satellite

 Basic level 94.

A satellite of mass m is circulating around the earth with constant angular velocity. If radius of the orbit is R 0 and mass of the earth M, the angular momentum about the centre of the earth is [MP PMT 1996; RPMT 2000] (a) m GMR 0

95.

(b) M GmR

0

(c) m

GM R0

(d) M

GM R0

A planet of mass m is moving in an elliptical path about the sun. Its maximum and minimum distances from the sun are r1 and r2 respectively. If M s is the mass of sun then the angular momentum of this planet about the center of sun will be (a)

2GM s (r1  r2 )

(b) 2GM sm

r1r2 (r1  r2 )

(c) m

2GM sr1r2 (r1  r2 )

(d) m

2GM sm (r1  r2 ) r1r2

Problems based on weightlessness in satellite

 Basic level 96.

97.

Reaction of weightlessness in a satellite is

[RPMT 2000]

(a) Zero gravity

(b) Centre of mass

(c) Zero reaction force by satellite surface

(d) None of these

A body suspended from a spring balance is placed in a satellite. Reading in balance is W1 when the satellite moves in an orbit of radius R. Reading in balance is W2 when the satellite moves in an orbit of radius 2 R . Then (a) W1  W2

(b) W1  W2

(c) W1  W2

(d) W1  2W2

genius PHYSICS 48 Gravitation 98.

An astronaut feels weightlessness because (a) Gravity is zero there (b) Atmosphere is not there (c) Energy is zero in the chamber of a rocket (d) The fictitious force in rotating frame of reference cancels the effect or weight

99.

Inside a satellite orbiting very close to the earth’s surface, water does not fall out of a glass when it is inverted. Which of the following is the best explanation for this (a) The earth does not exert any force on the water (b) The earth’s force of a attraction on the water is exactly balanced by the force created by the satellites motion (c) The water and the glass have the same acceleration, equal to g, towards the centre of the earth, and hence there is no relative motion between them (d) The gravitational attraction between the glass and the water balances the earth’s attraction on the water

100. To overcome the effect of weightlessness in an artificial satellite (a) The satellite is rotated its axis with compartment of astronaut at the centre of the satellite (b) The satellite is shaped like a wheel (c) The satellite is rotated around and around till weightlessness disappears (d) The compartment of astronaut is kept on the periphery of rotating wheel like satellite

Problems based on Kepler's laws

 Basic level 101. Which of the following astronomer first proposed that sun is static and earth rounds sun (a) Copernicus

(b) Kepler

(c) Galilio

[AFMC 2002]

(d) None

102. The period of a satellite in a circular orbit of radius R is T, the period of another satellite in a circular orbit of radius 4R is [CPMT 1982; MP PET/PMT 1998; AIIMS 2000; CBSE 2002]

(a) 4 T

(b) T / 4

(c) 8 T

(d) T / 8

103. Kepler’s second law is based on

[AIIMS 2002]

(a) Newton’s first law

(b) Newton’s second law

(c) Special theory of relativity

(d) Conservation of angular momentum

104. Two planets at mean distance d 1 and d 2 from the sun and their frequencies are n 1 and n 2 respectively then [Kerala (Med.) 2002] (a) n12 d 12  n 2 d 22

(b) n 22 d 23  n12 d 13

(c) n1 d 12  n 2 d 22

(d) n12 d 1  n 22 d 2

105. Earth needs one year to complete one revolution round the sun. If the distance between sun and earth is doubled then the period of revolution of earth will become (a) 2 2 yrs

(b) 8 yrs

[PM PMT 1997]

(c)

1 yrs 2

(d) 1 yrs

106. The eccentricity of earth’s orbit is 0.0167. The ratio of its maximum speed in its orbit to its minimum speed is (a) 2.507

(b) 1.033

(c) 8.324

[NCERT 1973]

(d) 1.000

107. For a planet around the sun in an elliptical orbit of semi – major and semi – minor axes a and b, respectively, and period T (A) The torque acting on the planet about the sun is non – zero (B) The angular momentum of the planet about the sun is constant (C) The areal velocity is ab / T (D) The planet moves with a constant speed around the sun (a) A, B

(b) B, C

(c) C, D

(d) D, A

108. A planet moves in an elliptical orbit around one of the foci. The ratio of maximum velocity v max and minimum velocity v min in terms of eccentricity e of the ellipse is given by (a)

1e 1e

(b)

e 1 e 1

(c)

1e 1e

(d)

e e 1

genius PHYSICS Gravitation 49 109. The satellites S 1 and S 2 describe circular orbits of radii r and 2r respectively around a planet. If the orbital angular velocity of S 1 is  , that of S 2 is (a)



(b)  2

2 2

(c)

 2

(d)

 2 3

 Advance level 110. Imagine a light planet revolving around a very massive star in a circular orbit of radius R with a period of revolution T. If the gravitational force of attraction between planet and star is proportional to R 5 / 2 , then T 2 is proportional to [IIT-JEE 1989; RPMT 1997]

(a) R

(b) R

3

(c)

7/2

R

(d) R 3 / 2

5/2

111. A binary star has stars of masses m and nm (where n is a numerical factor) having separation of their centres as r. If these stars revolve because of gravitational force of each other, the period of revolution is given by

(a)

2r 3 / 2

2r 1 / 2

(b)

1 2 2

(c)

1 2

 G(n  1)m   nm  

 Gnm    (n  1)m   

2r 3 2 GMn 3

(d)

2r 3 / 2 2   GMn  3 

2/3

Answers of Practice problems 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

c

d

b

a

c

d

a

b

c

d

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

c

a

c

b

c

a

b

d

a

d

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

b

d

d

b

a

d

a

d

b

c

31.

32.

33.

34.

35.

36.

37.

38.

39.

40.

d

c

d

b

d

a, c, d

d

d

b

c

41.

42.

43.

44.

45.

46.

47.

48.

49.

50.

b

c

c

c

d

c

d

a

c

a

51.

52.

53.

54.

55.

56.

57.

58.

59.

60.

b

c

c

c

b

b

a

c

a

c

61.

62.

63.

64.

65.

66.

67.

68.

69.

70.

a

c

b

a

a

b

c

b

b

a

71.

72.

73.

74.

75.

76.

77.

78.

79.

80.

c

a

d

a

c

a

b

c

b

c

81.

82.

83.

84.

85.

86.

87.

88.

89.

90.

b

c

b

b

c

c

a

a

b

d

91.

92.

93.

94.

95.

96.

97.

98.

99.

100.

c

b

d

a

c

c

a

d

c

d

101.

102.

103.

104.

105.

106.

107.

108.

109.

110.

a

c

d

b

a

b

b

c

a

b

111. a

genius PHYSICS 2 Elasticity

9.1 Interatomic Forces. The forces between the atoms due to electrostatic interaction between the charges of the atoms are called interatomic forces. These forces are electrical in nature and these are active if the distance between the two atoms is of the order of atomic size i.e. 10–10 metre. (1) Every atom is electrically neutral, the number of electrons (negative charge) orbiting around the nucleus is equal to the number of proton (positive charge) in the nucleus. So if two atoms are placed at a very large distance from each other then there will be a very small (negligible) interatomic force working between them. (2) When these two atoms are brought close to each other to a distance of the order of 10 –10 m, the distances between their positive nuclei and negative electron clouds get disturbed, and due to this, attractive interatomic force is produced between two atoms.

Repulsion

(3) This attractive force increases continuously with decrease in r and becomes maximum for one value of r called critical distance, represented by x (as shown in the figure). Beyond this the attractive force starts decreasing rapidly with further decrease in the value F of r. (4) When the distance between the two atoms becomes r0, the interatomic force will be zero. This distance r0 is called normal or equilibrium distance.

O

x

r0

(r0 = 0.74 Å for hydrogen). (5) When the distance between the two atoms further decreased, the interatomic force becomes repulsive in nature and increases very rapidly with decrease in distance between two atoms.

r Attractio n

U

O

r0

r

(6) The potential energy U is related with the interatomic force F by the following relation.

F

 dU dr

(i) When two atoms are at very large distance, the potential energy is negative and becomes more negative as r is decreased. (ii) When the distance between the two atoms becomes r0, the potential energy of the system of two atoms becomes minimum (i.e. attains maximum negative value). As the state of minimum potential energy is the state of equilibrium, hence the two atoms at separation r0 will be in a state of equilibrium. ( U 0  7 .2  10 19 Joule for hydrogen). (iii) When the distance between the two atoms is further decreased (i.e. r < r0) the negative value of potential energy of the system starts decreasing. It becomes zero and then attains positive value with further decrease in r (as shown in the figure).

genius PHYSICS Elasticity 3

9.2 Intermolecular Forces. The forces between the molecules due to electrostatic interaction between the charges of the molecules are called intermolecular forces. These forces are also called Vander Waal forces and are quite weak as compared to inter-atomic forces. These forces are also electrical in nature and these are active if the separation between two molecules is of the order of molecular size i.e.  10–9 m. (1) It is found that the force of attraction between molecules varies inversely as seventh power of the distance between them i.e.

Fatt 

1 r7

or

Fatt 

a r7

The negative sign indicates that the force is attractive in nature. (2) When the distance between molecules becomes less than r0, the forces becomes repulsive in nature and is found to vary inversely as ninth power of the distance between them i.e.

Frep 

1 r9

or

Frep 

b . r9

Therefore force between two molecules is given by F  Fatt  Frep 

a b  9 r7 r

The value of constants a and b depend upon the structure and nature of molecules. (3) Intermolecular forces between two molecules has the same general nature as shown in the figure for interatomic forces. (4) Potential Energy : Potential energy can be approximately expressed by the formula U 

A B  m n r r

A B represents repulsive contribution and term m represents the attractive contribution. n r r Constants A, B and numbers m and n are different for different molecules. For majority of solids n = 12 and m = 6. where the term

So potential energy can be expressed as U 

A B  6 12 r r

9.3 Comparison Between Inter atomic and Intermolecular Forces. (1) Similarities (i)

Both the forces are electrical in origin.

(ii) Both the forces are active over short distances. (iii) General shape of force-distance graph is similar for both the forces. (iv) Both the forces are attractive up to certain distance between atoms/molecules and become repulsive when the distance between them become less than that value. (2) Dissimilarities (i) Interatomic force depends upon the distance between the two atoms, whereas the intermolecular force depends upon the distance between the two molecules as well as their relative orientation. (ii) Interatomic forces are about 50 to100 times stronger than intermolecular forces.

genius PHYSICS 4 Elasticity (iii) The value of r0 for two atoms is smaller than the corresponding value for the molecules. Therefore one molecule is not restricted to attract only one molecule, but can attract many molecule. It is not so incase of atoms, since the atoms of one molecule cannot bind the atoms of other molecules.

9.4 States of Matter. The three states of matter differ from each other due to the following two factors. (1) The different magnitudes of the interatomic and intermolecular forces. (2) The extent of random thermal motion of atoms and molecules of a substance (which depends upon temperature).

Comparison Chart of Solid, Liquid and Gaseous States Property

Solid

Liquid

Gas

Shape

Definite

Not definite

Not definite

Volume

Definite

Definite

Not definite

Density

Maximum

Less than solids but more than gases.

Minimum

Incompressible

Less than gases but more than solids.

Compressible

Crystalline

Non-crystalline

Constant

Not constant

Not constant

K U

K >> U

Intermolecular force

Strongest

Less than solids but more than gases.

Weakest

Freedom of motion

Molecules vibrate about their mean position but cannot move freely.

Molecules have free motion.

limited

Molecules are free to move.

Effect of temperature

Matter remains in solid form below a certain temperature.

Liquids are found at temperatures more than that of solid.

These are found at temperatures greater than that of solids and liquids.

Compressibility Crystallinity Interatomic or intermolecular distance Relation between kinetic energy K and potential energy (U)

Note : 

The fourth state of matter in which the medium is in the form of positive and negative

ions, is known as plasma. Plasma occurs in the atmosphere of stars (including the sun) and in discharge tubes.

genius PHYSICS Elasticity 5

9.5 Types of Solids. A solid is that state of matter in which its constituent atoms or molecules are held strongly at the position of minimum potential energy and it has a definite shape and volume. The solids can be classified into two categories, crystalline and glassy or amorphous solids. Comparison chart of Crystalline and Amorphous Solids Crystalline solids

Amorphous or glassy solids

The constituent atoms, ions or molecules are arranged in a regular repeated three dimensional pattern, within the solid.

The constituent atoms, ions or molecules are not arranged in a regular repeated three dimensional pattern, within the solid.

Definite external geometric shape.

No regularity in external shape.

All the bonds in ions, or atoms or molecules are equally strong.

All the bonds are not equally strong.

They are anisotropic.

They are isotropic.

They have sharp melting point.

They don't have no sharp melting point.

They have a long-range order of atoms or ions or molecules in them.

They don’t have a long-range order.

They are considered true and stable solids.

They are not regarded as true and stable solids.

9.6 Elastic Property of Matter. (1) Elasticity : The property of matter by virtue of which a body tends to regain its original shape and size after the removal of deforming force is called elasticity. (2) Plasticity : The property of matter by virtue of which it does not regain its original shape and size after the removal of deforming force is called plasticity. (3) Perfectly elastic body : If on the removal of deforming forces the body regain its original configuration completely it is said to be perfectly elastic. A quartz fibre and phosphor bronze (an alloy of copper containing 4% to 10% tin, 0.05% to 1% phosphorus) is the nearest approach to the perfectly elastic body. (4) Perfectly plastic body : If the body does not have any tendency to recover its original configuration, on the removal of deforming force, it is said to be perfectly plastic. Paraffin wax, wet clay are the nearest approach to the perfectly plastic body. Practically there is no material which is either perfectly elastic or perfectly plastic and the behaviour of actual bodies lies between the two extremes. (5) Reason of elasticity : In a solids, atoms and molecules are arranged in such a way that each molecule is acted upon by the forces due to neighbouring molecules. These forces are known as intermolecular forces. For simplicity, the two molecules in their equilibrium positions (at intermolecular distance r = r0) (see graph in article 9.1) are shown by connecting them with a spring. In fact, the spring connecting the two molecules represents the intermolecular force between them. On applying the deforming forces, the molecules either come closer or go far apart from each other and restoring forces are developed. When the deforming force is removed, these

genius PHYSICS 6 Elasticity restoring forces bring the molecules of the solid to their respective equilibrium position (r = r0) and hence the body regains its original form. (6) Elastic limit : Elastic bodies show their property of elasticity upto a certain value of deforming force. If we go on increasing the deforming force then a stage is reached when on removing the force, the body will not return to its original state. The maximum deforming force upto which a body retains its property of elasticity is called elastic limit of the material of body. Elastic limit is the property of a body whereas elasticity is the property of material of the body. (7) Elastic fatigue : The temporary loss of elastic properties because of the action of repeated alternating deforming force is called elastic fatigue. It is due to this reason (i) Bridges are declared unsafe after a long time of their use. (ii) Spring balances show wrong readings after they have been used for a long time. (iii) We are able to break the wire by repeated bending. (8) Elastic after effect : The time delay in which the substance regains its original condition after the removal of deforming force is called elastic after effect. It is the time for which restoring forces are present after the removal of the deforming force it is negligible for perfectly elastic substance, like quartz, phosphor bronze and large for glass fibre.

9.7 Stress. When a force is applied on a body there will be relative displacement of the particles and due to property of elasticity an internal restoring force is developed which tends to restore the body to its original state. The internal restoring force acting per unit area of cross section of the deformed body is called stress. At equilibrium, restoring force is equal in magnitude to external force, stress can therefore also be defined as external force per unit area on a body that tends to cause it to deform. If external force F is applied on the area A of a body then, Stress 

Force F  Area A

Unit : N / m 2 (S.I.) , dyne / cm 2 (C.G.S.) Dimension : [ML1T 2 ] Stress developed in a body depends upon how the external forces are applied over it. On this basis there are two types of stresses : Normal and Shear or tangential stress (1) Normal stress : Here the force is applied normal to the surface. It is again of two types : Longitudinal and Bulk or volume stress (i) Longitudinal stress (a) It occurs only in solids and comes in picture when one of the three dimensions viz. length, breadth, height is much greater than other two. (b) Deforming force is applied parallel to the length and causes increase in length.

genius PHYSICS Elasticity 7

(c) Area taken for calculation of stress is area of cross section. (d) Longitudinal stress produced due to increase in length of a body under a deforming force is called tensile stress. (e) Longitudinal stress produced due to decrease in length of a body under a deforming force is called compressional stress. (ii) Bulk or Volume stress (a) It occurs in solids, liquids or gases. (b) In case of fluids only bulk stress can be found. (c) It produces change in volume and density, shape remaining same. (d) Deforming force is applied normal to surface at all points. (e) Area for calculation of stress is the complete surface area perpendicular to the applied forces. (f) It is equal to change in pressure because change in pressure is responsible for change in volume. (2) Shear or tangential stress : It comes in picture when successive layers of solid move on each other i.e. when there is a relative displacement between various layers of solid. A

(i) Here deforming force is applied tangential to one of the faces. (ii) Area for calculation is the area of the face on which force is applied.

F

Fixed face

(iii) It produces change in shape, volume remaining the same. Difference between Pressure and Stress Pressure

Stress

Pressure is always normal to the area.

Stress can be normal or tangential.

Always compressive in nature.

May be compressive or tensile in nature.

Sample problems based on Stress Problem 1.

Solution : (b)

A and B are two wires. The radius of A is twice that of B. they are stretched by the same load. Then the stress on B is [MP PMT 1993] (a) Equal to that on A

(b) Four times that on A

(c) Two times that on A

(d) Half that on A

Stress =

Force F  2 Area r

 Stress 

Problem 2.

1 r2

2



(Stress) B  rA      (2) 2  (Stress)B = 4 × (stress)A (Stress) A  rB 

[As F = constant]

One end of a uniform wire of length L and of weight W is attached rigidly to a point in the roof and a weight W1 is suspended from its lower end. If S is the area of cross-section of the wire, the stress in the wire at a height 3L/4 from its lower end is

genius PHYSICS 8 Elasticity (a) Solution : (c)

W1 S

(b)

W1  (W / 4 ) S

(c)

W1  (3W / 4 ) S

As the wire is uniform so the weight of wire below point P is

W1  W S

(d)

3W 4

P

3W  Total force at point P  W1  and area of cross-section = S 4

L

4

3W W1  Force 4   Stress at point P  Area S

Problem 3.

(b) Mg/2A

(c) 2Mg/A

(d) 4Mg/A

When the length of wire becomes double, its area of cross section will become half because volume of wire is constant (V  AL) . So the instantaneous stress =

Problem 4.

W1

On suspending a weight Mg, the length l of elastic wire and area of cross-section A its length becomes double the initial length. The instantaneous stress action on the wire is (a) Mg/A

Solution : (c)

3L

Force Mg 2 Mg .   Area A/2 A

A bar is subjected to equal and opposite forces as shown in the figure. PQRS is a plane making angle  with the cross-section of the bar. If the area of cross-section be ‘A’, then what is the tensile stress on PQRS Q

(a) F / A

F

(b) F cos / A (c) F

F

P R



/A

cos2

S

(d) F / A cos Solution : (c)

As tensile stress =

Normal force FN  Area AN

and here A N  ( A / cos  ) , FN  Normal force = F cos So, Tensile stress 

Problem 5.

F

A

F



F cos  F cos 2   A / cos  A

In the above question, what is the shearing stress on PQ (a) F / A cos 

(b) F sin 2 / 2A

(c) F / 2A sin 2

F sin θ cos θ F sin 2 Tangential force F sin θ    2A Area ( A/ cos θ ) A

Solution : (b)

Shear stress 

Problem 6.

In the above question, when is the tensile stress maximum (a)   0 o

(b)   30 o

(c)   45 o

(d)   90 o

F cos 2  . It will be maximum when cos 2   max . i.e. cos   1    0 o . A

Solution : (a)

Tensile stress 

Problem 7.

In the above question, when is the shearing stress maximum (a)   0 o

(d) F cos / A

(b)   30 o

(c)   45 o

(d)   90 o

genius PHYSICS Elasticity 9 Solution : (c)

Shearing stress 

F sin 2 . It will be maximum when sin 2  max i.e. sin 2  1  2  90 o  2A

  45 o .

9.8 Strain. The ratio of change in configuration to the original configuration is called strain. Being the ratio of two like quantities, it has no dimensions and units. Strain are of three types : (1) Linear strain : If the deforming force produces a change in length alone, the strain produced in the body is called linear strain or tensile strain. l

Change in length( l) Linear strain  Original length( l)

l

Linear strain in the direction of deforming force is called longitudinal strain and in a direction perpendicular to force is called lateral strain.

F

(2) Volumetric strain : If the deforming force produces a change in volume alone the strain produced in the body is called volumetric strain. Volumetric strain 

Change in volume( V ) Original v olume( V )

(V – V)

(3) Shearing strain : If the deforming force produces a change in the shape of the body without changing its volume, strain produced is called shearing strain. It is defined as angle in radians through which a plane perpendicular to the fixed surface of the cubical body gets turned under the effect of tangential force.



Note : 

x

F  L 



Fixed face

x L

When a beam is bent both compression strain as well as an

extension strain is produced.

Sample problems based on Strain Problem 8.

A cube of aluminium of sides 0.1 m is subjected to a shearing force of 100 N. The top face of the cube is displaced through 0.02 cm with respect to the bottom face. The shearing strain would be [MP PAT 1990] (a) 0.02

(b) 0.1

(c) 0.005

x 0 . 02 cm   0 . 002 L 0 . 1m

Solution : (d)

Shearing strain  

Problem 9.

A wire is stretched to double its length. The strain is (a) 2

(d) 0.002

(b) 1

(c) Zero

(d) 0.5

genius PHYSICS 10 Elasticity Solution : (b)

Strain 

Change in length 2L  L  1 Original length L

Problem 10. The length of a wire increases by 1% by a load of 2 kg-wt. The linear strain produced in the wire will be (a) 0.02 Solution : (c)

Strain 

(b) 0.001

(c) 0.01

(d) 0.002

Change in length 1 % of L L / 100 = 0.01   Original length L L

9.9 Stress-strain Curve.

Stress

If by gradually increasing the load on a vertically suspended metal wire, a graph is plotted between stress (or load) and longitudinal strain (or elongation) we get the curve as shown in figure. From this curve it is clear that : (1) When the strain is small (< 2%) (i.e., in region OP) stress is proportional to strain. This is the region where the so called Hooke’s law is obeyed. The point P is called limit of proportionality and slope of line OP gives the Young’s modulus Y of the material of the wire. If  is the angle of OP Elastic from strain axis then Y = tan . Plastic region Breaking region strength (2) If the strain is increased a little bit, i.e., in the region PE, the stress is B Elastic limit A C E not proportional to strain. However, the wire still regains its original length P after the removal of stretching force. This behaviour is shown up to point E known as elastic limit or yield-point. The region OPE represents the elastic  O Strain behaviour of the material of wire. (3) If the wire is stretched beyond the elastic limit E, i.e., between EA, the strain increases much more rapidly and if the stretching force is removed the wire does not come back to its natural length. Some permanent increase in length takes place. (4) If the stress is increased further, by a very small increase in it a very large increase in strain is produced (region AB) and after reaching point B, the strain increases even if the wire is unloaded and ruptures at C. In the region BC the wire literally flows. The maximum stress corresponding to B after which the wire begins to flow and breaks is called breaking or tensile strength. The region EABC represents the plastic behaviour of the material of wire. (5) Stress-strain curve for different materials.

O

Strain

C

The plastic region between E and C is small for brittle material and it will break soon after the elastic limit is crossed.

O

E P

Strain

Elastomers C C

The material of the wire have a good plastic range and such materials can be easily changed into different shapes and can be drawn into thin wires

Stress

E P

Ductile material

Stress

Stress

Brittle material

O

Strain

Stress strain curve is not a straight line within the elastic limit for elastomers and strain produced is much larger than the stress applied. Such materials have no plastic range and the breaking point lies very close to elastic limit. Example rubber

Sample problems based on Stress-strain curve

genius PHYSICS Elasticity 11

Problem 11. The stress-strain curves for brass, steel and rubber are shown in the figure. The lines A, B and C are for A

B

Stress

(a) Rubber, brass and steel respectively (b) Brass, steel and rubber

C

(c) Steel, brass and rubber respectively O

(d) Steel, rubber and brass Solution : (c)

Strain

From the graph tan  C  tan  B  tan  A  Y C  Y B  Y A Y Rubber  Y Brass  Y Steel

Problem 12. The strain stress curves of three wires of different materials are shown in the figure. P, Q and R are the elastic limits of the wires. The figure shows that P

(a) Elasticity of wire P is maximum Strain

(b) Elasticity of wire Q is maximum

Q

R

(c) Tensile strength of R is maximum O

(d) None of the above is true Solution : (d)

Stress

On the graph stress is represented on X- axis and strain Y-axis So from the graph Y  cot   Y P  Y Q  Y R

1 1  [where  is the angle from stress axis] tan  

[As  P   Q   R ]

We can say that elasticity of wire P is minimum and R is maximum.

9.10 Hooke’s law and Modulus of Elasticity.

i.e. stress  strain or

stress  constant  E strain

The constant E is called modulus of elasticity. (1) It’s value depends upon the nature of material of the body and the manner in which the body is deformed.

Stress

According to this law, within the elastic limit, stress is proportional to the strain.

O

Strain

(2) It's value depends upon the temperature of the body. (3) It’s value is independent of the dimensions (length, volume etc.) of the body. There are three modulii of elasticity namely Young’s modulus (Y), Bulk modulus (K) and modulus of rigidity () corresponding to three types of the strain.

9.11 Young's Modulus (Y). It is defined as the ratio of normal stress to longitudinal strain within limit of proportionality. Y 

Normal stress F / A FL   longitudin al strain l/L Al

If force is applied on a wire of radius r by hanging a weight of mass M, then

genius PHYSICS 12 Elasticity

Y 

MgL r 2 l

Important points (i) If the length of a wire is doubled, Then longitudinal strain = 

Young’s modulus =

change in length( l) final length  initial length 2L  L   1 initial length( L) Initial length L

stress  Y = stress strain

[As strain = 1]

So young’s modulus is numerically equal to the stress which will double the length of a wire. (ii) Increment in the length of wire

l

FL    As Y  Al   

FL r 2 Y

So if same stretching force is applied to different wires of same material, l 

L r2

[As F and Y are

constant] i.e., greater the ratio

L , greater will be the elongation in the wire. r2

(iii) Elongation in a wire by its own weight : The weight of the wire Mg act at the centre of gravity of the wire so that length of wire which is stretched will be L/2.  Elongation l 

FL Mg(L / 2) MgL L2 dg =   2Y AY AY 2 AY

[As mass (M) = volume (AL) × density (d)]

(iv) Thermal stress : If a rod is fixed between two rigid supports, due to change in temperature its length will change and so it will exert a normal stress (compressive if temperature increases and tensile if temperature decreases) on the supports. This stress is called thermal stress. As by definition, coefficient of linear expansion  

l L 

L F

F

l   L



thermal strain

So

thermal stress = Y 

[As Y = stress/strain]

And tensile or compressive force produced in the body = YA 

Note : 

In case of volume expansion Thermal stress = K Where K = Bulk modulus,  = coefficient of cubical expansion

(v) Force between the two rods : Two rods of different metals, having the same area of cross section A, are placed end to end between two massive walls as shown in figure. The L1 L2 first rod has a length L1, coefficient of linear expansion 1 and young’s 1 2 modulus Y1. The corresponding quantities for second rod are L2, 2 and Y2. If Y2 Y1 the temperature of both the rods is now raised by T degrees. Increase in length of the composite rod (due to heating) will be equal to

genius PHYSICS Elasticity 13

[As l = L  ]

l1  l 2  [L1 1  L 2 2 ] T

and due to compressive force F from the walls due to elasticity, L L F  FL  decrease in length of the composite rod will be equal to  1  2   As l  AY  Y Y A   2   1

as the length of the composite rod remains unchanged the increase in length due to heating must be equal L  F L to decrease in length due to compression i.e.  1  2   [L1 1  L 2 2 ] T A  Y1 Y 2  F

or

A[ L1 1  L 2 2 ]T  L1 L 2      Y1 Y 2 

(vi) Force constant of wire : Force required to produce unit elongation in a wire is called force constant of material of wire. It is denoted by k. 

k

F l

…..(i)

but from the definition of young’s modulus Y 

F/ A F YA   l/L l L

…..(ii)

YA L It is clear that the value of force constant depends upon the dimension (length and area of cross section) and material of a substance. (vii) Actual length of the wire : If the actual length of the wire is L, then under the tension T1, its length becomes L1 and under the tension T2, its length becomes L2. from (i) and (ii) k 

L1  L  l1  L1  L 

T1 k

……(i)

From (i) and (ii) we get L 

L2  L  l2  L 2  L 

and

T2 k

..…(ii)

L1 T2  L 2 T1 T2  T1

Sample problems based on Young's modulus Problem 13. The diameter of a brass rod is 4 mm and Young’s modulus of brass is 9  10 10 N / m 2 . The force required to stretch by 0.1% of its length is (a) 360 N Solution : (a)

[MP PET 1991; BVP 2003]

(c) 144   10 3 N

(b) 36 N

r  2  10 3 m, Y  9  10 10 N / m 2 , l  0.1% L 

As Y 

F L A l

 F  YA

(d) 36  10 5 N

l  0 . 001 L

l  9  10 10   (2  10 3 ) 2  0 .001  360 N L

Problem 14. A wire of length 2m is made from 10 cm3 of copper. A force F is applied so that its length increases by 2 mm. Another wire of length 8 m is made from the same volume of copper. If the force F is applied to it, its length will increase by [MP PET 2003] (a) 0.8 cm Solution : (d)

l

FL FL2  AY VY

(b) 1.6 cm l  L 2

(c) 2.4 cm [As V, Y and F are constant]

(d) 3.2 cm

genius PHYSICS 14 Elasticity 2

l2  L 2  8        16  l2 = 16 l1 = 16  2 mm = 32 mm = 3.2 cm l1  L1  2 2

Problem 15. A wire of length L and radius r is rigidly fixed at one end. On stretching the other end of the wire with a force F, the increase in its length is l. If another wire of same material but of length 2L and radius 2r is stretched with a force of 2F, the increase in its length will be [AIIMS 1980; MP PAT 1990; MP PET 1989, 92; MP PET/PMT 1988; MP PMT 1996, 2002; UPSEAT 2002]

(a) l Solution : (a)

l

(b) 2l

FL

 r2Y

(c)

2

l2 F L r  1  2 2  1   2  2     1 l1 F1 L1  r2  2



l 2

(d)

l 4

2

l 2  l1 i.e. the increment in length will be same.

Problem 16. Two wires A and B are of same materials. Their lengths are in the ratio 1 : 2 and diameters are in the ratio 2 : 1 when stretched by force FA and FB respectively they get equal increase in their lengths. Then the ratio FA/FB should be [Orissa JEE 2002] (a) 1 : 2 Solution : (d)

Y 

(b) 1 : 1

FL

r l

FA Y  A FB YB

 rA   rB

  

2

 lA   lB

  LB    LA

(d) 8 : 1

l L

 F  Y r 2

2

(c) 2 : 1

 2 2   1     1      8 1 1  2

Problem 17. A uniform plank of Young’s modulus Y is moved over a smooth horizontal surface by a constant horizontal force F. The area of cross-section of the plank is A. the compressive strain on the plank in the direction of the force is [Kerala (Engg.) 2002] (a) Solution : (a)

F AY

(b)

Compressive strain 

2F AY

Stress Young' s modulus

(c) 

1 F    2  AY 

(d)

3F AY

F/A F  Y AY

Problem 18. A wire is stretched by 0.01 m by a certain force F. Another wire of same material whose diameter and length are double to the original wire is stretched by the same force. Then its elongation will be [EAMCET (Engg.) 1995; CPMT 2001]

(a) 0.005 m Solution : (a)

l

FL

r Y 2

l2  L 2  l1  L1

(b) 0.01 m

 l   r1     r2

2

L

(c) 0.02 m

(d) 0.002 m

[As F and Y are constants]

r2

 l 0 . 01 1 1   (2)      l 2  1   0 . 005 m . 2 2 2 2  2

Problem 19. The length of an elastic string is a metres when the longitudinal tension is 4 N and b metres when the longitudinal tension is 5 N. The length of the string in metres when the longitudinal tension is 9 N is [EAMCET 2001]

(a) a – b Solution : (b)

(b) 5b – 4a

(c) 2b 

1 a 4

(d) 4a – 3b

Let the original length of elastic string is L and its force constant is k. When longitudinal tension 4N is applied on it

L

4 a k

.....(i)

genius PHYSICS Elasticity 15 L

and when longitudinal tension 5N is applied on it By solving (i) and (ii) we get k  Now L

when

longitudinal

5 b k

.....(ii)

1 and L  5a  4b ba

tension

9N

is

applied

on

elastic

string

then

its

length

=

9  5a  4 b  9(b  a)  5b  4 a k

Problem 20. The load versus elongation graph for four wires of the same material is shown in the figure. The thickest wire is represented by the line

[KCET (Engg./Med.) 2001] Load

(a) OD

D C B

(b) OC

A

(c) OB O

(d) OA Solution : (a)

Young's modulus Y 

Elongation

FL 1 l (As Y, L and F are constant) Al A

From the graph it is clear that for same load elongation is minimum for graph OD. As elongation (l) is minimum therefore area of cross-section (A) is maximum. So thickest wire is represented by OD.

