ALLEN PHYSICS HANDBOOK FOR NEET [1, 2019 ed.]

Citation preview

Physics Handbook Downloaded from Telegram

For more ​Click Here​ and Join.

ALLEN

Physics HandBook

C HAP TE R

ALLEN

Units, Dimension, Measurements and Practical Physics Systems of Units

Fundamental or base quantities

(ii) (iii)

Derived quantities

(iv)

e.g. Speed (=distance/time), volume, acceleration, force, pressure, etc.

Units of physical quantities

S.N. 1 2 3 4 5 6 7

Physical Qty. Mass Length Time Temperature Luminous intensity Electric current Amount of substance

SI Base Quantities and Units S I U n its

M ass

kilogra m

kg

T im e

se co n d

s

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\01_Unit & Dimension.p65

A

L e n gth

E

E lectric C urren t

am pe re

A

T h erm o d yn am ic T e m p era ture

kelvin

K

A m ou nt o f S ubsta n ce

m o le

m ol

L um inous Inte nsity

ca nde la

Cd

D efin itio n T h e m eter is the le ng th o f th e pa th tra vele d b y lig h t in va cuum durin g a tim e in te rva l o f 1/(2 99 , 7 92 , 4 58 ) o f a seco nd (1 9 8 3 ) T h e kilogra m is e qual to th e m ass o f the in tern atio n al p roto typ e o f th e k ilogra m (a platin um -iridium a lloy cy lin d e r) ke p t a t In tern atio n al B u rea u o f W eig h ts an d M ea su re s, a t S e vres, ne ar P a ris, Fra nc e. (1 8 8 9) T h e se co n d is th e dura tio n o f 9 , 1 9 2 , 6 31 , 77 0 pe rio ds o f the rad iatio n c orre spo nd in g to th e tran sition b etw e e n th e tw o h yp erfin e leve ls o f th e g ro u n d sta te of the ce sium -1 3 3 a to m (1 9 6 7 ) T h e a m p e re is th at c o n stan t cu rre nt w h ic h , if m a in ta in e d in tw o straig h t pa ralle l c o n du cto rs o f infin ite le ng th , of n egligible circular cro ss-section, an d p laced 1 m etre ap art in vacuum , w o uld prod uce be tw e en these con du ctors a fo rce e qual to 2 x 1 0 -7 N e w to n p e r m e tre o f le ng th . (1 9 4 8 ) The k elvin, is th e fra ctio n 1 /2 7 3 .1 6 of the th erm ody n am ic tem p erature o f the triple p o in t of w a ter. (1 9 6 7 ) T h e m ole is th e a m o unt o f substa nc e o f a sy stem , w h ic h co n ta in s a s m an y elem e nta ry e ntitie s as the re a re a to m s in 0.01 2 k ilo gram o f carb o n -1 2 . (1 9 7 1 ) T h e can dela is th e lum in o us in te nsity, in a give n dire ctio n, o f a so u rce th at em its m o n oc h ro m a tic rad iatio n of fre quen cy 5 4 0 x 1 0 1 2 h e rtz an d th at ha s a rad ia nt inten sity in th at directio n of 1/ 6 8 3 w a tt pe r stera d ian (1 9 7 9 ).

on

Sym bo l m

Symbol kg m s K Cd A mol

ss i

N am e m e te r

Name of Unit kilogram meter second kelvin candela ampere mole

Se

B a s e Q u a n tity

MKSA Length (m) Mass (kg) Time (s) Current (A)

20 19

e.g. Physical Quantity = Numerical Value × Unit

MKSQ Length (m) Mass (kg) Time (s) Charge (Q)

Fundamental Quantities in S.I. System and their units

LL E

The chosen reference standard of measurement in multiples of which, a physical quantity is expressed is called the unit of that quantity.

FPS Length (ft) Mass (pound) Time (s) –

N

The quantities which can be expressed in terms of the fundamental quantities are known as derived quantities.

CGS Length (cm) Mass (g) Time (s) –

0

e.g. : length, mass, time, etc.

MKS Length (m) Mass (kg) Time (s) –

(i)

-2

The quantities which do not depend upon other quantities for their complete definition are known as fundamental or base quantities.

1

Physics HandBook

CH APTER

Supplementary Units

Limitations of dimensional analysis

Dimensional Formula

•

In Mechanics the formula for a physical quantity depending on more than three other physical quantities cannot be derived. It can only be checked.

•

This method can be used only if the dependency is of multiplication type. The formulae containing exponential, trigonometrical and logarithmic functions can't be derived using this method. Formulae containing more than one term which are added or subtracted like s = ut +½ at2 also can't be derived.

•

The relation derived from this method gives no information about the dimensionless constants.

•

If dimensions are given, physical quantity may not be unique as many physical quantities have the same dimensions.

•

It gives no information whether a physical quantity is a scalar or a vector.

Relation which express physical quantities in terms of appropriate powers of fundamental units.

• • •

To check the dimensional correctness of a given physical relation To derive relationship between different physical quantities To convert units of a physical quantity from one system to another a

b

æM ö æL ö æT ö n1u1= n2u2 Þ n2=n1 ç 1 ÷ ç 1 ÷ ç 1 ÷ è M2 ø è L 2 ø è T2 ø

c

where u = MaLbTc

N

Use of dimensional analysis

0

Radian (rad) - for measurement of plane angle Steradian (sr) - for measurement of solid angle

-2

• •

ALLEN

20 19

LL E

SI PREFIXES

Prefix

Symbol

Power of 10

1018

d

E

10-1

deci

peta

P

10-2

centi

c

1012

tera

T

10-3

milli

m

10 9

giga

G

10-6

micro

m

10

mega

10

6

10 3

kilo

10 2

hecto

A

1

deca

on

exa

10

2

Symbol

15

M

10-9

nano

n

k

10-12

pico

p

h

10-15

femto

f

atto

a

da

Se

PREFIXES USED FOR DIFFERENT POWERS OF 10

Prefix

ss i

Power of 10

10

- 18

P hysical quan tity

Unit

P hys ical q uantity

U nit

A ngular acceleration

rad s- 2

F requen cy

h ertz

Mo ment o f in ertia

kg – m 2

Resistan ce

kg m 2 A- 2 s-3

Self in ductan ce

h en ry

Surface ten sio n

n ew ton/m joule K

-1

Magnetic flux

w eber

Universal gas con stant

Po le str en gth V iscosity

A –m po is e

Dipole mo ment Stefan co n stant

Reactance

oh m

Specific heat

J/kg°C

Strength of m agnetic field A stronom ical dis tance

Perm ittivity o f free space (e0) P erm eability o f free space (m0 )

n ewton A -1 m - 1

P lan ck's co nstan t

jo ule –se c

Parsec

En tro py

J/K

-1

mo l

co ulo mb–meter watt m - 2 K -4 co ulo mb 2 /N– m 2 weber/A-m

UNITS OF IMPORTANT PHYSICAL QUANTITIES

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\01_Unit & Dimension.p65

The magnitudes of physical quantities vary over a wide range. The CGPM recommended standard prefixes for magnitude too large or too small to be expressed more compactly for certain powers of 10.

E

Physics HandBook

C HAP TE R

ALLEN

DIMENSIONS OF IMPORTANT PHYSICAL QUANTITIES Physical quantity

Dimensions

Physical quantity

Momentum

M L T–

Capacitance

Calorie

M1 L 2 T –2

Modulus of rigidity

Latent heat capacity

M L T–

Magnetic permeability

1

1

0

1

2

2

Dimensions M–1 L–2 T4 A 2 M1 L– 1 T –2 M L T – A– 1

2

1

Self inductance

M L T – A–

2

Pressure

Coefficient of thermal conductivity

M L T– K –

1

Planck's constant

M L T–

1

3

1

2

1

2

1

3

2

1 1 2 M L– T – 1

2

Power

M L T–

Solar constant

M L T–

Impulse

M1 L 1 T –1

Magnetic flux

M1 L2 T –2 A– 1

M– L T A 1

0

2

Bulk modulus of elasticity

1 1 2 M L– T–

Potential energy

M L T–

Gravitational constant

M – L T–

1

1

2

2

3

2

M0 L1 T0

Light year

1

M–1 L–2 T3 K

0 1 2 0 M L– T A 1 1 2 M L– T –

Magnetic field intensity

0 1 0 1 M L– T A

Magnetic Induction

1 2 1 M T– A –

Electric Permittivity

1 3 4 2 M– L– T A

Electric Field Resistance

M1L 1T–3A- 1

2 3 2 ML T – A–

20 19

M1 L– 1 T–1

Coefficient of viscosity

0

Young modulus

LL E

Thermal resistance

Current density

1

0

3

-2

Hole mobility in a semi conductor

2

N

1

SETS OF QUANTITIES HAVING SAME DIMENSIONS

4.

[M 1 L 1 T – 2] [M 1 L – 1 T – 2 ]

Acceleration, g and gravitational field intensity. Surface tension, free surface energy (energy per unit area), force gradient, spring constant. Latent heat capacity and gravitational potential.

[ M 1 L0 T –2]

on

Thrust, force, weight, tension, energy gradient. Pressure, stress, Young's modulus, bulk modulus, shear modulus, modulus of rigidity, energy density. Angular momentum and Planck's constant (h).

A

5.

Dimensions

ss i

2. 3.

Quantities Strain, refractive index, relative density, angle, solid angle, phase, distance gradient, relative permeability, relative permittivity, angle of contact, Reynolds number, coefficient of friction, mechanical equivalent of heat, electric susceptibility, etc. Mass or inertial mass Mom entum and impulse.

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\01_Unit & Dimension.p65

6.

E

7. 8. 9.

10. 11.

Thermal capacity, Boltzmann constant, entropy. Work, torque, internal energy, potential energy, kinetic energy, mom ent of force, (q2 /C), (LI2 ), (qV), (V 2 C), (I 2 Rt),

12.

Se

S.N. 1.

V2 t , (VIt), (PV), (RT), (mL), (mc DT) R

Frequency, angular frequency, angular velocity, velocity gradient, radioactivity R 1 1 , , L RC LC

13.

ælö ç ÷ ègø

14.