Problem 21. A 5 m long aluminum wire (Y  7  10 10 N / m 2 ) of diameter 3 mm supports a 40 kg mass. In order to have the same elongation in a copper wire (Y  12  1010 N / m 2 ) of the same length under the same weight, the diameter should now be, in mm (a) 1.75 Solution : (c)

l

FL

r Y 2

(b) 2.0



4 FL

d 2 Y

(c) 2.3

(d) 5.0

[As r  d / 2 ]

If the elongation in both wires (of same length) are same under the same weight then d 2 Y  constant  d Cu   d Al

2

 Y Y Al 7  10 10   Al  d Cu  d Al  3  2 . 29 mm Y Cu Y Cu 12  10 10 

Problem 22. On applying a stress of 20  10 8 N / m 2 the length of a perfectly elastic wire is doubled. Its Young’s modulus will be (a) 40  10 8 N / m 2 Solution : (b)

[MP PET 2000]

(b) 20  10 8 N / m 2

(c) 10  10 8 N / m 2

(d) 5  10 8 N / m 2

When strain is unity then Young's modulus is equal to stress.

Problem 23. The dimensions of four wires of the same material are given below. In which wire the increase in length will be maximum when the same tension is applied [IIT-JEE 1981; NCERT 1976; CPMT 1983, 90; MP PMT 1992, 94, 97; MP PET/PMT 1998; MP PET 1989, 90, 99]

(a) Length 100 cm, diameter 1 mm

(b) Length 200 cm, diameter 2 mm

(c) Length 300 cm, diameter 3 mm

(d) Length 50 cm, diameter 0.5 mm

genius PHYSICS 16 Elasticity Solution : (d)

If same force is applied on four wires of same material then elongation in each wire depends on the length and diameter of the wire and given by l 

L d

2

and the ratio of

L d2

is maximum for (d) option.

Problem 24. The Young’s modulus of a wire of length L and radius r is Y N/m2. If the length and radius are reduced to L/2 and r/2, then its Young’s modulus will be (a) Y/2 Solution : (b)

(b) Y

[MP PMT 1985; MP PET 1997; KCET (Engg./Med.) 1999]

(c) 2Y

(d) 4Y

Young's modulus do not depend upon the dimensions of wire. It is constant for a given material of wire.

Problem 25. A fixed volume of iron is drawn into a wire of length L. The extension x produced in this wire by a constant force F is proportional to (a)

1 L

Solution : (c)

l

(b)

2

1 L

(c) L2

(d) L

FL FL2 FL2 for a fixed volume l  L2   AY ALY VY

Problem 26. A rod is fixed between two points at 20 o C . The coefficient of linear expansion of material of rod is 1 .1  10 5 / o C and Young’s modulus is 1.2  10 11 N / m 2 . Find the stress developed in the rod if

temperature of rod becomes 10 o C (a) 1 .32  10 7 N / m 2 Solution : (a)

Thermal stress

(b) 1 .10  10 15 N / m 2 (c) 1 .32  10 8 N / m 2

(d) 1 .10  10 6 N / m 2

F  Y    1.2  10 11  1.1  10 5  (20  10)  1.32  10 7 N / m 2 A

Problem 27. The coefficient of linear expansion of brass and steel are 1 and 2. If we take a brass rod of length L1 and steel rod of length L2 at 0 o C , their difference in length (L 2  L1 ) will remain the same at any temperature if [EAMCET (Med.) 1995]

(a)  1 L 2   2 L1 Solution : (d)

(b)  1 L22   2 L12

(c)  12 L1   22 L 2

(d)  1 L1   2 L 2

Difference in lengths of rods will remain same if expansion is same in both the rods. If expansion in first rod is l1  L1 1  and expansion in second rod is l 2  L 2 2  then L1 1   L 2 2  L1 1  L 2 2

Problem 28. The force required to stretch a steel wire of 1 cm2 cross-section to 1.1 times its length would be (Y  2  10 11 Nm 2 )

(a) 2  10 6 N Solution : (a)

L 2  1 .1 L1 F  YA

[MP PET 1992]

(b) 2  10 3 N

Strain 

(c) 2  10 6 N

(d) 2  10 7 N

L  L1 1 .1 L1  L1 l  2  = 0.1. L1 L1 L1

l  2  10 11  1  10 4  0.1  2  10 6 N . L

Problem 29. A two metre long rod is suspended with the help of two wires of equal length. One wire is of steel and its cross-sectional area is 0.1 cm2 and another wire is of brass and its cross-sectional area is 0.2 cm2. If a load

genius PHYSICS Elasticity 17 W is suspended from the rod and stress produced in both the wires is same then the ratio of tensions in them will be Steel

(a) Will depend on the position of W

Brass

T1

(b) T1 / T2  2

T2 2m

(c) T1 / T2  1 W

(d) T1 / T2  0.5 Solution : (d)

Stress = 

Tension  constant Area of cross-section

T1 T T A 0 .1 1   0 .5 .  2  1  1  T2 A2 0 .2 2 A1 A2

Problem 30. Three blocks, each of same mass m, are connected with wires W1 and W2 of same cross-sectional area a and Young’s modulus Y. Neglecting friction the strain developed in wire W2 is (a)

2 mg 3 aY

Solution : (a)

(c)

1 mg 3 aY

(d)

3 mg aY

W1

m

3 mg (b) 2 aY

W2

m

m

If the system moves with acceleration a and T is the tension in the string W2 then by comparing this condition from standard case T 

m1m 2 g m1  m 2

m1

T

P

In the given problem m 1  (m  m)  2m and m 2  m  Tension 

 Stress 

T

m . 2m . g 2  mg m  2m 3

m2

T 2 Stress 2 mg  mg and Strain   a 3a Young' s modulus 3 aY

Problem 31. A wire elongates by 1.0 mm when a load W is hanged from it. If this wire goes over a pulley and two weights W each are hung at the two ends, the elongation of the wire will be (a) 0.5 m Solution : (b)

(b) 1.0 mm

(c) 2.0 mm

(d) 4.0 mm

Elongation in the wire  Tension in the wire In first case T1  W and in second case T 2 

T l As 1  1  1  1  l2  l1  1.0mm l2 T2

2W  W W W W

T1

T2 T2

W W

W

genius PHYSICS 18 Elasticity

Problem 32. The Young’s modulus of three materials are in the ratio 2 : 2 : 1. Three wires made of these materials have their cross-sectional areas in the ratio 1 : 2 : 3. For a given stretching force the elongation's in the three wires are in the ratio (a) 1 : 2 : 3 Solution : (d)

l

(b) 3 : 2 : 1

(c) 5 : 4 : 3

(d) 6 : 3 : 4

FL 1 and for a given stretching force l  AY AY

Let three wires have young's modulus 2Y, 2Y and Y and their cross sectional areas are A, 2A and 3 A respectively.

l1 : l 2 : l 3 

1 1 1 1 1 1 1 1 1  : :  6:3:4 . : :  : : 2 4 3 A1 Y 1 A 2 Y 2 A 3 Y 3 A  2Y 2 A  2Y 3 A  Y

Problem 33. A light rod with uniform cross-section of 10 4 m 2 is shown in the adjoining figure. The rod consists of three different materials whose lengths are 0.1 m, 0.2 m and 0.15 m respectively and whose Young’s modulii are 2 .5  10 10 N / m 2 ,

A 0.1 m

Y = 2.5 × 1010 N/m2 B

4  10 10 N / m 2 and 1  1010 N / m 2 respectively. The displacement of 0.2 m

point B will be (a) 24  10 6 m

Y = 4 × 1010 N/m2 C Y = 1 × 1010 N/m2

0.15 m

(b) 9  10 6 m

D

(c) 4  10 6 m

10 kg

(d) 1  10 6 m Solution : (c)

Increment in the length AB =

MgL 10  10  0.1   4  10 6 m AY 10 4  2.5  10 10

 Displacement of point B = 4  10 6 m

Problem 34. In the above problem, displacement of point C will be (a) 24  10 6 m Solution : (b)

(b) 9  10 6 m

Increment in the length BC =

(c) 4  10 6 m

(d) 1  10 6 m

MgL 10  10  0.2  4  5  10 6 m AY 10  4  1010

 Displacement of point C = 4  10 6  5  10 6  9  10 6 m

Problem 35. In the above problem, the displacement of point D will be (a) 24  10 6 m Solution : (a)

(b) 9  10 6 m

Increment in the length CD 

(c) 4  10 6 m

(d) 1  10 6 m

MgL 10  10  0.15  4  15  10 6 m AY 10  1  1010

 Displacement of point D = 4  10 6  5  10 6 m  15  10 6  24  10 6 m .

Problem 36. Two blocks of masses m1 and m2 are joined by a wire of Young’s modulus Y via a massless pulley. The area of cross-section of the wire is S and its length is L. When the system is released, increase in length of the wire is (a)

m 1 m 2 gL YS (m 1  m 2 )

(b)

2m 1 m 2 gL YS (m 1  m 2 ) T m1

T m2

genius PHYSICS Elasticity 19

Solution : (b)

(c)

(m 1  m 2 )gL YS (m 1  m 2 )

(d)

4 m 1 m 2 gL YS (m 1  m 2 )

Tension in the wire T 

2m1m 2 T 2m1m 2 g g  stress in the wire   m1  m 2 S S (m1  m 2 )

2m 1 m 2 g 2m1m 2 gL l Stres s  l   L Y YS (m 1  m 2 ) YS (m1  m 2 )

 Strain

Problem 37. A steel wire of diameter d, area of cross-section A and length 2L is clamped firmly at two points A and B which are 2L metre apart and in the same plane. A body of mass m is hung from the middle point of wire such that the middle point sags by x lower from original position. If Young’s modulus is Y then m is given by (a)

1 YAx 2 2 gL2

(b)

1 YAL 2 2 gx 2

(c) (d) Solution : (c)

YAx

L

L x

3

gL3

m

YAL 3 gx 2

Let the tension in the string is T and for the equilibrium of mass m

2T sin   mg  T 

x mg mgL [As  is small then sin   ]  L 2 sin  2x L

A

Increment in the length l  AC  AB  L2  x 2  L  (L2  x 2 )1 / 2  L  x2   L  1  2   L  

As Young's modulus Y 

T L A l

1/2

   x2 1 x2  1  1  L 1     2 L2   2 L 

 T 



B x 2Tsin

T

T

 C m

YAl L

Substituting the value of T and l in the above equation we get

YAx 3 mgL YA x 2 m   . 2x L 2L gL3

Problem 38. Two wires of equal length and cross-section are suspended as shown. Their Young’s modulii are Y 2 respectively. The equivalent Young’s modulus will be

(a) Y1  Y2

Solution : (b)

(b)

Y1  Y 2 2

(c)

Y1Y 2 Y1  Y 2

(d)

Y1 Y 2

Y1

Y2

Let the equivalent young's modulus of given combination is Y and the area of cross section is 2A.

Y1 and

genius PHYSICS 20 Elasticity For parallel combination k1  k 2  k eq. L, A Y1

Y1 A Y 2 A Y 2 A   L L L

Y1  Y2  2Y ,  Y 

L, A Y2

L, 2A Y

Y1  Y 2 2

Problem 39. If a load of 9kg is suspended on a wire, the increase in length is 4.5 mm. The force constant of the wire is (a) 0.49  10 4 N / m Solution : (b)

Force constant k 

(b) 1.96  10 4 N / m

(c) 4.9  10 4 N / m

(d) 0.196  10 4 N / m

F mg 9  9.8  k  1.96  10 4 N / m   l l 4.5  10 3

Problem 40. One end of a long metallic wire of length L, area of cross-section A and Young’s modulus Y is tied to the ceiling. The other end is tied to a massless spring of force constant k. A mass m hangs freely from the free end of the spring. It is slightly pulled down and released. Its time period is given by (a) 2 Solution : (d)

m K

(b) 2

Force constant of wire k1 

mYA KL

(c) 2

mK YA

(d) 2

m(KL  YA ) KYA

F YA  and force constant of spring k 2  k (given) l L

Equivalent force constant for given combination

 Time period of combination T  2

L 1 kYA 1 1 1    k eq    YA k kL  YA k eq k1 k 2

m m (kL  YA )  2 k eq kYA

Problem 41. Two wires A and B have the same length and area of cross section. But Young’s modulus of A is two times the Young’s modulus of B. Then the ratio of force constant of A to that of B is (a) 1 Solution : (b)

(b) 2

Force constant of wire k 

YA k Y  A  A 2 L k B YB

(c)

1 2

(d)

2

[As L and A are same]

9.12 Work Done in Stretching a Wire. In stretching a wire work is done against internal restoring forces. This work is stored in the wire as elastic potential energy or strain energy. If a force F acts along the length L of the wire of cross-section A and stretches it by x then

Y 

stress F / A FL YA  F   .x strain x/L Ax L

So the work done for an additional small increase dx in length, dw  Fdx  Hence the total work done in increasing the length by l,

l



l



YA x . dx L

W  dW  Fdx  0

0

YA 1 YA 2 . x dx  l 0 L 2 L l



genius PHYSICS Elasticity 21

This work done is stored in the wire.  Energy stored in wire U 

1 YAl 2 L

2



YAl    As F  L   

1 Fl 2

Dividing both sides by volume of the wire we get energy stored in per unit volume of wire.

UV 

1 F l 1 1 1     stress strain   Y  (strain)2  (stress)2 2 2Y 2 A L 2

[As AL = volume of wire]

Total energy stored in wire (U)

Energy stored in per unit volume of wire (UV)

1 Fl 2

1 Fl 2 volume

1  stress  strain  volume 2

1  stress  strain 2

1  Y  (strain)2  volume 2

1  Y  (strain)2 2

1  (stress)2  volume 2Y

1  (stress)2 2Y

Note : 

If the force on the wire is increased from F1 to F2 and the elongation in wire is l then

energy stored in the wire U 

1 (F1  F2 ) l 2 2

 Thermal energy density = Thermal energy per unit volume = =

1  Thermal stress  strain 2

1 F l 1 1 = (Y  )( ) = Y  2 ( )2 2 2 2 AL

Sample problems based on Work done in Stretching a Wire Problem 42. A wire suspended vertically from one of its ends is stretched by attaching a weight of 200 N to the lower end. The weight stretches the wire by 1 mm, then the elastic energy stored in the wire is (a) 0.1 J Solution : (a)

(b) 0.2 J

Elastic energy stored in wire = U 

(c) 10 J

(d) 20 J

1 1 Fl   200  1  10  3  0 . 1 J 2 2

Problem 43. The graph shows the behaviour of a length of wire in the region for which the substance obeys Hooke’s law. P and Q represent (a) P = applied force, Q = extension

[AMU 2001] Q

(b) P = extension, Q = applied force (c) P = extension, Q = stored elastic energy (d) P = stored elastic energy, Q = extension Solution : (c)

P

The graph between applied force and extension will be straight line because in elastic range applied force  extension, but the graph between extension and stored elastic energy will be parabolic in nature.

genius PHYSICS 22 Elasticity As U 

1 2 kx or U  x 2 2

Problem 44. When a 4 kg mass is hung vertically on a light spring that obeys Hooke’s law, the spring stretches by 2 cms. The work required to be done by an external agent in the stretching this spring by 5 cms will be (g = 9.8 m/s2) [MP PMT 1995]

(a) 4.900 J Solution : (b)

(b) 2.450 J

(c) 0.495 J

(d) 0.245 J

When a 4 kg mass is hung vertically on a spring, it stretches by 2 cm k  Now work done in stretching this spring by 5 cms

U

F 4  9.8   1960 N / m x 2  10  2

1 2 1 kx   1960 (5  10  2 ) 2  2.45 J. 2 2

Problem 45. A rod of iron of Young’s modulus Y  2.0  10 11 N / m 2 just fits the gap between two rigid supports 1m apart. If the rod is heated through 100 o C the strain energy of the rod is (   18  10 6 o C 1 and area of cross-section A  1 cm 2 ) (a) 32.4 J Solution : (a)

U 

(b) 32.4 mJ

(c) 26.4 J

(d) 26.4 mJ

l    Thermal strain     L  

1 1  Y  (strain) 2  volume   Y (  ) 2  A  L 2 2

1  (2  10 11 )  (18  10  6  100 ) 2  1  10  4  1  324  10 1 = 32.4 J. 2

Problem 46. Which of the following cases will have the greatest strain energy (F is the stretching force, A is the area of cross section and s is the strain) (a) F = 10 N, A = 1 cm2, s = 10–3 (b) (c) F = 10 N, A = Solution: (b)

Strain energy 

1 cm2, s = 10–4 2

F = 15 N, A = 2 cm2, s = 10–3 (d) F = 5 N, A = 3 cm2, s = 10–3

1 1 F 1  stress × strain × volume   × strain × AL   F  strain × L 2 2 A 2

For wire (a)

U

1  10  10  3  L  5  10  3 L ; 2

For wire (c)

U

1 1  10  10  4  L  0 . 5  10  3 L ; For wire (d) U   5  10  3  2 . 5  10  3 L 2 2

For wire (b) U 

1  15  10  3  L  7 . 5  10  3 L 2

For a given length wire (b) will have greatest strain energy.

9.13 Breaking of Wire. When the wire is loaded beyond the elastic limit, then strain increases much more rapidly. The maximum stress corresponding to B (see stress-strain curve) after which the wire begin to flow and breaks, is called breaking stress or tensile strength and the force by application of which the wire breaks is called the breaking force. (i) Breaking force depends upon the area of cross-section of the wire i.e., Breaking force  A 

Breaking force = P × A

Here P is a constant of proportionality and known as breaking stress. (ii) Breaking stress is a constant for a given material and it does not depends upon the dimension (length or thickness) of wire.

A F

genius PHYSICS Elasticity 23

(iii) If a wire of length L is cut into two or more parts, then again it's each part can hold the same weight. Since breaking force is independent of the length of wire. (iv) If a wire can bear maximum force F, then wire of same material but double thickness can bear maximum force 4F because Breaking force  r2. (v) The working stress is always kept lower than that of a breaking stress. So that safety factor =

breaking stress may have large value. working stress

(vi) Breaking of wire under its own weight. Breaking force = Breaking stress  Area of cross section [As mass = volume  density = ALd]

Weight of wire = Mg = ALdg = PA 

Ldg  P 

L

P dg

This is the length of wire if it breaks by its own weight.

Sample problems based on Breaking of Wire Problem 47. A wire of diameter 1 mm breaks under a tension of 1000 N. Another wire of same material as that of the first one, but of diameter 2 mm breaks under a tension of (a) 500 N Solution : (d)

(b) 1000 N

(c) 10000 N

(d) 4000 N

Breaking force  area of cross-section (r 2 )  d 2 F2  d 2  F1  d 1

2

 F2  2mm       F2  1000  4  4000 N . 1000  1mm   2

Problem 48. In steel, the Young’s modulus and the strain at the breaking point are 2  10 11 Nm 2 and 0.15 respectively. The stress at the breaking point for steel is therefore (a) 1 .33  10 11 Nm 2 Solution : (d)

Y 

Stress Strain

(b) 1 .33  10 12 Nm 2 (c) 7 .5  10 13 Nm 2

(d) 3  10 10 Nm 2

 Stress = Y × Strain  2  10 11  0.15  0.3  10 11  3  10 10 N / m 2

Problem 49. To break a wire, a force of 10 6 N / m 2 is required. If the density of the material is 3  10 3 kg / m 3 , then the length of the wire which will break by its own weight will be (a) 34 m Solution : (a)

(b) 30 m

(c) 300 m

Length of the wire which will break by its own weight L 

[Roorkee 1979]

(d) 3 m

P 10 6 100    33.3 m ≃ 34 m. dg 3  10 3  10 3

Problem 50. The wires A and B shown in the figure are made of the same material and have radii rA and rB respectively. The block between them has a mass m. When the force F is mg/3, one of the wires break (a) A will break before B if rA = rB

A

(b) A will break before B if rA < 2rB (c) Either A or B may break if rA = 2rB (d) The lengths of A and B must be known to predict which wire will break

m B F

genius PHYSICS 24 Elasticity Solution :(a,b,c) When force F 

mg is applied at the lower end then 3

Stress in wire B 

F



rB2

mg 3rB2

and stress in wire A =

(i) if rA  rB  r (Let) then stress in wire B 

mg 3r

2

F  mg

rA2

and

mg  mg 4 mg  3 2  3  rA2  rA

stress in wire A =

4 mg . 3 r 2

i.e. stress in wire A > stress in wire B so the A will break before B (ii) if rB  r , (let) then rA  2r Stress in wire B =

mg 3r

2

and Stress in wire A 

4 mg 3 (2r)

2



mg 3r 2

i.e. stress in wire A = stress in wire B. It means either A or B may break. (iii) If rA  2rB then stress in A will be more than B. i.e. A will break before B.

Problem 51. A body of mass 10 kg is attached to a wire 0.3 m long. Its breaking stress is 4 .8  10 7 N / m 2 . The area of cross-section of the wire is 10 6 m 2 . What is the maximum angular velocity with which it can be rotated in the horizontal circle (a) 1 rad/sec Solution : (c)

(b) 2 rad/sec

(c) 4 rad/sec

(d) 8 rad/sec

Breaking force = centrifugal force Breaking stress × area of cross-section = m  2 l

4.8  10 7  10 6  10   2  0.3   2  16    4rad / sec

Problem 52. Two block of masses 1 kg and 4 kg are connected by a metal wire going over a smooth pulley as shown in the figure. The breaking stress of the metal is 3 .18  10 10 N / m 2 . The minimum radius of the wire so it will not break is (a) 1  10 5 m (b) 2  10 5 m (c) 3  10 5 m

1kg

(d) 4  10 5 m Solution : (d)

Tension in the wire T 

4 kg

2m 1 m 2 2 1 4  10  T  16 N g  T 14 m1  m 2

Breaking force = Breaking stress × Area of cross-section Tension in the wire = 3.18  10 10  r 2

16  3.18  10 10  r 2  r 

9.14 Bulk Modulus.

16 3 .18  10 10  3 .14

 4  10 5 m.

genius PHYSICS Elasticity 25

When a solid or fluid (liquid or gas) is subjected to a uniform pressure all over the surface, such that the shape remains the same, then there is a change in volume. Then the ratio of normal stress to the volumetric strain within the elastic limits is called as Bulk modulus. This is denoted by K.

K K

Normal stress volumetric strain

(V – V)

F/ A  pV   V / V V

where p = increase in pressure; V = original volume; V = change in volume The negative sign shows that with increase in pressure p, the volume decreases by V i.e. if p is positive, V is negative. The reciprocal of bulk modulus is called compressibility. 1 V  K pV

C = compressibility =

S.I. unit of compressibility is N–1m2 and C.G.S. unit is dyne–1 cm2. Gases have two bulk moduli, namely isothermal elasticity E and adiabatic elasticity E . (1) Isothermal elasticity (E) : Elasticity possess by a gas in isothermal condition is defined as isothermal elasticity. For isothermal process, PV = constant

(Boyle’s law)

Differentiating both sides PdV + VdP = 0  PdV = – VdP P



dP stress  E  (dV / V ) strain

E = P

i.e., Isothermal elasticity is equal to pressure. (2) Adiabatic elasticity (E) : Elasticity possess by a gas in adiabatic condition is defined as adiabatic elasticity. For adiabatic process, PV  = constant

(Poisson’s law)

Differentiating both sides, P  V  1 dV  V  dP  0   PdV  VdP  0

P

dP stress  E    dV  strain    V 

E =  P



i.e., adiabatic elasticity is equal to  times pressure.

Note : 

Ratio of adiabatic to isothermal elasticity

[where   E E



P P

 1

i.e., adiabatic elasticity is always more than isothermal elasticity.

9.15 Density of Compressed Liquid.

Cp Cv

]

 E > E

genius PHYSICS 26 Elasticity If a liquid of density  , volume V and bulk modulus K is compressed, then its density increases.



As density

m V



so

But by definition of bulk modulus K  

From (i) and (ii)

 







 V V

 VP V P    V V K

…..(ii)

    P   K

[As  =   –  ]

P 

1    As K  C   

    1   [1  CP] K  

or

…..(i)

9.16 Fractional Change in the Radius of Sphere. A solid sphere of radius R made of a material of bulk modulus K is surrounded by a liquid in a cylindrical container. A massless piston of area A floats on the surface of the liquid. Volume of the spherical body V 

m

4 3 R 3

V R 3 V R R 1 V  R 3 V



Bulk modulus K   V

…..(i)

P V

V P mg   V K AK



Substituting the value of

mg    As P  A   

…..(ii)

V R 1 mg from equation (ii) in equation (i) we get  V R 3 AK

Sample problems based on Bulk modulus Problem 53. When a pressure of 100 atmosphere is applied on a spherical ball of rubber, then its volume reduces to 0.01%. The bulk modulus of the material of the rubber in dyne/cm2 is (a) 10  1012 Solution : (c)

(b) 100  1012

(c) 1  1012

1 atm  10 5 N / m 2 100 atm  10 7 N / m 2 and V  0.01%V 

K

(d) 20  1012

V  0 . 0001 V

10 7 Dyne P .   1  10 11 N / m 2  1  10 12 0 .0001 V / V cm 2

Problem 54. Coefficient of isothermal elasticity E and coefficient of adiabatic elasticity E are related by (  C p / Cv ) [MP PET 2000]

(a) E   E

(b) E   E

(c) E   / E

(d) E   2 E

genius PHYSICS Elasticity 27 Solution : (b)

Adiabatic elasticity    isothermal elasticity  E   E .

Problem 55. A uniform cube is subjected to volume compression. If each side is decreased by 1%,then bulk strain is [EAMCET (Engg.) 1995; DPMT 2000]

(a) 0.01 Solution : (d)

(b) 0.06

(c) 0.02

(d) 0.03

Volume of cube V  L3  Percentage change in V  3  (percentage change in L)= 3 (1%)  3%

V  3% of V  Volumetric strain 

V 3   0 . 03 V 100

Problem 56. A ball falling in a lake of depth 200 m shows 0.1% decrease in its volume at the bottom. What is the bulk modulus of the material of the ball (a) 19 .6  10 8 N / m 2 Solution : (a)

K

[AFMC 1997]

(b) 19 .6  10 10 N / m 2 (c) 19 .6  10 10 N / m 2

(d) 19.6  10 8 N / m 2

P hdg 200  10 3  9.8  19.6  10 8 N / m 2   V / V V / V 0.001

Problem 57. The ratio of the adiabatic to isothermal elasticities of a triatomic gas is (a) Solution : (b)

3 4

4 3

(b)

[MP PET 1991]

(c) 1

(d)

For triatomic gas   4 / 3  Ratio of adiabatic to isothermal elasticity  

5 3

4 . 3

Problem 58. A gas undergoes a change according to the law P  P0 e V . The bulk modulus of the gas is (a) P Solution : (b)

P  Po e V 

(b) PV dP  Po e V   P dV

(c) P

(d)

PV



[As P  Po e V ]

dP dP V  P V   P V dV dV / V 

K  P V

Problem 59. The ratio of two specific heats of gas C p / Cv for argon is1.6 and for hydrogen is 1.4. Adiabatic elasticity of argon at pressure P is E. Adiabatic elasticity of hydrogen will also be equal to E at the pressure (a) P Solution : (b)

(b)

8 P 7

(c)

7 P 8

(d) 1.4 P

Adiabatic elasticity =  (pressure) For Argon (E ) Ar  1 .6 P and for Hydrogen (E ) H 2  1 .4 P  According to problem (E ) H 2  ( ) Ar  1.4 P   1.6 P  P  

16 8 P  P. 14 7

Problem 60. The pressure applied from all directions on a cube is P. How much its temperature should be raised to

maintain the original volume ? The volume elasticity of the cube is  and the coefficient of volume expansion is



(a) Solution : (a)

P



(b)

P



(c)

P



Change in volume due to rise in temperature V  V 

(d)

 P

genius PHYSICS 28 Elasticity  volumetric strain 

V    V

But bulk modulus   

stress P P    strain   

9.17 Modulus of Rigidity. Within rigidity of Shearing  Shearing

limits of proportionality, the ratio of tangential stress to the shearing strain is called modulus of the material of the body and is denoted by , i.e. x F A stress R Q Q'  strain 

L 

In this case the shape of a body changes but its volume remains unchanged.

Fixed face P

S

Consider a cube of material fixed at its lower face and acted upon by a tangential force F at its upper surface having area A. The shearing stress, then, will be Shearing stress 

F|| A



F A

This shearing force causes the consecutive horizontal layers of the cube to be slightly displaced or sheared relative to one another, each line such as PQ or RS in the cube is rotated through an angle  by this shear. The shearing strain is defined as the angle  in radians through which a line normal to a fixed surface has turned. For small values of angle, Shearing strain    So



QQ ' x  PQ L

shear stress F / A F   shear strain  A

Only solids can exhibit a shearing as these have definite shape.

9.18 Poisson’s Ratio. When a long bar is stretched by a force along its length then its length increases and the radius decreases as shown in the figure. Lateral strain : The ratio of change in radius to the original radius is called lateral strain. Longitudinal strain : The ratio of change in length to the original length is called longitudinal strain. The ratio of lateral strain to longitudinal strain is called Poisson’s ratio ( ). i.e.

 

Lateral strain Longitudin al strain

 

 dr / r dL / L

L

r

r – dr L + dL

Negative sign indicates that the radius of the bar decreases when it is stretched. Poisson’s ratio is a dimensionless and a unitless quantity.

9.19 Relation Between Volumetric Strain, Lateral Strain and Poisson’s Ratio. If a long bar have a length L and radius r then volume V  r 2 L

F

genius PHYSICS Elasticity 29

Differentiating both the sides dV  r 2 dL   2rL dr Dividing both the sides by volume of bar

dV r 2 dL  2rL dr dL dr    2 2 2 V L r r L r L



Volumetric strain = longitudinal strain + 2(lateral strain)



dV dL dL dL   2  (1  2 ) V L L L

or

 

dr / r dr dL    As   dL / L  r   L   

1 dV  1 2  AdL 

[where A = cross-section of bar]

Important points (i) If a material having  = – 0.5 then 

dV dL =0  [1  2 ] V L

Volume = constant or K =  i.e., the material is incompressible.

(ii) If a material having  = 0, then lateral strain is zero i.e. when a substance is stretched its length increases without any decrease in diameter e.g. cork. In this case change in volume is maximum. (iii) Theoretical value of Poisson’s ratio  1    0.5 . (iv) Practical value of Poisson’s ratio 0    0.5

9.20 Relation between Y, k,  and . Moduli of elasticity are three, viz. Y, K and  while elastic constants are four, viz, Y, K,  and . Poisson’s ratio  is not modulus of elasticity as it is the ratio of two strains and not of stress to strain. Elastic constants are found to depend on each other through the relations : Y  3 K(1  2 ) and Y  2(1   ) Eliminating  or Y between these, we get Y 

9 K 3 K  2 and   3K   6 K  2

Sample problems based on relation between Y, k,  and  Problem 61. Minimum and maximum values of Poisson’s ratio for a metal lies between (a) –  to + 

(b) 0 to 1

(c) –  to 1

[Orissa JEE 2003]

(d) 0 to 0.5

Solution : (d)

Problem 62. For a given material, the Young’s modulus is 2.4 times that of rigidity modulus. Its Poisson’s ratio is [EAMCET 1990; RPET 2001]

(a) 2.4 Solution : (d)

(b) 1.2

(c) 0.4

(d) 0.2

Y  2(1   )  2.4  2(1   )  1.2  1      0.2

Problem 63. There is no change in the volume of a wire due to change in its length on stretching. The Poisson’s ratio of the material of the wire is (a) + 0.50

(b) – 0.50

(c) + 0.25

(d) – 0.25

genius PHYSICS 30 Elasticity Solution : (b)

dV dL dL dL   2  (1  2 ) =0 V L L L

1  2  0    

[As there is no change in the volume of the wire]

1 2

Problem 64. The values of Young’s and bulk modulus of elasticity of a material are 8  10 10 N / m 2 and 10  10 10 N / m 2 respectively. The value of Poisson’s ratio for the material will be (a) 0.25 Solution : (c)

(b) – 0.25

(c) 0.37

(d) – 0.37

Y  3 K(1  2 )  8  10 10  3  10  10 10 (1  2 )    0.37

Problem 65. The Poisson’s ratio for a metal is 0.25. If lateral strain is 0.0125, the longitudinal strain will be (a) 0.125 Solution : (b)

 

(b) 0.05

Lateral strain Longitudin al strain

(c) 0.215

 Longitudinal strain 

(d) 0.0125

Lateral strain





0 . 0125  0.05 0 . 25

Problem 66. The ‘’ of a material is 0.20. If a longitudinal strain of 4.0  10 3 is caused, by what percentage will the volume change (a) 0.48% Solution : (c)

(b) 0.32%

(c) 0.24%

(d) 0.50%

Longitudinal strain  4  10 3 or 0.4% Lateral strain    0.4%  0.2  0.4%  0.08%

 Volumetric strain = longitudinal strain – 2 lateral strain  0.4  2  (0.08 ) = 0.24%  Volume will change by 0.24%.

9.21 Torsion of Cylinder. If the upper end of a cylinder is clamped and a torque is applied at the lower end the cylinder gets twisted by angle . Simultaneously shearing strain  is produced in the cylinder. P

r

(i) The angle of twist  is directly proportional to the distance from the fixed end of the cylinder.

Q l



At fixed end   0 o and at free end  = maximum.

O

(ii) The value of angle of shear  is directly proportional to the radius of the cylindrical shell.



A

B

At the axis of cylinder  = 0 and at the outermost shell  = maximum. (iii) Relation between angle of twist () and angle of shear () AB = r = l



r l

(iv) Twisting couple per unit twist or torsional rigidity or torque required to produce unit twist. C

 r 4 2l

 C  r4  A2

(v) Work done in twisting the cylinder through an angle  is W 

 r 4  2 1 C 2  2 4l

Sample problems based on Torsion

genius PHYSICS Elasticity 31

Problem 67. Mark the wrong statement

[MP PMT 2003]

(a) Sliding of molecular layer is much easier than compression or expansion (b) Reciprocal of bulk modulus of elasticity is called compressibility (c) It is difficult to twist a long rod as compared to small rod (d) Hollow shaft is much stronger than a solid rod of same length and same mass Solution : (c)

Problem 68. A rod of length l and radius r is joined to a rod of length l / 2 and radius r / 2 of same material. The free end of small rod is fixed to a rigid base and the free end of larger rod is given a twist of , the twist angle at the joint will be [RPET 1997] (a)  / 4 Solution : (d)

(b)  / 2

(c) 5 / 6

(d) 8 / 9

If torque  is applied at the free end of larger rod and twist  is given to it then twist at joint is  1 and twist at the upper end (fixed base)  2 4

 

r 4 (  1 ) 2l



(   1 ) 



r 2 2(l / 2)

   1   2 

( 1  0 ) 8

[As  2  0]

8  8 1   1  91  8   1 



8 . 9

Problem 69. The upper end of a wire of radius 4 mm and length 100 cm is clamped and its other end is twisted through an angle of 30 o . Then angle of shear is

[NCERT

1990;

MP

PMT

1996]

(a) 12o Solution : (b)

L  r

(b) 0.12o

 

(c) 1 .2 o

(d) 0.012 o

r 4  10 3  30 o   0.12 o L 1

Problem 70. Two wires A and B of same length and of the same material have the respective radii r1 and r2. Their one end is fixed with a rigid support, and at the other end equal twisting couple is applied. Then the ratio of the angle of twist at the end of A and the angle of twist at the end of B will be [AIIMS 1980] (a)

Solution : (c)

r12

(b)

r22

1   2 

r14 1 2l1



r24 2 2l 2

r22

(c)

r12

r   1   2  2  r1

  

r24 r14

(d)

r14 r24

4

Problem 71. The work done in twisting a steel wire of length 25 cm and radius 2mm through 45 o will be (  8  10 10 N / m 2 )

(a) 2.48 J Solution : (a)

W

(b) 3.1 J

(c) 15.47 J

r 4  2 1 3 .14  8  10 10  (2  10 3 ) 4  ( / 4 ) 2  C 2   2.48 J 2 4l 4  25  10  2

(d) 18.79 J

genius PHYSICS 32 Elasticity

9.22 Interatomic Force Constant. Behaviour of solids with respect to external forces is such that if their atoms are connected to springs. When an external force is applied on a solid, this distance between its atoms changes and interatomic force works to restore the original dimension. The ratio of interatomic force to that of change in interatomic distance is defined as the interatomic force F constant. K  r It is also given by K  Y  r0 [Where Y = Young's modulus, r0 = Normal distance between the atoms of wire] Unit of interatomic force constant is N/m and Dimension MT–2

Note : 

The number of atoms having interatomic distance r0 in length l of a wire, N = l/r0.

 The number of atoms in area A of wire having interatomic separation r0 is N  A / r02 .

Sample problems based on Interatomic Force Constant Problem 72. The mean distance between the atoms of iron is 3  10 10 m and interatomic force constant for iron is 7 N/m. The Young’s modulus of elasticity for iron is (a) 2 .33  10 5 N / m 2 Solution : (d)

Y 

[JIPMER 2002]

(b) 23 .3  10 10 N / m 2 (c) 233  1010 N / m 2

(d) 2 .33  10 10 N / m 2

k 7   2 .33  10 10 N / m 2 . ro 3  10 10

Problem 73. The Young’s modulus for steel is Y  2  10 11 N / m 2 . If the inter-atomic distance is 3.2Å, the inter atomic force constant in N/ Å will be (b) 6.4  10 9

(a) 6.4  10 9 Solution : (b)

(d) 3.2  10 9

(c) 3.2  10 9

k  Y  r0  2  10 11  3.2  10 10  6.4  10 1 N / m = 6.4  10 9 N / Å .

9.23 Elastic Hysteresis. When a deforming force is applied on a body then the strain does not change simultaneously with stress rather it lags behind the stress. The lagging of strain behind the stress is defined as elastic hysteresis. This is the reason why the values of strain for same stress are different while increasing the load and while decreasing the load. Hysteresis loop : The area of the stress-strain curve is called the hysteresis loop and it is numerically equal to the work done in loading the material and then unloading it. For B

Stress

Stress

For A

O

Strain

O

Strain

If we have two tyres of rubber having different hysteresis loop then rubber B should be used for making the car tyres. It is because of the reason that area under the curve i.e. work done in case of rubber B is lesser and hence the car tyre will not get excessively heated and rubber A should be used to absorb vibration of the machinery because of the large area of the curve, a large amount of vibrational energy can be dissipated.

genius PHYSICS Elasticity 33

9.24 Factors Affecting Elasticity. (1) Hammering and rolling : Crystal grains break up into smaller units by hammering and rolling. This result in increase in the elasticity of material. (2) Annealing : The metals are annealed by heating and then cooling them slowly. Annealing results in decrease in the elasticity of material. (3) Temperature : Intermolecular forces decreases with rise in temperature. Hence the elasticity decreases with rise in temperature but the elasticity of invar steel (alloy) does not change with change of temperature. (4) Impurities : Due to impurities in a material elasticity can increase or decrease. The type of effect depends upon the nature of impurities present in the material.

9.25 Important Facts About Elasticity. (1) The body which requires greater deforming force to produce a certain change in dimension is more elastic. Example : Ivory and steel balls are more elastic than rubber. (2) When equal deforming force is applied on different bodies then the body which shows less deformation is more elastic. Example : (i) For same load, more elongation is produced in rubber wire than in steel wire hence steel is more elastic than rubber. (ii) Water is more elastic than air as volume change in water is less for same applied pressure. (iii) Four identical balls of different materials are dropped from the same height then after collision balls rises upto different heights. The order of their height can be given by hivory > hsteel > hrubber > hclay because Yivory > Ysteel > Yrubber > Yclay. (3) The value of moduli of elasticity is independent of the magnitude of the stress and strain. It depends only on the nature of material of the body. (4) For a given material there can be different moduli of elasticity depending on the type of stress applied and resulting strain. Name of substance

Young’s modulus (Y) 1010N/m2

Bulk modulus (K) 1010N/m2

Modulus of rigidity () 1010N/m2

Aluminium

6.9

7.0

2.6

Brass

9.0

6.7

3.4

Copper

11.0

13.0

4.5

Iron

19.0

14.0

4.6

Steel

20.0

16.0

8.4

Tungsten

36.0

20.0

15.0

Diamond

83.0

55.0

34.0

Water

–

0.22

–

Glycerin

–

0.45

–

Air

–

1.01

–

genius PHYSICS 34 Elasticity (5) The moduli of elasticity has same dimensional formula and units as that of stress since strain is dimensionless.  Dimensional formula ML1 T 2 while units dyne/cm2 or Newton/m2. (6) Greater the value of moduli of elasticity more elastic is the material. But as Y  (1/l), K  (1/V) and   (1/) for a constant stress, so smaller change in shape or size for a given stress corresponds to greater elasticity. (7) The moduli of elasticity Y and  exist only for solids as liquids and gases cannot be deformed along one dimension only and also cannot sustain shear strain. However K exist for all states of matter viz. solid, liquid or gas. (8) Gases being most compressible are least elastic while solids are most i.e. the bulk modulus of gas is very low while that for liquids and solids is very high. Ksolid > Kliquid > Kgas (9) For a rigid body l, V or  = 0 so Y, K or  will be , i.e. elasticity of a rigid body is infinite. Diamond and carborundum are nearest approach to rigid bodies. (10) In a suspension bridge there is a stretch in the ropes by the load of the bridge. Due to which length of rope changes. Hence Young’s modulus of elasticity is involved. (11) In an automobile tyre as the air is compressed, volume of the air in tyre changes, hence the bulk modulus of elasticity is involved. (12) In transmitting power, an automobile shaft is sheared as it rotates, so shearing strain is set up, hence modulus of rigidity is involved. (13) The shape of rubber heels changes under stress, so modulus of rigidity is involved.

9.26 Practical Applications of Elasticity. (i) The metallic parts of machinery are never subjected to a stress beyond elastic limit, otherwise they will get permanently deformed. (ii) The thickness of the metallic rope used in the crane in order to lift a given load is decided from the knowledge of elastic limit of the material of the rope and the factor of safety. (iii) The bridges are declared unsafe after long use because during its long use, a bridge under goes quick alternating strains continuously. It results in the loss of elastic strength. (iv) Maximum height of a mountain on earth can be estimated from the elastic behaviour of earth. At the base of the mountain, the pressure is given by P = hg and it must be less than elastic limit (K) of earth’s supporting material. K > P > hg

 h

K g

or

h m ax 

K g

(v) In designing a beam for its use to support a load (in construction of roofs and bridges), it is advantageous to increase its depth rather than the breadth of the beam because the depression in rectangular beam. l b Wl 3   4Ybd

d

 3

To minimize the depression in the beam, it is designed as I-shaped

W

genius PHYSICS Elasticity 35

girder. (vi) For a beam with circular cross-section depression is given by  

WL 3 12  r 4 Y

(vii) A hollow shaft is stronger than a solid shaft made of same mass, length and material. Torque required to produce a unit twist in a solid shaft  solid 

 r 4

and torque required to produce a unit twist in a hollow shaft  hollow  From (i) and (ii),

.…..(i)

2l

 (r24  r14 )

…..(ii)

2l

 hollow r24  r14 (r22  r12 )(r22  r12 )    solid r4 r4

…..(iii)

Since two shafts are made from equal volume  r 2 l   (r22  r12 )l  r 2  r22  r12 Substituting this value in equation (iii) we get,

 hollow r22  r12   1  hollow > solid  solid r2

i.e., the torque required to twist a hollow shaft is greater than the torque necessary to twist a solid shaft of the same mass, length and material through the same angle. Hence, a hollow shaft is stronger than a solid shaft.

Problems based on Interatomic and Intermolecular forces 1.

2.

In solids, inter-atomic forces are

[DCE 1999]

(a) Totally repulsive

(b) Totally attractive

(c)

(d) None of these

Combination of (a) and (b)

The potential energy U between two molecules as a function of the distance X between them has been shown in the figure. The two molecules are [CPMT 1986, 88, 91] (a) Attracted when x lies between A and B and are repelled when X lies between B and C

U + ve

(b) Attracted when x lies between B and C and are repelled when X lies between A and B

0

(c) Attracted when they reach B

– ve

(d) Repelled when they reach B 3.

X

A

B

C

The nature of molecular forces resembles with the nature of the (a) Gravitational force

(b) Nuclear force

(c) Electromagnetic force

(d) Weak force

Problems based on Stress 4.

The ratio of radius of two wire of same material is 2 : 1. Stretched by same force, then the ratio of stress is (a) 2 : 1

5.

(c)

1:4

(d) 4 : 1

If equal and opposite forces applied to a body tend to elongate it, the stress so produced is called (a) Tensile stress

6.

(b) 1 : 2

[PET 1991]

(b) Compressive stress

(c) Tangential stress

(d) Working stress

A vertical hanging bar of length l and mass m per unit length carries a load of mass M at the lower end, its upper end is clamped to a rigid support. The tensile force at a distance x from support is (a) Mg + mg(l – x)

(b) Mg

(c) Mg + mgl

(d) (M  m )g

x l

genius PHYSICS 36 Elasticity 7.

8.

One end of a uniform rod of mass m1 and cross-sectional area A is hung from a ceiling. The other end of the bar is supporting mass m2. The stress at the midpoint is (a)

g(m 2  2m 1 ) 2A

(b)

g(m 2  m1 ) 2A

(c)

g (2 m 2  m 1 ) 2A

(d)

g(m 2  m1 ) A

m1

m2

A uniform bar of square cross-section is lying along a frictionless horizontal surface. A horizontal force is applied to pull it from one of its ends then F

(a) The bar is under same stress throughout its length (b) The bar is not under any stress because force has been applied only at one end (c) The bar simply moves without any stress in it (d) The stress developed reduces to zero at the end of the bar where no force is applied

Problems based on Strain 9.

Which one of the following quantities does not have the unit of force per unit area (a) Stress

(b) Strain

(c) Young’s modulus of elasticity 10.

(b) Shearing strain

[EAMCET 1980]

(c) Longitudinal strain

(d) Metallic strain

(b) Longitudinal

(c) Volume

(d) Transverse

(c)

(d) Liquids

The longitudinal strain is only possible in (a) Gases

13.

Pressure

When a spiral spring is stretched by suspending a load on it, the strain produced is called (a) Shearing

12.

(d)

The reason for the change in shape of a regular body is (a) Volume stress

11.

[MP PMT 1992]

(b) Fluids

Solids

The face EFGH of the cube shown in the figure is displaced 2 mm parallel to itself when forces of 5  10 5 N each are applied on the lower and upper faces. The lower face is fixed. The strain produced in the cube is E

(a) 2

H

(b) 0.5

G

(c) 0.05

C

D A

(d) 1.2  10 8 14.

F

B

4 cm

Forces of 10 5 N each are applied in opposite direction on the upper and lower faces of a cube of side 10 cm, shifting the upper face parallel to itself by 0.5 cm. If the side of the cube were 20 cm, the displacement would be E

(a) 1 cm

D

(b) 0.5 cm (c) 0.25 cm

A

(d) 0.125 cm

Problems based on Stress strain curve

F C

B

genius PHYSICS Elasticity 37 15.

The stress versus strain graphs for wires of two materials A and B are as shown in the figure. If YA and YB are the Young’s modulii of the materials, then [Kerala (Engg.) 2001] Y

(a) YB  2Y A

(c)

Stress

(b) Y A  YB

YB  3Y A

60o 30o

O

(d) Y A  3YB 16.

A

Strain

B X

The graph is drawn between the applied force F and the strain (x) for a thin uniform wire. The wire behaves as a liquid in the part [CPMT 1988] F

b

a

(a) ab

d

(b) bc c

(c) cd

O

17.

X

x

(d) oa

The diagram shows stress v/s strain curve for the materials A and B. From the curves we infer that A Stress

(a) A is brittle but B is ductile (b) A is ductile and B is brittle

B

(c) Both A and B are ductile Strain

(d) Both A and B are brittle 18.

The figure shows the stress-strain graph of a certain substance. Over which region of the graph is Hooke’s law obeyed A Stress

(a) AB (b) BC

D

C

B

(c) CD Which one of the following is the Young’s modulus (in N/m2) for the wire having the stress-strain curve shown in the figure Stress (107 N/m2)

19.

Strain

E

(d) ED

(a) 24  10 11 (b) 8.0  10 11 (c) 10  10 11

10 8 6 4 2 O

2

(d) 2.0  10 11

4

6 8  10–4

Strain

Problems based on Young's Modulus 20.

The adjacent graph shows the extension (l) of a wire of length 1m suspended from the top of a roof at one end with a load W connected to the other end. If the cross sectional area of the wire is 10 –6m2, calculate the young’s modulus of the material of the wire (a) 2  10 (b) 2  10

11

(c)

N /m

2

N / m2

3  10 12 N / m 2

(d) 2  10 13 N / m 2 21.

l(10 – 4) m

[IIT-JEE (Screening) 2003] 11

4 3 2 1 20 40 60 80 W(N)

In the Young’s experiment, if length of wire and radius both are doubled then the value of Y will become

[RPET 2003]

genius PHYSICS 38 Elasticity (a) 2 times 22.

(b) 4 times

A rubber cord catapult has cross-sectional area

(c) Remains same 25mm2

(d) Half

and initial length of rubber cord is 10cm. It is stretched to 5cm. and then

released to project a missile of mass 5gm. Taking Yrubber  5  10 8 N / m 2 velocity of projected missile is (a) 20 ms–1 23.

(b) 100 ms–1

(c) 250 ms–1

[CPMT 2002]

(d) 200 ms–1

Consider the following statements Assertion (A) : Stress is the internal force per unit area of a body. Reason (R) : Rubber is more elastic than steel. Of these statements

[AIIMS 2002]

(a) Both A and R are true and the R is a correct explanation of the A (b) Both A and R are true but the R is not a correct explanation of the A (c) A is true but the R is false (d) Both A and R are false (e) A is false but the R is true 24.

The area of cross-section of a steel wire (Y  2.0  10 11 N / m 2 ) is 0.1 cm2. The force required to double its length will be [MP PET 2002]

(a) 2  10 N 12

25.

(d) 2  10 N

(b) Y A  t

YL t A

(c)

(d) Y AL t

(b) Glass

(c)

Steel

(d) Copper

There are two wires of same material and same length while the diameter of second wire is 2 times the diameter of first wire, then ratio of extension produced in the wires by applying same load will be (a) 1 : 1

28.

2  10 N

6

Which one of the following substances possesses the highest elasticity [MP PMT 1992; RPMT 1999; RPET 2000; MH CET (Med.) 2001] (a) Rubber

27.

(c)

10

A metal bar of length L and area of cross-section A is clamped between two rigid supports. For the material of the rod, its Young’s modulus is Y and coefficient of linear expansion is . If the temperature of the rod is increased by t o C , the force exerted by the rod on the supports is [MP PMT 2001] (a) Y AL t

26.

(b) 2  10 N 11

(b) 2 : 1

(c) 1 : 2

(d) 4 : 1

Consider the following statements Assertion (A) : Rubber is more elastic than glass. Reason (R) : The rubber has higher modulus of elasticity than glass. Of these statements

[AIIMS 2000]

(a) Both A and R are true and the R is a correct explanation of the A (b) Both A and R are true but the R is not a correct explanation of the A (c) A is true but the R is false (d) Both A and R are false (e) A is false but the R is true 29.

30.

31.

The longitudinal extension of any elastic material is very small. In order to have an appreciable change, the material must be in the form of (a) Thin block of any cross section

(b) Thick block of any cross section

(c)

(d) Short thin wire

Long thin wire

In suspended type moving coil galvanometer, quartz suspension is used because (a) It is good conductor of electricity

(b) Elastic after effects are negligible

(c) Young’s modulus is greater

(d) There is no elastic limit

You are given three wires A, B and C of the same length and cross section. They are each stretched by applying the same force to the ends. The wire A is stretched least and comes back to its original length when the stretching force is removed. The wire B is stretched more than A and also comes back to its original length when the stretching force is removed. The wire C is stretched most and remains stretched even when stretching force is removed. The greatest Young’s modulus of elasticity is possessed by the material of wire (a) A

32.

(b) B

(c) C

(d) All have the same elasticity

The ratio of diameters of two wires of same material is n : 1. The length of wires are 4 m each. On applying the same load, the increase in length of thin wire will be

genius PHYSICS Elasticity 39 (a) n2 times 33.

(b) Increase by (c) Decrease by (d)

35.

36.

37.

(d) None of the above

Vlg Y r 2

Vrg Y l 2 Vg Y r

Vg Y

If the ratio of lengths, radii and Young’s modulii of steel and brass wires in the figure are a, b and c respectively. Then the corresponding ratio of increase in their lengths would be (a)

2a 2c b

(b)

3a 2b 2 c

(c)

2 ac b2

Steel M Brass 2M

3c (d) 2 ab 2 A uniform heavy rod of weight W, cross sectional area A and length L is hung from a fixed support. Young’s modulus of the material of the rod is Y. If lateral contraction is neglected, the elongation of the rod under its own weight is 2WL WL WL (a) (b) (c) (d) Zero AY AY 2 AY

A constant force F0 is applied on a uniform elastic string placed over a smooth horizontal surface as shown in figure. Young’s modulus of string is Y and area of cross-section is S. The strain produced in the string in the direction of force is (a)

F0 Y S

(b)

F0 SY

(c)

F0 2 SY

(d)

F0 Y 2S

F0

A uniform rod of length L has a mass per unit length  and area of cross section A. The elongation in the rod is l due to its own weight if it is suspended from the ceiling of a room. The Young’s modulus of the rod is (a)

38.

(c) 2n times

A wire of radius r, Young’s modulus Y and length l is hung from a fixed point and supports a heavy metal cylinder of volume V at its lower end. The change in length of wire when cylinder is immersed in a liquid of density  is in fact (a) Decrease by

34.

(b) n times

2 gL2 Al

(b)

gL2 2 Al

(c)

2gL Al

(d)

gl 2 AL

AB is an iron wire and CD is a copper wire of same length and same cross-section. BD is a rod of length 0.8 m. A load G = 2kg-wt is suspended from the rod. At what distance x from point B should the load be suspended for the rod to remain in a horizontal position (YCu  11 .8  10 10 N / m 2 , Y Fe  19 .6  10 10 N / m 2 ) C

A T1

(a) 0.1 m (b) 0.3 m (c) 0.5 m (d) 0.7 m 39.

O

T2 D

B x G

A slightly conical wire of length L and end radii r1 and r2 is stretched by two forces F, F applied parallel to length in opposite directions and normal to end faces. If Y denotes the Young’s modulus, then extension produced is

genius PHYSICS 40 Elasticity (a)

(a)

3

(b)

1

(d)

(b) FKA

(c)

F K

(d)

FLY r1r2

FK AL

Y

1 2

O

3 (d) 2 The force constant of a wire does not depend on (b) Radius of the wire

30o

X

Displacement

(c) Length of the wire

(d) None of the above

A metal wire of length L, area of cross-section A and Young’s modulus Y behaves as a spring. The equivalent spring constant will be (a)

44.

FL r1r2 Y

3

(a) Nature of the material 43.

(c)

The value of force constant between the applied elastic force F and displacement will be

(c)

42.

FL r1 Y

The force constant of wire is K and its area of cross-section is A. If the force F is applied on it, then the increase in its length will be (a) KA

41.

(b)

Force

40.

FL r12 Y

Y AL

(b)

YA L

(c)

YL A

(d)

L AY

A highly rigid cubical block A of small mass M and side L is fixed rigidly onto another cubical block B of the same dimensions and modulus of rigidity  such that the lower face of A completely covers the upper face of B. The lower face of B is rigidly held on a horizontal surface. A small force is applied perpendicular to one of the sides faces of A. After the force is withdrawn, block A execute small oscillations the time period of which is given by (a) 2 ML

(b) 2

M L

(c)

2

ML



(d) 2

M

L

Problems based on Stretching a wire 45.

A wire of length L and cross-sectional area A is made of a material of Young’s modulus Y. It is stretched by an amount x. The work done is [MP PET 1996; BVP 2003] (a)

46.

YxA 2L

Yx 2 A L

(c)

Yx 2 A 2L

(d)

2 Yx 2 A L

Two wires of same diameter of the same material having the length l and 2l. If the force F is applied on each, the ratio of the work done in the two wires will be [MP PET 1989] (a) 1 : 2

47.

(b)

(b) 1 : 4

(c) 2 : 1

(d) 1 : 1

If the potential energy of a spring is V on stretching it by 2 cm, then its potential energy when it is stretched by 10 cm will be [CPMT 1976]

(a) V/25 48.

(c) V/5

(d) 25V

The strain energy stored in a body of volume V due to shear S and shear modulus  is (a)

49.

(b) 5V

S 2V 2

(b)

SV 2 2

(c)

S 2V



(d)

K is the force constant of a spring. The work done in increasing its extension from l1 to l2 will be (a) K(l2  l1 )

(b)

K (l2  l1 ) 2

(c)

K(l22  l12 )

(d)

1 S 2 V 2 [MP PET 1995; MP PMT 1996]

K 2 2 (l2  l1 ) 2

Problems based on Breaking of wire 50.

The breaking stress of a wire depends upon (a) Length of the wire

(b) Radius of the wire

[AIIMS 2002]

(c) Material of the wire

(d) Shape of the cross section

genius PHYSICS Elasticity 41 51.

An aluminium rod has a breaking strain of 0.2%. The minimum cross sectional area of the rod, in m2, in order to support a load of 104 N is (Y  7  10 9 N / m 2 ) (a) 1.4  10 4

52.

(b) Double

(c) Four times

(d) One-fourth

A heavy mass is attached to a thin wire and is whirled in a vertical circle. The wire is most likely to break (a) When the mass is at the highest point (c) When the wire is horizontal

54.

(d) 7.1  10 5

(c) 1.4  10 3

A cable is replaced by another one of the same length and material but of twice the diameter. The maximum load that the new wire can support without exceeding the elastic limit, as compared to the load that the original wire could support, is (a) Half

53.

(b) 7.1  10 4

(b) When the mass is at the lowest point (d) At an angle of cos–1 (1/3) from the upward vertical

A heavy uniform rod is hanging vertically from a fixed support. It is stretched by its own weight. The diameter of the rod is (a) (b) (c) (d)

Smallest at the top and gradually increases down the rod Largest at the top and gradually decreases down the rod Uniform everywhere Maximum in the middle

Problems based on Bulk modulus 55.

The isothermal bulk modulus of a gas at atmospheric pressure is (a) 1 mm of Hg

56.

(c) C p Cv

(d) 1 / C p Cv

(b) 2  10 8

(c) 10 9

(d) 2  10 9

(b) 4  10 5 cc

(c) 0.025 cc

(d) 0.004 cc

An ideal gas of mass m, volume V, pressure p and temperature T undergoes a small change in state at constant temperature. Its C adiabatic exponent i.e., p is . The bulk modulus of the gas at the constant temperature process called isothermal process is Cv m p pV (d) T T An ideal gas of mass m, volume V, pressure p and temperature T undergoes a small change under a condition that heat can neither enter into it from outside nor can it leave the system. Such a process is called adiabatic process. The bulk modulus of the gas

(a) p 60.

(b) C p / Cv

The compressibility of water is 4  10–5 per unit atmospheric pressure. The decrease in volume of 100 cubic centimetre of water under a pressure of 100 atmosphere will be [MP PMT 1990] (a) 0.4 cc

59.

(d) 2.026  10 5 N / m 2

If a rubber ball is taken at the depth of 200 m in a pool. Its volume decreases by 0.1%. If the density of the water is 1  10 3 kg / m 3 and g = 10 m/s2, then the volume elasticity in N/m2 will be [MP PMT 1991] (a) 10 8

58.

(c) 1.013  10 5 N / m 2

The specific heat at constant pressure and at constant volume for an ideal gas are Cp and Cv and its adiabatic and isothermal elasticities are E and E respectively. The ratio of E to E is [MP PMT 1989; MP PET 1992] (a) Cv / C p

57.

(b) 13.6 mm of Hg

[AIIMS 2000; KCET (Engg./Med.) 1999]

(b)  p

(c)

C      p  is  C v   m p pV (d) T T An ideal gas whose adiabatic exponent is  is expanded according to the law p= V where  is a constant. For this process the bulk modulus of the gas is p (a) p (b) (c) p (d) (l – )p  1 c.c. of water is taken from the top to the bottom of a 200 m deep lake. What will be the change in its volume if K of water is

(a) p 61.

62.

(b)  p

(c)

(b) 8.8  10 2 c.c.

(c) 8.8  10 4 c.c.

2.2  10 9 N / m 2

(a) 8.8  10 6 c.c.

Problems based on Modulus of rigidity 63.

Modulus of rigidity of a liquid

(d) 8.8  10 1 c.c

genius PHYSICS 42 Elasticity (a) Non zero constant 64.

(b) Infinite

(c) Zero

(d) Cannot be predicted

The Young’s modulus of the material of a wire is 6  10 12 N / m 2 and there is no transverse strain in it, then its modulus of rigidity will be (a) 3  10 12 N / m 2

(b) 2  10 12 N / m 2

(c) 10 12 N / m 2

(d) None of the above

Problems based on relation between Y, , K and  65.

66. 67. 68.

The value of Poisson’s ratio lies between 1 3 1 (a) –1 to (b)  to  2 4 2

70.

71.

(c)



1 to 1 2

Which of the following will be  if Y = 2.4 (a) –1 (b) 0.2 (c) 0.1 Which is correct relation (a) Y <  (b) Y >  (c) Y =  The relationship between Young’s modulus Y, bulk modulus K and modulus of rigidity  is (a) Y 

69.

[AIIMS 1985; MP PET 1986; DPMT 2002]

9K   3K

(b)  

9 yK Y  3K

The Poisson’s ratio cannot have the value (a) 0.7 (b) 0.2 Which of the following relations is true 9Y (a) 3Y = K(1 – ) (b) K  Y 

9K 3  K

(c) Y 

(d) 1 to 2 [RPET 2001]

(d) – 0.25 [RPET 2001]

(d)  = +1 [MP PET 1991; MP PMT 1997]

3K 9  K

(d) Y 

[EAMCET 1989]

(c) 0.1

(d) 0.5 [CPMT 1984]

(c)  = (6K + ) Y

0 .5 Y  

(d)  



The wrong relation for modulus of rigidity () is (a)  

Shearing stress Shearing strain

(b) Unit of  is N / m 2

(c)  

Y 2(1   )

(d)  

Y 2(1   )

Problems based on Torsion 72.

A rod of 2m length and radius 1 cm is twisted at one end by 0.8 rad with respect to other end being clamped. The shear strain developed in its rod will be [RPET 1997] (a) 0.002

73.

(d) 0.016

(b) 0 .9 o

(c) 9 o

(d) 90 o

The end of a wire of length 0.5m and radius 10–3m is twisted through 0.80 radian. The shearing strain at the surface of wire will be (a) 1.6  10 3

75.

(c) 0.008

The upper end of a wire 1 metre long and 2 mm in radius is clamped. The lower end is twisted through an angle of 45 o . The angle of shear is [MP PMT 1990] (a) 0.09 o

74.

(b) 0.004

(b) 1.6  10 3

(c) 16  10 3

(d) 16  10 6

Two cylinders A and B of the same material have same length, their radii being in the ratio of 1 : 2 respectively. The two are joined in series. The upper end of A is rigidly fixed. The lower end of B is twisted through an angle , the angle of twist of the cylinder A is fig. (a)

15  16

(b)

16  15

(c)

16  17

(d)

17  16

Problems based on Interatomic force constant 76.

If the interatomic spacing in a steel wire is 3.0Å and Ysteel  20  10 10 N / m 2 , then force constant is (a) 6  10 2 N / Å

77.

(b) 6  10 9 N / Å

(c)

4  10 5 N / Å

(d) 6  10 5 N / Å

The Young’s modulus of a metal is 1.2  10 11 N / m 2 and the inter-atomic force constant is 3.6  10 9 N / Å . The mean distance between the atoms of the metal is (a) 2Å

(b) 3 Å

(c) 4.5 Å

(d) 5 Å

genius PHYSICS Elasticity 43 78.

The interatomic distance for a metal is 3  10 10 m . If the interatomic force constant is 3.6  10 9 N / Å , then the Young’s modulus in N / m 2 will be (a) 1.2  10 11

(c) 10 .8  10 19

(b) 4.2  10 11

(d) 2.4  10 10

Miscellaneous problems 79.

80.

A particle of mass m is under the influence of a force F which varies with the displacement x according to the relation F  kx  F0 in which k and F0 are constants. The particle when disturbed will oscillate (a) About x = 0, with   k / m

(b) About x = 0,with   k / m

(c) About x = F0/k with   k / m

(d) About x = F0/k with   k / m

The extension in a string obeying Hooke’s law is x. The speed of sound in the stretched string is v. If the extension in the string is increased to 1.5x, the speed of sound will be [IIT 1996] (a) 1.22 v

81.

(b) 0.61 v

(c) 1.50 v

(d) 0.75 v

Railway lines and girders for buildings, are I shaped, because (a) The bending of a girder is inversely proportional to depth, hence high girder bends less (b) The coefficient of rigidity increases by this shape (c) Less volume strain is caused (d) This keeps the surface smooth

82.

If Young’s modulus for a material is zero, then the state of material should be (a) Solid

83.

84.

(b) Solid but powder

(a) Increases with temperature rise

(b) Decreases with temperature rise

(c) Does not depend on temperature

(d) None of the above

For the same cross-sectional area and for a given load, the ratio of depressions for the beam of square cross-section and circular cross-section is (c) 3 : 

(d) 1 : 

m  2 L2 AY

(b)

m  2 L2 2 AY

(c)

m  2 L2 3 AY

(d)

2 m  2 L2 AY

A steel wire is suspended vertically from a rigid support. When loaded with a weight in air, it extends by la and when the weight is immersed completely in water, the extension is reduced to lw. Then the relative density of the material of the weight is (a)

87.

(b)  : 1

A uniform rod of mass m, length L, area of cross-section A is rotated about an axis passing through one of its ends and perpendicular to its length with constant angular velocity  in a horizontal plane. If Y is the Young’s modulus of the material of rod, the increase in its length due to rotation of rod is (a)

86.