(VI), (I2 R), (V 2/R), Power

12

æmö ,ç ÷ è k ø

1 2

æL ,ç èR

ö ÷ , (RC), ( LC ) , time ø

[M 0 L 0 T 0] [M 1 L 0 T 0 ] [M 1 L 1 T – 1]

[ M 1 L2 T –1] [ M 0 L1 T –2]

[ M 0 L2 T –2] [ ML 2 T – 2K – 1 ] [M 1 L 2 T – 2] [M 0 L 0 T – 1] [ M 0 L 0 T 1] [ M L 2 T – 3]

3

CH APTER

6.67 × 10 –11 N m 2 kg –2

Speed of light in vacuum (c)

3 × 10 8 ms –1

Permeability of vacuum (m 0 )

4p × 10 –7 H m –1

Permittivity of vacuum (e 0 )

8.85 × 10 –12 F m –1

Planck constant (h)

6.63 × 10 –34 Js

Atom ic mass unit (am u)

1.66 × 10 –27 kg

Energy equivalent of 1 amu

931.5 MeV 9.1 × 10

Trigonometric functions sinq, cosq, tanq etc and their arrangement s q are dimensionless.

•

Dimensions of differential é dny ù

•

MeV 6.02 × 10

Faraday constant (F)

9.648 × 10 C mol

Stefan–Boltzmann constant (s)

5.67× 10 –8 W m –2 K –4

Wien constant (b)

2.89× 10

Rydberg constant (R ¥ )

1.097× 10 m –1

Triple point for water

273.16 K (0.01°C)

mol

23

–1

4

–3

•

•

22.4 L = 22.4× 10 –3 m 3

PRACTICAL PHYSICS

For a Number Less than 1

All accurately known digits in measurement plus the first uncertain digit together form significant figure. Ex. 0.108 ® 3SF, 40.000 ® 5SF, 1.23 × 10-19 ® 3SF, 0.0018 ® 2SF

Significant Digits The product or quotient will be reported as having as many significant digits as the number involved in the operation with the least number of significant digits. For example : 0.000170 × 100.40 = 0.017068 Another example : 2.000 × 104 / 6.0 × 10–3 = 0.33 × 107

4

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\01_Unit & Dimension.p65

Significant Figures

Se

A

Any zero to the right of a non-zero digit is significant. All zeros between decimal point and first non-zero digit are not significant.

For example : 3.0 × 800.0 = 2.4 × 103 The sum or difference can be no more precise than the least precise number involved in the mathematical operation. Precision has to do with the number of positions to the RIGHT of the decimal. The more position to the right of the decimal, the more precise the number. So a sum or difference can have no more indicated positions to the right of the decimal as the number involved in the operation with the LEAST indicated positions to the right of its decimal. For example : 160.45 + 6.732 = 167.18 (after rounding off) Another example : 45.621 + 4.3 – 6.41 = 43.5 (after rounding off) Rules for rounding off digits : 1. If the digit to the right of the last reported digit is less than 5 round it and all digits to its right off. 2. If the digit to the right of the last reported digit is greater than 5 round it and all digits to its right off and increased the last reported digit by one. 3. If the digit to the right of the last reported digit is a 5 followed by either no other digits or all zeros, round it and all digits to its right off and if the last reported digit is odd round up to the next even digit. If the last reported digit is even then leave it as is.

on

For a number greater than 1 • All non-zero digits are significant. • All zeros between two non-zero digits are significant. Location of decimal does not matter. • If the numbe is without decimal part, then the terminal or trailing zeros are not significant. • Trailing zeros in the decimal part are significant.

Independent quantities may be taken as fundamental quantities in a new system of units.

ss i

Rules for Counting Significant Figures

We can't add or subtract two physical quantities of different dimensions.

20 19

LL E

mol –1

Dimensions of integrals é ydx ù = [ yx ] úû ëê ò

–1

mK

7

éyù

coefficients ê n ú = ê n ú ë dx û ë x û

kg º 0.511

–31

Avogadro constant (N A )

Molar volum e of ideal gas (NTP)

•

-2

Electron rest mass (m e )

KEY POINTS

0

Gravitational constant (G)

ALLEN

N

SOME FUNDAMENTAL CONSTANTS

Physics HandBook

E

Physics HandBook

C HAP TE R

ALLEN

For example if we wish to round off the following number to 3 significant digits : 18.3682 The answer is : 18.4. Another example : Round off 4.565 to three significant digits.

Error in Product and Division A physical quantity X depend upon Y & Z as X = Ya Zb then maximum possible fractional error in X. DX DY DZ = a + b X Y Z

The answer would be 4.56.

Rounding off 6.75 ® 6.8,

6.84 ® 6.8, 6.65 ® 6.6,

6.85 ® 6.8, x=

6.95 ® 7.0

Order of magnitude Power of 10 required to represent a quantity 49 = 4.9 × 101 » 101 Þ order of magnitude =1 51 = 5.1 × 101 » 102 Þ order of magnitude = 2

The quotient rule is not applicable if the numerator and denominator are dependent on each other. e.g if R =

XY . We cannot apply quotient rule X+Y

EN

0.051 =5.1 × 10-2 » 10-1order of magnitude = -1

é æ Da ö Dx æ Db ö ù am = ± êm ç ÷ + n ç ÷ ú n then è ø è b øû x a ë b

1 1 1 = + . Differentiating both R X Y

as follows

(1) random and (2) systematic errors.

the sides, we get

r

R

x

2

=-

dX

X

2

-

dY

Y2

.

+

y

X

Y2

io n

The smallest value of a physical quantity which can be measured accurately with an instrument is called the least count of the measuring instrument.

Vernier Callipers

Least count = 1MSD – 1 VSD (MSD ® main scale division, VSD ® Vernier scale division)

Least Count Error :– If the instrument has known least count, the absolute error is taken to be equal to the least count unless otherwise stated.

A node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\01_Unit & Dimension.p65

E

R

2

Se

A measurement with relatively small random error is said to have high precision. A measurement with small random error and small systematic error is said to have high accuracy.

Least count

=

dR

ss

2. Systematic errors occur due to error in the procedure, or miscalibration of the instrument etc. Such errors have same size and sign for all the measurements. Such errors can be determined.

2

-

20

Thus

LL

1. Random errors appear randomly because of operator, fluctuations in external conditions and variability of measuring instruments. The effect of random error can be some what reduced by taking the average of measured values. Random errors have no fixed sign or size.

-2

Whenever an experiment is performed, two kinds of errors can appear in the measured quantity.

to find the error in R. Instead we write the equation

19

Errors

0

6.87® 6.9,

Error in Power of a Quantity

Relative error =

A.

absolute error in a measurement size of the measurement

0

1

2

3

4

5

6

14

15

Systematic errors :

They have a known sign. The systematic error is removed before beginning calculations. Bench error and zero error are examples of systematic error.

Propagation of combination of errors Error in Summation and Difference : x = a + b then Dx = ± (Da+Db)

Ex. A vernier scale has 10 parts, which are equal to 9 parts of main scale having each path equal to 1 mm then least count = 1 mm –

9 mm = 0.1 mm 10

[Q 9 MSD = 10 VSD]

5

Physics HandBook

CH APTER

ALLEN

Zero Error

Main scale

0

0

Main scale

1

5

10

Main scale

1

0

5

0

0

0

10

Vernier scale with positive zero error

Vernier scale without zero error

1

5

10

Vernier scale with negative zero error

(i)

(ii)

0

Ratchet

Circular (Head) scale

Spindle

5 10

Thimble

LL

Linear (Pitch) Scale Sleeve

0

-2

pitch total no. of divisions on circular scale

19

Least count =

20

Screw Gauge

EN

The zero error is always subtracted from the reading to get the corrected value. If the zero error is positive, its value is calculated as we take any normal reading. Negative zero error = – [Total no. of vsd – vsd coinciding] ×L.C.

io n

Ex. The distance moved by spindle of a screw gauge for each turn of head is 1mm. The edge of the humble is

Se

A

Positive Zero Error

(2 division error) i.e., +0.002 cm

Circular scale

Negative Zero Error

(3 division error) i.e., –0.003 cm Circular scale

0 Main scale reference line

6

10 5 0 95 90 85

0

zero of the circular scale is above the zero of main scale

Main scale reference line

15 10 5 0 95 90

Zero of the circular scale is below the zero of main scale

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\01_Unit & Dimension.p65

Zero Error

If there is no object between the jaws (i.e. jaws are in contact), the screwgauge should give zero reading. But due to extra material on jaws, even if there is no object, it gives some excess reading. This excess reading is called Zero error.

1mm = 0.01 mm 100

ss

provided with a angular scale carrying 100 equal divisions. The least count =

E

Physics HandBook

C HAP TE R

ALLEN

Basic Mathematics used in Physics Quadratic Equation - b ± b - 4ac 2a b ; a

(1–x)n = 1 – nx +

c a For real roots, b2 – 4ac ³ 0 For imaginary roots, b2 – 4ac < 0

Product of roots x1x2 =

log mn = n log m log2 = 0.3010

Componendo and dividendo theorem

Geometrical progression-GP

a, ar, ar2, ar3, ...... here, r = common ratio

nth term, an = a.rn–1

Note: (i) 1+2+3+4+5....+ n =

6 2 é n ( n + 1) ù 3 2 2 3 ê ú (iii) 1 +2 +3 +...+ n = 2 ëê ûú

A node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\02_Basic Maths & Vector.p65

E

n ( n + 1)(2n + 1)

Sum of ¥ terms S¥ =

[where r < 1 ]

Se

(ii) 12+22+32+...+ n2=

n(n + 1) 2

a 1- r

io n

n [2a+(n–1)d] 2 nth term, an = a + (n – 1)d

Sum of n terms Sn =

ss

a, a+d, a+2d, a+3d, .....a+(n – 1)d here d = common difference

a(1 - r n ) 1-r

Sum of n terms Sn =

LL

Arithmetic progression-AP

-2

m =log m–log n n logem = 2.303 log10m log 3 = 0.4771

log

p a p+q a+b = = then q b p-q a -b

19

log mn = log m + log n

If x m2 now for mass m1, m1 g – T = m1a for mass m2,T – m2 g = m2 a

a f1

m1

f1

Acceleration = a =

m2 Tension = T =

Fig. 1(a) : F.B.D. representation of action and reaction forces.