(d) None of the above

The elasticity of invar

(a)  : 3 85.

(c) Gas

la lw

(b)

la la  lw

(c)

la la  lw

(d)

lw la

The twisting couple per unit twist for a solid cylinder of radius 4.9 cm is 0.1 N-m. The twisting couple per unit twist for a hollow cylinder of same material with outer and inner radii of 5 cm and 4 cm respectively, will be (b) 0.64  10–1 N-m

(a) 0.64 N-m

(c) 0.64  10–2 N-m

(d) 0.64  10–3 N-m

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

c

b

c

c

a

a

c

b

b

b

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

a

c

c

c

d

b

b

d

d

a

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

genius PHYSICS 44 Elasticity c

c

c

d

b

c

d

d

c

b

31.

32.

33.

34.

35.

36.

37.

38.

39.

40.

a

a

a

b

c

c

b

b

c

c

41.

42.

43.

44.

45.

46.

47.

48.

49.

50.

b

d

b

d

c

a

d

d

d

c

51.

52.

53.

54.

55.

56.

57.

58.

59.

60.

b

c

b

a

c

b

d

a

a

b

61.

62.

63.

64.

65.

66.

67.

68.

69.

70.

a

c

c

a

a

b

b

a

a

d

71.

72.

73.

74.

75.

76.

77.

78.

79.

80.

c

b

a

a

c

b

b

a

c

a

81.

82.

83.

84.

85.

86.

87.

a

b

c

c

c

b

b

genius PHYSICS 48 Surface Tension

10.1 Intermolecular Force. The force of attraction or repulsion acting between the molecules are known as intermolecular force. The nature of intermolecular force is electromagnetic. The intermolecular forces of attraction may be classified into two types. Cohesive force

Adhesive force

The force of attraction between molecules of same substance is called the force of cohesion. This force is lesser in liquids and least in gases.

The force of attraction between the molecules of the different substances is called the force of adhesion.

Ex. (i) Two drops of a liquid coalesce into one when brought in mutual contact.

Ex. (i) Adhesive force enables us to write on the blackboard with a chalk.

(ii) It is difficult to separate two sticky plates of glass welded with water.

(ii) A piece of paper sticks to another due to large force of adhesion between the paper and gum molecules.

(iii) It is difficult to break a drop of mercury into small droplets because of large cohesive force between the mercury molecules.

(iii) Water wets the glass surface due to force of adhesion.

Note : 

Cohesive or adhesive forces are inversely proportional to the eighth power of distance between

the molecules.

10.2 Surface Tension. The property of a liquid due to which its free surface tries to have minimum surface area and behaves as if it were under tension some what like a stretched elastic membrane is called surface tension. A small liquid drop has spherical shape, as due to surface Imaginary line tension the liquid surface tries to have minimum surface area and for a given volume, the sphere has minimum surface area. Surface tension of a liquid is measured by the force acting per unit length on either side of an imaginary line drawn on the free surface of liquid, the direction of this force being perpendicular to the line and tangential to the free surface of liquid. So if F is the force acting on one side of imaginary line of length L, then T = (F/L) (1) It depends only on the nature of liquid and is independent of the area of surface or length of line considered. (2) It is a scalar as it has a unique direction which is not to be specified. (3) Dimension : [MT – 2].

(Similar to force constant)

(4) Units : N/m (S.I.) and Dyne/cm [C.G.S.] (5) It is a molecular phenomenon and its root cause is the electromagnetic forces.

genius PHYSICS Surface Tension 49

10.3 Force Due to Surface Tension. If a body of weight W is placed on the liquid surface, whose surface tension is T. If F is the minimum force required to pull it away from the water then value of F for different bodies can be calculated by the following table. Body Needle

Figure

Force

F

(Length = l ) F = 2l T + W

Hollow disc

T T F = T F × 2l

F = 2 (r1 + r2)T + W

(Inner radius = r1 Outer radius = r2)

F

F = 2 (r + r)T + W

Thin ring

F = 4rT + W

(Radius = r)

F

Circular plate or disc

F = 2rT + W

(Radius = r)

F

Square frame

F = 8l T + W

(Side = l )

F

Square plate

F = 4l T + W

genius PHYSICS 50 Surface Tension

10.4 Examples of Surface Tension. (1) When mercury is split on a clean glass plate, it forms globules. Tiny globules are spherical on the account of surface tension because force of gravity is negligible. The bigger globules get flattened from the middle but have round shape near the edges, figure

(3) When a molten metal is poured into water from a suitable height, the falling stream of metal breaks up and the detached portion of the liquid in small quantity acquire the spherical shape. Molten metal Water

(2) When a greased iron needle is placed gently on the surface of water at rest, so that it does not prick the water surface, the needle floats on the surface of T T water despite it being heavier because the weight of needle is balanced by the vertical components of the forces of surface tension. If the mg water surface is pricked by one end of the needle, the needle sinks down. (4) Take a frame of wire and dip it in soap solution and take it out, a soap film will be formed in the frame. Place a loop of wet thread gently on the film. It will remain in the form, we place it on the film according to figure. Now, piercing the film with a pin at any point inside the loop, Thread It immediately takes the circular loop form as shown in figure.

(5) Hair of shaving brush/painting brush when dipped in water spread out, but as soon as it is taken out, its hair stick together.

(6) If a small irregular piece of camphor is floated on the surface of pure water, it does not remain steady but dances about on the surface. This is because, irregular shaped camphor dissolves unequally and decreases the surface tension of the water locally. The unbalanced forces make it move haphazardly in different directions.

(7) Rain drops are spherical in shape because each drop tends to acquire minimum surface area due to surface tension, and for a given volume, the surface area of sphere is minimum.

(8) Oil drop spreads on cold water. Whereas it may remain as a drop on hot water. This is due to the fact that the surface tension of oil is less than that of cold water and is more than that of hot water.

10.5 Factors Affecting Surface Tension. (1) Temperature : The surface tension of liquid decreases with rise of temperature. The surface tension of liquid is zero at its boiling point and it vanishes at critical temperature. At critical temperature, intermolecular forces for liquid and gases becomes equal and liquid can expand without any restriction. For small temperature differences, the variation in surface tension with temperature is linear and is given by the relation Tt  T0 (1   t)

where T t , T0 are the surface tensions at t o C and 0 o C respectively and  is the temperature coefficient of surface tension. Examples : (i) Hot soup tastes better than the cold soup. (ii) Machinery parts get jammed in winter. (2) Impurities : The presence of impurities either on the liquid surface or dissolved in it, considerably affect the force of surface tension, depending upon the degree of contamination. A highly soluble substance like sodium chloride when dissolved in water, increases the surface tension of water. But the sparingly soluble substances like phenol when dissolved in water, decreases the surface tension of water.

genius PHYSICS Surface Tension 51

10.6 Applications of Surface Tension. (1) The oil and grease spots on clothes cannot be removed by pure water. On the other hand, when detergents (like soap) are added in water, the surface tension of water decreases. As a result of this, wetting power of soap solution increases. Also the force of adhesion between soap solution and oil or grease on the clothes increases. Thus, oil, grease and dirt particles get mixed with soap solution easily. Hence clothes are washed easily. (2) The antiseptics have very low value of surface tension. The low value of surface tension prevents the formation of drops that may otherwise block the entrance to skin or a wound. Due to low surface tension, the antiseptics spreads properly over wound. (3) Surface tension of all lubricating oils and paints is kept low so that they spread over a large area. (4) Oil spreads over the surface of water because the surface tension of oil is less than the surface tension of cold water. (5) A rough sea can be calmed by pouring oil on its surface. (6) In soldering, addition of ‘flux’ reduces the surface tension of molten tin, hence, it spreads.

10.7 Molecular Theory of Surface Tension. The maximum distance upto which the force of attraction between two molecules is appreciable is called molecular range ( 10 9 m ) . A sphere with a molecule as centre and radius equal to molecular range is called the sphere of influence. The liquid enclosed between free surface (PQ) of the liquid and an imaginary plane (RS) at a distance r (equal to molecular range) from the free surface of the liquid form a liquid film. To understand the tension acting on the free surface of a liquid, let us consider four liquid molecules like A, B, C and D. Their sphere of influence are shown in the figure. (1) Molecule A is well within the liquid, so it is attracted equally in all directions. Hence the net force on this molecule is zero and it moves freely inside the liquid. D

(2) Molecule B is little below the free surface of the liquid and it is also attracted equally in all directions. Hence the resultant force on it is also zero.

P R

C

B

Q S

A (3) Molecule C is just below the upper surface of the liquid film and the part of its sphere of influence is outside the free liquid surface. So the number of molecules in the upper half (attracting the molecules upward) is less than the number of molecule in the lower half (attracting the molecule downward). Thus the molecule C experiences a net downward force.

(4) Molecule D is just on the free surface of the liquid. The upper half of the sphere of influence has no liquid molecule. Hence the molecule D experiences a maximum downward force. Thus all molecules lying in surface film experiences a net downward force. Therefore, free surface of the liquid behaves like a stretched membrane.

Sample problems based on Surface tension Problem 1.

A wooden stick 2m long is floating on the surface of water. The surface tension of water 0.07 N/m. By putting soap solution on one side of the sticks the surface tension is reduced to 0.06 N/m. The net force on the stick will be [Pb. PMT 2002] (a) 0.07 N

(b) 0.06 N

(c) 0.01 N

(d) 0.02 N

genius PHYSICS 52 Surface Tension Solution : (d)

Force on one side of the stick F1  T1  L

 0.07  2  0.14 N

and force on other side of the stick F2  T2  L  0.06  2  0.12 N So net force on the stick  F1  F2  0.14  0.12  0.02 N

Problem 2.

A thin metal disc of radius r floats on water surface and bends the surface downwards along the perimeter making an angle  with vertical edge of disc. If the disc displaces a weight of water W and surface tension of water is T, then the weight of metal disc is [AMU (Med.) 1999] (a) 2 rT + W

Solution : (c)

(b) 2 rT cos – W

(c) 2 rT cos + W

(d) W – 2 rT cos

Weight of metal disc = total upward force = upthrust force + force due to surface tension

T

T





= weight of displaced water + T cos  (2 r) = W + 2 rT cos 

Problem 3.

A 10 cm long wire is placed horizontally on the surface of water and is gently pulled up with a force of 2  10 2 N to keep the wire in equilibrium. The surface tension in Nm–1 of water is (a) 0.1 N/m

Solution : (a)

(d) 0.002 N/m

F 2  10 2   0 .1 N /m 2l 2  10  10  2

There is a horizontal film of soap solution. On it a thread is placed in the form of a loop. The film is pierced inside the loop and the thread becomes a circular loop of radius R. If the surface tension of the loop be T, then what will be the tension in the thread (a) R 2 / T

Solution : (d)

(c) 0.001 N/m

Force on wire due to surface tension F  T  2l

T 

Problem 4.

(b) 0.2 N/m

(b) R 2 T

(c) 2RT

(d) 2RT

Suppose tension in thread is F, then for small part l of thread

l  R and 2F sin / 2  2Tl  2TR

F cos /2  /2 F

TR  TR    2TR (sin  / 2   / 2) sin  / 2  / 2

F 

l  /2 /2

F sin/2

Problem 5.

F sin/2

A liquid is filled into a tube with semi-elliptical cross-section as shown in the figure. The ratio of the surface tension forces on the curved part and the plane part of the tube in vertical position will be (a)

 (a  b )

b

4b

(b)

2 a b

(c)

a 4b

(d) Solution : (a)

2 × T × l F cos /2  /2 F

a

 (a  b ) 4b

From the figure Curved part = semi perimeter 

 (a  b ) 2

and the plane part = minor axis = 2b

2b

a

 (a+b) 2

genius PHYSICS Surface Tension 53  Force on curved part = T 

 (a  b ) 2

and force on plane part = T  2b  Ratio 

Problem 6.

 (a  b ) 4b

A liquid film is formed over a frame ABCD as shown in figure. Wire CD can slide without friction. The mass to be hung from CD to keep it in equilibrium is (a)

Tl g

(b)

2 Tl g

A

B

D

C

Liquid film

l

g (c) 2 Tl

(d) T  l Solution : (b)

Weight of the body hung from wire (mg) = upward force due to surface tension (2Tl )  m 

2Tl g

10.8 Surface Energy. The molecules on the liquid surface experience net downward force. So to bring a molecule from the interior of the liquid to the free surface, some work is required to be done against the intermolecular force of attraction, which will be stored as potential energy of the molecule on the surface. The potential energy of surface molecules per unit area of the surface is called surface energy. Unit : Joule/m2 (S.I.) erg/cm2 (C.G.S.) Dimension : [MT–2] If a rectangular wire frame ABCD, equipped with a sliding wire LM dipped in soap solution, a film is formed over the frame. Due to the surface tension, the film will have a tendency to shrink and thereby, the sliding wire LM will be pulled in inward direction. However, the sliding C L L' D wire can be held in this position under a force F, which is equal and opposite to the force acting on the sliding wire LM all along its length due F l T × 2l to surface tension in the soap film. If T is the force due to surface tension per unit length, then F = T  2l

B

M M' A x

Here, l is length of the sliding wire LM. The length of the sliding wire has been taken as 2l for the reason that the film has got two free surfaces. Suppose that the sliding wire LM is moved through a small distance x, so as to take the position L' M' . In this process, area of the film increases by 2l  x (on the two sides) and to do so, the work done is given by W = F  x = (T  2l)  x = T  (2lx) = T  A  W = T  A

[A = Total increase in area of the film from both the sides]

If temperature of the film remains constant in this process, this work done is stored in the film as its surface energy. From the above expression T 

W or T = W A

[If A = 1]

genius PHYSICS 54 Surface Tension i.e. surface tension may be defined as the amount of work done in increasing the area of the liquid surface by unity against the force of surface tension at constant temperature.

10.9 Work Done in Blowing a Liquid Drop or Soap Bubble. (1) If the initial radius of liquid drop is r1 and final radius of liquid drop is r2 then W = T  Increment in surface area W = T  4 [r22  r12 ]

[drop has only one free surface]

(2) In case of soap bubble W = T  8 [r22  r12 ]

[Bubble has two free surfaces]

10.10 Splitting of Bigger Drop. When a drop of radius R splits into n smaller drops, (each of radius r) then surface area of liquid increases. Hence the work is to be done against surface tension. Since the volume of liquid remains constant therefore

4 4 R 3  n r 3  R 3  nr 3 3 3

Work done = T  A = T [Total final surface area of n drops – surface area of big drop] = T [n 4r 2  4R 2 ]

Various formulae of work done 4T[nr 2  R 2 ]

4Tr 2n 2 / 3 [n1 / 3  1]

4R 2 T [n1 / 3  1]

R

1 1  4TR 3    r R 

If the work is not done by an external source then internal energy of liquid decreases, subsequently temperature decreases. This is the reason why spraying causes cooling. By conservation of energy, Loss in thermal energy = work done against surface tension JQ = W 1 1  JmS   4TR 3    r R 

 

J

4 1 1  R 3 d S   4R 3 T    3 r R 

 Decrease in temperature

 

3T JSd

[As m = V  d =

4  R3 d ] 3

1 1  r  R   

where J = mechanical equivalent of heat, S = specific heat of liquid, d = density of liquid.

10.11 Formation of Bigger Drop. If n small drops of radius r coalesce to form a big drop of radius R then surface area of the liquid decreases. Amount of surface energy released = Initial surface energy – final surface energy E  n4r 2 T  4R 2 T R

Various formulae of released energy 4T [nr 2  R 2 ]

4R 2T (n1 / 3  1)

4Tr 2n 2 / 3 (n1 / 3  1)

1 1  4TR 3    r R 

genius PHYSICS Surface Tension 55

(i) If this released energy is absorbed by a big drop, its temperature increases and rise in temperature can 3 T 1 1   be given by   JSd  r R  (ii) If this released energy is converted into kinetic energy of a big drop without dissipation then by the law of conservation of energy. 1 1 1  mv 2  4R 3 T    2 r R 



v

6T d



1 4  1 1  R 3 d v 2  4R 3 T     2 3  r R 



v2 

6 T 1 1   d  r R 

1 1     r R

Sample problems based on Surface energy Problem 7.

Two small drops of mercury, each of radius R, coalesce to form a single large drop. The ratio of the total surface energies before and after the change is [AIIMS 2003] (a) 1 : 21 / 3

Solution : (b)

(b) 21 / 3 : 1

As R  n1 / 3 r  2 1 / 3 r  R 2  2 2 / 3 r 2   r2 Initial surface energy 2(4 r 2 T )   2   R2 Final surface energy (4R 2 T ) 

Problem 8.

(c) 2 : 1

r

2

R2

(d) 1 : 2

 22 / 3

   2  2 2 / 3 = 21/3  

Radius of a soap bubble is increased from R to 2R work done in this process in terms of surface tension is [CPMT 1991; RPET 2001; BHU 2003]

(a) 24R 2 S



(b) 48R 2 S

  8S [(2 R)

(c) 12R 2 S

(d) 36R 2 S

Solution : (a)

W  8T R 22  R12

Problem 9.

The work done in blowing a soap bubble of 10cm radius is (surface tension of the soap solution is

2

 (R)2 ]  24R 2 S

3 N /m ) 100 [MP PMT 1995; MH CET 2002]

(a) 75.36  10 4 J Solution : (a)

(b) 37.68  10 4 J

W  8R 2 T  8 (10  10  2 ) 2

(c) 150 .72  10 4 J

(d) 75.36 J

3  75 . 36  10  4 J 100

Problem 10. A drop of mercury of radius 2mm is split into 8 identical droplets. Find the increase in surface energy. (Surface tension of mercury is 0.465 J/m2) (a) 23.4J Solution : (a)

(b) 18.5J

(c) 26.8J

(d) 16.8J

Increase in surface energy  4R 2 T (n 1 / 3  1)  4 (2  10 3 )2 (0 .465 )(8 1 / 3  1) = 23 .4  10 6 J = 23.4 J

Problem 11. The work done in increasing the size of a soap film from 10cm  6cm to 10cm  11cm is 3  10 4 J . The surface tension of the film is (a) 1 .5  10 2 Nm 1 Solution : (b)

[MP PET 1999; MP PMT 2000; AIIMS 2000; JIPMER 2001, 02]

(b) 3 .0  10 2 Nm 1 (c) 6.0  10 2 Nm 1

A1  10  6  60 cm 2  60  10 4 m 2 , A 2  10  11  110 cm 2  110  10 4 m 2

As the soap film has two free surfaces W  T  2 A

(d) 11 .0  10 2 Nm 1

genius PHYSICS 56 Surface Tension  W  T  2  ( A2  A1 )  T 

W 2  50  10

4



3  10 4 2  50  10  4

 3  10  2 N/m

Problem 12. A film of water is formed between two straight parallel wires of length 10cm each separated by 0.5cm. If their separation is increased by 1 mm while still maintaining their parallelism, how much work will have to be done (Surface tension of water  7.2  10 2 N / m )

[Roorkee 1986; MP PET

2001]

(a) 7.22  10 6 J Solution : (b)

(b) 1.44  10 5 J

(d) 5.76  10 5 J

(c) 2.88  10 5 J

As film have two free surfaces W  T  2A

W  T  2l  x F

l

 7.2  10 2  2  0.1  1  10 3

 1.44  10 5 J x

Problem 13. If the work done in blowing a bubble of volume V is W, then the work done in blowing the bubble of volume 2V from the same soap solution will be (a) W/2

(b)

[MP PET 1989]

(c)

2W

1/3

Solution : (d)

As volume of the bubble V 

4  3  R 3  R    3  4 

3

2W

 3  V1 / 3  R 2     4 

(d)

3

4 W

2/3

V 2/3  R2  V 2/3

Work done in blowing a soap bubble W  8R 2 T  W  R 2  V 2 / 3 V W  2   2 W1  V1

  

2/3

 2V     V 

2/3

 (2) 2 / 3  (4 )1 / 3  W2 

3

4W

Problem 14. Several spherical drops of a liquid of radius r coalesce to form a single drop of radius R. If T is surface tension and V is volume under consideration, then the release of energy is

1 1  (a) 3 VT    r R Solution : (b)

1 1  (b) 3 VT    r R

1 1  (c) VT    r R

1  1 (d) VT  2  2  R  r

1 1  1 1  4  1 1  Energy released = 4TR 3     3 R 3 T     3 VT    r R  r R  3  r R 

10.12 Excess Pressure. Due to the property of surface tension a drop or bubble tries to contract and so compresses the matter enclosed. This in turn increases the internal pressure which prevents further contraction and equilibrium is achieved. So in equilibrium the pressure inside a bubble or drop is greater than outside and the difference of pressure between two sides of the liquid surface is called excess pressure. In case of a drop excess pressure is provided by hydrostatic pressure of the liquid within the drop while in case of bubble the gauge pressure of the gas confined in the bubble provides it.

genius PHYSICS Surface Tension 57

Excess pressure in different cases is given in the following table : Plane surface

Concave surface P

P=0

P = 0

P 

Convex surface

P  P

2T R

Drop

P  P

Bubble in air

P

P 

4T R

P 

P

P 

2T P   hdg R

Liquid surface of unequal radii

P

Note : 

 1 1  P  T     R1 R 2 

2T R

Cylindrical liquid surface

R

P

2T R

Bubble in liquid

Bubble at depth h below the free surface of liquid of density d

h

2T R

T R

Liquid film of unequal radii

P

 1 1  P  2T     R1 R 2 

Excess pressure is inversely proportional to the radius

of bubble (or drop), i.e., pressure inside a smaller bubble (or drop) is higher than inside a larger bubble (or drop). This is why when two bubbles of different sizes are put in communication with each other, the air will rush from smaller to larger bubble, so that the smaller will shrink while the larger will expand till the smaller bubble reduces to droplet.

genius PHYSICS 58 Surface Tension

Sample problems based on Excess pressure Problem 15. The pressure inside a small air bubble of radius 0.1mm situated just below the surface of water will be equal to (Take surface tension of water 70  10 3 Nm 1 and atmospheric pressure  1.013  10 5 Nm  2 ) [AMU (Med.) 2002]

(b) 1.027  10 3 Pa

(a) 2.054  10 3 Pa Solution : (c)

(c) 1.027  10 5 Pa

Pressure inside a bubble when it is in a liquid  Po 

(d) 2.054  10 5 Pa

2T 70  10 3  1 .013  10 5  2   1.027  10 5 Pa. 3 R 0 .1  10

Problem 16. If the radius of a soap bubble is four times that of another, then the ratio of their excess pressures will be [AIIMS 2000]

(a) 1 : 4 Solution : (a)

(b) 4 : 1

(c) 16 : 1

Excess pressure inside a soap bubble P 

(d) 1 : 16

P1 r 4T   2 1:4 r P2 r1

Problem 17. Pressure inside two soap bubbles are 1.01 and 1.02 atmospheres. Ratio between their volumes is [MP PMT 1991]

(b) (102)3 : (101)3

(a) 102 : 101 Solution : (c)

(c) 8 : 1

Excess pressure P  Pin  Pout  1.01atm  1atm and volume of air bubble V 



V1  P2  V2  P1

3

(d) 2 : 1

 0.01atm and similarly P2  0.02atm

4 3 1 r V  r 3  3 (P)3

[as P 

1 1 or r  ] r P

 8  0 . 02  2        1  0 . 01  1  3

3

Problem 18. The excess pressure inside an air bubble of radius r just below the surface of water is P1. The excess pressure inside a drop of the same radius just outside the surface is P2. If T is surface tension then (a) P1  2P2 Solution : (b)

(b) P1  P2

(c) P2  2P1

Excess pressure inside a bubble just below the surface of water P1  and excess pressure inside a drop P2 

2T r

(d) P2  0, P1  0 2T r

P1  P2

10.13 Shape of Liquid Meniscus. We know that a liquid assumes the shape of the vessel in which it is contained i.e. it can not oppose permanently any force that tries to change its shape. As the effect of force is zero in a direction perpendicular to it, the free surface of liquid at rest adjusts itself at right angles to the resultant force. When a capillary tube is dipped in a liquid, the liquid surface becomes curved near the point of contact. This curved surface is due to the resultant of two forces i.e. the force of cohesion and the force of adhesion. The curved surface of the liquid is called meniscus of the liquid. If liquid molecule A is in contact with solid (i.e. wall of capillary tube) then forces acting on molecule A are

genius PHYSICS Surface Tension 59

(i) Force of adhesion Fa (acts outwards at right angle to the wall of the tube). (ii) Force of cohesion Fc (acts at an angle 45o to the vertical). Resultant force FN depends upon the value of Fa and Fc. If resultant force FN make an angle  with Fa. Then tan  

Fc sin 135 o Fa  Fc cos 135

o



Fc 2 Fa  Fc

By knowing the direction of resultant force we can find out the shape of meniscus because the free surface of the liquid adjust itself at right angle to this resultant force. If Fc 

Fc 

2 Fa

tan = 

  = 90o

i.e. the resultant force acts vertically downwards. Hence the liquid meniscus must be horizontal. Fa

Fc 

2 Fa

2 Fa

tan  = positive   is acute angle

tan  = negative   is obtuse angle

i.e. the resultant force directed outside the liquid. Hence the liquid meniscus must be concave upward.

i.e. the resultant force directed inside the liquid. Hence the liquid meniscus must be convex upward.

Fa

A

A

Fa



 Fc

FN

FN

45°

A



Fc

45 °

Fc

FN

Example: Pure water in silver coated capillary tube.

Example: Water in glass capillary tube.

Example: Mercury in glass capillary tube.

10.14 Angle of Contact. Angle of contact between a liquid and a solid is defined as the angle enclosed between the tangents to the liquid surface and the solid surface inside the liquid, both the tangents being drawn at the point of contact of the liquid with the solid.  < 90o Fa 

 = 90o

Fc 2



Fa 

 > 90o

Fc



2

Fa 

Fc



2

concave meniscus.

plane meniscus.

convex meniscus.

Liquid wets the solid surface

Liquid does not wet the solid surface.

Liquid does not wet the solid surface.

genius PHYSICS 60 Surface Tension

Important points (i) Its value lies between 0o and 180o

  0 o for pure water and glass,   8 o for tap water and glass,   90 o for water and silver   138 o for mercury and glass,   160 o for water and chromium (ii) It is particular for a given pair of liquid and solid. Thus the angle of contact changes with the pair of solid and liquid. (iii) It does not depends upon the inclination of the solid in the liquid. (iv) On increasing the temperature, angle of contact decreases. (v) Soluble impurities increases the angle of contact. (vi) Partially soluble impurities decreases the angle of contact.

10.15 Capillarity. If a tube of very narrow bore (called capillary) is dipped in a liquid, it is found that the liquid in the capillary either ascends or descends relative to the surrounding liquid. This phenomenon is called capillarity. The root cause of capillarity is the difference in pressures on two sides of (concave and convex) curved surface of liquid. Examples of capillarity : (i) Ink rises in the fine pores of blotting paper leaving the paper dry. (ii) A towel soaks water. (iii) Oil rises in the long narrow spaces between the threads of a wick. (iv) Wood swells in rainy season due to rise of moisture from air in the pores. (v) Ploughing of fields is essential for preserving moisture in the soil. (vi) Sand is drier soil than clay. This is because holes between the sand particles are not so fine as compared to that of clay, to draw up water by capillary action.

10.16 Ascent Formula. When one end of capillary tube of radius r is immersed into a liquid of density d which wets the sides of the capillary tube (water and capillary tube of glass), the shape of the liquid meniscus in the tube becomes concave upwards. R = radius of curvature of liquid meniscus. T = surface tension of liquid P = atmospheric pressure

2T Pressure at point A = P, Pressure at point B = P  R

R  r

A A(P) C(P)

B

h

y



B D(P) E(P ) (P – 2T/R)

Pressure at points C and D just above and below the plane surface of liquid in the vessel is also P (atmospheric pressure). The points B and D are in the same horizontal plane in the liquid but the pressure at these points is different. In order to maintain the equilibrium the liquid level rises in the capillary tube upto height h.

genius PHYSICS Surface Tension 61

Pressure due to liquid column = pressure difference due to surface tension 

hdg 



h

2T R 2T 2 T cos   Rdg rdg

r    As R  cos    

Important points (i) The capillary rise depends on the nature of liquid and solid both i.e. on T, d,  and R. (ii) Capillary action for various liquid-solid pair. Meniscus

Angle of contact

Level

 < 90o

Rises

Plane

 = 90o

No rise no fall

Convex

 > 90o

Fall

Glass

Concave Water Silver

Water Glass

Mercury

(iii) For a given liquid and solid at a given place

h

1 r

[As T, , d and g are constant]

i.e. lesser the radius of capillary greater will be the rise and vice-versa. This is called Jurin’s law. (iv) If the weight of the liquid contained in the meniscus is taken into consideration then more accurate ascent formula is given by h

2T cos  r  rdg 3

r

(v) In case of capillary of insufficient length, i.e., L < h, the liquid will neither overflow from the upper end like a fountain nor will it tickle along the vertical sides of the tube. The liquid after reaching the upper end will increase the radius of its meniscus without changing nature such that : hr = Lr  L < h

r' h

L

r'>r

(vi) If a capillary tube is dipped into a liquid and tilted at an angle  from vertical, then the vertical height of liquid column remains same whereas the length of liquid column (l) in the capillary tube increases.

R h

Water



l

h

genius PHYSICS 62 Surface Tension

h cos

h = l cos  or l 

(vii) It is important to note that in equilibrium the height h is independent of the shape of capillary if the radius of meniscus remains the same. That is why the vertical height h of a liquid column in capillaries of different shapes and sizes will be same if the radius of meniscus remains the same.

h

Sample problems based on Capillarity Problem 19. Water rises to a height of 10cm in a capillary tube and mercury falls to a depth of 3.5cm in the same capillary tube. If the density of mercury is 13.6 gm/cc and its angle of contact is 135o and density of water is 1 gm/cc and its angle of contact is 0 o , then the ratio of surface tensions of the two liquids is (cos 135 o  0.7) [MP PMT 1988; EAMCET (Med.) 2003]

(a) 1 : 14 Solution : (b)

h



(b) 5 : 34

2T cos  rdg



hW T cos  W d Hg  W h Hg T Hg cos  Hg d W

(c) 1 : 5

(d) 5 : 27

[as r and g are constants]

T T 10 cos 0 o 13 . 6 10  0 .7 20 5  W .  W    3 .5 T Hg cos 135 1 T Hg 3 .5  13 .6 136 34

Problem 20. Water rises in a vertical capillary tube upto a height of 2.0 cm. If the tube is inclined at an angle of 60 o with the vertical, then upto what length the water will rise in the tube (a) 2.0 cm

(b) 4.0 cm

(c)

4

cm

[UPSEAT 2002]

(d) 2 2 cm

3 Solution : (b)

The height upto which water will rise l 

2cm h   4 cm . [h = vertical height,  = angle with cos 60 cos 

vertical]

Problem 21. Two capillary tubes of same diameter are kept vertically one each in two liquids whose relative densities are 0.8 and 0.6 and surface tensions are 60 and 50 dyne/cm respectively. Ratio of heights of liquids in the two tubes

Solution : (d)

h1 is h2

[MP PMT 2002]

(a)

10 9

h

2T cos  rdg



h1  T1   d 2    h 2  T2   d 1

(b)

3 10

(c)

10 3

(d)

9 10

[If diameter of capillaries are same and taking value of  same for both liquids]   60   0 .6   36  9       .      50   0 .8   40  10

Problem 22. A capillary tube of radius R is immersed in water and water rises in it to a height H. Mass of water in the capillary tube is M. If the radius of the tube is doubled, mass of water that will rise in the capillary tube will now be [RPMT 1997; RPET 1999; CPMT 2002] (a) M Solution : (b)

(b) 2M

(c) M/2

Mass of the liquid in capillary tube M = V = (r2h)

 M  r 2h  r

(d) 4M [As h 

1 ] r

genius PHYSICS Surface Tension 63 So if radius of the tube is doubled, mass of water will becomes 2M, which will rise in capillary tube.

Problem 23. Water rises to a height h in a capillary at the surface of earth. On the surface of the moon the height of water column in the same capillary will be (a) 6h Solution : (a)

h

(b)

2T cos  rdg

h 

[MP PMT 2001]

1 h 6

(c) h

1 g

(d) Zero

[If other quantities remains constant]

hmoon g  earth = 6  hmoon  6 h h earth g moon

[As gearth= 6gmoon]

Problem 24. Water rises upto a height h in a capillary on the surface of earth in stationary condition. Value of h increases if this tube is taken

Solution : (d)

(a) On sun

(b) On poles

(c) In a lift going upward with acceleration

(d) In a lift going downward with acceleration

h

1 . In a lift going downward with acceleration (a), the effective acceleration decreases. So h increases. g

Problem 25. If the surface tension of water is 0.06 N/m, then the capillary rise in a tube of diameter 1mm is (  0 o ) [AFMC 1998]

(a) 1.22 cm Solution : (b)

h h

(b) 2.44 cm

(c) 3.12 cm

(d) 3.86 cm

1 2T cos  , [ =0, r  mm  0 . 5  10  3 m , T  0.06 N / m , d = 10 3 kg / m 3 , g = 9.8 m/s2 ] 2 rdg

2  0.06  cos  0.5  10 3  10 3  9.8

 0.0244 m  2.44 cm

Problem 26. Two capillaries made of same material but of different radii are dipped in a liquid. The rise of liquid in one capillary is 2.2cm and that in the other is 6.6cm. The ratio of their radii is (a) 9 : 1 Solution : (c)

As h 

(b) 1 : 9

1 r



(c) 3 : 1

[MP PET 1990]

(d) 1 : 3

r h h1 r2 6.6 3 or 1  2    h 2 r1 r2 h1 2 .2 1

Problem 27. The lower end of a capillary tube is at a depth of 12cm and the water rises 3cm in it. The mouth pressure required to blow an air bubble at the lower end will be X cm of water column where X is (a) 3 Solution : (d)

(b) 9

(c) 12

[CPMT 1989]

(d) 15

The lower end of capillary tube is at a depth of 12 + 3 = 15 cm from the free surface of water in capillary tube. So, the pressure required = 15 cm of water column.

Problem 28. The lower end of a capillary tube of radius r is placed vertically in water. Then with the rise of water in the capillary, heat evolved is (a)  Solution : (b)

 2r 2 h 2 J

dg

(b) 

r 2h 2 dg 2J

(c) 

r 2 h 2dg 2J

(d) 

r 2 h 2dg

When the tube is placed vertically in water, water rises through height h given by h  Upward force  2r  T cos Work done by this force in raising water column through height h is given by

 rhdg W  (2rT cos  )h  (2rh cos  )T  (2rh cos  )   2 cos 

 2 2   r h dg 

J

2T cos  rdg

genius PHYSICS 64 Surface Tension However, the increase in potential energy E p of the raised water column  mg

h 2

where m is the mass of the raised column of water m  r 2 hd 2 2  hg  r h dg So, E P  (r 2 hd )   2  2 

Further, W  E p 

r 2 h 2 dg 2

The part (W  E P ) is used in doing work against viscous forces and frictional forces between water and glass surface and appears as heat. So heat released =

W  E p J



r 2 h 2 dg 2J

Problem 29. Water rises in a capillary tube to a certain height such that the upward force due to surface tension is balanced by 75  10 4 N force due to the weight of the liquid. If the surface tension of water is

6  10 2 N / m , the inner circumference of the capillary must be (a) 1 .25  10 2 m Solution : (d)

(b) 0 .50  10 2 m

(c) 6 .5  10 2 m

(d) 12 .5  10 2 m

Weight of liquid = upward force due to surface tension

75  10 4  2rT Circumference 2r 

75  10 4 75  10 4   0.125 = 12.5  10 2 m T 6  10  2

10.17 Shape of Drops. Whether the liquid will be in equilibrium in the form of a drop or it will spread out; depends on the relative strength of the force due to surface tension at the three interfaces. TLA

TLA = surface tension at liquid-air interface, TSA = surface tension at solid-air interface. TSL = surface tension at solid-liquid interface,  = angle of contact between the liquid and solid.

Liquid TSL

Solid

Air TSA  O

For the equilibrium of molecule TSL + TLA cos = TSA or cos  

TSA  TSL TLA

Air

…..(i) TSL Solid

 O

TLA TSA

Special Cases TSA > TSL, cos is positive i.e. 0 o    90 o . This condition is fulfilled when the molecules of liquid are strongly attracted to that of solid. Example : (i) Water on glass. (ii) Kerosene oil on any surface. TSA < TSL, cos is negative i.e. 90 o    180 o . This condition is fulfilled when the molecules of the liquid are strongly attracted to themselves and relatively weakly to that of solid. Example : (i) Mercury on glass surface. (ii) Water on lotus leaf (or a waxy or oily surface)

genius PHYSICS Surface Tension 65 (TSL + TLA cos) > TSA In this condition, the molecule of liquid will not be in equilibrium and experience a net force at the interface. As a result, the liquid spreads. Example : (i) Water on a clean glass plate.

10.18 Useful Facts and Formulae. (1) Formation of double bubble : If r1 and r2 are the radii of smaller and larger bubble and P0 is the 4T 4T atmospheric pressure, then the pressure inside them will be P1  P0  and P2  P0  . r1 r2 Now as r1  r2  P1  P2 1 1  So for interface P  P1  P2  4 T     r1 r2 

r2

…..(i)

r

P

r1 P1

P2

As excess pressure acts from concave to convex side, the interface will be concave towards the smaller bubble and convex towards larger bubble and if r is the radius of interface.

P  From (i) and (ii)

4T r

…..(ii)

1 1 1   r r1 r2

 Radius of the interface r 

r1 r2 r2  r1

(2) Formation of a single bubble (i) Under isothermal condition two soap bubble of radii ‘a’ and ‘b’ coalesce to form a single bubble of radius ‘c’. If the external pressure is P0 then pressure inside bubbles 4T  4T  4T     Pa   P0   , Pb   P0   and Pc   P0   a  b  c    

and volume of the bubbles

Va 

b

 a   b   c  Pa Va  Pb Vb  Pc Vc

Pa Va  Pb Vb  Pc Vc

RT a

…..(i)

RT b

RT c



4T  4 3   4T  4 3   4T  4 3    P0  a   3 a    P0  b   3 b    P0  c   3 c           4 T (a 2  b 2  c 2 )  P0 (c 3  a 3  b 3 )

 Surface tension of the liquid T 

PV    As PV  RT , i.e .,   RT   

[As temperature is constant, i.e., Ta  Tb  Tc ]

Substituting the value of pressure and volume 

c

4 3 4 4 a , Vb  b 3 , Vc  c 3 3 3 3

Now as mass is conserved 

a

P0 (c 3  a 3  b 3 ) 4 (a 2  b 2  c 2 )

genius PHYSICS 66 Surface Tension (ii) If two bubble coalesce in vacuum then by substituting P0  0 in the above expression we get

a2  b 2  c 2  0  c 2  a2  b 2 Radius of new bubble  c  a 2  b 2

or can be expressed as r  r12  r22 .

(3) The difference of levels of liquid column in two limbs of u-tube of unequal radii r1 and r2 is 2 T cos  h  h1  h 2  dg

h

1 1      r1 r2 

h1 h2

(4) A large force (F) is required to draw apart normally two glass plate enclosing a thin water film because the thin water film formed between the two glass plates will have concave surface all around. Since on the concave side of a liquid surface, pressure is more, work will have to be done in drawing the plates apart.

F

2 AT where T= surface tension of water film, t= thickness of film, A = area of film. t

(5) When a soap bubble is charged, then its size increases due to outward force on the bubble. (6) The materials, which when coated on a surface and water does not enter through that surface are known as water proofing agents. For example wax etc. Water proofing agent increases the angle of contact. (7) Values of surface tension of some liquids. Liquid

Surface tension Newton/metre

Mercury

0.465

Water

0.075

Soap solution

0.030

Glycerine

0.063

Carbon tetrachloride

0.027

Ethyl alcohol

0.022

Sample problems (Miscellaneous) Problem 30. The radii of two soap bubbles are r1 and r2. In isothermal conditions, two meet together in vacuum. Then the radius of the resultant bubble is given by (a) R  (r1  r2 ) / 2 Solution : (c)

(b) R  r1 (r1r2  r2 )

[RPET 1999; MP PMT 2001; EAMCET 2003]

(c) R 2  r12  r22

(d) R  r1  r2

Under isothermal condition surface energy remain constant  8r12 T  8r22 T  8R 2 T  R 2  r12  r22

Problem 31. Two soap bubbles of radii r1 and r2 equal to 4cm and 5cm are touching each other over a common surface S 1S 2 (shown in figure). Its radius will be

[MP PMT 2002] S1

(a) 4 cm (b) 20 cm (c) 5 cm (d) 4.5 cm

5 cm

4 cm S2

genius PHYSICS Surface Tension 67 Solution : (b)

r2 r1 54   20 cm 54 r2  r1

Radius of curvature of common surface of double bubble r 

Problem 32. An air bubble in a water tank rises from the bottom to the top. Which of the following statements are true [Roorkee 2000]

(a) Bubble rises upwards because pressure at the bottom is less than that at the top (b) Bubble rises upwards because pressure at the bottom is greater than that at the top (c) As the bubble rises, its size increases (d) As the bubble rises, its size decreases Solution : (b, c)

Problem 33. The radii of two soap bubbles are R1 and R2 respectively. The ratio of masses of air in them will be R3 (a) 13 R2

Solution : (c)

 P (c)   P 

R3 (b) 23 R1

4T R1 4T R2

  3  R1  R3  2 

 P (d)   P 

4T R2 4T R1

  3  R2  R3  1 

From PV = RT.

At a given temperature, the ratio masses of air

1 2

  P  P1 V1    P2 V2   P  

 4T  4  R 3  P  1 R1  3   4  4 T  R 3   P  2 R 2  3 

  3  R1 . 4 T  R 23  R2  4T R1

Problem 34. On dipping one end of a capillary in liquid and inclining the capillary at an angles 30 o and 60 o with the vertical, the lengths of liquid columns in it are found to be l1 and l2 respectively. The ratio of l1 and l2 is (a) 1 : 3 Solution : (a)

l1 

(b) 1 : 2

(c)

2 :1

(d)

3 :1

l cos  2 cos 60 o 1/2 h h 1    and l 2  = 1: 3 l2 cos  1 cos  1 cos  2 cos 30 o 3 /2

Problem 35. A drop of water of volume V is pressed between the two glass plates so as to spread to an area A. If T is the surface tension, the normal force required to separate the glass plates is (a)

Solution : (b)

TA 2 V

(b)

2 TA 2 V

Force required to separate the glass plates F 

(c)

4 TA 2 V

(d)

TA 2 2V

2 AT A 2TA 2 2TA 2    . t A ( A  t) V

Problems based on Cohesive and adhesive force 1.

Mercury does not wet glass, wood or iron because (a) Cohesive force is less than adhesive force

[MP PMT 1995; MP PET 1997]

(b) Cohesive force is greater than adhesive force

genius PHYSICS 68 Surface Tension (c) Angle of contact is less than 90 o

2.

The force of cohesion is (a) Maximum in solids

3.

[CPMT 1996]

(b) Maximum in liquid

(c) Same in different matters

(d)

Maximum in gases

What enables us to write on the black board with chalk (a) Gravity

4.

(d) Cohesive force is equal to adhesive force

(b) Cohesion

(c) Adhesion

(d) None of the above

Intermolecular forces decrease rapidly as the distance between the molecules increases and do so much more (a) Slowly than demanded by the inverse square law of the distance (b) Rapidly than anticipated through the inverse square law of the distance (c) According to inverse square law (d) It actually remains the same for all the distances

Problems based on Surface tension 5.

The spherical shape of rain-drop is due to [CPMT 1976, 90; CPMT 2001; NCERT 1982; AIIMS 1998; MHCET 2000; DCE 1999; AFMC 1999, 2001]

(a) Density of the liquid

6.

(a) 4 C

(b) 25 C

10.

(d) 750 dynes

(c) 750 dynes

(b) 4TL

(c) 8TL

(d) 10TL

Ball pen and fountain pen depend respectively upon the principle of (a) Surface tension and viscosity

(b) Surface tension and gravity

(c) Gravitation and surface tension

(d) Surface tension and surface tension

Which graph represents the variation of surface tension with temperature over small temperature ranges for water (b)

(c)

S.T.

(d)

S.T.

Temp

Temp

S.T.

Temp

Temp

The material of a wire has a density of 1.4 g per cm3. If it is not wetted by a liquid of surface tension 44 dyne per cm, then the maximum radius of the wire which can float on the surface of the liquid is (a)

1 cm 7

(b) 0.7 cm

(c)

10 cm 14

(d)

10 cm 28

A water drop of 0.05cm3 is squeezed between two glass plates and spreads into area of 40cm2. If the surface tension of water is 70 dyne/cm then the normal force required to separate the glass plates from each other will be (a) 90 N

13.

(d) 75 C

A square frame of side L is dipped in a liquid. On taking it out, a membrane is formed. If the surface tension of the liquid is T, the force acting on the frame will be [MP PMT 1990]

S.T.

12.

[MP PMT/PET 1998] o

o

(b) 60 dynes

(a)

11.

(d) Gravity

Force necessary to pull a circular plate of 5cm radius from water surface for which surface tension is 75 dynes/cm, is [MP PMT 1991]

(a) 2TL

9.

(c) 50 C

o

(a) 30 dynes

8.

(c) Atmospheric pressure

At which of the following temperatures, the value of surface tension of water is minimum o

7.

(b) Surface tension

(b) 45 N

(c) 22.5 N

(d) 450 N

The main difference between a stretched membrane and the liquid surface is (a) The liquid surface has a tendency to contract but the stretched membrane does not (b) The surface tension does not depend on area but on the tension of the stretched membrane does (c) The surface tension increases with increases in area (d) Surface tension increases irregularly with temperature

14.

On bisecting a soap bubble along a diameter, the force due to surface tension on any of its half part will be

genius PHYSICS Surface Tension 69 (a) 4RT

15.

16.

(b)

4 R T

(c)

T 4 R

(d)

2T R

The addition of soap changes the surface tension of water to  1 and that of sugar changes it to  2 . Then (a)  1   2

(b)  1   2

(c)  1   2

(d) It is not possible to predict the above

A hollow disc of aluminum whose external and internal radii are R and r respectively, is floating on the surface of a liquid whose surface tension is T. The maximum weight of disc can be (a) 2 (R + r) T

(b) 2 (R – r) T

(c) 4 (R + r) T

(d) 4 (R – r) T

Problems based on Surface energy 17.

8000 identical water drops are combined to form a big drop. Then the ratio of the final surface energy to the initial surface energy of all the drops together is (a) 1 : 10

18.

(c) 1 : 20

(d) 1 : 25

8 mercury drops coalesce to form one mercury drop, the energy changes by a factor of (a) 1

19.

(b) 1 : 15

(b) 2

(c) 4

[DCE 2000]

(d) 6

Which of the following statements are true in case when two water drops coalesce and make a bigger drop

[Roorkee 1999]

(a) Energy is released (b) Energy is absorbed (c) The surface area of the bigger drop is greater than the sum of the surface areas of both the drops (d) The surface area of the bigger drop is smaller than the sum of the surface areas of both the drops

20.

An oil drop of radius 1cm is sprayed into 1000 small equal drops of same radius. If the surface tension of oil drop is 50 dyne/cm then the work done is [RPET 1990] (a) 18 ergs

21.

(b) 2W

(c) 4W

(d) 2

1 W 3

(b) N 2 / 3

(c)

N1 / 3

(d) N 0

The work done in increasing the volume of a soap bubble of radius R and surface tension T by 700% will be (a) 8R 2T

24.

(d) 18000 ergs

A liquid drop of radius R is broken up into N small droplets. The work done is proportional to (a) N

23.

(c) 1800 ergs

If work W is done in blowing a bubble of radius R from a soap solution, then the work done in blowing a bubble of radius 2R from the same solution is [MP PET 1990] (a) W/2

22.

(b) 180 ergs

(b) 24 R 2T

(c)

48R 2T

(d) 8R 2T 2 / 3

1000 drops of water all of same size join together to form a single drop and the energy released raises the temperature of the drop. Given that T is the surface tension of water, r the radius of each small drop,  the density of liquid, J the mechanical equivalent of heat. What is the rise in the temperature (a) T/Jr

(b) 10T/Jr

(c) 100T/Jr

(d) None of these

Problems based on Excess pressure 25.

Two bubbles A and B (A > B) are joined through a narrow tube. Then

[UPSEAT

2001;

Kerala

(Med.)

2002]

(a) The size of A will increase (c) The size of B will increase until the pressure equals

26.

(b)

The size of B will increase

(d) None of these

Excess pressure of one soap bubble is four times more than the other. Then the ratio of volume of first bubble to another one is [CPMT 1997; MH CET 2000]

genius PHYSICS 70 Surface Tension (a) 1 : 64

27.

(b) 1 : 4

(b) 68.66 dyne/cm

(c) 137 dyne/cm

(d) 150 dyne/cm

An air bubble of radius r in water is at a depth h below the water surface at some instant. If P is atmospheric pressure, d and T are density and surface tension of water respectively, the pressure inside the bubble will be [Roorkee 1990] (a) P  h dg 

29.

(d) 1 : 2

The pressure of air in a soap bubble of 0.7cm diameter is 8mm of water above the pressure outside. The surface tension of the soap solution is [MP PET 1991; MP PMT 1997] (a) 100 dyne/cm

28.

(c) 64 : 1

4T r

(b) P  h dg 

2T r

(c)

P  h dg 

2T r

(d) P  h dg 

4T r

A soap bubble is very slowly blown at the end of a glass tube by a mechanical pump which supplies a fixed volume of air every minute whatever the pressure against which it is pumping. The excess pressure P inside the bubble varies with time as shown by which graph P

P

(a)

P

(b)

P

(c)

(d)

t

t

t

t

Problems based on Angle of contact 30.

A liquid does not wet the sides of a solid, if the angle of contact is [MP PAT 1990; AFMC 1988, MNR 1998, KCET 1998, Haryana CEE 1998; RPMT 1999; 2003]

(a) Zero

31.

(b) Obtuse (More than 90o)

34.

36.

(c) Plane

(d) Uncertain

The angle of contact between glass and mercury is (b) 30 o

[MP PMT 1987]

(c) 90 o

(d) 135 o

When the temperature is increased the angle of contact of a liquid (a) Increases

(b) Decreases

(c) Remains the same

(d) First increases and then decreases

For those liquids which do not wet the solid surface, the ratio of cohesive force and adhesive force will be (a) Greater than

35.

[MP PET/PMT 1988]

(b) Concave

(a) 0 o

33.

(d) 90o0

The meniscus of mercury in the capillary tube is (a) Convex

32.

(c) Acute (Less than 90o)

1 2

(b) Greater than

2

(c) Lesser than

1 2

(d) Lesser than

2

The water proofing agent makes an angle of contact (a) From acute angle to obtuse angle

(b) From obtuse angle to acute angle

(c) From obtuse angle to right angle

(d) From acute angle to right angle

A glass plate is partly dipped vertically in the mercury and the angle of contact is measured. If the plate is inclined, then the angle of contact will (a) Increase

(b) Remain unchanged

(c) Increase or decrease

(d) Decrease

Problems based on Capillarity 37.

The surface tension for pure water in a capillary tube experiment is (a)

38.

g 2 hr

(b)

2 hr g

(c)

[MH CET 2002]

rg 2h

If capillary experiment is performed in vacuum then for a liquid there (a) It will rise (b) Will remain same (c) It will fall

(d)

hrg 2

(d) Rise to the top

genius PHYSICS Surface Tension 71 39.

A surface tension experiment with a capillary tube in water is repeated in an artificial satellite. Which is revolving around the earth, water will rise in the capillary tube upto a height of [Roorkee 1992] (a) 0.1 m (b) 0.2 m (c) 0.98 m (d) Full length of the capillary tube

40.

When a capillary is dipped in water, water rises to a height h. If the length of the capillary is made less than h, then [MP PAT 1990]

(a) The water will come out (c) The water will not rise

41.

A long cylindrical glass vessel has a small hole of radius ‘r’ at its bottom. The depth to which the vessel can be lowered vertically in the deep water bath (surface tension T) without any water entering inside is [MP PMT 1990] (a) 4T/rg

42.

(b) The water will not come out (d) The water will rise but less than height of capillary

(b) 3T/rg

(c) 2T/rg

(d) T/rg

Water rises to a height of 10cm in capillary tube and mercury falls to a depth of 3.112cm in the same capillary tube. If the density

of mercury is 13.6 and the angle of contact for mercury is 135 o , the ratio of surface tension of water and mercury is[MP PET/PMT 1988] (a) 1 : 0.15 (b) 1 : 3 (c) 1 : 6 (d) 1.5 : 1

43.

Water can rise to a height h in a capillary tube lowered vertically into water. If the height of tube above the surface of water be l and l < h, then water will rise in the capillary to a height (a) h (b) l (c) l – h (d) l + h

44.

The height upto which water will rise in a capillary tube will be

45.

(a) Maximum when water temperature is 4 o C

(b) Maximum when water temperature is 0 o C

(c) Minimum when water temperature is 4 o C

(d) Same at all temperatures

The exact expression for surface tension of liquid which rises up in the capillary tube is (a) T  rhdg / 2

(b) T  rhdg / 2 cos 

(c) T 

r(h  r / 3)dg 2

(d) T 

r(h  r / 3)dg 2 cos 

46.

If a wax coated capillary tube is dipped in water, then water in it will (a) Rise up (b) Depress (c) Sometimes rise and sometimes fall (d) Rise up and come out as a fountain

47.

Capillaries made from various materials but having the same bore are dipped in the same liquid, then (a) Liquid will not rise in any of them (b) Liquid will rise in all upto same height (c) Liquid will not rise in all upto same height (d) Liquid will rise in all and height of liquid columns will be inversely proportional to the density of material used

48.

A straight capillary tube is immersed in water and the water rises to 5cm. If the capillary is bent as shown in figure then the height of water column will be h

(a) 5cm (b) Less than 5cm (c) Greater than 5cm (d) 4 cos 

49.

Water rises in a capillary tube through a height h. If the tube is inclined to the liquid surface at 30 o , the liquid will rise in the tube upto its length equal to (a)

h 2

(b) h

(c) 2h

(d) 4h

Problems (Miscellaneous) 50.

If a water drop is kept between two glass plates, then its shape is

(a)

51.

(b)

(c)

(d) None of these

When two soap bubbles of radius r1 and r2 (r2  r1 ) coalesce, the radius of curvature of common surface is

[MP PMT 1996]

genius PHYSICS 72 Surface Tension (a) r2  r1

52.

(b)

r1r2 r2  r1

(d) r2  r1

(b) 600  ergs

(c) 1000  ergs

(d) 1200  ergs

(c)

(d)

In the above question, the radius of the bigger drop will be (a)

54.

(c)

Two soap bubbles of radius 1cm and 2cm coalesce to form a single drop under isothermal conditions. The total energy possessed by them if surface tension is 30 dyne cm–1, will be (a) 400  ergs

53.

r2  r1 r1r2

3 cm

(b)

5 cm

7 cm

8 cm

In a U-tube the radii of two columns are respectively r1 and r2 and if a liquid of density d filled in it has level difference of h then the surface tension of the liquid is (a) T 

hdg r2  r1

(b) T 

(r2  r1 )hdg 2

(c) T 

(r1  r2 )hdg 2

(d) T 

hdg (r1r2 ) 2 r2  r1

r1

r2 h

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

b

a

c

b

b

d

d

c

c

b

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

a

b

b

a

c

a

c

c

a, d

c

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

c

c

b

d

a

a

b

b

b

b

31.

32.

33.

34.

35.

36.

37.

38.

39.

40.

a

d

b

b

a

b

d

a

d

b

41.

42.

43.

44.

45.

46.

47.

48.

49.

50.

c

c

b

c

d

b

c

a

c

c

51.

52.

53.

54.

c

d

b

d

genius PHYSICS Surface Tension 69

Problems based on Cohesive and adhesive force 1.

2.

Mercury does not wet glass, wood or iron because

(b) Cohesive force is greater than adhesive force

(c) Angle of contact is less than 90 o

(d) Cohesive force is equal to adhesive force

The force of cohesion is (a) Maximum in solids

3.

[CPMT 1996]

(b) Maximum in liquid

(c) Same in different matters

(d)

Maximum in gases

What enables us to write on the black board with chalk (a) Gravity

4.

[MP PMT 1995; MP PET 1997]

(a) Cohesive force is less than adhesive force

(b) Cohesion

(c) Adhesion

(d) None of the above

Intermolecular forces decrease rapidly as the distance between the molecules increases and do so much more (a) Slowly than demanded by the inverse square law of the distance (b) Rapidly than anticipated through the inverse square law of the distance (c) According to inverse square law (d) It actually remains the same for all the distances

Problems based on Surface tension 5.

The spherical shape of rain-drop is due to [CPMT 1976, 90; CPMT 2001; NCERT 1982; AIIMS 1998; MHCET 2000; DCE 1999; AFMC 1999, 2001]

(a) Density of the liquid 6.

(a) 4 C

(b) 25 C

10.

[MP PMT/PET 1998]

(d) 75 C o

o

(b) 60 dynes

(d) 750 dynes

(c) 750 dynes

A square frame of side L is dipped in a liquid. On taking it out, a membrane is formed. If the surface tension of the liquid is T, the force acting on the frame will be [MP PMT 1990] (b) 4TL

(c) 8TL

(d) 10TL

Ball pen and fountain pen depend respectively upon the principle of (a) Surface tension and viscosity

(b) Surface tension and gravity

(c) Gravitation and surface tension

(d) Surface tension and surface tension

Which graph represents the variation of surface tension with temperature over small temperature ranges for water (a)

(b) S.T.

(c)

S.T.

(d)

S.T.

Temp

Temp

11.

(d) Gravity

Force necessary to pull a circular plate of 5cm radius from water surface for which surface tension is 75 dynes/cm, is [MP PMT 1991]

(a) 2TL 9.

(c) 50 C

o

(a) 30 dynes 8.

(c) Atmospheric pressure

At which of the following temperatures, the value of surface tension of water is minimum o

7.

(b) Surface tension

S.T.

Temp

Temp

cm3.

The material of a wire has a density of 1.4 g per If it is not wetted by a liquid of surface tension 44 dyne per cm, then the maximum radius of the wire which can float on the surface of the liquid is (a)

1 cm 7

(b) 0.7 cm

(c)

10 cm 14

(d)

10 cm 28

genius PHYSICS 70 Surface Tension 12.

A water drop of 0.05cm3 is squeezed between two glass plates and spreads into area of 40cm2. If the surface tension of water is 70 dyne/cm then the normal force required to separate the glass plates from each other will be (a) 90 N

13.

(b) 45 N

(c) 22.5 N

(d) 450 N

The main difference between a stretched membrane and the liquid surface is (a) The liquid surface has a tendency to contract but the stretched membrane does not (b) The surface tension does not depend on area but on the tension of the stretched membrane does (c) The surface tension increases with increases in area (d) Surface tension increases irregularly with temperature

14.

On bisecting a soap bubble along a diameter, the force due to surface tension on any of its half part will be (a) 4RT

15.

16.

(b)

4 R T

(c)

T 4 R

(d)

2T R

The addition of soap changes the surface tension of water to  1 and that of sugar changes it to  2 . Then (a)  1   2

(b)  1   2

(c)  1   2

(d) It is not possible to predict the above

A hollow disc of aluminum whose external and internal radii are R and r respectively, is floating on the surface of a liquid whose surface tension is T. The maximum weight of disc can be (a) 2 (R + r) T

(b) 2 (R – r) T

(c) 4 (R + r) T

(d) 4 (R – r) T

Problems based on Surface energy 17.

8000 identical water drops are combined to form a big drop. Then the ratio of the final surface energy to the initial surface energy of all the drops together is (a) 1 : 10

18.

(c) 1 : 20

(d) 1 : 25

8 mercury drops coalesce to form one mercury drop, the energy changes by a factor of (a) 1

19.

(b) 1 : 15 (b) 2

(c) 4

[DCE 2000]

(d) 6

Which of the following statements are true in case when two water drops coalesce and make a bigger drop

[Roorkee 1999]

(a) Energy is released (b) Energy is absorbed (c) The surface area of the bigger drop is greater than the sum of the surface areas of both the drops (d) The surface area of the bigger drop is smaller than the sum of the surface areas of both the drops 20.

An oil drop of radius 1cm is sprayed into 1000 small equal drops of same radius. If the surface tension of oil drop is 50 dyne/cm then the work done is [RPET 1990] (a) 18 ergs

21.

(a) W/2 22.

(d) 18000 ergs

(b) 2W

(c) 4W

(d) 2

1 W 3

(b) N 2 / 3

(c)

N1 / 3

(d) N 0

The work done in increasing the volume of a soap bubble of radius R and surface tension T by 700% will be (a) 8R 2T

24.

(c) 1800 ergs

A liquid drop of radius R is broken up into N small droplets. The work done is proportional to (a) N

23.

(b) 180 ergs

If work W is done in blowing a bubble of radius R from a soap solution, then the work done in blowing a bubble of radius 2R from the same solution is [MP PET 1990]

(b) 24 R 2T

(c)

48R 2T

(d) 8R 2T 2 / 3

1000 drops of water all of same size join together to form a single drop and the energy released raises the temperature of the drop. Given that T is the surface tension of water, r the radius of each small drop,  the density of liquid, J the mechanical equivalent of heat. What is the rise in the temperature

genius PHYSICS Surface Tension 71 (a) T/Jr

(b) 10T/Jr

(c) 100T/Jr

(d) None of these

Problems based on Excess pressure 25.

Two bubbles A and B (A > B) are joined through a narrow tube. Then

[UPSEAT

2001;

Kerala

(Med.)

2002]

(a) The size of A will increase

(b)

(c) The size of B will increase until the pressure equals 26.

The size of B will increase

(d) None of these

Excess pressure of one soap bubble is four times more than the other. Then the ratio of volume of first bubble to another one is [CPMT 1997; MH CET 2000]

(a) 1 : 64 27.

(b) 1 : 4

(b) 68.66 dyne/cm

(c) 137 dyne/cm

(d) 150 dyne/cm

An air bubble of radius r in water is at a depth h below the water surface at some instant. If P is atmospheric pressure, d and T are density and surface tension of water respectively, the pressure inside the bubble will be [Roorkee 1990] (a) P  h dg 

29.

(d) 1 : 2

The pressure of air in a soap bubble of 0.7cm diameter is 8mm of water above the pressure outside. The surface tension of the soap solution is [MP PET 1991; MP PMT 1997] (a) 100 dyne/cm

28.

(c) 64 : 1

4T r

(b) P  h dg 

2T r

(c)

P  h dg 

2T r

(d) P  h dg 

4T r

A soap bubble is very slowly blown at the end of a glass tube by a mechanical pump which supplies a fixed volume of air every minute whatever the pressure against which it is pumping. The excess pressure P inside the bubble varies with time as shown by which graph P

P

(a)

P

(b)

P

(c)

(d)

t

t

t

t

Problems based on Angle of contact 30.

A liquid does not wet the sides of a solid, if the angle of contact is [MP PAT 1990; AFMC 1988, MNR 1998, KCET 1998, Haryana CEE 1998; RPMT 1999; 2003]

(a) Zero 31.

(b) Obtuse (More than 90o)

34.

(c) Plane

(d) Uncertain

The angle of contact between glass and mercury is (b) 30 o

[MP PMT 1987]

(c) 90 o

(d) 135 o

When the temperature is increased the angle of contact of a liquid (a) Increases

(b) Decreases

(c) Remains the same

(d) First increases and then decreases

For those liquids which do not wet the solid surface, the ratio of cohesive force and adhesive force will be (a) Greater than

35.

[MP PET/PMT 1988]

(b) Concave

(a) 0 o 33.

(d) 90o0

The meniscus of mercury in the capillary tube is (a) Convex

32.

(c) Acute (Less than 90o)

1 2

(b) Greater than

2

(c) Lesser than

1 2

(d) Lesser than

The water proofing agent makes an angle of contact (a) From acute angle to obtuse angle

(b) From obtuse angle to acute angle

(c) From obtuse angle to right angle

(d) From acute angle to right angle

2

genius PHYSICS 72 Surface Tension 36.

A glass plate is partly dipped vertically in the mercury and the angle of contact is measured. If the plate is inclined, then the angle of contact will (a) Increase

(b) Remain unchanged

(c) Increase or decrease

(d) Decrease

Problems based on Capillarity 37.

The surface tension for pure water in a capillary tube experiment is (a)

38. 39.

40.

g 2 hr

(b)

2 hr g

(c)

[MH CET 2002]

rg 2h

(d)

hrg 2

If capillary experiment is performed in vacuum then for a liquid there (a) It will rise (b) Will remain same (c) It will fall (d) Rise to the top A surface tension experiment with a capillary tube in water is repeated in an artificial satellite. Which is revolving around the earth, water will rise in the capillary tube upto a height of [Roorkee 1992] (a) 0.1 m (b) 0.2 m (c) 0.98 m (d) Full length of the capillary tube When a capillary is dipped in water, water rises to a height h. If the length of the capillary is made less than h, then [MP PAT 1990]

41.

42.

43.

44.

(a) The water will come out (b) The water will not come out (c) The water will not rise (d) The water will rise but less than height of capillary A long cylindrical glass vessel has a small hole of radius ‘r’ at its bottom. The depth to which the vessel can be lowered vertically in the deep water bath (surface tension T) without any water entering inside is [MP PMT 1990] (a) 4T/rg (b) 3T/rg (c) 2T/rg (d) T/rg Water rises to a height of 10cm in capillary tube and mercury falls to a depth of 3.112cm in the same capillary tube. If the density

of mercury is 13.6 and the angle of contact for mercury is 135 o , the ratio of surface tension of water and mercury is[MP PET/PMT 1988] (a) 1 : 0.15 (b) 1 : 3 (c) 1 : 6 (d) 1.5 : 1 Water can rise to a height h in a capillary tube lowered vertically into water. If the height of tube above the surface of water be l and l < h, then water will rise in the capillary to a height (a) h (b) l (c) l – h (d) l + h The height upto which water will rise in a capillary tube will be (a) Maximum when water temperature is 4 o C

45.