R = 2T =

Pulley system

Case – I :

Case – II a For mass m1 : m1 T = m1 a For mass m2 : m2g – T = m2 a Acceleration:

T m1

m2

a

m2 g m1 m2 g and T = (m1 + m2 ) (m1 + m2 )

m2

a

19

a=

T

T

EN

A single fixed pulley changes the direction of force only and in general, assumed to be massless and frictionless. SOME CASES OF PULLEY

4m1 m2 g (m1 + m2 )

0

m2 F m1 + m2

-2

f1=m2a Þ action of m1 on m2: f1 =

LL

FRAME OF REFERENCE

Inertial frames of reference : A reference frame which is either at rest or in uniform motion along the straight line. A non–accelerating frame of reference is called an inertial frame of reference.

io n

•

2m1 m2 2 ´ Pr oduct of masses g g= (m1 + m2 ) Sum of two masses

Reaction at the suspension of pulley :

For body m2 :

a

(m1 - m2 ) net pulling force g= total mass to be pulled (m1 + m2 )

20

F

ALLEN

All the fundamental laws of physics have been formulated in respect of inertial frame of reference. Non–inertial frame of reference : An accelerating frame of reference is called a non–inertial frame of reference. Newton's laws of motion are not directly applicable in such frames, before application we must add pseudo force.

ss

•

Se

Pseudo force:

is mass of the particle or body. The direction of pseudo force must be opposite to the direction of acceleration of the non–inertial frame. When we draw the free body diagram of a mass, with respect to an inertial frame of reference we apply only the real forces (forces which are actually acting on the mass). But when the free body diagram is drawn from a non– r r inertial frame of reference a pseudo force (in addition to all real forces) has to be applied to make the equation F = ma

to be valid in this frame also.

r

åF

real

r r r r + Fpseudo = ma (where ar is acceleration of object in non inertial reference frame) & Fpseudo = -ma a r

(where a 0 is acceleration of non inertial reference frame).

24

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\04_NLM & Friction.p65

A

The force on a body due to acceleration of non–inertial frame is called fictitious or apparent or pseudo force and r r r is given by F = - ma0 , where a0 is acceleration of non–inertial frame with respect to an inertial frame and m

E

Physics HandBook

CH APTER

Graph between applied force and force of friction block

f

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Friction force (f)

applied force

Limiting friction

f=fL Kinetic friction

Applied force F

45°

F=fL = µ SN

EN Static friction coefficient ms =

w

limiting friction

Kinetic friction coefficient

w

fk r , f = - ( m k N ) vˆ rel N k Angle of Friction (l)

19

•

20

mk =

N

io n

LL

l

Applied force

ss

f

W

Se

A node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\04_NLM & Friction.p65

E

w

N

r r , 0 £ fs £ m s N , fs = - Fapplied

nt lta su d N Re f an of

Fr ictional fo rces ar ise due t o molecular interactions. In some cases friction acts as a supporting force and in some cases it acts as opposing force.

s max

( fs ) max = m s N =

FRICTION

Friction is the force of two surfaces in contact, or the force of a medium acting on a moving object. (i.e. air on aircraft.)

(f )

Cause of Friction: Friction arises on account of strong atomic or molecular forces of attraction between the two surfaces at the point of actual contact. Types of friction

fS m s N = = mS N N Angle of repose : The maximum angle of an inclined plane for which a block remains stationary on the plane. tan l =

w

N fs

Friction

Static friction (No relative motion between objects)

-2

•

0

(a) If the lift moving with constant velocity v upwards or downwards. In this case there is no accelerated motion hence no pseudo force experienced by observer inside the lift. So apparent weight W´=Mg=Actual weight. (b) If the lift is accelerated upward with constant acceleration a. Then forces acting on the man w.r.t. observed inside the lift are (i) Weight W=Mg downward (ii) Fictitious force F0=Ma downward. So apparent weight W´=W+F0=Mg+Ma=M(g+a) (c) If the lift is accelerated downward with acceleration a g . Then as in Case (c). Apparent weight W´ =M(g–a) is negative, i.e., the man will be accelerated upward and will stay at the ceiling of the lift.

fri ct io n

Man in a Lift

St at ic

ALLEN

Kinetic friction (There is relative motion between objects)

q si n Mg q R

Mg

Mgcosq

tanq R=m s

25

Physics HandBook

C HAP TE R

Normal constraint : displacements, velocities & accelerations of both objects should be same along C.N.

Dependent Motion of Connected Bodies Method I : Method of constraint equations

å x i = constant Þ å

ALLEN

v2

dx i d2 x = 0 Þ å 2i = 0 dt dt

r

For n moving bodies we have x1, x2,...xn

r

No. of constraint equations = no. of strings

v1 a2

a1

q

e.g. a2 = a1 tan q & v2 = v1 tan q

KEY POINTS Aeroplanes always fly at low altitudes because according to Newton's III law of motion as aeroplane displaces air & at low altitude density of air is high.

•

Rockets move by pushing the exhaust gases out so they can fly at low & high altitude.

EN

The sum of scalar products of tension forces applied by connecting links of constant length and displacement of corresponding contact points equal to zero. r r r r r r åT × x = 0 Þ åT × v =0 Þ åT ×a =0

•

Pulling (figure I) is easier than pushing (figure II) on a rough horizontal surface because normal reaction is less in pulling than in pushing. F

x2

T T

•

a2

F

m

m

Fig. II

Fig. I

While walking on ice, one should take small steps to avoid slipping. This is because smaller step increases the normal reaction and that ensure smaller friction.

io n

LL

2T 1 a1 2

Here 2a2 = a1

q

20

T

x1

m

19

m

q

-2

•

0

Method II : Method of virtual work :

26

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\04_NLM & Friction.p65

Se

A

ss

IMPORTANT NOTES

E

Physics HandBook

CH APTER

ALLEN

CIRCULAR MOTION Definition of Circular Motion w

r r r r dv d r r dw r r dr Acceleration a = = (w ´ r ) = ´ r +w´ dt dt dt dt r r r r r r = a ´ r + w ´ v = a t + aC

w

Tangential acceleration: a t =

Radius Vector :

ér r r r æ dv ö ù ê a t = component of a along v= ( a × vˆ ) vˆ = çè dt ÷ø vˆ ú ë û

EN

The vector joining the centre of the circle and the center of the particle performing circular motion is called radius vector. It has constant magnitude and variable direction. It is directed outwards.

Centripetal acceleration : a C = wv =

No. of revolutions described by particle per sec. is its frequency. Its unit is revolutions per second (r.p.s.) or revolutions per minute (r.p.m.) Time Period (T) :

r r aC × v = 0

w

Magnitude of net acceleration :

LL

It is time taken by particle to complete one revolution.

q

s

w

Dq Average angular velocity w = (a scalar quantity) Dt

w

Instantaneous angular velocity

2

r d|v| = 0 = at dt

If a is contant, then following equations hold (i) Dq = qf – qi

(ii) wf = wi + at

(iii) q = wit +

1 2 at 2

(iv) w2f = w2i + 2aq

dq w= (a vector quantity) dt

(v) q = wf t 2p =2pf or 2pn T

w

For uniform angular velocity w =

w

Angular displacement

q = wt

w ® Angular frequency

n or f = frequency

Relation between w and v

v w= r

w

w

Se

r

A node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\05_Circular motion.p65

E

For uniform circular motion

r

arc length s = r radius

io n

1 n

2

æ v2 ö æ dv ö a = a2C + a2t = ç ÷ + ç ÷ è r ø è dt ø

ss

T=

r v2 = w 2 r or a C = w 2 r ( -ˆr ) r

20

Frequency (n) :

-2

w

19

r r r r k × v = 0 & r & v always in same plane.

Angle q =

dv = ar dt

0

When a particle moves in a plane such that its distance from a fixed (or moving) point remains constant then its motion is called as circular motion with respect to that fixed point. That fixed point is called centre and the distance is called radius of circular path.

w

r r r v =w´r

In vector form velocity

(vi) q =

1 2 at 2

(wi + wf )t 2

æ wf - wi ö (vii) a = ç t ÷ø è

27

Physics HandBook

C HAP TE R

Curvilinear Motion :

Maximum speed of in circular motion :

®

at

On unbanked road : v max = ms Rg

v

•

On banked road :

aNet æ m + tan q ö v max = ç s Rg = tan ( q + f) Rg è 1 - m s tan q ÷ø

aN r r dv a Net = ; dt

v min = Rg tan ( q - f) ; v min £ v car £ v max

r r r components of a Net along v = a t

where f = angle of friction = tan–1µs; q = angle of banking w

r r d|v| aT = ; aN is responsible for change of direction dt

N

19

v2 v2 ÞR= R aN

mg

®

v1 aN R2

LL

2

®

v2

R1 >R2; Radius of curvature doesn't remains constant R is a property of curves, not of the particle

28

Se

A

(If a bee follows this path instead of the particle then its radius of curvature will be the same)

io n

1

R1

ss

aN

q

-2

Radius of Curvature : aN =

v2 rg

0

Bending of cyclist : tan q =

EN

r r r components of a Net perpendicular to v = aN

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\05_Circular motion.p65

q

• ®

20

P

ALLEN

E

Physics HandBook

CH APTER

ALLEN

C

Circular motion in vertical plane A. Condition to complete vertical circle u ³ 5gR

B

If u = 5gR then Tension at C is equal to 0 and tension at A is equal to 6mg

R

Velocity at B: v B = 3gR

u

A C

Velocity at C: v C = gR

B q

2

mv - mg cos q R

EN R

Velocity can be zero but T never be zero between A & B.

R

LL u - 2gR ; q is from vertical line 3gR

q

mgsinq

mg

B

A

io n

Tension will be zero in between B to C & the angle where T = 0

T

20

2gR < u < 5gR

Particle crosses the point B but not complete the vertical circle.

cos q =

V

mgcos q

q

2

mgcosq

mg

19

mv 2 Because T is given by T = mg cos q + R

q

u

A

C

C. Condition for leaving path :

V

mgsinq

B. Condition for pendulum motion (oscillating condition) u £ 2gR (in between A to B)

T

-2

From B to C : T =

mv 2 R

0

From A to B : T = mgcos q +

v

C

T=0,v¹ 0

R

q B

ss

Note : After leaving the circle, the particle will follow a parabolic path.

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\05_Circular motion.p65

E

•

•

Average angular velocity is a scalar phy sical qu an it y whereas instantaneous angular velocity is a vector physical quanity. Small Angular displacement r dq is a vector quantity, but large angular displacement q is scalar quantity. r r r r dq1 + dq2 = dq2 + dq1 r r r r But q1 + q2 ¹ q2 + q1

Se

A

* T is maximum at the bottom & minimum at the top.