(c) Minimum when water temperature is 4 o C (d) Same at all temperatures The exact expression for surface tension of liquid which rises up in the capillary tube is (a) T  rhdg / 2

46.

47.

48.

(b) Maximum when water temperature is 0 o C

(b) T  rhdg / 2 cos 

(c) T 

r(h  r / 3)dg 2

(d) T 

r(h  r / 3)dg 2 cos 

If a wax coated capillary tube is dipped in water, then water in it will (a) Rise up (b) Depress (c) Sometimes rise and sometimes fall (d) Rise up and come out as a fountain Capillaries made from various materials but having the same bore are dipped in the same liquid, then (a) Liquid will not rise in any of them (b) Liquid will rise in all upto same height (c) Liquid will not rise in all upto same height (d) Liquid will rise in all and height of liquid columns will be inversely proportional to the density of material used A straight capillary tube is immersed in water and the water rises to 5cm. If the capillary is bent as shown in figure then the height of water column will be (a) 5cm

h

(b) Less than 5cm (c) Greater than 5cm (d) 4 cos  49.

Water rises in a capillary tube through a height h. If the tube is inclined to the liquid surface at 30 o , the liquid will rise in the tube upto its length equal to

genius PHYSICS Surface Tension 73 (a)

h 2

(b) h

(c) 2h

(d) 4h

Problems (Miscellaneous) 50.

If a water drop is kept between two glass plates, then its shape is

(a)

51.

(b)

r2  r1 r1r2

(c)

r1r2 r2  r1

(d) r2  r1

(b) 600  ergs

(c) 1000  ergs

(d) 1200  ergs

(c)

(d)

In the above question, the radius of the bigger drop will be (a)

54.

(b)

[MP PMT 1996]

Two soap bubbles of radius 1cm and 2cm coalesce to form a single drop under isothermal conditions. The total energy possessed by them if surface tension is 30 dyne cm–1, will be (a) 400  ergs

53.

(d) None of these

When two soap bubbles of radius r1 and r2 (r2  r1 ) coalesce, the radius of curvature of common surface is (a) r2  r1

52.

(c)

3 cm

(b)

5 cm

7 cm

8 cm

In a U-tube the radii of two columns are respectively r1 and r2 and if a liquid of density d filled in it has level difference of h then the surface tension of the liquid is (a) T 

hdg r2  r1

(b) T 

(r2  r1 )hdg 2

(c) T 

(r1  r2 )hdg 2

(d) T 

hdg (r1r2 ) 2 r2  r1

r1

r2 h

genius PHYSICS

74 74 Surface Tension

Answer Sheet (Practice problems) 1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

b

a

c

b

b

d

d

c

c

b

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

a

b

b

a

c

a

c

c

a, d

c

21.

22.

23.

24.

25.

26.

27.

28.

29.

30.

c

c

b

d

a

a

b

b

b

b

31.

32.

33.

34.

35.

36.

37.

38.

39.

40.

a

d

b

b

a

b

d

a

d

b

41.

42.

43.

44.

45.

46.

47.

48.

49.

50.

c

c

b

c

d

b

c

a

c

c

51.

52.

53.

54.

c

d

b

d

genius PHYSICS Fluid Mechanics 109

Problems based on Pressure

 Basic level 1.

2.

The pressure at the bottom of a tank containing a liquid does not depend on (a) Acceleration due to gravity

(b) Height of the liquid column

(c) Area of the bottom surface

(d) Nature of the liquid

When a large bubble rises from the bottom of a lake to the surface. Its radius doubles. If atmospheric pressure is equal to that of column of water height H, then the depth of lake is [AIIMS 1995; AFMC 1997] (a) H

3.

(b) 2H

(b) 10 m

(c) 15 m

(d) 20 m

(b) Decreases by 2%

(c) Remains unchanged

(d) Sometimes increases and sometimes decreases

Two stretched membranes of area 2 (a) 1 : 1

6.

(d) 8H

The value of g at a place decreases by 2%. The barometric height of mercury (a) Increases by 2%

5.

(c) 7H

The volume of an air bubble becomes three times as it rises from the bottom of a lake to its surface. Assuming atmospheric pressure to be 75 cm of Hg and the density of water to be 1/10 of the density of mercury, the depth of the lake is (a) 5 m

4.

[Kerala (Engg.) 2002]

cm2 and 3 cm2 are placed

(b) 2 : 3

in a liquid at the same depth. The ratio of pressures on them is (c) 3 : 2

(d) 22 : 32

Three identical vessels are filled to the same height with three different liquids A, B and C ( A   B   C ) . The pressure at the base will be (a) Equal in all vessels

7.

(b) Maximum in vessel A

(c) Maximum in vessel B

(d) Maximum in vessel C

Three identical vessels are filled with equal masses of three different liquids A, B and C ( A   B   C ) . The pressure at the base will be (a) Equal in all vessels

8.

(c) Maximum in vessel B

(d) Maximum in vessel C

A barometer kept in a stationary elevator reads 76 cm. If the elevator starts accelerating up the reading will be (a) Zero

9.

(b) Maximum in vessel A (b) Equal to 76 cm

(c) More than 76 cm

(d) Less than 76 cm

A closed rectangular tank is completely filled with water and is accelerated horizontally with an acceleration a towards right. Pressure is (i) maximum at, and (ii) minimum at (a) (i) B (ii) D (b) (i) C (ii) D (c) (i) B (ii) C (d) (i) B (ii) A

10.

D

B

C

a

A beaker containing a liquid is kept inside a big closed jar. If the air inside the jar is continuously pumped out, the pressure in the liquid near the bottom of the liquid will (a) Increases increase

11.

A

(b) Decreases

(c) Remain constant

(d) First

decrease

and

then

A barometer tube reads 76 cm of mercury. If the tube is gradually inclined at an angle of 60o with vertical, keeping the open end immersed in the mercury reservoir, the length of the mercury column will be (a) 152 cm

(b) 76 cm

(c) 38 cm

(d) 38 3 cm

genius PHYSICS 110 Fluid Mechanics

 Advance level 12.

A ring shaped tube contains two ideal gases with equal masses and molar masses M 1  32 and M 2  28 . The gases are separated by one fixed partition and another movable stopper S which can move freely without friction inside the ring. The angle  in degrees is M1 M2

(a) 192



(b) 291

S

(c) 129 (d) 219 13.

The height to which a cylindrical vessel be filled with a homogeneous liquid, to make the average force with which the liquid presses the side of the vessel equal to the force exerted by the liquid on the bottom of the vessel, is equal to (a) Half of the radius of the vessel

(b)

(c) One-fourth of the radius of the vessel

(d) Three-fourth of the radius of the vessel

A vertical U-tube of uniform inner cross section contains mercury in both sides of its arms. A glycerin (density = 1.3 g/cm3) column of length 10 cm is introduced into one of its arms. Oil of density 0.8 gm/cm3 is poured into the other arm until the upper surfaces of the oil and glycerin are in the same horizontal level. Find the length of the oil column, Density of mercury = 13.6 g/cm3 Glycerin e

14.

Radius of the vessel

(a) 10.4 cm (b) 8.2 cm

Oil 10 cm

h

(c) 7.2 cm Mercury

(d) 9.6 cm 15.

There are two identical small holes of area of cross-section a on the opposite sides of a tank containing a liquid of density  . The difference in height between the holes is h. Tank is resting on a smooth horizontal surface. Horizontal force which will has to be applied on the tank to keep it in equilibrium is (a) gha (b)

2 gh a

h

(c) 2 agh (d) 16.

gh a

Two communicating vessels contain mercury. The diameter of one vessel is n times larger than the diameter of the other. A column of water of height h is poured into the left vessel. The mercury level will rise in the right-hand vessel (s = relative density of mercury and  = density of water) by (a)

n 2h (n  1)2 s

17.

(c)

h (n  1)2 s

(d)

h n2s

Water

h

h (b) 2 (n  1) s

Mercury

A triangular lamina of area A and height h is immersed in a liquid of density  in a vertical plane with its base on the surface of the liquid. The thrust on the lamina is (a)

1 A gh 2

(b)

1 A gh 3

(c)

1 A gh 6

(d)

2 A gh 3

genius PHYSICS Fluid Mechanics 111

Problems based on Pascal's law 18.

Pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel. This law was first formulated by (a) Bernoulli

19.

(b) Archimedes

(c) Boyle

(d) Pascal

A piston of cross-section area 100 cm2 is used in a hydraulic press to exert a force of 10 7 dynes on the water. The cross-sectional area of the other piston which supports an object having a mass 2000 kg. is (a) 100 cm2

(b) 109cm2

(c) 2 × 104cm2

(d) 2 × 1010cm2

Problems based on Archimedes principle 20.

A block of steel of size 5 cm × 5 cm × 5 cm is weighed in water. If the relative density of steel is 7, its apparent weight is [AFMC 1997]

(a) 6 × 5 × 5 × 5 gf 21.

22.

(b) 4 × 4 × 4 × 7 gf

(c) 5 × 5 × 5 × 7 gf

(d) 4 × 4 × 4 × 6 gf

A body is just floating on the surface of a liquid. The density of the body is same as that of the liquid. The body is slightly pushed down. What will happen to the body [AIIMS 1980] (a) It will slowly come back to its earlier position

(b) It will remain submerged, where it is left

(c) It will sink

(d) It will come out violently

A uniform rod of density  is placed in a wide tank containing a liquid of density 0 (0  ) . The depth of liquid in the tank is half the length of the rod. The rod is in equilibrium, with its lower end resting on the bottom of the tank. In this position the rod makes an angle  with the horizontal (a) sin  

23.

1 2

(b) sin  

1 0 . 2 

(c)

sin  

 / 0

(d) sin   0 / 

A cork is submerged in water by a spring attached to the bottom of a bowl. When the bowl is kept in an elevator moving with acceleration downwards, the length of spring (a) Increases

24.

0 / 

(b) Decreases

(c) Remains unchanged

(d) None of these

A cubical block of wood 10 cm on a side floats at the interface between oil and water with its lower surface horizontal and 4 cm below the interface. The density of oil is 0.6 gcm 1 . The mass of block is 6 cm

Block

(a) 706 g (b) 607 g

4 cm

(c) 760 g (d) 670 g 25.

A solid sphere of density  ( > 1) times lighter than water is suspended in a water tank by a string tied to its base as shown in fig. If the mass of the sphere is m then the tension in the string is given by

 1   mg (a)     (b)  mg (c)

mg  1

(d) (  1) mg 26.

A spherical ball of radius r and relative density 0.5 is floating in equilibrium in water with half of it immersed in water. The work done in pushing the ball down so that whole of it is just immersed in water is : (where  is the density of water) (a)

27.

5 r 4 g 12

(b) 0.5 rg

(c)

4 3 r g 3

(d)

2 4 r g 3

A hollow sphere of volume V is floating on water surface with half immersed in it. What should be the minimum volume of water poured inside the sphere so that the sphere now sinks into the water (a) V / 2

(b) V / 3

(c)

V/4

(d) V

genius PHYSICS 112 Fluid Mechanics 28.

A rectangular block is 5 cm × 5 cm × 10cm in size. The block is floating in water with 5 cm side vertical. If it floats with 10 cm side vertical, what change will occur in the level of water? (a) No change

(b) It will rise

(c) It will fall 29.

30.

(d) It may rise or fall depending on the density of block

A ball whose density is 0.4 ×

103

(a) 9 cm

(b) 6 cm

33.

34.

(c) 4.5 cm

(d) 2.25 cm

1 1 of its body immersed in water and B floats with of its 2 4 volume above the water level. The ratio of the density of A to that of B is

(b) 2 : 3

(c) 3 : 4

(d) 1 : 2

A boat carrying steel balls is floating on the surface of water in a tank. If the balls are thrown into the tank one by one, how will it affect the level of water (a) It will remain unchanged

32.

falls into water from a height of 9 cm . To what depth does the ball sink

Two solids A and B float in water. It is observed that A floats with

(a) 4 : 3 31.

kg/m3

(b)

It will rise

(c) It will fall

(d)

Two pieces of metal when immersed in a liquid have equal upthrust on them; then (a) Both pieces must have equal weights

(b) Both pieces must have equal densities

(c) Both pieces must have equal volumes

(d) Both are floating to the same depth

A wooden cylinder floats vertically in water with half of its length immersed. The density of wood is (a) Equal of that of water

(b) Half the density of water

(c) Double the density of water

(d) The question is incomplete

If W be the weight of a body of density  in vacuum then its apparent weight in air of density  is (a)

W

  (b) W  1   



(c)

W



  (d) W  1    



Problems based on Density 35.

A block of ice floats on a liquid of density 1.2in a beaker then level of liquid when ice completely melt (a) Remains same

36.



(b)  

2

1   2

(b)  

2



   0 1  

 0 gy  B 

   0 [1   d ]

1   2 2 1  2

(c)



2 1  2 1   2

(d)  

1  2 1   2

1   2 2 1  2

(c)



2 1  2 1   2

(d)  

1  2 1   2

 gy   (b)    0 1  0  B  

(c)



Β



 0 hgy 

   0 1 

 

 B  (d)    0 1    0 gy  

(b)    0 [1   d ]

(c)

   0 d

(d)    0 / d

Three liquids of densities d, 2d and 3 d are mixed in equal volumes. Then the density of the mixture is (a) d

41.

1   2

With rise in temperature, density of a given body changes according to one of the following relations (a)

40.



The density  of water of bulk modulus B at a depth y in the ocean is related to the density at surface  0 by the relation (a)

39.

(d) (A), (B) or (c)

If two liquids of same volume but different densities  1 and  2 are mixed, then density of mixture is given by (a)

38.

(c) Lowers

If two liquids of same masses but densities  1 and  2 respectively are mixed, then density of mixture is given by (a)

37.

(b) rises

[IIT-JEE 1994]

(b) 2d

(c) 3d

(d) 5d

Three liquids of densities d, 2d and 3 d are mixed in equal proportions of weights. The relative density of the mixture is

genius PHYSICS Fluid Mechanics 113 (a)

11d 7

(b)

18 d 11

(c)

13 d 9

23 d 18

(d)

Problems based on Streamlined & turbulent flow 42.

Stream-line flow is more likely for liquids with

[Pb. CET 1997]

(a) Low density and low viscosity (c) High viscosity and low density 43.

44.

45.

46.

(b)

High viscosity and high density

(d) Low viscosity and high density

In a laminar flow the velocity of the liquid in contact with the walls of the tub is (a) Zero

(b) Maximum

(c) In between zero and maximum

(d) Equal to critical velocity

In a turbulent flow, the velocity of the liquid molecules in contact with the walls of the tube is (a) Zero

(b) Maximum

(c) Equal to critical velocity

(d) May have any value

Which of the following is NOT the characteristic of turbulent flow (a) Velocity more than the critical velocity

(b) Velocity less than the critical velocity

(c) Irregular flow

(d) Molecules crossing from one layer to another

The Reynolds number of a flow is the ratio of (a) Gravity to viscous force

(b) Gravity force to pressure force

(c) Inertia forces to viscous force

(d) Viscous forces to pressure forces

Problems based on Equation of Continuity 47.

Water is flowing through a tube of non-uniform cross-section ratio of the radius at entry and exit end of the pipe is 3 : 2. Then the ratio of velocities at entry and exit of liquid is [RPMT 2001] (a) 4 : 9

48.

49.

(b) 9 : 4

(c) 8 : 27

(d) 1 : 1

Water is flowing through a horizontal pipe of non-uniform cross-section. At the extreme narrow portion of the pipe, the water will have (a) Maximum speed and least pressure

(b) Maximum pressure and least speed

(c) Both pressure and speed maximum

(d) Both pressure and speed least

A liquid flows in a tube from left to right as shown in figure. A 1 and A 2 are the cross-sections of the portions of the tube as shown. Then the ratio of speeds v1 / v 2 will be (a) A1 / A 2 (b) A 2 / A1

50.

(c)

A 2 / A1

(d)

A1 / A 2

In a streamline flow (a) The speed of a particle always remains same (b) The velocity of a particle always remains same (c) The kinetic energies of all the particles arriving at a given point are the same (d) The moments of all the particles arriving at a given point are the same

A1

A2 v1

v2

genius PHYSICS 114 Fluid Mechanics 51.

Water coming out of the mouth of a tap and falling vertically in streamline flow forms a tapering, column, i.e., the area of crosssection of the liquid column decreases as it moves down. Which of the following is the most accurate explanation for this (a) As the water moves down, its speed increases and hence its pressure decreases. It is then compressed by the atmosphere (b) Falling water tries to reach a terminal velocity and hence reduces the area of crosssection to balance upward and downward forces (c) The mass of water flowing past any cross-section must remain constant. Also, water is almost incompressible. Hence, the rate of volume flow must remain constant. As this is equal to velocity × area, the area decreases as velocity increases (d) The surface tension causes the exposed surface area of the liquid to decrease continuously

Problems based on Equation of Bernoulli's Theorem 52.

53.

An application of Bernoulli's equation for fluid flow is found in (a) Dynamic lift of an aeroplane

(b) Viscosity meter

(c) Capillary rise

(d) Hydraulic press

The Working of an atomizer depends upon (a) Bernoulli's theorem

54.

(b) Boyle's law

[MP PMT 1992]

(c) Archimedes principle

(d) Newton's law of motion

The pans of a physical balance are in equilibrium. Air is blown under the right hand pan; then the right hand pan will (a) Move up

55.

[IIT-JEE (Screening) 1994]

(b) Move down

(c) Move erratically

(d) Remain at the same level

According to Bernoulli's equation

P 1 v2 h  constant g 2 g The terms A, B and C are generally called respectively:

56.

(a) Gravitational head, pressure head and velocity head

(b) Gravity, gravitational head and velocity head

(c) Pressure head, gravitational head and velocity head

(d) Gravity, pressure and velocity head

At what speed the velocity head of a stream of water be equal to 40 cm of Hg (a) 1032.6 cm/sec

57.

(b) 432.6 cm/sec

(c) 632.6 cm/sec

(d) 832.6 cm/sec

The weight of an aeroplane flying in air is balanced by (a) Upthrust of the air which will be equal to the weight of the air having the same volume as the plane (b) Force due to the pressure difference between the upper and lower surfaces of the wings, created by different air speeds on the surface (c) Vertical component of the thrust created by air currents striking the lower surface of the wings (d) Force due to the reaction of gases ejected by the revolving propeller

58.

In this figure, an ideal liquid flows through the tube, which is of uniform cross-section. The liquid has velocities v A and v B , and pressure PA and PB at points A and B respectively A

(a) v A  v B (b) v B  v A

B

(c) PA = PB (d) PB > PA 59.

A liquid flows through a horizontal tube. The velocities of the liquid in the two sections, which have areas of cross-section A1 and

A 2 , are v1 and v2 respectively. The difference in the levels of the liquid in the two vertical tubes is h h A1

A2

v1

v2

genius PHYSICS Fluid Mechanics 115 (a) The volume of the liquid flowing through the tube in unit time is A1v1 (b) v 2  v1  2 gh (c) v 22  v12  2 gh (d) The energy per unit mass of the liquid is the same in both sections of the tube 60.

A sniper fires a rifle bullet into a gasoline tank making a hole 53.0 m below the surface of gasoline. The tank was sealed at 3.10 atm. The stored gasoline has a density of 660 kgm–3. The velocity with which gasoline begins to shoot out of the hole is (a) 27 .8 ms 1

61.

(b) 41 .0 ms 1

(c) 9.6 ms 1

(d)

19 .7 ms 1

An L-shaped tube with a small orifice is held in a water stream as shown in fig. The upper end of the tube is 10.6 cm above the surface of water. What will be the height of the jet of water coming from the orifice? Velocity of water stream is 2.45 m/s (a) Zero (b) 20.0 cm (c) 10.6 cm

2.45 m/s

(d) 40.0 cm 62.

To get the maximum flight, a ball must be thrown as

(a)

63.

v

v

(c)

(d) Any of (a), (b) and (c)

v

Fig. represents vertical sections of four wings moving horizontally in air. In which case is the force upwards

(a) 64.

(b)

(b)

(c)

(d)

An L-shaped glass tube is just immersed in flowing water such that its opening is pointing against flowing water. If the speed of water current is v, then (a) The water in the tube rises to height

v2 2g

(b) The water in the tube rises to height

g 2v 2

h

v

(c) The water in the tube does not rise at all (d) None of these

Problems based on Velocity of Efflux 65.

A tank is filled with water up to a height H. Water is allowed to come out of a hole P in one of the walls at a depth D below the surface of water. Express the horizontal distance x in terms of H and D (a) x  D(H  D) (b) x  (c)

D(H  D) 2

D H

x  2 D(H  D)

x

(d) x  4 D(H  D) 66.

A cylindrical vessel of 90 cm height is kept filled upto the brim. It has four holes 1, 2, 3, 4 which are respectively at heights of 20 cm, 30 cm, 45 cm and 50 cm from the horizontal floor PQ. The water falling at the maximum horizontal distance from the vessel comes from [CPMT 1989]

P

4 3 2 1 Q

genius PHYSICS 116 Fluid Mechanics (a) Hole number 4 (b) Hole number 3 (c) Hole number 2 (d) Hole number 1 67.

A rectangular vessel when full of water takes 10 minutes to be emptied through an orifice in its bottom. How much time will it take to be emptied when half filled with water (a) 9 minute

68.

A a

2 [ H1  H 2 ] g

2h g

(b)

2 gh

(c)

2 gh (H 1  H 2 )

(d)

A a

g [ H1  H 2 ] 2

(b)

2h D . g d

(c)

2h d . g D

(d)

2h  d    g Dd

A large tank is filled with water to a height H. A small hole is made at the base of the tank. It takes T1 time to decrease the height of water to

H



(  1) ; and it takes T 2 time to take out the rest of water. If T1  T2 , then the value of  is

(a) 2 71.

(d) 3 minute

A streamlined body falls through air from a height h on the surface of a liquid. If d and (D > d) represents the densities of the material of the body and liquid respectively, then the time after which the body will be instantaneously at rest, is (a)

70.

(c) 5 minute

A vessel of area of cross-section A has liquid to a height H. There is a hole at the bottom of vessel having area of cross-section a. The time taken to decrease the level from H 1 to H 2 will be (a)

69.

(b) 7 minute

(b) 3

(c) 4

(d) 2 2

A tank is filled upto a height h with a liquid and is placed on a platform of height h from the ground. To get maximum range x m a small hole is punched at a distance of y from the free surface of the liquid. Then (a)

x m  2h

y

h

(b) x m  1.5h (c)

h

y h

xm

(d) y  0.75 h

Problems based on Viscosity 72.

73.

74.

75.

Velocity of water in a river is (a) Same everywhere

(b) More in the middle and less near its banks

(c) Less in the middle and more near its banks

(d) Increase from one bank to other bank

As the temperature of water increases, its viscosity (a) Remains unchanged

(b) Decreases

(c) Increases

(d) Increases or decreases depending on the external pressure

The coefficient of viscosity for hot air is (a) Greater than the coefficient of viscosity for cold air

(b) Smaller than the coefficient of viscosity for cold air

(c) Same as the coefficient of viscosity for cold air

(d) Increases or decreases depending on the external pressure

A good lubricant should have (a) High viscosity

76.

[CBSE PMT 1988]

(b) Low viscosity

(c) Moderate viscosity

(d) High density

The relative velocity of two consecutive layers is 8 cm/s. If the perpendicular distance between the layers is 0.1 cm, then the velocity gradient will be (a) 8sec–1

(b) 80 sec–1

(c) 0.8 sec–1

(d) 0.08 sec–1

genius PHYSICS Fluid Mechanics 117 77.

We have three beakers A, B and C containing glycerine, water and kerosene respectively. They are stirred vigorously and placed on a table. The liquid which comes to rest at the earliest is (a) Glycerine

(b) Water

(c) Kerosene

(d) All of them at the same time

Problems based on Stoke's law and Terminal velocity 78.

A lead shot of 1mm diameter falls through a long column of glycerine. The variation of its velocity v. with distance covered is represented by [AIIMS 2003] (a)

(b) v

v

Distance

79.

(c)

(d)

v

v

Distance

Distance

Distance

covered covered covered A small spherical solid ball is droppedcovered from a great height in a viscous liquid. Its journey in the liquid is best described in the diagram given below by the [CPMT 1988] A B

Velocity (v)

(a) Curve A (b) Curve B

C D

(c) Curve C

Time (t)

(d) Curve D 80.

A small drop of water falls from rest through a large height h in air; the final velocity is (a)  h

(b)  h

(c)  (1 / h)

(d) Almost independent of h

Problems based on Poiseuille's law 81.

The rate of flow of liquid in a tube of radius r, length l, whose ends are maintained at a pressure difference P is V  where

 is coefficient of the viscosity and Q is

(a) 8 82.

(b)

1 8

QP r 4 l

[DCE 2002]

(c) 16

(d)

1 16

In Poiseuilli's method of determination of coefficient of viscosity, the physical quantity that requires greater accuracy in measurement is [EAMCET 2001]

83.

(a) Pressure difference

(b) Volume of the liquid collected

(c) Length of the capillary tube

(d) Inner radius of the capillary tube

Two capillary tubes of the same length but different radii r1 and r2 are fitted in parallel to the bottom of a vessel. The pressure head is P. What should be the radius of a single tube that can replace the two tubes so that the rate of flow is same as before (a) r1  r2

84.

(c) r14  r24

(d) None of these

Under a constant pressure head, the rate of flow of liquid through a capillary tube is V. If the length of the capillary is doubled and the diameter of the bore is halved, the rate of flow would become (a) V / 4

85.

(b) r12  r22

(b) 16 V

(c) V / 8

(d) V / 32

Two capillaries of same length and radii in the ratio 1 : 2 are connected in series. A liquid flows through them in streamlined condition. If the pressure across the two extreme ends of the combination is 1 m of water, the pressure difference cross first capillary is (a) 9.4 m

(b) 4.9 m

(c) 0.49 m

(d) 0.94 m

genius PHYSICS 118 Fluid Mechanics 86.

Water flows in a streamlined manner through a capillary tube of radius a, the pressure difference being P and the rate of flow Q. If the radius is reduced to a/2 and the pressure increased to 2P, the rate of flow becomes (a) 4 Q

(b) Q

(c)

Q 4

(d)

Q 8

Physics classes by Pradeep Kshetrapal

FORMULA BANK

THERMODYNAMICS

genius PHYSICS Simple Harmonic Motion 1

15.1 Periodic Motion. A motion, which repeat itself over and over again after a regular interval of time is called a periodic motion and the fixed interval of time after which the motion is repeated is called period of the motion. Examples : (i) Revolution of earth around the sun (period one year) (ii) Rotation of earth about its polar axis (period one day) (iii) Motion of hour’s hand of a clock (period 12-hour) (iv) Motion of minute’s hand of a clock (period 1-hour) (v) Motion of second’s hand of a clock (period 1-minute) (vi) Motion of moon around the earth (period 27.3 days)

15.2 Oscillatory or Vibratory Motion. Oscillatory or vibratory motion is that motion in which a body moves to and fro or back and forth repeatedly about a fixed point in a definite interval of time. In such a motion, the body is confined with in welldefined limits on either side of mean position. Oscillatory motion is also called as harmonic motion. Example : (i) The motion of the pendulum of a wall clock. (ii) The motion of a load attached to a spring, when it is pulled and then released. (iii) The motion of liquid contained in U- tube when it is compressed once in one limb and left to itself. (iv) A loaded piece of wood floating over the surface of a liquid when pressed down and then released executes oscillatory motion.

15.3 Harmonic and Non-harmonic Oscillation. Harmonic oscillation is that oscillation which can be expressed in terms of single harmonic function (i.e. sine or cosine function). Example : y  a sin  t or y  a cos  t Non-harmonic oscillation is that oscillation which can not be expressed in terms of single harmonic function. It is a combination of two or more than two harmonic oscillations. Example : y  a sin  t  b sin 2 t

15.4 Some Important Definitions. (1) Time period : It is the least interval of time after which the periodic motion of a body repeats itself. S.I. units of time period is second. (2) Frequency : It is defined as the number of periodic motions executed by body per second. S.I unit of frequency is hertz (Hz). (3) Angular Frequency : Angular frequency of a body executing periodic motion is equal to product of frequency of the body with factor 2. Angular frequency  = 2  n S.I. units of  is Hz [S.I.]  also represents angular velocity. In that case unit will be rad/sec.

genius PHYSICS 2 Simple Harmonic Motion (4) Displacement : In general, the name displacement is given to a physical quantity which undergoes a change with time in a periodic motion. Examples : (i) In an oscillation of a loaded spring, displacement variable is its deviation from the mean position. (ii) During the propagation of sound wave in air, the displacement variable is the local change in pressure (iii) During the propagation of electromagnetic waves, the displacement variables are electric and magnetic fields, which vary periodically. (5) Phase : phase of a vibrating particle at any instant is a physical quantity, which completely express the position and direction of motion, of the particle at that instant with respect to its mean position. In oscillatory motion the phase of a vibrating particle is the argument of sine or cosine function involved to represent the generalised equation of motion of the vibrating particle.

y  a sin   a sin( t   0 )

here,    t   0 = phase of vibrating particle.

(i) Initial phase or epoch : It is the phase of a vibrating particle at t = 0. In    t   0 , when t = 0;    0

here,  0 is the angle of epoch.

(ii) Same phase : Two vibrating particle are said to be in same phase, if the phase difference between them is an even multiple of  or path difference is an even multiple of ( / 2) or time interval is an even multiple of (T / 2) because 1 time period is equivalent to 2 rad or 1 wave length () (iii) Opposite phase : When the two vibrating particles cross their respective mean positions at the same time moving in opposite directions, then the phase difference between the two vibrating particles is 180 o Opposite phase means the phase difference between the particle is an odd multiple of  (say , 3, 5,  3 7…..) or the path difference is an odd multiple of  (say , ,.......) or the time interval is an odd multiple of 2 2 (T / 2). (iv) Phase difference : If two particles performs S.H.M and their equation are

y 1  a sin( t  1 ) and y 2  a sin( t   2 ) then phase difference   ( t   2 )  ( t  1 )   2  1

15.5 Simple Harmonic Motion. Simple harmonic motion is a special type of periodic motion, in which a particle moves to and fro repeatedly about a mean position under a restoring force which is always directed towards the mean position and whose magnitude at any instant is directly proportional to the displacement of the particle from the mean position at that instant. Restoring force  Displacement of the particle from mean position. F –x F = – kx Where k is known as force constant. Its S.I. unit is Newton/meter and dimension is [MT –2].

15.6 Displacement in S.H.M..

genius PHYSICS Simple Harmonic Motion 3

The displacement of a particle executing S.H.M. at an instant is defined as the distance of particle from the mean position at that instant. As we know that simple harmonic motion is defined as the projection of uniform circular motion on any diameter of circle of reference. If the projection is taken on y-axis. Y

then from the figure y  a sin  t

y  a sin

P

N

2 t T

y X

a  =t

O

M

X

y  a sin 2 n t y  a sin( t   )

Y

where a = Amplitude,  = Angular frequency, t = Instantaneous time, T = Time period, n = Frequency and  = Initial phase of particle If the projection of P is taken on X-axis then equations of S.H.M. can be given as

x  a cos ( t   )  2  x  a cos  t   T  x  a cos (2n t   )

Important points (i) y  a sin  t

when the time is noted from the instant when the vibrating particle is at mean

position. (ii) y  a cos  t

when the time is noted from the instant when the vibrating particle is at extreme

position. (iii) y  a sin( t   ) when the vibrating particle is  phase leading or lagging from the mean position. (iv) Direction of displacement is always away from the equilibrium position, particle either is moving away from or is coming towards the equilibrium position. (v) If t is given or phase ( ) is given, we can calculate the displacement of the particle. If t 

T  (or   ) 2 4

Similarly if t 

then from equation y  a sin

2 2 T    a sin    a t , we get y  a sin T T 4 2

T (or    ) then we get y  0 2

Sample problems based on Displacement Problem 1.

A simple harmonic oscillator has an amplitude A and time period T. The time required by it to travel from [CBSE 1992; SCRA 1996] x  A to x  A / 2 is (a) T / 6

Solution : (a)

(b) T / 4

(c) T / 3

(d) T / 2

Because the S.H.M. starts from extreme position so y  a cos  t form of S.H.M. should be used. A 2  2  A cos t  cos  cos t  t T /6 2 T 3 T

genius PHYSICS 4 Simple Harmonic Motion

Problem 2.

A mass m = 100 gms is attached at the end of a light spring which oscillates on a friction less horizontal table with an amplitude equal to 0.16 meter and the time period equal to 2 sec. Initially the mass is released from rest at t = 0 and displacement x = – 0.16 meter. The expression for the displacement of the mass at any time (t) is [MP PMT 1995] (a) x  0.16 cos ( t)

Solution : (b)

2 t  x  0.16 cos( t) T

(b) 0.02 sec

(c) 0.1 sec

(d) 0.2 sec

By comparing the given equation with standard equation y  a sin( t   )

  100 so T 

Problem 4.

[As a = – 0.16 meter, T = 2 sec]

The motion of a particle executing S.H.M. is given by x  0.01 sin 100 (t  .05) . Where x is in meter and time t is in seconds. The time period is (a) 0.01 sec

Solution : (b)

(d) x  0.16 cos( t   )

(c) x  0.16 cos( t   )

Standard equation for given condition x  a cos

Problem 3.