KEY POINTS

•

Relative Angular Velocity Relative angular velocity of a particle 'A' w.r.t. other moving particle B is the angular velocity of the position vector of A w.r.t. B. That means it is the rate at which position vector of 'A' w.r.t. B rotates at that instant

w AB

A

vB

r

q2 vBsin q2

q1

vA

vAsinq1

B

Relative velocity of A w.r.t. (v AB ) ^ B perpendicular to line AB = = rAB seperation between A and B

here (vAB)^ = vA sinq1 + vBsinq2 \ w AB =

v A sin q1 + v B sin q2 r

29

Physics HandBook

C HAP TE R

ALLEN

WORK, ENERGY & POWER WORK DONE r r W = ò dW = ò F.dr = ò Fdr cos q r r [where q is the angle between F & dr ] rr w For constant force W = F.d = Fd cos q w For Unidirectional force

W = ò dW = ò Fdx = Area between F–x curve and x-axis.

EN

NATURE OF WORK DONE Although work done is a scalar quantity, yet its value may be positive, negative or even zero

F

F

q

q

S

q

S

(q = 90°)

(q < 90°)

S

LL

f

mg

S

F

S

N

Se

A A

30

f

B

F

A

f=friction force

As f = F, hence S = 0

WORK DONE BY VARIABLE FORCE A force varying with position or time is known as the variable force B ® r dS ˆ ˆ ˆ F = Fxi + Fyj + Fzk q ® F r A dS = dxiˆ + dyjˆ + dzkˆ

WAB

F=2.5 N

mg = 100 N

Work done by gravity (q =180°)

r r = ò F × dS

Motion under gravity (q =0°)

ss

fmax=10N

mg

mg

Motion of particle on circular path (uniform) (q =90°)

io n

Work done by friction force (q =180°)

20

S

B

S

19

(q >90°)

Work done by friction force on block A (q =0°)

Calculation of work done from force– displacement graph : Total work done,

F P2

x2

W = å Fdx

P1

x1

xB

= ò Fxdx + xA

yB

ò Fydy +

yA

zB

ò Fzdz

zA

= Area of P1P2NM

O

M

x1

dx

N x x2

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\06_Work, energy & power.p65

F

0

Positive work

Zero work

-2

Negative work

E

Physics HandBook

CH APTER

ALLEN

POTENTIAL ENERGY

Kinetic energy

Kinetic energy is a frame dependent quantity.

Work energy theorem (W = DKE )

dU dx

Conservative Forces •

Work done in a round trip is zero.

•

Central forces, spring forces etc. are conservative forces

F=–

dU Þ dU = –Fdx Þ dx

U2

x2

U1

x1

ò dU = - ò Fdx

Þ DU = –WC

l Potential energy may be positive or negative or even zero

Se

When only a conservative force acts within a system, the kinetic energy and potential energy can change into each other. However, their sum, the mechanical energy of the system, doesn't change. Work done is completely recoverable.

A •

l If force varies with only one dimension (say along x-axis) then

LL

Work done does not depend upon path.

•

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\06_Work, energy & power.p65

r ¶U ˆ ¶U ˆ ¶U ˆ ijk F = -ÑU = - grad(U) = – ¶x ¶y ¶z

•

•

E

-2

Change in potential energy DU = - ò F(x)dx

19

For conservative force F ( x ) = -

EN

Change in kinetic energy = work done by all force

io n

•

1 1 rr mv 2 = m(v.v) 2 2

20

K=

0

The energy possessed by a body by virtue of its motion is called kinetic energy.

ss

•

l The energy possessed by a body by virtue of its position or configuration in a conservative force field. l Potential energy is a relative quantity. l Potential energy is defined only for conservative force field. l Potential energy of a body at any position in a conservative force field is defined as the external work done against the action of conservative force in order to shift it from a certain reference point (PE = 0) to the present position. l Potential energy of a body in a conservative force field is equal to the work done by the conservative force in moving the body from its present position to reference position. l At a certain reference position, the potential energy of the body is assumed to be zero or the body is assumed to have lost the capacity of doing work. l Relationship between conservative force field and potential energy :

r r r r If F is a conservative force then Ñ ´ F = 0 r (i.e. curl of F is zero)

Work done depends upon path.

•

Work done in a round trip is not zero.

•

r Attraction forces

Non–conservative Forces

•

Repulsion forces U+ve U-ve

i)

Potential energy is positive, if force field is repulsive in nature

Forces are velocity–dependent & retarding in nature e.g. friction, viscous force etc.

ii)

Potential energy is negative, if force field is attractive in nature

•

Work done against a non–conservative force may be dissipated as heat energy.

l If r ­ (separation between body and force centre), U ­, force field is attractive or vice–versa.

•

Work done is not recoverable.

l If r ­, U ¯, force field is repulsive in nature.

31

Physics HandBook

C HAP TE R

Potential energy curve and equilibrium

It is a curve which shows the change in potential energy with the position of a particle.

Neutral equilibrium After a particle is slightly displaced from its equilibrium position if no force acts on it then the equilibrium is said to be neutral equilibrium. Point H corresponds to neutral equilibrium Þ U = constant ;

G

Law of conservation of Mechanical energy

F

x

F

position of particle

Stable Equilibrium :

After a particle is slightly displaced from its equilibrium position if it tends to come back towards equilibrium then it is said to be in stable equilibrium. dU At point A : slope is negative so F is positive dx

dU is positive. so F is negative dx

Power

• Power is a scalar quantity with dimension M 1L2T–3 • SI unit of power is J/s or watt • 1 horsepower = 746 watt = 550 ft–lb/sec. Average power : Pav= W/t

dU =0 dx

At point B : it is the point of stable equilibrium. At point B : U = Umin ,

d2 U dU = 0 & 2 = positive dx dx

Se

Unstable equilibrium :

A

After a particle is slightly displaced from its equilibrium position, if it tends to move away from equilibrium position then it is said to be in unstable equilibrium. At point D : slope

dU is positive so F is dx

negative ; At point G : slope

dU is negative dx

so F is positive At point E : it is the point of unstable equilibrium; At point E negative

32

U=U max,

ss

At equilibrium F = -

io n

LL

At point C : slope

w

in spring U ( x ) = 1 kx 2 2 Mass and energy are equivalent and are related by E 2 = mc

d2 U dU = 0 and = dx dx2

-2

F

19

F

0

B

Total mechanical (kinetic + potential) energy of a system remains constant if only conservative forces are acting on the system of particles or the work done by all other forces is zero. From work energy theorem W = DKE Proof : For internal conservative forces Wint = –DU So W=W ext +W int = 0 + W int =–DUÞ–DU=DKE ÞD(KE+U) =0 ÞKE+U=constant w Spring force F=–kx, Elastic potential energy stored

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\06_Work, energy & power.p65

H

C

20

A

d2 U dU =0, =0. dx dx2

EN

potential energy (U)

E D

ALLEN

E

Physics HandBook

CH APTER

ALLEN r r dW F.dr r r = = F.v Instantaneous power : P = dt dt

KEY POINTS •

fig.(b)

kine ti c ener gy

fig.(c)

and

work

e n e r g y

•

If vr = constant then

instantaneous power dW P= =tanq dt

mechanical

t2

energy may not

average power

be valid every

P=Pav=

W 2- W 1 D W = t 2- t 1 Dt

•

r r d r dv r dm F = ( mv ) = m +v dt dt dt r r dm rr 2 dm F=v then P = F.v = v dt dt

In rotatory motion :

dq = tw P=t dt

Efficiency

Output Energy h= Input Energy

time.

move

elliptical orbits.

The gravitational fo rce

on

t he

comet due to sun is not normal to th e

co met' s

velocity but the work done by the gravitational force is

z ero

in

ss

complete round trip

because

gravitational force

Se

A node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\06_Work, energy & power.p65

E

Co mets

around the sun in

io n

•

t1

LL

•

time time

-2

For a system of varying mass

of conservation of

19

•

because principle

w1

EN

time dt W= Pdt

simultaneously

20

power

work

w2

q

potential

0

fig.(a)

A body may gain

is a conservative force. •

Wo rk d on e by static friction may be

pos it ive

be ca us e st at ic friction may acts alon g direct ion

th e of

mo ti on o f an object.

33

Physics HandBook

C HAP TE R

ALLEN

COLLISIONS & CENTRE OF MASS CENTRE OF MASS OF SOME COMMON OBJECTS

Centre of mass : Centre of mass of system is the point associated with the system which have same acceleration as the acceleration of point mass (of same mass as that of system) would have under the application of same external force. Centre of mass of system of discrete particles

Body

Uniform Ring

Shape of body

Position of centre of mass

Centre of ring

CM

y

m3 (x3 ,y3 ,z3 )

Uniform Rod

LL

co-ordinates of centre of mass :

Centre of mass of continuous distribution of particles

A

r

(0,0,0)

Se

dm

Triangular plane lamina

x

z

x cm =

1 1 1 z dm x dm , y cm = ò y dm and z cm = Mò M M

ò

x, y, z are the co-ordinate of the COM of the dm mass.

The centre of mass after removal of a part of a body Original mass (M) – mass of the removed part (m) = {original mass (M)}+{–mass of the removed part (m)} The formula changes to: x CM =

34

CM

ss

y

Solid sphere/ hollow sphere

CM

io n

1 1 1 Sm i x i , y cm = Smi y i & z cm = Smi z i M M M

r 1 r R CM = ò rdm M

Centre of disc

-2

Total mass of the body : M = m1 + m2 + ..... + mn then r r r r m1 r1 + m2 r2 + m3 r3 + ... 1 r R CM = = Smi ri M m1 + m2 + m3 + ...

x cm =

CM

x

(0,0,0)

z

Uniform Disc

Mx - mx ¢ My - my ¢ Mz - mz ¢ ; y CM = ; y CM = M-m M-m M-m

0

mn(xn,yn,zn)

rn

Plane lamina in the form of a square or rectangle or parallelogram

Hollow/solid cylinder

Centre of rod

Centre of sphere

Point of intersection of the medians of the triangle i.e. centroid

CM

CM

CM

Point of intersection of diagonals

Middle point of the axis of cylinder

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\07_COM & Collision, Momentum.p65

r3

19

r2

20

r1

m2(x2,y2,z2)

EN

m1 (x1 ,y1,z1 )

E

MOTION OF CENTRE OF MASS

Position of centre of mass

Shape of body

For a system of particles, velocity of centre of mass

r r r dR CM m1 v1 + m2 v 2 + ... r v CM = = m1 + m2 + .... dt

y

Half ring

y cm =

CM ycm

R

2R p

Similarly acceleration

x

r r m a + m2 a2 + ... r d r a CM = ( v CM ) = 1 1 m1 + m2 + .... dt

y

y cm =

CM ycm

q

R

Rsinq q

Law of conservation of linear momentum Linear momentum of a system of particles is equal to the product of mass of the system with velocity of its centre of mass.

y cm =

CM ycm x

y

Sector of a disc (plate)

y cm =

CM ycm

R

q

2Rsinq 3q

r r r If Fext. = 0 then Mv CM = constant

If no external force acts on a system the velocity of its centre of mass remains constant, i.e., velocity of centre of mass is unaffected by internal forces.