(b) x  0.16 cos( t)

2





2  0.02 sec 100

Two equations of two S.H.M. are x  a sin( t   ) and y  b cos( t   ) . The phase difference between the two is (a)

Solution : (c)

0o

[MP PMT 1985]

(b) 

(c)

o

90o

(d)

180o

x  a sin( t   ) and y  b cos( t   ) = b sin( t     / 2) Now the phase difference = ( t   

 2

)  ( t   )   / 2  90 o

15.7 Velocity in S.H.M.. Velocity of the particle executing S.H.M. at any instant, is defined as the time rate of change of its displacement at that instant. In case of S.H.M. when motion is considered from the equilibrium position y  a sin  t dy  a cos  t dt

so

v



v  a cos  t

or

v  a 1  sin 2  t

or

v   a2  y 2

……(i) [As sin t = y/a] …..(ii)

Important points (i) In S.H.M. velocity is maximum at equilibrium position. From equation (i)

vmax  a

when

from equation (ii)

vmax  a

when

cos  t =1

i.e.

=t=0

i.e

 t 

y 0

(ii) In S.H.M. velocity is minimum at extreme position. cos  t = 0

From equation (i)

vmin  0

when

From equation (ii)

v min  0

when y = a

 2

genius PHYSICS Simple Harmonic Motion 5

(iii) Direction of velocity is either towards or away from mean position depending on the position of particle.

Sample problems based on Velocity Problem 5.

A body is executing simple harmonic motion with an angular frequency 2 rad/sec. The velocity of the body at 20 mm displacement. When the amplitude of motion is 60 mm is [AFMC 1998] (a) 40 mm/sec

(b) 60 mm/sec

(c) 113 mm/sec

(d) 120 mm/sec

Solution : (c)

v   a2  y 2  2 (60)2  (20)2 = 113 mm/sec

Problem 6.

A body executing S.H.M. has equation y  0.30 sin(220 t  0.64 ) in meter. Then the frequency and maximum velocity of the body is (a) 35 Hz , 66 m / s

Solution : (a)

(b) 45 Hz , 66 m / s

(c) 58 Hz , 113m / s

(d) 35 Hz , 132m / s

By comparing with standard equation y  a sin( t   ) we get a  0.30;   220  2n  220  n  35 Hz so v max  a  0.3  220  66 m / s

Problem 7.

A particle starts S.H.M. from the mean position. Its amplitude is A and time period is T. At the time when its speed is half of the maximum speed. Its displacement y is (a) A/2

(b) A / 2

(c) A 3 / 2

a a2   a2  y 2   a2  y 2  y  2 4

3A 2

(d) 2 A / 3 v m ax a  ] 2 2

Solution : (c)

v   a2  y 2 

Problem 8.

A particle perform simple harmonic motion. The equation of its motion is x  5 sin(4 t 

[As v 

displacement. If the displacement of the particle is 3 units then its velocity is (a) 2 / 3

(b) 5 / 6

(c) 20

 6

) . Where x is its [MP PMT 1994]

(d) 16

[As  = 4, a = 5, y = 3]

Solution : (d)

v   a2  y 2  4 5 2  3 2 = 16

Problem 9.

A simple pendulum performs simple harmonic motion about x = 0 with an amplitude (A) and time period (T). The speed of the pendulum at x  (a)

Solution : (a)

A 3

(b)

T

v   a2  y 2  v 

2 T

A will be 2

 A T

A2 

A 3 A2  4 T

[MP PMT 1987]

(c)

A 3 2T

(d)

3 2 A T

[As y = A/2]

Problem 10. A particle is executing S.H.M. if its amplitude is 2 m and periodic time 2 seconds. Then the maximum velocity of the particle will be (a) 6 Solution : (c)

(b) 4

v m ax  a  a

(c) 2

(d) 

2 2 2  v max  2 T 2

Problem 11. A S.H.M. has amplitude ‘a’ and time period T. The maximum velocity will be (a) Solution : (d)

4a T

v m ax  a 

(b) a 2 T

2a T

(c) 2 a T

[MP PMT 1985]

(d)

2a T

genius PHYSICS 6 Simple Harmonic Motion

Problem 12. A particle executes S.H.M. with a period of 6 second and amplitude of 3 cm its maximum speed in cm/sec is [AIIMS 1982]

(a)  / 2 Solution : (b)

v m ax  a  a

(b) 

(c) 2

(d) 3

2 2 3  v max   T 6

Problem 13. A body of mass 5 gm is executing S.H.M. about a point with amplitude 10 cm. Its maximum velocity is 100 cm/sec. Its velocity will be 50 cm/sec, at a distance (a) 5 Solution : (c)

(b) 5 2

[CPMT 1976]

(c) 5 3

(d) 10 2

v max  a  100 cm / sec and a  10 cm so   10 rad / sec .  v   a2  y 2  50 = 10 10 2  y 2  y  5 3

15.8 Acceleration in S.H.M.. The acceleration of the particle executing S.H.M. at any instant, is defined as the rate of change of its velocity at that instant. So acceleration A  dv  d (a cos t) dt

dt

A   2 a sin  t

……(i)

A   2 y

……(ii)

[As y  a sin  t ]

Important points (i) In S.H.M. as Accelerati on   2 y is not constant. So equations of translatory motion can not be applied. (ii) In S.H.M. acceleration is maximum at extreme position. sin  t  maximum  1 i.e. at t 

From equation (i) A m ax   2 a when

T or t   4 2

From equation (ii) | A m ax |   2 a when y  a (iii) In S.H.M. acceleration is minimum at mean position From equation (i) A min  0

when sin  t  0 i.e. at t  0 or t 

From equation (ii) A min  0

when y  0

T or  t   2

(iv) Acceleration is always directed towards the mean position and so is always opposite to displacement i.e.,

A  y

15.9 Comparative Study of Displacement, Velocity and Acceleration. Displacement

y  a sin  t

Velocity

v  a cos t  a sin( t  ) 2

Acceleration

A  a sin  t  a sin( t   )



2

2

From the above equations and graphs we can conclude that.

+a y

0

T

2T

Displacement y = a sin t

0

T

2T

Velocity v = a cos t

0

T

Acceleration 2 2T A = – a cos t

–a +a v – a +a2 a – a2

1 T 2

T

3 T 2

2T

Time

genius PHYSICS Simple Harmonic Motion 7

(i) All the three quantities displacement, velocity and acceleration show harmonic variation with time having same period. (ii) The velocity amplitude is  times the displacement amplitude (iii) The acceleration amplitude is  2 times the displacement amplitude (iv) In S.H.M. the velocity is ahead of displacement by a phase angle  / 2 (v) In S.H.M. the acceleration is ahead of velocity by a phase angle  / 2 (vi) The acceleration is ahead of displacement by a phase angle of  (vii) Various physical quantities in S.H.M. at different position : Physical quantities

Equilibrium position (y = 0)

Extreme Position (y =  a)

Displacement y  a sin  t

Minimum (Zero)

Maximum (a)

Velocity v   a2  y 2

Maximum (a )

Minimum (Zero)

Acceleration A   2 y

Minimum (Zero)

Maximum (  2 a )

15.10 Energy in S.H.M.. A particle executing S.H.M. possesses two types of energy : Potential energy and Kinetic energy (1) Potential energy : This is an account of the displacement of the particle from its mean position. The restoring force F  ky against which work has to be done



U   dw  

So  potential Energy



x

0

Fdx 

y

1

 ky dy  2 k y

2

0

U

1 m 2y 2 2

[As  2  k / m ]

U

1 m  2 a 2 sin 2  t 2

[As y  a sin  t ]

Important points (i) Potential energy maximum and equal to total energy at extreme positions

1 2 1 when y  a ;  t   / 2 ; t  T / 4 ka  m 2a2 2 2 (ii) Potential energy is minimum at mean position U max 

when y  0 ;  t  0 ; t  0

U min  0

(2) Kinetic energy : This is because of the velocity of the particle Kinetic Energy

K

1 mv 2 2

K

1 ma 2 2 cos 2  t 2

[As v  a cos  t ]

genius PHYSICS 8 Simple Harmonic Motion

1 [As v   a 2  y 2 ] m  2 (a 2  y 2 ) 2 (i) Kinetic energy is maximum at mean position and equal to total energy at mean position. K

1 when y  0 ; t  0 ;  t  0 m 2a2 2 (ii) Kinetic energy is minimum at extreme position. K max 

when y  a ; t  T / 4 ,  t   / 2

K min  0

(3) Total energy : Total mechanical energy = Kinetic energy + Potential energy

1 1 1 m  2 (a 2  y 2 )  m  2 y 2  m  2 a 2 2 2 2 Total energy is not a position function i.e. it always remains constant. E

(4) Energy position graph : Kinetic energy (K)  Potential Energy (U) =

Energy

1 m  2 (a 2  y 2 ) 2

U

1 m 2y 2 2

K

1 Total Energy (E) = m  2 a 2 2

y =– a

y =+ a

y=0

It is clear from the graph that (i) Kinetic energy is maximum at mean position and minimum at extreme position (ii) Potential energy is maximum at extreme position and minimum at mean position (iii) Total energy always remains constant. (5) Kinetic Energy Potential Energy

K

1 1 1 m  2 a 2 cos 2  t  m  2 a 2 (1  cos 2 t)  E(1  cos ' t) 2 4 2

U

1 1 1 m  2 a 2 sin 2  t  m  2 a 2 (1  cos 2 t)  E(1  cos  ' t) 2 2 4

where  '  2 and E  1 m  2 a 2 i.e. in S.H.M., kinetic energy and potential energy vary periodically with double the frequency of S.H.M. (i.e. with time period T '  T / 2 ) From the graph we note that potential energy or kinetic energy completes two vibrations in a time during which S.H.M. completes one vibration. Thus the frequency of potential energy or kinetic energy double than that of S.H.M.

Energy

2

Total energy 

1 2

2 2

m a E

KE  PE 

0

1 2

T T

1 2 1

2 Time

2

2

2

m  a cos t 2 2

2

m  a sin t

Sample problems based on Energy Problem 14. A particle is executing simple harmonic motion with frequency f. The frequency at which its kinetic energy changes into potential energy is (a) f / 2

(b) f

(c) 2 f

(d) 4 f

Solution : (c)

Problem 15. When the potential energy of a particle executing simple harmonic motion is one-fourth of the maximum value during the oscillation, its displacement from the equilibrium position in terms of amplitude ‘a’ is

genius PHYSICS Simple Harmonic Motion 9 [CBSE 1993; MP PMT 1994; MP PET 1995, 96; MP PMT 2000]

(a) a / 4 Solution : (c)

(b) a / 3

According 

to

(c) a / 2

problem

potential

(d) 2a / 3 energy

=

1 maximum 4

Energy

1 1 1 a2  y a/2 m  2 y 2   m  2a2   y 2  2 4 2 4 

Problem 16. A particle of mass 10 grams is executing S.H.M. with an amplitude of 0.5 meter and circular frequency of 10 radian/sec. The maximum value of the force acting on the particle during the course of o scillation is [MP PMT 2000]

(a) 25 N Solution : (d)

(b) 5 N

(c) 2.5 N

(d) 0.5 N

Maximum force = mass  maximum acceleration = m  2 a  10  10 3 (10) 2 (0 .5) = 0.5 N

Problem 17. A body is moving in a room with a velocity of 20 m/s perpendicular to the two walls separated by 5 meters. There is no friction and the collision with the walls are elastic. The motion of the body is

Solution : (b)

(a) Not periodic

(b) Periodic but not simple harmonic

(c) Periodic and simple harmonic

(d) Periodic with variable time period

[MP PMT 19

Since there is no friction and collision is elastic therefore no loss of energy take place and the body strike again and again with two perpendicular walls. So the motion of the ball is periodic. But here, there is no restoring force. So the characteristics of S.H.M. will not satisfied.

Problem 18. Two particles executes S.H.M. of same amplitude and frequency along the same straight line. They pass one another when going in opposite directions. Each time their displacement is half of their amplitude. The phase difference between them is (a) 30o Solution : (d)

(b) 60o

(c) 90o

(d) 120o

Let two simple harmonic motions are y  a sin  t and y  a sin( t   ) In the first case

a  a sin  t  sin  t  1 / 2 2

In the second case 



cos  t 

3 2

a  a sin ( t   ) 2

1    [sin  t. cos   cos  t sin  ]  1   1 cos   3 sin   2 2 2 2  

 1  cos   3 sin   (1  cos  )2  3 sin 2   (1  cos  ) 2  3(1  cos 2  )

cos   1 or cos   1 / 2

By solving we get

 0

i.e.

or   120 o

Problem 19. The acceleration of a particle performing S.H.M. is 12 cm/sec2 at a distance of 3 cm from the mean position. Its time period is (a) 0.5 sec Solution : (d)

A   2y   

(b) 1.0 sec A  y

(c) 2.0 sec

(d) 3.14 sec

2 2 12     3.14  2 ; but T   2 3

Problem 20. A particle of mass 10 gm is describing S.H.M. along a straight line with period of 2 sec and amplitude of 10 cm. Its kinetic energy when it is at 5 cm. From its equilibrium position is (a) 37 .5 2 erg

(b) 3.75 2 erg

(c) 375 2 erg

(d) 0 .375 2erg

genius PHYSICS 10 Simple Harmonic Motion Solution : (c)

Kinetic energy 

1 2 1 4 2 m  2 (a 2  y 2 )  10 (10 2  5 2 )  375 ergs . 2 2 4

Problem 21. The total energy of the body executing S.H.M. is E. Then the kinetic energy when the displacement is half of the amplitude is (a) E / 2 Solution : (c)

[RPET 1996]

(b) E / 4

Kinetic energy 

(c) 3E / 4

2  1 3E  m  2 (a 2  y 2 )  1 m  2  a 2  a   3  1 m  2 a 2  =   2 4 2 4  4 2  

(d)

3E / 4

[As y 

a ] 2

Problem 22. A body executing simple harmonic motion has a maximum acceleration equal to 24 m/sec2 and maximum velocity equal to 16 meter/sec. The amplitude of simple harmonic motion is (a) Solution : (a)

32 meters 3

(b)

3 meters 32

(c)

1024 meters 9

Maximum acceleration  2 a  24

…..(i)

and maximum velocity a  16

….(ii)

Dividing (i) by (ii)



[MP PMT 1995]

(d)

64 meters 9

3 2

Substituting this value in equation (ii) we get a  32 / 3meter

Problem 23. The displacement of an oscillating particle varies with time (in seconds) according to the equation. y(cm )  sin

t

1   . The maximum acceleration of the particle approximately 2 2 3

(a) 5.21 cm/sec2 Solution : (d)

(b) 3.62 cm/sec2

(c) 1.81 cm/sec2

(d) 0.62 cm/sec2

By comparing the given equation with standard equation, y  a sin( t   ) We find that a  1 and    / 4   2   3 . 14  2   0 . 62 cm / sec 2 Now maximum acceleration   2 a     4  4  

Problem 24. The potential energy of a particle executing S.H.M. at a distance x from the mean position is proportional to [Roorkee 1992]

(a)

x

(b) x

(c) x 2

(d) x 3

Solution : (c)

Problem 25. The kinetic energy and potential energy of a particle executing S.H.M. will be equal, when displacement is (amplitude = a) (a) a / 2 Solution : (c)

[MP PMT 1987; CPMT 1990]

(b) a 2

(c) a / 2

According to problem Kinetic energy = Potential energy 

(d)

a 2 3

1 1 m  2 (a 2  y 2 )  m  2 y 2 2 2

 a2  y 2  y 2  y  a / 2

Problem 26. The phase of a particle executing S.H.M. is (a) Maximum velocity

 when it has 2

(b) Maximum acceleration

(c)

Maximum energy

(d)

Solution : (b, d) Phase  / 2 means extreme position. At extreme position acceleration and displacement will be maximum.

genius PHYSICS Simple Harmonic Motion 11

Problem 27. The displacement of a particle moving in S.H.M. at any instant is given by y  a sin  t . The acceleration after time t 

T is (where T is the time period) 4

(a) a

[MP PET 1984]

(c) a 2

(b) a

(d)  a 2

Solution : (d)

Problem 28. A particle of mass m is hanging vertically by an ideal spring of force constant k, if the mass is made to oscillate vertically, its total energy is (a) Maximum at extreme position

(b) Maximum at mean position

(c) Minimum at mean position

(d)

Same at all position

Solution : (d)

15.11 Time Period and Frequency of S.H.M.. For S.H.M. restoring force is proportional to the displacement Fy

or F  ky

…(i) A   2 y

For S.H.M. acceleration of the body  Restoring force on the body

Time period (T ) 

or

Frequency (n) 

2



…(ii)

F  mA  m  2 y

m k

 2

1 1  T 2

…(iii)

k m

From (i) and (iii) ky  m  2 y    

where k is a force constant.

k m

In different types of S.H.M. the quantities m and k will go on taking different forms and names. In general m is called inertia factor and k is called spring factor. Thus

T  2

or

n

Inertia factor Spring factor Spring factor Inertia factor

1 2

In linear S.H.M. the spring factor stands for force per unit displacement and inertia factor for mass of the body executing S.H.M. and in Angular S.H.M. k stands for restoring torque per unit angular displacement and inertial factor for moment of inertia of the body executing S.H.M. For linear S.H.M. T  2 m  k

or

n

1 2

Acceleration Dispalcement

m Force/Disp lacement 

1 2

 2

A y

15.12 Differential Equation of S.H.M.. For S.H.M. (linear)

Acceleration  – (Displacement) A  y

m  Displacement  2 m  Accelerati on

Displacement  2 Acceleration

y A

genius PHYSICS 12 Simple Harmonic Motion or

A   2 y

or

d 2y   2 y dt 2 d 2y  ky  0 dt 2

[As  

or

m

For angular S.H.M.

2   c and d    2  0

k ] m

dt 2

where  2 

c [As c = Restoring torque constant and I = Moment of inertia] I

Sample problems based on Differential equation of S.H.M. Problem 29. A particle moves such that its acceleration a is given by a  bx . Where x is the displacement from equilibrium position and b is a constant. The period of oscillation is

[NCERT 1984; CPMT 1991; MP PMT 1994; MNR 1995]

(a) 2 b

(b)

2

(c)

b Solution : (b)

We know that Acceleration =   2 (displacement) and

2 b

d 2y dt 2

b

a  bx (given in the problem)

Comparing above two equation  2  b    b  Time period T 

Problem 30. The equation of motion of a particle is

(d) 2 

2





2 b

 ky  0 where k is a positive constant. The time period of the

motion is given by (a) Solution : (c)

2 k

(b) 2k

Standard equation m So, T  2

d 2y dt 2

(c)

2 k

(d) 2 k

 ky  0 and in a given equation m =1 and k = k

2 m  k k

15.13 Simple Pendulum. An ideal simple pendulum consists of a heavy point mass body suspended by a weightless, inextensible and perfectly flexible string from a rigid support about which it is free to oscillate. But in reality neither point mass nor weightless string exist, so we can never construct a simple pendulum strictly according to the definition. S Let mass of the bob = m



Length of simple pendulum = l Displacement of mass from mean position (OP) = x When the bob is displaced to position P, through a small angle  from the vertical. Restoring force acting on the bob

F  mg sin 

T l

y O

P

mg sin  mg mg cos

genius PHYSICS Simple Harmonic Motion 13

(When  is small sin  ~  

or

F  mg 

or

F  mg



F  mg   k (Spring factor) x l

So time period

T  2

OP x Arc = = ) l l Length

x l

Inertia factor Spring factor

m  2 mg / l

 2

l g

Important points (i) The period of simple pendulum is independent of amplitude as long as its motion is simple harmonic. But if  is not small, sin    then motion will not remain simple harmonic but will become oscillatory. In this situation if 0 is the amplitude of motion. Time period T  2

  0 2  l  1 2 0  1  sin  .......  T 1     0   g 16  22  2   

(ii) Time period of simple pendulum is also independent of mass of the bob. This is why (a) If the solid bob is replaced by a hollow sphere of same radius but different mass, time period remains unchanged. (b) If a girl is swinging in a swing and another sits with her, the time period remains unchanged. (iii) Time period T  l where l is the distance between point of suspension and center of mass of bob and is called effective length. (a) When a sitting girl on a swinging swing stands up, her center of mass will go up and so l and hence T will decrease. (b) If a hole is made at the bottom of a hollow sphere full of water and water comes out slowly through the hole and time period is recorded till the sphere is empty, initially and finally the center of mass will be at the center of the sphere. However, as water drains off the sphere, the center of mass of the system will first move down and then will come up. Due to this l and hence T first increase, reaches a maximum and then decreases till it becomes equal to its initial value. (iv) If the length of the pendulum is comparable to the radius of earth then T  2

(a) If l  R , then

1 1  l R

(b) If l  R( )1 / l  1 / R

so T  2

so T  2

l g

R 6 .4  10 6  2  84 .6 minutes g 10

and it is the maximum time period which an oscillating simple pendulum can have (c) If l  R

so

T  2

R  1hour 2g

1 1 1  g   R l

genius PHYSICS 14 Simple Harmonic Motion (v) If the bob of simple pendulum is suspended by a wire then effective length of pendulum will increase with the rise of temperature due to which the time period will increase. l  l0 (1    )

(If  is the rise in temperature, l0  initial length of wire, l = final length of

wire)

1 l  (1    )1 / 2  1    l0 2

T  T0

So

T 1  1    T0 2

i.e.

T 1    T 2

(vi) If bob a simple pendulum of density  is made to oscillate in some fluid of density  (where  vwater

> vair

5000 m/s > 1500 m/s > 330 m/s As for sound vwater > vAir while for light vw < vA. Water is rarer than air for sound and denser for light. The concept of rarer and denser media for a wave is through the velocity of propagation (and not density). Lesser the velocity, denser is said to be the medium and vice-versa. (2) Newton's formula : He assumed that when sound propagates through air temperature remains K P  constant.(i.e. the process is isothermal) vair = As K = E = P ; E = Isothermal elasticity; P =   Pressure. By calculation vair = 279 m/sec. However the experimental value of sound in air is 332 m/sec which is greater than that given by Newton's formula. (3) Laplace correction : He modified Newton's formula assuming that propagation of sound in air as adiabatic process. v=

k



=

E

(As k = E    = Adiabatic elasticity)



v = 1.41 × 279= 331.3 m/s (4) Effect of density : v =

(  Air  1.41 )

1 P v   

P  RT . Velocity of sound is independent of the pressure of gas   M provided the temperature remains constant. ( P   when T = constant) (5) Effect of pressure : v 

(6) Effect of temperature : v 

RT M

 v  T in K 

When the temperature change is small then vt = v0 (1 +  t) v0 = velocity of sound at 0°C, vt = velocity of sound at t°C ,  = temp-coefficient of velocity of sound. Value of  = 0.608

m/s = 0.61 (Approx.) o C

Temperature coefficient of velocity of sound is defined as the change in the velocity of sound, when temperature changes by 1°C. (7) Effect of humidity : With increase in humidity, density of air decreases. So with rise in humidity velocity of sound increases. This is why sound travels faster in humid air (rainy season) than in dry air (summer) at the same temperature. (8) Effect of wind velocity : Because wind drifts the medium (air) along its direction of motion therefore the velocity of sound in a particular direction is the algebraic sum of the velocity of sound and the component of wind velocity in that direction. Resultant velocity of sound along A SL = v + w cos. Wind (w) S

wcos

 Sound (v)

L

genius PHYSICS 6 Wave Motion (9) Sound of any frequency or wavelength travels through a given medium with the same velocity. (v = constant) For a given medium velocity remains constant. All other factors like phase, loudness pitch, quality etc. have practically no effect on sound velocity. (10) Relation between velocity of sound and root mean square velocity. vsound =

RT M

and vrms =

3 RT M

so

v rms = v sound

3



or

vsound = [  /3]1/2 vrms.

(11) There is no atmosphere on moon, therefore propagation of sound is not possible there. To do conversation on moon, the astronaut uses an instrument which can transmit and detect electromagnetic waves.

16.6 Reflection and Refraction of Waves. When sound waves are incident on a boundary between two media, a part of incident waves returns back into the initial medium (reflection) while the remaining is partly absorbed and partly transmitted into the second medium (refraction) In case of reflection and refraction of sound (1) The frequency of the wave remains unchanged that means

Incident wave

i = r = t =  = constant i r

(2) The incident ray, reflected ray, normal and refracted ray all lie in the same plane. (3) For reflection angle of incidence (i) = Angle of reflection (r)

t

Reflected wave

Transmitted wave

sin i v i (4) For refraction  sin t v t

(5) In reflection from a denser medium or rigid support, phase changes by 180° and direction reverses if incident wave is y = A1 sin (t  kx ) then reflected wave becomes y = Ar sin ( t  kx   ) = – Ar sin ( t  kx ) . (6) In reflection from a rarer medium or free end, phase does not change and direction reverses if incident wave is y = AI sin (  t  kx ) then reflected wave becomes y = Ar sin ( t  kx ) d Source

(7) Echo is an example of reflection. If there is a sound reflector at a distance d from the source then time interval 2d between original sound and its echo at the site of source will be t  v

Detecto r

16.7 Reflection of Mechanical Waves Medium Reflection from rigid end/denser medium Reflection from free end/rarer medium

Longitudinal wave Compression as rarefaction and vice-versa Compression as compression and rarefaction as rarefaction

Transverse wave

Change in direction

Crest as crest and Trough as trough

Reversed

Crest as trough and trough as crest

No change

Phase change 

Time change T 2

 2

Zero

Zero

Zero

16.8 Progressive Wave. (1) These waves propagate in the forward direction of medium with a finite velocity.

Path change

genius PHYSICS Wave Motion 7

(2) Energy and momentum are transmitted in the direction of propagation of waves without actual transmission of matter. (3) In progressive waves, equal changes in pressure and density occurs at all points of medium. (4) Various forms of progressive wave function. (i) y = A sin ( t – kx) (ii) y = A sin ( t –

2



where y = displacement A = amplitude  = angular frequency n = frequency k = propagation constant T = time period  = wave length v = wave velocity t = instantaneous time x = position of particle from origin

x)

x t (iii) y = A sin 2     T  

(iv) y = A sin

2



(vt – x)

x  (v) y = A sin   t   v 

Important points (a) If the sign between t and x terms is negative the wave is propagating along positive X-axis and if the sign is positive then the wave moves in negative X-axis direction. (b) The coefficient of sin or cos functions i.e. Argument of sin or cos function i.e. ( t - kx) = Phase. (c) The coefficient of t gives angular frequency  = 2 n 

2 = vk. T

(d) The coefficient of x gives propagation constant or wave number k 

2





 v

(e) The ratio of coefficient of t to that of x gives wave or phase velocity. i.e. v =

.  . k

(f) When a given wave passes from one medium to another its frequency does not change. (g) From v = n  v   n = constant 

v 1 1 .  v 2 2

(5) Some terms related to progressive waves (i) Wave number ( n ) : The number of waves present in unit length is defined as the wave number ( n ) =

1 .  Unit = meter–1 ; Dimension = [L–1]. (ii) Propagation constant (k) : k = k

 v



Phase difference between particles  = Distance between them x

Angular velocity 2 and k =  2  Wave velocity 

(iii) Wave velocity (v) : The velocity with which the crests and troughs or compression and rarefaction     . travel in a medium, is defined as wave velocity v = = n  2 T k

genius PHYSICS 8 Wave Motion (iv) Phase and phase difference : Phase of the wave is given by the argument of sine or cosine in the 2 equation of wave. It is represented by  (x , t)  (vt  x ) .



(v) At a given position (for fixed value of x) phase changes with time (t).

d 2v 2 2 2  d =    Time difference. .dt  Phase difference = T dt  T T (vi) At a given time (for fixed value of t) phase changes with position (x). d 2 2 2   d   dx  Phase difference =  Path difference dx     Time difference =

T × Path difference 

Sample problems based on Progressive wave Problem 1.

The speed of a wave in a certain medium is 960 m/sec. If 3600 waves pass over a certain point of the medium in 1 minute, the wavelength is [MP PMT 2000] (a) 2 meters

(b) 4 meters 3600 Hz . 60

So  

(c) 8 meters

(d) 16 meters

v 960   16 meters. n 60

Solution : (d)

v = 960 m/s ; n =

Problem 2.

A simple harmonic progressive wave is represented by the equation y = 8 sin 2  (0.1x – 2t) where x and y are in cm and t is in seconds. At any instant the Phase difference between two particles separated by 2.0 cm in the x-direction is [MP PMT 2000] (a) 18°

Solution : (d)

(b) 36°

(d) 72°

 x   t x  2 t  given by comparing with standard equation y = a sin 2     y = 8 sin 2   10 T      10 cm So Phase Difference =

Problem 3.

(c) 54°

2



× path difference =

2 2  2   180  = 72° 10 5

The frequency of sound wave is n and its velocity is v if the frequency is increased to 4n the velocity of the wave will be [MP PET 2000] (a) v

(b) 2v

(c) 4v

(d) v/4

Solution : (a)

Wave velocity does not depends on the frequency. It depends upon the Elasticity and inertia of the medium.

Problem 4.

The displacement of a particle is given by x = 3 sin (5  t) + 4 cos (5  t ) The amplitude of particle is [MP PMT 1999]

(a) 3 Solution : (c)

(b) 4

(c) 5

(d) 7

Standard equation : x = a sin  t + b cos  t





x  a 2  b 2 sin  t  tan 1 b / a Given equation x  3 sin 5 t   4 cos(5 t)



x  9  16 sin 5  t  tan 1 4 / 3



x  5 sin( 5 t  tan 1 (4 / 3))

Problem 5.

The equation of a transverse wave travelling on a rope is given by y  10 sin  (0.01 x  2.00 t) where y and x are in cm and t in seconds. The maximum transverse speed of a particle in the rope is about [MP PET 1999] (a) 63 cm/sec

(b) 75 cm/s

(c) 100 cm/sec

(d) 121 cm/sec

genius PHYSICS Wave Motion 9 Solution : (a)

Standard eq. of travelling wave y = A sin (k x  t) By comparing with the given equation y = 10 sin (0.01  x  2 t ) A = 10 cm,  = 2  Maximum particle velocity = A  = 2  × 10 = 63 cm/sec

Problem 6.

In a wave motion y = a sin (k x   t), y can represents (a) Electric Field

(b) magnetic field

(c) Displacement

(d) Pressure

Solution : (a,b,c,d)

Problem 7.

Find the ratio of the speed of sound in nitrogen gas to that of helium gas, at 300 k is (a)

Solution : (c)

v

1 2

(b)

(c)

3 5

7/5 4 3 . = . 5 / 3 28 5

The displacement x (in metres) of a particle performing simple harmonic motion is related to time t (in

(a) 0.5 Hz (b) 1.0 Hz From the given equation, coefficient of t =   4 

Problem 9.

Solution : (a)

4 5

M

  seconds) as x = 0.05 cos  4 t   . The frequency of the motion will be 4  Solution : (d)

(d)

 RT

 N 2 M He vN   v He  He M N 2

Problem 8.

2 3

n=

 2



(c) 1.5 Hz

[MP PMT / PET 1998]

(d) 2.0 Hz

4  2 Hz 2

  A wave is represented by the equation Y = 7 sin  7 t  0 .04 x   x is in meters and t is in seconds. 3  The speed of the wave is [MP PET 1996] 49 m/s (a) 175 m/sec (b) 49  m/s (c) (d) 0.28  m / s  Standard equation y = A sin  t  kx   0  In a given equation   7  , k  0.04 v

 k



7 = 175 m/sec . 04 

Problem 10. A wave is represented by the equation y = 0.5 sin (10 t + x)m. It is a travelling wave propagating along the Solution : (a)

x direction with velocity. (a) 10 m/s (b) 20 m/s v =  / k  10 / 1  10 m / s

[Roorkee 1995]

(c) 5 m/s

(d) None of these

Problem 11. A transverse progressive wave on a stretched string has a velocity of 10 ms–1 and a frequency of 100 Hz.

Solution : (d)

The phase difference between two particles of the string which are 2.5 cm apart will be (a)  / 8 (b)  / 4 (c) 3 / 8 (d)  / 2 10  v / n   0 .1 m  10 cm 100 2  2 Phase difference = × path difference  × 2.5 = 10 2 

Problem 12. In a stationary wave, all particles are (a) At rest at the same time twice in every period of oscillation (b) At rest at the same time only once in every period of oscillation

[MP PMT 1994]

genius PHYSICS 10 Wave Motion (c) Never at rest at the same time (d) Never at rest at all Solution : (a) Problem 13. The path difference between the two waves

2 x  2x    y1  a1 sin   t     is  and y 2  a 2 cos   t        (a) Solution : (b)

 2



(b)

      2  2

(c)

[MP PMT 1994]

2        2

(d)

2



 

2x  2x    y1  a1 sin   t      ; y 2  a2 sin t    2   

 2x   2x    Phase difference =   t        t        2     2  Path difference =

    × Phase difference     2 2 2 

x  Problem 14. A plane wave is described by the equation y  3 cos   10 t   . The maximum velocity of the particles 4

Solution : (a)

of the medium due to this wave is (a) 30 (b) 3  / 2 Maximum velocity = A  = 3 × 10 = 30

2

(c) 3/4

(d) 40



Problem 15. A wave represented by the given equation y = A sin (10  x  15  t  ) where x is in meter and t is in 3 second. The expression represents (a) A wave travelling in the positive x-direction with a velocity of 1.5 m/sec (b) A wave travelling in the negative x-direction with a velocity of 1.5 m/sec (c) A wave travelling in the negative x-direction with a wavelength of 0.2 m (d) A wave travelling in the positive x-direction with a wavelength of 0.2 m

Solution : (b, c) By comparing with standard equation Y = A sin k x   t   / 3  K = 10  ,   15  We know that : v 

 k

= 1.5 m/sec;  

2 = 0.2 meter. k

x Problem 16. A transverse wave is described by the equation Y  y 0 sin 2   ft   The maximum particle velocity is 



four times the wave velocity if (a)   Solution : (b)

y 0

(b)  

4

y 0 2

(c)    y 0

(d)   2y 0

Maximum particle velocity = 4 wave velocity A = 4 f  y0 2  f  4 f



y 0 2

Problem 17. The equation of a wave travelling in a string can be written as y = 3 cos  (100 t – x) Its wavelength is [MP PMT 1991, 94, 97; MNR 1985]

(a) 100 cm Solution : (b)

(b) 2 cm

y = A cos (  t  k x)  standard equation

(c) 5 cm

(d) None of these

genius PHYSICS Wave Motion 11 y = 3 cos (100  t   x ) – given equation So K =  and  

2  2 cm k

Problem 18. A plane wave is represented by x = 1.2 sin (314 t + 12.56 y) where x and y are distances measured along in x and y direction in meter and t is time in seconds. This wave has

[MP PET 1991]

(a) A wave length of 0.25 m and travels m + ve x-direction (b) A wavelength of 0.25 m and travels in + ve y-direction (c) A wavelength of 0.5 m and travels in – ve y-direction (d) A wavelength of 0.5 m and travels in – ve x-direction Solution : (c)

From given equation k = 12.56



2 0.5 m direction = – y k

Problem 19. A wave is reflected from a rigid support. The change in phase on reflection will be (a)  / 4

(b)  / 2

(d) 2

(c) 

Solution : (c)

 Problem 20. The equation of displacement of two waves are given as y1 = 10 sin  3  t   ; 

3

y2 = 5

[sin 3  t  3 cos 3 t]

Then what is the ratio of their amplitudes (a) 1 : 2 Solution : (c)

(b) 2 : 1

y2 = 5 [sin 3  t  3 cos 3  t]

=5

[AIIMS 1997]

(c) 1 : 1

(d) None of these

    1  3 sin  3 t   = 10 sin  3  t   3 3   

So, A1 = 10 and A2 = 10

Problem 21. The equation of a wave travelling on a string is y = 4 sin



x  8 t   if x and y are in cm, then velocity of 2 8

wave is

Solution : (d)

[MP PET 1990]

(a) 64 cm/sec in – x direction

(b) 32 cm/sec in – x direction

(c) 32 cm/sec in +x direction

(d) 64 cm/sec in +x direction

   .x y = 4 sin  4 t  16  

  4 , k   / 16 v

 k



4  64 cm / sec in + x direction.  / 16

Problem 22. The equation of wave is y = 2 sin  (0.5 x  200 t) where x and y are expressed in cm and t in sec. The wave velocity is (a) 100 cm/sec Solution : (d)

v

 k



[MP PMT 1986]

(b) 200 cm/sec

(c) 300 cm/sec

(d) 400 cm/sec

200  = 400 cm/sec 0 .5 

16.9 Principle of Superposition. The displacement at any time due to any number of waves meeting simultaneously at a point in a medium is the vector sum of the individual displacements due each one of the waves at that point at the same time.

genius PHYSICS 12 Wave Motion    If y 1 , y 2, y 3 ………. are the displacements at a particular time at a particular position, due to individual     waves, then the resultant displacement. y  y1  y 2  y 3  .......... ... Examples (i) Radio waves from different stations having different frequencies cross the antenna. But our T.V/Radio set can pick up any desired frequency. (ii) When two pulses of equal amplitude on a string approach each other [fig. (A)], then on meeting, they superimpose to produce a resultant pulse of zero amplitude [fig (B)]. After crossing, the two pulses travel independently as shown in [fig (C)] as if nothing had happened.

(A) (B)

(C)

Important applications of superposition principle : (a) Interference of waves

(b) Stationary waves

(c) Beats.

16.10 Interference of Sound Waves. When two waves of same frequency, same wavelength, same velocity (nearly equal amplitude) moves in the same direction, Their superimposition results in the interference. Due to interference the resultant intensity of sound at that point is different from the sum of intensities due to each wave separately. This modification of intensity due to superposition of two or more waves is called interference. Let at a given point two waves arrives with phase difference  and the equation of these waves is given by y1 = a1 sin  t , y2 = a2 sin (  t   ) then by the principle of superposition    y  y 1  y 2  y = A sin  t    where A  a1 2  a 2 2  2a1 a 2 cos  and tan  =

a2 sin  a1  a 2 cos 

and since Intensity  A 2 . So I = a1 2  a 2 2  2 a1 , a 2 cos   I  I1  I 2  2 I1 I 2 cos 

Important points (1) Constructive interference : Intensity will be maximum when   0, 2  , 4 ,...... 2 n ; where n = 0,1,2 …….. when x = 0, , 2, ……….. n; where n = 0, 1 ……. Imax = I1 + I2 + 2 I1 I 2 =



I1  I 2



2

  A1  A 2 2

It means the intensity will be maximum at those points where path difference is an integral multiple of wavelength  . These points are called points of constructive interference or interference maxima. (2) Destructive interference : Intensity will be minimum when    , 3 ,5 ...... ( 2 n  1) ; where n= 1, 2, 3 …….. when x = /2, 3/2, ……….. (2n-1); where n = 1, 2, 3…….. Imin = I1+I2 – 2

I1 I 2  Imin =



I1  I 2



2

  A1 ~ A 2 

2

genius PHYSICS Wave Motion 13

(3) All maxima are equally spaced and equally loud. Same is also true for minima. Also interference  3 5  maxima and minima are alternate as for maximum x  0, ,2 .... etc . and for minimum x  , , ....... etc . 2 2 2

I (4) max  I min

I I

1

 I2

1

~

I2

 

2 2



 A1  A 2 2  A1 ~ A 2 2

with

I1 A1 2  I2 A2 2

(5) If I1  I 2  I0 then Imax = 4 I o and Imin = 0



I1  I 2



2

exceeds

the

sum

of

Imax  ( I1  I2 )2

I

(6) In interference the intensity in maximum

2 I1 I 2

individual

intensities (I1 + I2) by an amount 2 I1 I 2 while in minima



I1 ~

amount 2 I1 I 2 .

I2



2

lacks I1  I 2  by

the same

I1  I2

2 I1 I 2 Imin  ( I1 ~

Phase Diff.0



2

Path Diff. 0

/



2

3

4

5

6

3/2 2 5/2 3

I2 )2

 x

Hence in interference energy is neither created nor destroyed but is redistributed.

16.11 Standing Waves or Stationary Waves. When two sets of progressive wave trains of same type (both longitudinal or both transverse) having the same amplitude and same time period/frequency/wavelength travelling with same speed along the same straight line in opposite directions superimpose, a new set of waves are formed. These are called stationary waves or standing waves. Characteristics of standing waves : (1) The disturbance confined to a particular region between the starting point and reflecting point of the wave. (2) There is no forward motion of the disturbance from one particle to the adjoining particle and so on, beyond this particular region. (3) The total energy associated with a stationary wave is twice the energy of each of incident and reflected wave. But there is no flow or transference of energy along the stationary wave. (4) There are certain points in the medium in a standing wave, which are permanently at rest. These are called nodes. The distance between two consecutive nodes is

 2

(5) Points of maximum amplitude is known as antinodes. The distance between two consecutive antinodes is also  / 2 . The distance between a node and adjoining antinode is  / 4 . (6)The medium splits up into a number of segments. Each segment is vibrating up and down as a whole. (7) All the particles in one particular segment vibrate in the same phase. Particles in two consecutive segments differ in phase by 180°. (8) All the particles except those at nodes, execute simple harmonic motion about their mean position with the same time period. (9) The amplitude of vibration of particles varies from zero at nodes to maximum at antinodes. (10) Twice during each vibration, all the particles of the medium pass simultaneously through their mean position. (11) The wavelength and time period of stationary waves are the same as for the component waves. (12) Velocity of particles while crossing mean position varies from maximum at antinodes to zero at nodes.

genius PHYSICS 14 Wave Motion (13) In standing waves, if amplitude of component waves are not equal. Resultant amplitude at nodes will be minimum (but not zero). Therefore, some energy will pass across nodes and waves will be partially standing.

16.12 Standing Waves on a String. When a string under tension is set into vibration, transverse harmonic waves propagate along its length. When the length of string is fixed, reflected waves will also exist. The incident and reflected waves will superimpose to produce transverse stationary waves in a string Incident wave y1 = a sin

2

Reflected wave y 2  a sin

 2



x=L

x=0

(vt + x)

y1

vt  x       a sin 2 vt  x 

y2



According to superposition principle : y = y1 + y2 = 2 a cos General formula for wavelength   of the string.

2vt



sin

2x



2L where n = 1,2,3, … correspond to 1st , 2nd, 3rd modes of vibration n

(1) First normal mode of vibration n1 

v

1 v  n1  2L 2L



1

T m

A

N

This mode of vibration is called the fundamental mode and the frequency is called fundamental frequency. The sound from the note so produced is called fundamental note or first harmonic. (2) Second normal mode of vibration : n 2 

v

2



v 2v   2(n1 ) L 2L

L

This is second harmonic or first over tone. (3) Third normal mode of vibration : n3 

v

3



3v  3n1 2L

N

L 2L 3 L , , .......... .. L n n n

For first mode of vibration

x=0,x=L

For second mode of vibration

x = 0, x =

L ,x  L 2

[Three nodes]

For third mode of vibration

x=0,x=

L 2L ,x  ,x L 3 3

[Four nodes]

Position of antinodes : x 

[Two nodes]

2  1L L 3L 5L , , .......... . 2n 2n 2n 2n

For first mode of vibration x = L/2 For second mode of vibration x =

2

A

N

L = 2

This is third harmonic or second overtone. Position of nodes : x  0,

1

N

A

N

N

L 3L , 4 4

16.13 Standing Wave in a Closed Organ Pipe.

[One antinode] [Two antinode]

A

N

A

N

A

N

genius PHYSICS Wave Motion 15

Organ pipes are the musical instrument which are used for producing musical sound by blowing air into the pipe. Longitudinal stationary waves are formed on account of superimposition of incident and reflected longitudinal waves. Equation of standing wave y = 2a cos General formula for wavelength  

2vt



sin

2x



A

N

4L 2n  1

(1) First normal mode of vibration : n1 

L

1 2

v 4L

This is called fundamental frequency. The note so produced is called fundamental note or first harmonic. (2) Second normal mode of vibration : n 2 

v

2



A

N

3v  3 n1 4L

N

L

A

3 2 4

This is called third harmonic or first overtone. (3) Third normal mode of vibration : n 3 

N

5v  5n1 4L

N

A

N

A

A

This is called fifth harmonic or second overtone. Position of nodes : x = 0,

2L 4L 6L 2nL , , ......... (2n  1) (2n  1) (2n  1) (2n  1)

For first mode of vibration x = 0 For second mode of vibration For third mode of vibration Position of antinode : x =

[One node]

x = 0, x = x = 0, x 

2L 3

[Two nodes]

2L 4 L , 5 5

[Three nodes]

L 3L 5L , , ........., L 2n  1 2n  1 2n  1

For first mode of vibration

x=L

[One antinode]

For second mode of vibration

x=

L ,x  L 3

[Two antinode]

For third mode of vibration

x=

L 3L , ,L 5 5

[Three antinode]

16.14 Standing Waves in Open Organ Pipes. General formula for wavelength

2L  n

where n = 1,2,3 ………

(1) First normal mode of vibration : n 1 

N

A

v

1



L

v 2L

A

1 2

N A

N A L = 2

A

genius PHYSICS 16 Wave Motion This is called fundamental frequency and the note so produced is called fundamental note or first harmonic. (2) Second normal mode of vibration n 2 

v

2



v  v   2   2n1  n 2  2n1 L  2L 

This is called second harmonic or first overtone. A

v

3v (3) Third normal mode of vibration n 3  , n3  3n1  3 2L

N

N

A

L

This is called third harmonic or second overtone.

N A

A

3 3 2

Important points (i) Comparison of closed and open organ pipes shows that fundamental note in open organ pipe v  v     n1   has double the frequency of the fundamental note in closed organ pipe  n1  . 2 L 4 L   

Further in an open organ pipe all harmonics are present whereas in a closed organ pipe, only alternate harmonics of frequencies n1 ,3n1 , 5 n1 , ...... etc are present. The harmonics of frequencies 2n1, 4n1, 6n1 …… are missing. Hence musical sound produced by an open organ pipe is sweeter than that produced by a closed organ pipe. (ii) Harmonics are the notes/sounds of frequency equal to or an integral multiple of fundamental frequency (n). Thus the first, second, third, harmonics have frequencies n1 ,2n1 ,3n1 , 0.6 r

……… (iii) Overtones are the notes/sounds of frequency twice/thrice/ four times the fundamental frequency (n) eg. 2 n,3n,4 n ……. and so on. (iv) In organ pipe an antinode is not formed exactly at the open end rather it is formed a little distance away from the open end outside it. The distance of antinode from the open end of the pipe is known as end correction.

16.15 Vibration of a String. Fundamental frequency n  General formula n p 

p 2L

1 2L

T m

T m

Fixed end

N

A

N

Fixed end

Plucking

L = Length of string, T = Tension in the string m = Mass per unit length (linear density), p = mode of vibration

Important points (1) As a string has many natural frequencies, so when it is excited with a tuning fork, the string will be in resonance with the given body if any of its natural frequencies concides with the body.

genius PHYSICS Wave Motion 17

(2) (i) n 

1 L

if T and m are constant (ii) n  T

if L and m are constant (iii) n 

1

if T and L are

m

constant (3) If M is the mass of the string of length L, m  So n 

1 2L

T 1  m 2L

T 1  M/L 2

M L

T 1 T 1 =  ML 2 L r 2  2 Lr

T



where m = r2 (r = Radius,  = Density)

16.16 Comparative Study of Stretched Strings, Open Organ Pipe and Closed Organ Pipe S. No.

Parameter

Stretched string

Open organ pipe

(1)

Fundamental frequency or 1st harmonic

n1 

(2)

Frequency of 1st overtone or 2nd harmonic

n 2  2n1

n 2  2n1

Missing

(3)

Frequency of 2nd overtone or 3rd harmonic

n 3  3n1

n 3  3n1

n 3  3n1

(4)

Frequency ratio of overtones

2 : 3 : 4…

2 : 3 : 4…

3 : 5 : 7…

(5)

Frequency ratio of harmonics

1 : 2 : 3 : 4…

1 : 2 : 3 : 4…

1 : 3 : 5 : 7…

(6)

Nature of waves

Transverse stationary

Longitudinal stationary

Longitudinal stationary

v 2l

n1 

v 2l

Closed organ pipe n1 

v 4l

16.17 Beats. When two sound waves of slightly different frequencies, travelling in a medium along the same direction, superimpose on each other, the intensity of the resultant sound at a particular position rises and falls regularly with time. This phenomenon of regular variation in intensity of sound with time at a particular position is called beats.

Important points (1) One beat : If the intensity of sound is maximum at time t = 0, one beat is said to be formed when intensity becomes maximum again after becoming minimum once in between. (2) Beat period : The time interval between two successive beats (i.e. two successive maxima of sound) is called beat period. (3) Beat frequency : The number of beats produced per second is called beat frequency. (4) Persistence of hearing : The impression of sound heard by our ears persist in our mind for 1/10th of a second. If another sound is heard before 1/10 second is over, the impression of the two sound mix up and our mind cannot distinguish between the two. So for the formation of distinct beats, frequencies of two sources of sound should be nearly equal (difference of frequencies less than 10) (5) Equation of beats : If two waves of equal amplitudes 'a' and slightly different frequencies n1 and n2 travelling in a medium in the same direction are. y1 = a sin 1 t  a sin 2 n1 t ; y 2  a sin  2 t  a sin 2 n 2 t    By the principle of super position : y  y 1  y 2 y = A sin  (n1  n 2 ) t

where A = 2a cos  (n1  n 2 ) t = Amplitude of resultant wave.

genius PHYSICS 18 Wave Motion (6) Beat frequency : n = n1 ~ n2. (7) Beat period : T=

1 1  Beat frequency n1 ~ n 2

16.18 Determination of Unknown Frequency. Let n2 is the unknown frequency of tuning fork B, and this tuning fork B produce x beats per second with another tuning fork of known frequency n1. As number of beat/sec is equal to the difference in frequencies of two sources, therefore n2 = n1  x The positive/negative sign of x can be decided in the following two ways : By loading

By filing

If B is loaded with wax so its frequency decreases

If B is filed, its frequency increases

If number of beats decreases n2 = n1 + x

If number of beats decreases n2 = n1 – x

If number of beats Increases n2 = n1 – x

If number of beats Increases n2 = n1 + x

If number of beats remains unchanged n2 = n1 + x

If number of beats remains unchanged n2 = n1 – x

If number of beats becomes zero n2 = n1 + x

If number of beats becomes zero n2 = n1 – x

If A is loaded with wax its frequency decreases

If A is filed, its frequency increases

If number of beats decreases n2 = n1 – x

If number of beats decreases n2 = n1 + x

If number of beats increases n2 = n1+ x

If number of beats Increases n2 = n1 – x

If number of beats remains unchanged n2 = n1 – x

If number of beats remains unchanged n2 = n1 + x

If number of beats becomes zero n2 = n1 – x

If no of beats becomes zero n2 = n1 + x

Sample problems based on Superposition of waves Problem 23. The stationary wave produced on a string is represented by the equation y = 5 cos 

x 

 sin (40  t) where x  3 

and y are in cm and t is in seconds. The distance between consecutive nodes is (b)  cm

(a) 5 cm Solution : (c)

(c) 3 cm

(d) 40 cm

By comparing with standard equation of stationary wave y = a cos We get

2x



2 x



=

sin

2  vt



 x   = 6; Distance between two consecutive nodes =  3 cm 3 2

Problem 24. On sounding tuning fork A with another tuning fork B of frequency 384 Hz, 6 beats are produced per second. After loading the prongs of A with wax and then sounding it again with B, 4 Beats are produced per second what is the frequency of the tuning fork A. (a) 388 Hz

(b) 80 Hz

(c) 378 Hz

Solution : (c) 390

390

256

384

378 YA

378 YB

(d) 390 Hz

genius PHYSICS Wave Motion 19 Probable frequency of A is 390 Hz and 378 Hz and After loading the beats are decreasing from 6 to 4 so the original frequency of A will be n2 = n 1 – x = 378 Hz.

Problem 25. Two sound waves of slightly different frequencies propagating in the same direction produces beats due to [MP PET 2000]

(a) Interference

(b) Diffraction

(c) Polarization

(d) Refraction

Solution : (a)

Problem 26. Beats are produced with the help of two sound waves on amplitude 3 and 5 units. The ratio of maximum to minimum intensity in the beats is (a) 2 : 1 2

Solution : (d)

[MP PMT 1999]

(b) 5 : 3

(c) 4 : 1

(d) 16 : 1

2

 53  Imax  A1  A2   = 16:1  =     Imin 5 3  A1  A2 

Problem 27. Two tuning forks have frequencies 380 and 384 hertz respectively. When they are sounded together, they produce 4 beats. After hearing the maximum sound, how long will it take to hear the minimum sound [MP PMT/PET 1998]

(a) 1/2 sec Solution : (c)

(b) 1/4 sec

(c) 1/8 sec

(d) 1/16 sec

Beats period = Time interval between two minima T=

1 1  sec n1  n2 4

Time interval between maximum sound and minimum sound = T/2 = 1/8 sec

Problem 28. Two tuning fork A and B give 4 beats per second when sounded together. The frequency of A is 320 Hz. When some wax is added to B and it is sounded with A, 4 beats per second are again heard. The frequency of B is (a) 312 Hz Solution : (c)

(b) 316 Hz

(c) 324 Hz

(d) 328 Hz

Since there is no change in beats. Therefore the original frequency of B is n2 = n 1 + x = 320 + 4 = 324

Problem 29. 41 forks are so arranged that each produces 5 beat/sec when sounded with its near fork. If the frequency of last fork is double the frequency of first fork, then the frequencies of the first and last fork respectively [MP PMT 1997]

(a) 200, 400 Solution : (a)

(b) 205, 410

(c) 195, 390

(d) 100, 200

Let the frequency of first tuning fork = n and that of last = 2n n, n + 5, n + 10, n + 15 …… 2n this forms A.P. Formula of A.P l = a + (N – 1) r where l = Last term, a = First term, N = Number of term, r = Common difference 2n = n + (41 – 1) 5 2n = n + 200 n = 200

and

2n = 400

Problem 30. In stationary waves, antinodes are the points where there is (a) Minimum displacement and minimum pressure change (b) Minimum displacement and maximum pressure change (c) Maximum displacement and maximum pressure change (d) Maximum displacement and minimum pressure change Solution : (d)

At Antinodes displacement is maximum but pressure change is minimum.

[MP PMT 1996]

genius PHYSICS 20 Wave Motion

Problem 31. The equation y = 0.15 sin 5x cos 300 t, describes a stationary wave. The wavelength of the stationary wave is [MP PMT 1995]

(a) Zero meter Solution : (b)

(b) 1.256 meter

By comparing with standard equation 

(c) 2.512 meter

2x



5x   

Problem 32. The equation of a stationary wave is y = 0.8 cos 

(d) 0.628 meter

2   = 1.256 meter 5

x 

 sin 200  t where x is in cm. and t is in sec. The  20 

separation between consecutive nodes will be (a) 20 cm Solution : (a)

(b) 10 cm

Standard equation y = A cos

2x



sin

(c) 40 cm

2vt



By comparing this equation with given equation. Distance Between two nodes =

(d) 30 cm

2x x    = 40 cm  20

 = 20 cm. 2

Problem 33. Which of the property makes difference between progressive and stationary waves (a) Amplitude Solution : (c)

(b) Frequency

(c) Propagation of energy (d) Phase of the wave

In stationary waves there is no transfer of energy.

Problem 34. If amplitude of waves at distance r from a point source is A, the amplitude at a distance 2r will be [MP PMT 1985]

(a) 2A Solution : (c)

I  A2 and I 

(b) A

1 r

2

so r 

(c) A/2 1 ; A

r1 A  2  A2 = A1 r2 A1

(d) A/4  r1   r2

 1  = A   = A/2 2 

Problem 35. If two waves of same frequency and same amplitude respectively on superimposition produced a resultant disturbance of the same amplitude the wave differ in phase by (a)  Solution : (b)

(b) 2 / 3

(c)  / 2

[MP PMT 1990]

(d) zero

A  A1 2  A 2 2  2 A1 A 2 cos  A2 = A2 + A2 + 2A2 cos  [A1 = A2 = A given] cos  = – 1/2    120  

2 3

Problem 36. The superposition takes place between two waves of frequency f and amplitude a. The total intensity is directly proportional to (a) a Solution : (d)

2 I  a1  a2 

[MP PMT 1986]

(b) 2a

(c)

2a2

(d)

4a2

[As a1 = a2 = a]

I  4a2

Problem 37. The following equation represent progressive transverse waves z1 = A cos (  t – kx) z2 = A cos (  t + kx)

[MP PET 1993]

genius PHYSICS Wave Motion 21 z3 = A cos (  t + ky) z4 = A cos ( 2  t – 2ky) A stationary wave will be formed by superposing (a) z1 and z2 Solution : (a)

(b) z1 and z4

(c) z2 and z3

(d) z3 and z4

The direction of wave must be opposite and frequencies will be same then by superposition, standing wave formation takes place.

Problem 38. When two sound waves with a phase difference of  / 2 and each having amplitude A and frequency  are superimposed on each other, then the maximum amplitude and frequency of resultant wave is [MP PMT 1989] (a)

A

; / 2

(b)

2 Solution : (d)

A

;

(c)

2A;

2

Resultant Amplitude =

a12  a2 2  2a1a2 cos  =

 2

A 2  A 2  2 A 2 cos

(d)

 2

=

2 A; 

2A

and frequency remains same = .

Problem 39. There is a destructive interference between the two waves of wavelength  coming from two different paths at a point. To get maximum sound or constructive interference at that point, the path of one wave is to be increased by [MP PET 1985] (a)  / 4 Solution : (b)

(b)  / 2

(c)

3 4

Destructive interference means the path difference is 2n  1 

(d) 

 2

If it is increased by  / 2 Then new path difference (2n – 1)

   = n 2 2

which is the condition of constructive interference.

Problem 40. The tuning fork and sonometer wire were sounded together and produce 4 beats/second when the length of sonometer wire is 95 cm or 100 cm. The frequency of tuning fork is (a) 156 Hz Solution : (a)

Frequency n 

(b) 152 Hz

1 l

 As n 

1 2l

(c) 148 Hz

[MP PMT 1990]

(d) 160 Hz

T m

If n is the frequency of tuning fork. n+4 

1 1  n–4   (n + 4 ) 95 = (n – 4) 100  n = 156 Hz. 95 100

Problem 41. A tuning fork F1 has a frequency of 256 Hz and it is observed to produce 6 beats/second with another tuning fork F2. When F2 is loaded with wax. It still produces 6 beats/second with F1. The frequency of F2 before loading was [MP PET 1990] (a) 253 Hz Solution : (b)

(b) 262 Hz

(c) 250 Hz

(d) 259 Hz

No of beats does not change even after loading then n2 = n1 + x = 256 + 6 = 262 Hz.

16.19 Doppler Effect. Whenever there is a relative motion between a source of sound and the listener, the apparent frequency of sound heard by the listener is different from the actual frequency of sound emitted by the source.

genius PHYSICS 22 Wave Motion When the distance between the source and listener is decreasing the apparent frequency increases. It means the apparent frequency is more than the actual frequency of sound. The reverse is also true. General expression for apparent frequency n' =

v  v m   v L n v  v m   v S 

S

Here n = Actual frequency; vL = Velocity of listener; vS = Velocity of source vm = Velocity of medium and v = Velocity of sound wave Sign convention : All velocities

L

v vL

vS

along the direction S to L are taken as positive and all velocities along the direction L to S are taken as  v  vL   n negative. If the medium is stationary vm = 0 then n' =   v  vS  Special cases : (1) Source is moving towards the listener, but the listener at rest n' 

v .n v  vS

(2) Source is moving away from the listener but the listener is at rest n' 

v .n v  vS

v  vL n v v  vL (4) Source is at rest and listener is moving towards the source n'  .n v  v  vL   n (5) Source and listener are approaching each other n'    v  vS  (3) Source is at rest and listener is moving away from the source n' 

 v  vL  n (6) Source and listener moving away from each other n'    v  vS  (7) Both moves in the same direction with same velocity n' = n, i.e. there will be no Doppler effect because relative motion between source and listener is zero. (8) Source and listener moves at right angle to the direction of wave propagation. n' = n It means there is no change in frequency of sound heard if there is a small displacement of source and listener at right angle to the direction of wave propagation but for a large displacement the frequency decreases because the distance between source of sound and listener increases.

Important points seen.

(i) If the velocity of source and listener is equal to or greater than the sound velocity then Doppler effect is not

(ii) Doppler effect gives information regarding the change in frequency only. It does not says about intensity of sound. (iii) Doppler effect in sound is asymmetric but in light it is symmetric.

16.20 Some Typical Features of Doppler’s Effect in Sound. (1) When a source is moving in a direction making an angle  w.r.t. the listener : The apparent frequency heard by listener L at rest L (Listener at rest) nv When source is at point A is n   v  v s cos  vs cos As source moves along AB, value of  increases, cos decreases, n  goes on decreasing.



s A

 vs

C

vs B

vs cos

genius PHYSICS Wave Motion 23

At point C,   90 o , cos   cos 90 o  0 , n  n . At point B, the apparent frequency of sound becomes n  

nv v  v s cos 

(2) When a source of sound approaches a high wall or a hill with a constant velocity v s , the reflected sound propagates in a direction opposite to that of direct sound. We can assume that the source and observer are approaching each other with same velocity i.e. v s  v L 

 v  vL n     v  vs

  n 

(3) When a listener moves between two distant sound sources : Let v L be the velocity of listener (v  v L )n away from S 1 and towards S 2 . Apparent frequency from S 1 is n   v (v  v L )n and apparent frequency heard from S L is n   v 2nv L  Beat frequency  n   n   v  (4) When source is revolving in a circle and listener L is on one side v s  r so n max 

nv nv and n m in  v  vs v  vs

O

S

r

(5) When listener L is moving in a circle and the source is on one side v L  r so n max 

(v  v L )n (v  v L )n and n min  v v



(6) There will be no change in frequency of sound heard, if the source is situated at the centre of the circle along which listener is moving.

O r

L

(7) Conditions for no Doppler effect : (i) When source (S) and listener (L) both are at rest. (ii) When medium alone is moving. (iii) When S and L move in such a way that distance between S and L remains constant. (iv) When source S and listener L, are moving in mutually perpendicular directions.

Sample problems based on Doppler effect Problem 42. A source of sound of frequency 90 vibration/sec is approaching a stationary observer with a speed equal to 1/10 the speed of sound. What will be the frequency heard by the observer (a) 80 vibration/sec Solution : (c)

n' 

(b) 90 vibration/sec

(c) 100 vibration/sec

10 10  90 v v n  100 vibration/sec .n  n'  .n  n'  v 9 9 v  vs v 10

[MP PMT 2000]

(d) 120 vibration/sec vs(+) S

vL = 0 L

Problem 43. A source of sound of frequency 500 Hz is moving towards an observer with velocity 30 m/s. The speed of the sound is 330 m/s. The frequency heard by the observer will be (a) 550 Hz Solution : (a)

n' 

(b) 458.3 Hz

v 330 .n  n'  .500  n' = 550 Hz v  vs 330  30

[MP PET 2000]

(c) 530 Hz vs(+) S

(d) 545.5 Hz vL = 0 L

genius PHYSICS 24 Wave Motion

Problem 44. A motor car blowing a horn of frequency 124 vibration/sec moves with a velocity 72 km/hr towards a tall wall. The frequency of the reflected sound heard by the driver will be (velocity of sound in air is 330 m/s) [MP PET 1997]

(a) 109 vibration/sec Solution : (c)

(b) 132 vibration/sec

(c) 140 vibration/sec

(d) 248 vibration/sec

In the given condition source and listener are at the same position i.e. (car) for given condition

n' 

330  20 v  vcar .n = 140 vibration/sec .n  330  20 v  vcar

Problem 45. The driver of car travelling with a speed 30 meter/sec. towards a hill sounds a horn of frequency 600 Hz.

If the velocity of sound in air is 330 m/s the frequency of reflected sound as heard by the driver is [MP PMT 1996 (a) 720 Hz Solution : (a)

(b) 555.5 Hz

(c) 550 Hz

This question is same as that of previous one so n' 

(d) 500 Hz

v  vcar .n  720 Hz v  vcar

Problem 46. The source of sound s is moving with a velocity 50 m/s towards a stationary observer. The observer

measures the frequency of the source as 1000 Hz. What will be the apparent frequency of the source when it is moving away from the observer after crossing him ? The velocity of sound in the medium is 350 m/s [MP PMT (a) 750 Hz Solution : (a)

(b) 857 Hz

(c) 1143 Hz

When source is moving towards the stationary listener.

(d) 1333 Hz vs(+)

v 350 n'  n  1000  .n  n  857 .14 v  vs 350  50

vL = 0

S

When source is moving away from the stationary observer n' ' 

L

350 v  857 = 750 Hz = 350  50 v  vs

Problem 47. A source and listener are both moving towards each other with speed v/10 where v is the speed of sound. If the frequency of the note emitted by the source is f, the frequency heard by the listener would be nearly [MP PMT 1994]

(a) 1.11 f Solution : (b)

 v  vL n'    v  vs

(b) 1.22 f

v  v   10  n  n'   v   v  10 

  n   

(c) f  n' 

(d) 1.27 f

11 f = 1.22 f. 9

Problem 48. A man is watching two trains, one leaving and the other coming in with equal speed of 4 m/s. If they sound

their whistles, each of frequency 240 Hz, the number of beats heard by the man (velocity of sound in air = 320 m/s) will be equal to [MP PET 1999; CPMT 1997; NCERT 1984] (a) 6 (b) 3 (c) 0 (d) 12

Solution : (a)

App. Frequency due to train which is coming in n1  App. Frequency due to train which is leaving n2 

v .n v  vs

v .n v  vs

1   1  So number of beats n1 – n2 =   320  240  316 324 

 n1 – n2 = 6

Problem 49. At what speed should a source of sound move so that observer finds the apparent frequency equal to half of the original frequency (a) v / 2 (b) 2v

[RPMT 1996]

(c) v / 4

(d) v

genius PHYSICS Wave Motion 25 Solution : (d)

n' 

v .n v  vs



n v  .n 2 v  vs

 vs = v