Impulse – Momentum theorem

Impulse of a force is equal to the change of momentum

LL

x

4R 3p

-2

Half disc (plate)

r r d ( Mv CM ) From Newton's second law Fext. = dt

19

y

EN

x

0

Segement of a ring

20

Body

Physics HandBook

C HAP TE R

ALLEN

Hollow hemisphere

CM

ycm

R

x

y cm

io n

y

R = 2

t2

r

r

ò Fdt = Dp t1

ss

Force time graph area momentum.

y

Collision of bodies

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\07_COM & Collision, Momentum.p65

A

CM

E

ycm

R

Se

Solid hemisphere

y cm =

3R 8

y cm =

h 3

x

The event or the process, in which two bodies either coming in contact with each other or due to mutual interaction at distance apart, affect each others motion (velocity, momentum, energy or direction of motion) is defined as a collision.

y

Hollow cone

h

CM

ycm

In collision •

The particles come closer before collision and after collision they either stick together or move away from each other.

•

The particles need not come in contact with each other for a collision.

•

The law of conservation of linear momentum is necessarily applicable in a collision, whereas the law of conservation of mechanical energy is not.

x

y

Solid cone

h

gives change in

y cm =

CM ycm x

h 4

35

Physics HandBook

C HAP TE R

ALLEN

T Y P ES O F CO LL I S IO N On the basis of kinetic energy

On the basis of direction

Two dimensional collision or Oblique collision

Elastic collision

In-elastic collision

The collision, in which the particles move along the same plane at different angles before and after collision, is defined as oblique collision.

A collision is said to be elastic, if the total kinetic energy before and after collision remains the same

A collision is said to be inelastic, if the total kinetic energy does not remains constant

one-dimensional collision or Head on collision The collision, in which the particles move along the same straight line before and after the collision, is defined as one dimensional collision.

Perfectly In-elastic collision

The collision, in which particles have no tendancy to regain its shape is called perfectly inelastic collision. In this collision, velocity of particles along the line of collision becomes equal after collsion.

u1

B

u2

m2 m1 Before collision

Head on elastic collision

A

B

Se

Head on inelastic collision of two particles

v2

ss

Þ u1 – u2 = v2 – v1

B

io n

LL 1 1 1 1 m1 u12 + m2 u 22 = m1 v 12 + m2 v 22 2 2 2 2 Rate of separation = Rate of approach i.e. e=1

v1

m1 m2 After collision

Collision

(i) Linear momentum is conserved m1u1 + m2u2 = m1v1 + m2v2 (ii) KE is not conserved but initial KE is equal to final KE

(iii)

A

19

A

20

Head on collision

-2

Value of e is 1 for elastic collision, 0 for perfectly inelastic collision and 0 < e < 1 for inelastic collision.

0

velocity of separation along line of impact v 2 - v1 = velocity of approach along line of impact u1 - u2

A

Let the coefficient of restitution for collision is e

(i) Momentum is conserved m1u1 + m2 u2 = m1 v1 + m2 v 2 ...(i)

(ii) Kinetic energy is not conserved.

v 2 - v1 (iii) According to Newton's law e = u - u ...(ii) 1 2

By solving eq. (i) and (ii) :

æ (1 + e ) m2 ö = m1 u1 + m 2 u 2 - m2 e ( u1 - u 2 ) æ m - em2 ö v1 = ç 1 ÷ u1 + ç m + m ÷ u2 m1 + m2 + m m è 1 è 1 2 ø 2 ø

m1 u1 + m2 u2 - m1e ( u2 - u1 ) æ (1 + e ) m1 ö æ m - em1 ö v2 = ç 2 u2 + ç ÷ u1 = ÷ m1 + m2 è m1 + m2 ø è m1 + m2 ø

Elastic Collision (e=1) • If the two bodies are of equal masses : m1 = m2 = m, v1 = u2 and v2 = u1 Thus, if two bodies of equal masses undergo elastic collision in one dimension, then after the collision, the bodies will exchange their velocities.

36

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\07_COM & Collision, Momentum.p65

e=-

EN

Coefficient of restitution (Newton's law)

E

Physics HandBook

C HAP TE R

ALLEN

• If the mass of a body is negligible as compared to other. If m1>> m2 and u2 = 0 then v 1 = u1 ,

v 2 = 2u1 when a heavy body A collides against a light body B at rest, the body A should keep on moving with same velocity and the body B will move with velocity double that of A. If m 2 >> m1 and u2 = 0 then

v 2 = 0 , v 1 = - u1 When light body A collides against a heavy body B at rest, the body A should start moving with same speed just in opposite direction while the body B should practically remains at rest. w

Loss in kinetic energy in inelastic collision

DK =

m1m2 (1 - e2 )|u1 - u2 |2 2(m1 + m2 )

Oblique Collision

EN

Conserving the momentum of system in directions along normal (x axis in our case) and tangential (y axis in our case) m1u1cosa1 + m2u2cosa2 = m1v1cosb1 + m2v2cosb2

Before collision

b2

x

b1

19

a1 u1

m1

a2

v1

After collision

20

m1

v2

m2

-2

u2

m2

0

y

LL

Since no force is acting on m1 and m2 along the tangent (i.e. y–axis) the individual momentum of m1 and m2 remains conserved. m1u1sina1 = m1v1sinb1 & m2u2sina2 = m2v2sinb2

Rocket propulsion :

E

u

Se

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\07_COM & Collision, Momentum.p65

A

æ dm ö Thrust force on the rocket= v r çè dt ÷ø

ss

io n

v 2 cos b2 - v 1 cos b1 By using Newton's experimental law along the line of impact e = u cos a - u cos a 1 1 2 2

Velocity of rocket at any instant

v

At t=0 v=u m=m0

æ m0 ö v=u–gt+vrln çè m ÷ø

At t=t m=m v=v

exhaust velocity =v r

KEY POINTS r

åm r

r =0

•

Sum of mass moments about centre of mass is zero. i.e.

•

A quick collision between two bodies is more violent then slow collision, even when initial and final velocities are equal because the rate of change of momentum determines that the impulsive force small or large.

•

Heavy water is used as moderator in nuclear reactors as energy transfer is maximum if m1; m2

•

Impulse-momentum theorem is equivalent to Newton's second law of motion.

•

For a system, conservation of linear momentum is equivalent to Newton's third law of motion.

i i / cm

37

Physics HandBook

C HAP TE R

ALLEN

ROTATIONAL MOTION Rigid body is defined as a system of particles in which distance between each pair of particles remains constant (with respect to time) that means the shape and size do not change, during the motion. Eg : Fan, Pen, Table, stone and so on.

DY BO D I RIG

Type of Motion of rigid body

L NA O I N TAT RO OTIO M

rotation.

I = ò dm r 2

19

-2

Radius of Gyration (K) K has no meaning without axis of rotation. I = MK2 K is a scalar quantity

LL

Perpendicular axis Theorems : Iz = Ix + Iy (body lies on the x-y plane)

Moment of inertia of system of particle axis

discrete body

r1

m1

r2

dm

axis continuous body

r3 m3

ss

y

x

(Valid only for 2-dimensional body) Parallel axis Theorem : I = ICM + Md2

I CM

2

I =ICM+Md

CM

I = r2 dm

I = m1r12 + m2r22 + m3r32 + .....

38

o

d

r

m2

z

Se

A

The moment of inertia of a particles with respect to an axis of rotation is equal to the product of mass of the particle and square of distance from rotational axis. I = mr2 r = perpendicular distance from axis of rotation

io n

r

(for all type of bodies) ICM = moment of inertia about the axis Passing through the centre of mass

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\08_Rotational motion.p65

m

·

I M

20

Radius of gyaration : K =

axis

0

For Rigid Bodies : Moment of inertia of a rigid body about any axis of

Moment of Inertia The virtue by which a body revolving about an axis opposes the change in rotational motion is known as moment of inertia.

·

Combined Translational and Rotational Motion

Pure Rotational Motion

EN

Pure Translational Motion

E

Physics HandBook

CH APTER

ALLEN

MOMENT OF INERTIA OF SOME REGULAR BODIES Position of the axis Figure Moment of of rotation Inertia (I) (a) About an axis perpendicular to the plane and passes through the centre

z

A

CM

Mass = M Radius = R

d

I

Z

x

y'

I1

ss

Se Iz

R2

M

2

MR2 4

R 2

5 MR2 4

5 R 2

3 MR2 2

3 R 2

M 2 é R1 + R22 ùû 2ë

R12 + R22 2

y'

CM O I CM

CM

R

x'

x

R

3 R 2

19

io n

LL

A node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\08_Rotational motion.p65

E

M = Mass R1= Internal Radius R2 = Outer Radius

1 MR2 2

R

R M

R

(a) About an axis passing through the centre and perpendicular to the plane of disc

20

R

(b) About a diametric axis

(c) About an axis tangential to the rim and lying in the plane of the disc (d) About an axis tangential to the rim & perpendicular to the plane of disc

3 MR2 2

O

(a) About an axis passing through the centre and perpendicular to the plane of disc

2R

R

M

y

R1

2

2MR2

CM

(d) About an axis tangential to the rim and lying in the plane of ring

R2

R

EN

(c) About an axis tangential to the rim and perpendicular to the plane of the ring

(3) Annular disc

1 MR2 2

x'

(b) About the diametric axis

M = Mass R = Radius

R

B

y

(2) Circular Disc

MR2

0

(1) Circular Ring

Radius of gyration (K)

-2

Shape of the body

R1

39

Shape of the body

C HAP TE R

Position of the axis of rotation

Figure

R2

(b) About a diameteric axis

R1

M

I

(a) About its diametric axis which passes through its centre of mass

Moment of Inertia (I)

Radius of gyration (K)

M 2 é R1 + R22 ùû 4ë

R12 + R22

2 MR2 5

2 R 5

2

R

M

EN

(4) Solid Sphere

ALLEN

M = Mass R = Radius

(b) About a tangent to the Sphere

M

7 MR2 5

I

2 MR2 3

2 R 3

io n

LL

M

20

R

(5) Hollow (a) About diametric axis Sphere passing through centre of mass (Thin spherical Shell)

7 R 5

19

R

-2

I

0

Physics HandBook

R

M = Mass R = Radius L = Length

40

5 MR2 3

5 R 3

R

table tennis ball

(a) About its geometrical axis which is parallel to its length

(b) About an axis which is perpendicular to its length and passes through its centre of mass

M

MR2

MR2 ML2 + 2 12 L

R

R 2 L2 + 2 12

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\08_Rotational motion.p65

(6) Hollow Cylinder

I

Se

(b) About a tangent to the surface

A

M = Mass R = Radius Thickness negligible

ss

table tennis ball

E

Position of the axis of rotation

Figure

(c) About an axis perpendicular to its length and passing through one end of the cylinder

Radius of gyration (K)

MR2 ML2 + 2 3

R2 L2 + 2 3

MR2 2

R

L

(a) About its geometrical axis, which is along its length

I R

M

2

EN

(7) Solid Cylinder M = Mass R = Radius L = Length

Moment of Inertia (I)

(b) About an axis tangential to the cylinderical surface and parallel to its geometrical axis

I

3 MR2 2

M

19

20

ML2 MR2 + 12 4

(9) Rectangular Plate

M = Mass a = Length b = Breadth

ss

(b) About an axis passing through one end and perpendicular to length of the rod

(a) About an axis passing through centre of mass and perpendicular to side b in its plane (b) About an axis passing through centre of mass and perpendicular to side a in its plane.

ML2 12

L 12

L

Se

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\08_Rotational motion.p65

E

Mass = M Length = L

L2 R2 + 12 4

L

(a) About an axis passing through centre of mass and perpendicular to its length

A

Thickness is negligible w.r.t. length

io n

LL

(c) About an axis passing through the centre of mass and perpendicular to its length

(8) Thin Rod

3 R 2

-2

R

0

Shape of the body

Physics HandBook

CH APTER

ALLEN

ML2 3

L 3

L

b

b

Mb 2 12

2 3

Ma 2 12

2 3

a

b

a

a

41

Physics HandBook Shape of the body

C HAP TE R

Position of the axis of rotation

Figure

Moment of Inertia (I)

(

(c) About an axis passing throught centre of mass and perpendicular to plane (10) Cube

ALLEN

M a2 + b 2

Radius of gyration (K)

)

a2 + b 2 12

12

b a

About an axis passes through centre of mass and perpendicular to face

a

Ma2 6

6

IC =

r

C

A

0

Line of action of force F

P q

rsinq

Se

of force from the axis of

r/ 2

r

O

t=rFsinq Direction of torque can be determined by using right hand thumb rule.

ROTATIONAL EQUILIBRIUM

If a rigid body is in rotational equilibrium under the action of several coplanar forces, the resultant torque of all the forces about any axis perpendicular to the plane containing the forces must be zero. In the figure a body is shown under the action of several external coplanar forces F1, F2, … …Fi, and Fn. v åtP = 0

Here P is a point in the plane of the forces about which we calculate torque of all the external forces acting on the body. The flexibility available in selection of the point P provides us with advantages that we can select such a point about which torques of several unknown forces will become zero or we can make as many number of equations as desired by selecting

42

Fn

F2 y

T1

C

l /4 400

T2

x

D

A

Fi

F1

B l/2

200

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\08_Rotational motion.p65

distance of line of action

mr 2 2

io n

Magnitude of torque = Force × perpendicular

ss

TORQUE

19

-2

C

r Torque about point : rt = rr ´ F

rotation.

r

I C = mr 2

r

LL

(12) Sector of a uniform disk of mass m

About an axis Passing through center and perpendicular to the plane containing the arc About an axis Passing through center and perpendicular to the plane containing the sector.

20

Mass = M Side a (11) Uniform thin rod bent into shape of an arc of mass m

EN

a

E

Physics HandBook

CH APTER

ALLEN

several different points. The first situation yields to a simpler equation to be solved and second situation though does not give independent equation, which can be used to determine additional unknowns yet may be used to check the solution. The above condition reveals that a body cannot be in rotational equilibrium under the action of a single force unless the line of action passes through the mass center of the body. A case of particular interest arises where only three coplanar forces are involved and the body is in rotational equilibrium. It can be shown that if a body is in rotational equilibrium under the action of three forces, the lines of action of the three forces must be either concurrent or parallel. This condition provides us with a graphical technique to analyze rotational equilibrium.

Equilibrium of Rigid Bodies

A rigid body is said to be in equilibrium, if it is in translational as well as rotational equilibrium both. To analyze such problems conditions for both the equilibriums must be applied.

Rotation about fixed axis not passing through mass center

r F1

r Fn

0

r aC on circular path of radius rP / C . To deal with this kind of motion, we have

-2

EN

In this kind of rotation the axis of rotation remains fixed and does not passes through the mass center. Rotation of door is a common example of this category. Doors are hinged about their edges; therefore their axis of rotation does not pass through the mass center. In this kind of rotation motion the mass center executes circular motion about the axis of rotation. r In the figure, free body diagram and kinetic diagram of a body R r r P rotating about a fixed axis through point P is shown. It is easy to conceive maC P Fi that as the body rotates its mass center moves on a circular path of radius r I pa r rP / C . The mass center of the body is in translation motion with acceleration C C

LL

we can write the following equation also.

)

r r St P = I P a

io n

Pure Rotation about P

20

(

19

to make use of both the force and the torque equations. r r r r r Translation of mass center SFi = MaC = M a ´ rC / P - M w 2 rC / P r r Centroidal Rotation St C = I C a r r r r I P = MrP2 / C + I C and aC / P = a ´ rC / P - w 2 rC / P Making use of parallel axis theorem

ANGULAR MOMENTUM (MOMENT OF LINEAR MOMENTUM)

E

r

ss

l According to Newtons Second Law’s for rotatory r r dL r = Ia . motion t = dt

Se

p= mv

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\08_Rotational motion.p65

A

Angular momentum of a particle about a given axis is the product of its linear momentum and perpendicular distance of line of action of linear momentum vector from the axis r r r of rotation, L = r ´ p

l Angular Impulse = Change in angular momentum.

q

q

b= rs i nq

Magnitude of Angular momentum = Linear momentum × Perpendicular distance of line of action of momentum from the axis of rotation L = mv × r sinq Direction of angular momentum can be used by using right hand thumb rule.

l If a large torque acts on a body for a small time r then, angular impulse= tdt

Conservation of Angular Momentum Angular momentum of a particle or a system remains constant if t ext = 0 about that point or axis of rotation. If t = 0 then

DL = 0 Þ L= constant Dt

Þ Lf = Li

or I1w1 = I2w2

43

Physics HandBook

C HAP TE R

ALLEN

Examples of Conservation of Angular Momentum

l If a person skating on ice folds his arms then his M.I. decreases and 'w' increases.

l A diver jumping from a height folds his arms and legs (I decrease) in order to increase no. of rotation in air by increasing 'w'.

wi

Ii

wf

If

Ii>If w i< w f

19

-2

0

EN

l If a person moves towards the centre of rotating platform then 'I' decrease and 'w' increase.

LL

Kinetic Energy of Rotation KE R =

Other forms

l

If external torque acting on a body is equal to zero (t = 0), L=constant

l

Rotational Work : Wr = tq (If torque is constant)

l

l

Wr =

dW dq =t = tw dt dt

K µw

ò tdq (If torque is variable)

q1

The work done by torque = Change in kinetic energy of rotation. W =

Instantaneous power =

1 Kµ , I

q2

Se

A

ss

l

Average power Pav =

1 2 1 2 1 Iw - Iw = I(w 22 - w12 ) 2 2 2 1 2

DW Dt

COMBINED TRANSLATIONAL AND ROTATIONAL MOTION OF A RIGID BODY

When a body perform translatory motion as well as rotatory motion then it is known as rolling.

Pure rolling

In Pure Rolling

w CM

vCM =Rw CM R

(i) If the velocity of point of contact with respect to the surface is zero then it is known as pure rolling. Contact point

44

vCM

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\08_Rotational motion.p65

1 2 L2 1 Iw = = Lw 2 2I 2

1 2 Iw 2

io n

K=

20

ROTATIONAL KINETIC ENERGY

E

(ii)

Physics HandBook

CH APTER

ALLEN

If a body is performing rolling then the velocity of any point of the body with respect to the surface r r r r is given by v = v CM + wCM ´ R A VCM F

VCM

C C

A

VCM VCM

wR

VCM

VCM E

VCM=wR VCM=0

F

E

C

VCM

VCM

A

wR

2VCM VCM VCM

VCM

C

2VCM

wR

VCM VCM

wR

Only T.M.

VCM=wR P VCM-w R=0

B

Only R.M.

Pure rolling motion

EN

B

F

VCM+w R=2VCM

2 VCM

VF =

VB = 0

Rolling Kinetic Energy under pure rolling

Velocity at a point on rim of sphere

19

Rolling body

v 2 + R 2w 2 + 2vRw cos q

VCM

v

q

LL

v net =

2 VCM

q

-2

VE =

20

VA = 2VCM

0

Only Translatory motion + Only Rotatory Motion = Rolling motion. For pure rolling above body

surface

io n

Rw

Rolling Kinetic Energy

ss

æ v2 ö 1 2 1 1 1 mv + mK2 ç 2 ÷ E = mv 2 + Iw2 = 2 2 2 2 èR ø

For pure rolling v = Rw

E

q 2

Se

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\08_Rotational motion.p65

A

v net = 2v cos

Rolling Kinetic Energy E =

1 2æ K2 ö mv ç 1 + 2 ÷ 2 R ø è

Etranslation : Erotation : ETotal = 1 :

K2 K2 : 1 + R2 R2 2

Body

K2 R2

E trans 1 = 2 E rotation KR2

E trans 1 = 2 E total 1 + KR2

K E rotation 2 = R K2 E total 1 + R2

Ring

1

1

1/2

1/2

Disc

1/2

2

2/3

1/3

Solid sphere

2/5

5/2

5/7

2/7

Spherical shell

2/3

3/2

3/5

2/5

Solid cylinder

1/2

2

2/3

1/3

1

1

1/2

1/2

Hollow cylinder

45

Physics HandBook

C HAP TE R

ALLEN

General Plane Motion: Rotation about axis in translation motion Rotation of bodies about an axis in translation motion can be dealt with either as superposition of translation of mass center and centroidal rotation or assuming pure rotation about the instantaneous axis of rotation. In the figure is shown the free body diagram and kinetic diagram of a body in general plane motion. n r r Translation of mass center å Fi = MaC i =0 n

r

åt

Centroidal Rotation

i -1

C

r F1

r Fi

r F2

r I Ca

r MaC

C

r Fn

r = IC a

This kind of situation can also be dealt with considering it rotation about IAR. It gives sometimes quick solutions, especially when IAR is known and forces if acting at the IAR are not required to be found.

EN

Rolling Motion on an inclined plane

Rolling body

-2

19

S

Applying Conservation of energy

ss

1 2 1 2 mv + Iw 2 2

Linear accleration on reaching the lowest point a

l

A

2 1 2 1 2æ v ö mgh = 2 mv + 2 mK ç 2 ÷ èR ø

1 2æ K2 ö mgh = 2 mv ç 1 + R2 ÷ ...(1) è ø

h = s sinq

l

...(2)

from (1) & (2)

VRolling =

46

2gh K2 1+ 2 R

=

2gs sin q K2 1+ R2

l

=

g sin q 1 + K2 / R2

Time taken to reach the lowest point of the plane is

Se

mgh =

l

q

io n

LL

Inclined plane

20

height

VCM

0

m

t=

2s(1 + K 2 / R 2 ) g sin q

K2 R2

Least, will reach first

K2 R2

Maximum, will reach last

K2 R2

equal, will reach together

When ring, disc, hollows sphere, solid sphere rolls on same inclined plane then vS > vD > vH > vR aS > aD > aH > aR tS < tD < tH < tR

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\08_Rotational motion.p65

w

E

Physics HandBook

CH APTER

ALLEN

For a pure rolling body after one full rotation

A2

A3 A4

A1

A5 2p R displacement of lowermost point =2pR distance = 8R

0

20

and parallel to the original is expressed by the following equation. r r r r Lo = rC ´ ( Mv C ) + I C w

19

-2

EN

Angular Momentum in general plane motion Angular momentum of a body in plane motion can also be written similar y r to torque equation or kinetic energy as sum of angular momentum about the vC w axis due to translation of mass center and angular momentum of centroidal C rotation about centroidal axis parallel to the original axis. r Consider a rigid body of mass M in plane motion. At the instant shown its mass rC r r O x center has velocity v and it is rotating with angular velocity w about an axis r perpendicular to the plane of the figure. It angular momentum Lo about an axis passing though the origin

io n

LL

The first term of the above equation represent angular momentum due to translation of the mass center and the second term represents angular momentum in centroidal rotation. Angular momentum in rotation about fixed axis y Consider a body of mass M rotating with angular velocity w about a r fixed axis perpendicular to plane of the figure passing through point P r

O

vC

rC / P

w C

x

The above equation reveals that the angular momentum of a rigid body in plane motion can also be expressed in a single term due to rotation about the instantaneous axis of rotation. Angular momentum in pure centroidal rotation In pure centroidal rotation, mass center remains at rest, therefore w angular momentum due to translation of the mass center vanishes. C r r LC = I C w

A node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\08_Rotational motion.p65

E

Se

the body about the fixed rotational axis. r r LP = I P w

ss

P. Making use of the parallel axis theorem I P = MrC2 / P + I C and r r r r equation vC = w ´ rC / P we can express the angular momentum LP of

Eccentric Impact

In eccentric impact the line of impact which is the common normal drawn at the point of impact does not passes through mass center of at least one of the colliding bodies. It involves change in state of rotation motion of either or both the bodies. Consider impact of two A and B such that the mass center CB of B does Line of CB not lie on the line of impact as shown in figure. If we assume bodies to Impact be frictionless their mutual forces must act along the line of impact. A B The reaction force of A on B does not passes through the mass center CA of B as a result state of rotation motion of B changes during the impact.

47

Physics HandBook

C HAP TE R

ALLEN

20

0

o

1 3

M l2 for Io, we have

3mv B¢ + M lw ¢ = 3mv o (1) The velocity of the lower end of the rod before the impact was zero and immediately after the impact it becomes lw' towards right. Employing these facts we can express the coefficient of restitution according to eq.

e=

v Qn ¢ - v Pn ¢ v pn - v Qn ®

lw ¢ - v B¢ = ev o

(2)

From eq. (1) and (2), we have

48

Velocity of the ball immediately after the impact

v B¢ =

Angular velocity of the rod immediately after the impact

w¢ =

(3m - eM ) v o 3m + M 3 (1 + e ) mv o (3m + M ) l

Ans. Ans.

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\08_Rotational motion.p65

Substituting

Se

A

ss

io n

LL

19

-2

EN

Problems of Eccentric Impact Problems of eccentric impact can be divided into two categories. In one category both the bodies under going eccentric impact are free to move. No external force act on either of them. There mutual forces are responsible for change in their momentum and angular momentum. In another category either or both of the bodies are hinged. Eccentric Impact of bodies free to move Since no external force acts on the two body system, we can use principle of conservation of linear momentum, principle of conservation of angular momentum about any point and concept of coefficient of restitution. The coefficient of restitution is defined for components of velocities of points of contacts of the bodies along the line of impact. While applying principle of conservation of angular momentum care must be taken in selecting the point about which we write the equation. The point about which we write angular momentum must be at rest relative to the selected inertial reference frame and as far as possible its location should be selected on line of velocity of the mass center in order to make zero the first term involving moment of momentum of mass center. Eccentric Impact of hinged bodies When either or both of the bodies are hinged the reaction of the hinge during the impact act as external force on the two body system, therefore linear momentum no longer remain conserved and we cannot apply principle of conservation of linear momentum. When both the bodies are hinged we cannot also apply conservation of angular momentum, and we have to use impulse momentum principle on both the bodies separately in addition to making use of coefficient of restitution. But when one of the bodies is hinged and other one is free to move, we can apply conservation of angular momentum about the hinge. O Ex. A uniform rod of mass m and length l is suspended from a fixed support and can rotate freely in the vertical plane. A small ball of mass m moving horizontally with velocity vo strikes elastically the lower end of the rod as shown in the figure. Find the angular velocity of the rod and velocity of the ball immediately after the impact. v Sol. The rod is hinged and the ball is free to move. External forces acting on w' O O the rod ball system are their weights and reaction from the hinge. Weight of the ball as well as the rod are finite and contribute negligible impulse during the impact, but impulse of reaction of the hinge during impact is considerable and cannot be neglected. Obviously linear momentum of the system is not vo v' conserved. The angular impulse of the reaction of hinge about the hinge is zero. Therefore angular momentum of the system about the hinge is Immediately after Before the impact conserved. Let velocity of the ball after the impact becomes v'B and angular the impact velocity of the rod becomes w'. We denote angular momentum of the ball and the rod about the hinge before the impact by L and L and B1 R1 after the impact by L and L . B2 R2 Applying conservation of angular momentum about the hinge, we have r r r r LB 1 + LR1 = LB 2 + LR 2 ® mv o l + 0 = mv B¢ l + I o w ¢

E

A

E

io n

ss

CH APTER

0

-2

19

20

EN

LL

Se

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\08_Rotational motion.p65

ALLEN Physics HandBook

IMPORTANT NOTES

49

Physics HandBook

C HAP TE R

ALLEN

SIMPLE HARMONIC MOTION Periodic Motion

Oscillatory Motion

Any motion which repeats itself after regular interval of time (i.e. time period) is called periodic motion or harmonic motion. Example: (i) Motion of planets around the sun. (ii) Motion of the pendulum of wall clock.

The motion of body is said to be oscillatory or vibratory motion if it moves back and forth (to and fro) about a fixed point after regular interval of time.

Simple Harmonic Motion (SHM) Simple harmonic motion is the simplest form of vibratory or oscillatory mo tion . In whi ch restoring force is directly proporiontal to distance from mean.

The fixed point about which the body oscillates is called mean position or equilibrium position.

EN

Example: (i) Vibration of the wire of 'Sitar'. (ii) Oscillation of the mass suspended from spring.

Some Basic Terms in SHM

2p 1 where w is angular = n w frequency and n is frequency. Frequency (n or f)

It is given by T =

•

The number of oscillations per second is defined as frequency. It is given by n =

•

50

w 1 = T 2p

Phase Phase of a vibrating particle at any instant is the state of the vibrating particle regarding its displacement and direction of vibration at that particular instant. In the equation x = A sin (wt + f), (wt+f) is the phase of the particle.

0

-2

19

20

Two vibrating particles are said to be in same phase if the phase difference between them is an even multiple of p, i.e. Df = 2np where n = 0, 1, 2, 3,....

Two vibrating particle are said to be in opposite phase if the phase difference between them is an odd multiple of p i.e., Df = (2n + 1)p where n = 0, 1, 2, 3,....

io n

•

Angular frequency (w) :The rate of change of phase angle of a particle with respect to time is defined as its angular frequency.

w

w=

For linear SHM

(F µ – x) : F = m w

k m

d2 x = –kx = –mw2x where w = dt 2

k m

For angular SHM ( t µ -q ) : t=I

d2q = Ia = – kq = – mw2q where w = dt2

w

Displacement x = A sin (wt + f),

w

Angular displacement q =q0 sin(wt+ f)

k m

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\09_SHM_Oscillation.p65

A

•

The difference of total phase angles of two particles executing SHM with respect to the mean position is known as phase difference.

ss

•

The phase angle at time t = 0 is known as initial phase or epoch.

Se

•

Mean Position The point at which the restoring force on the particle is zero and potential energy is minimum, is known as its mean position. Restoring Force The force acting on the particle which tends to bring the particle towards its mean position, is known as restoring force. Restoring force always acts in a direction opposite to that of displacement. Displacement is measured from the mean position. Amplitude The maximum (positive or negative) value of displacement of particle from mean position is defined as amplitude. Time period (T) The minimum time after which the particle keeps on repeating its motion is known as time period. The smallest time taken to complete one oscillation or vibration is also defined as time period.

LL

•

E

Physics HandBook

CH APTER

dx = Aw cos(wt + f) = w A 2 - x2 dt

w

Velocity v =

w

Angular velocity

w

d2 x Acceleration a = 2 =– Aw2 sin(wt + f) = –w2 x dt

w

Angular acceleration

Average energy in SHM (i) The time average of P.E. and K.E. over one cycle is

dq = q0wcos(wt+f) dt

(c) < TE>t =

w

Potential energyU= kx2 = mw2A2sin2 (wt + f)

w

1 Total energy E = K+U= mw2A2 = constant 2

w

1 2 kA + U0 2

Spring block system

T = 2p

k

k

m k

m

w

time

A

20

m 1 1 1 1 = + + where k eff k eff k1 k 2 k 3

K.E.

Parallel combination of springs

Se

K,U node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\09_SHM_Oscillation.p65

E

ET

w

Note : Total energy of a particle in S.H.M. is same at all instant and at all displacement.

(ii)

To tal ener gy depends upon mass, amplitude and frequency of vibration of the particle executing S.H.M.

k1

k3

k2

T = 2p

k2 k3 m

m k eff

m

where keff = k1 + k2 + k3

P.E.

(i)

k1

Series combination of springs T = 2p

w

Umax or

m k

ss

displacement

Kmax or

T=2p

io n

TE

m2

m k

m1 m2 where µ=reduced mass = m + m 1 2

1 k A2 2

LL

1 k A2 2

m1

w

T = 2p

19

m

TE

1 kA2 6

-2

Kinetic energy K = mv2 = mw2A2cos2(wt+ f) 1 2

(b) < PE>x = U 0 +

EN

(c) x =

w

1 2

1 2 kA 4

1 2 kA + U0 2

1 2 kA 3

(a) x =

d q = – q0w2sin(wt+f)=–w2q dt2 1 2

(b) t =

(ii) The position average of P.E. and K.E. between x = – A to x=A

2

1 2

1 2 kA 4

(a) t =

0

ALLEN

Time period of simple pendulum

L

w

Time period T = 2p

L g

Second pendulum Time period = 2 seconds, Length » 1 meter (on earth's surface)

51

Physics HandBook

w

C HAP TE R

Time period of Physical pendulum k2 +l I T = 2p = 2p l mgl g

ALLEN

In accelerating cage l

a

cm

where Icm = mk2 w

a

Time period of Conical pendulum q

h l cos q T = 2p = 2p g g

h

a

l m

I where k = torsional constant of the wire k

geff =

T = 2p

I=moment of inertia of the body about the vertical axis

l g+a

T = 2p

l g-a

KEY POINTS

T = 2p

(g

g2 + a 2

l

2

+ a2 )

1

SHM is the projection of uniform circular motion along one of the diameters of the circle.

•

The periodic time of a hard spring is less as compared to that of a soft spring because the spring constant is large for hard spring.

•

For a system executing SHM, the mechanical energy remains constant.

•

Maximum kinetic energy of a particle in SHM may be greater than mechanical energy as potential energy of a system may be negative.

•

The frequency of oscillation of potential energy and kinetic energy is twice as that of displacement or velocity or acceleration of a particle executing S.H.M.

w

Spring cut into two parts :

k1

Se

A

l1 m æ m ö æ n ö Here l = n l 1 = ç ÷ø l , l 2 = çè m + n ÷ø l è m n + 2

l1

k2

20

l2

But kl=k1l1= k2l2 Þ k1 =

IMPORTANT NOTES

(m + n) (m + n) k; k 2 = k m n node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\09_SHM_Oscillation.p65

l

ss

k

io n

LL

19

•

52

2

-2

T=2p

geff = g–a

0

geff = g + a

Time period of Torsional pendulum

EN

w

E

A

E

io n

ss

CH APTER

0

-2

19

20

EN

LL

Se

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\09_SHM_Oscillation.p65

ALLEN Physics HandBook

IMPORTANT NOTES

53

Physics HandBook

C HAP TE R

ALLEN

WAVE MOTION & DOPPLER'S EFFECT A wave is a disturbance that propagates in space, transports energy and momentum from one point to another without the transport of matter. CLASSIFICATION OF WAVES Medium Necessity

Vibration of medium particle

Stationary (standing) waves 1-D (Waves on strings)

Transverse waves

Dimension

Longitudinal waves

2-D (Surface waves or ripples on water) 3-D (Sound or light waves)

If particle vibrates in pependiuclar direction to the wave motion then its called transverse wave. In strings, mechanical waves are always transverse. In gases and liquids, mechanical waves are always longitudinal because fluids cannot sustain shear. Partially transverse waves are possible on a liquid surface because surface tension provide some rigidity on a liquid surface. These waves are called as ripples as they are combination of transverse & longitudinal. In solids mechanical waves (may be sound) can be either transverse or longitudinal depending on the mode of excitation. In longitudinal wave motion, oscillatory motion of the medium particles produce regions of compression (high pressure) and rarefaction (low pressure).

0

-2

PLANE PROGRESSIVE WAVES

io n

•

2p = wave propagation constant l (Wave is moving along x-axis and paticle is moving along y axis) If coefficient of x & t are of opposite sign than wave move in +ve x-direction Ex. y = A sin ( wt - kx ) wave is moving in +x direction and particle is moving in y direction

Wave equation : y = A sin ( wt - kx ) where k =

LL

•

19

•

Non-mechanical waves (EM waves)

Progressive waves

20

•

Propagation of energy

EN

• • • •

Mechanical (Elastic) waves

Simillarly Ex. z = A sin ( wt - ky ) here wave move in +y direction and particle move in z direction

¶2 y 1 ¶2 y = ¶x 2 v 2 ¶t 2

dx w dx w = Q wt - kx= constant Þ = dt k dt k

A

Wave velocity (phase velocity) v w = •

•

•

Particle velocity v p =

æ dy ö dy = Aw cos ( wt - kx ) v p = - v w ´ slope = - v w ç ÷ è dx ø dt

¶2 y = -w 2 A sin ( wt - kx ) = -w 2 y ¶t 2 For particle 1 : vp ¯ and ap ¯ For particle 2 : vp ­ and ap ¯ For particle 3 : vp ­ and ap ­ For particle 4 : vp ¯ and ap ­ Relation between phase difference, path difference & time difference Particle acceleration : a p =

0

p T/2 l

54

2p T

Df Dl DT = = 2p T l

y 1 2

3

4

t

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\10_Wave motion & Doppler's effect.p65

ss

Differential equation :

Se

•

E

ENERGY IN WAVE MOTION •

WAVE FRONT Spherical wave front (source ®point source)

•

Cylindrical wave front (source ®linear source)

•

Plane wave front (source ® point / linear source at very large distance)

KE 1 æ Dm ö 2 = ç ÷ vp volume 2 è volume ø 1 1 = rv 2p = rw2 A 2 cos2 ( wt - kx ) 2 2 2

•

TE = rw 2 A2 cos2 ( wt - kx ) volume

[i.e. Average total energy / volume]

INTENSITY OF WAVE

• Power : P = (energy density) (volume/ time)

[where S = Area of cross-section]

y ( r,t ) =

Power 1 = rw 2 A 2 v Intensity : I = area of cross-section 2

A node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\10_Wave motion & Doppler's effect.p65

Due to cylindrical source I µ

•

y ( r, t ) =

1 r

A

r r sin ( wt - k × r )

r Due to plane source I = constant

r r y ( r,t ) = A sin ( wt - k × r )

INTERFERENCE OF WAVES y1 = A1sin(wt–kx), y2 = A2sin(wt–kx+f0) y = y1 + y2 = A sin (wt – kx + f)

Se

T = tension in the string.

E

•

r r A sin ( wt - k × r ) r

ss

T where m = mass/length and m

KEY POINTS

1 r2

io n

Speed of transverse wave on string :

v=

Due to point source I µ

LL

•

•

19

æ 1 2 2ö P = èç rw A ø÷ ( Sv ) 2

0

1 2 2 rw A 2

-2

• Pressure energy density u =

EN

•

PE 1 1 æ dy ö = rv 2 ç ÷ = rw 2 A 2 cos 2 ( wt - kx ) è dx ø volume 2 2

20

•

Physics HandBook

CH APTER

ALLEN

• A wave can be represent ed by fu ncti on y=f(kx ± wt) because it satisfy the differential

• Longitudinal waves can be produced in solids, liquids and gases because bulk modulus of elasticity is present in all three.

A

f0

1 æ ¶2 y ö w ¶2 y = 2 ç 2 ÷ where v = . equation 2 k v è ¶t ø ¶x

• A pulse whose wave function is given by y=4 / [(2x +5t)2+2] propagates in –x direction as this wave function is of the form y=f (kx + wt) which represent a wave travelling in –x direction.

A2

f A1

where

A=

and

tan f =

As

I µ A2

So

I = I1 + I2 + 2 I1 I2 cos f0

A12 + A 22 + 2A1 A 2 cos f0 A 2 sin f0 A1 + A 2 cos f 0

55

Physics HandBook •

(

I1 + I2

)

Rarer

Denser

yi=Aisin(w t-k1x)

yt=A tsin(wt-k2x)

2

For destructive interference [Minimum Intensity] f0 = (2n+1)p or path difference = (2n+1) where n = 0, 1, 2, 3, .... Imin =

•

ALLEN

• For constructive interference [Maximum intensity] f0 = 2np or path difference = nl where n = 0, 1, 2, 3, .... Imax =

•

C HAP TE R

(

l 2

I1 - I2

yr=-A rsin(wt+k1x)

•

)

Denser

2

Rarer

yi=Aisin(wt-k 1x)

yt=A tsin(w t-k 2x)

I max - I min Degree of hearing= I + I ´ 100 max min

EN

yr=A rsin(wt+k 1x)

REFLECTION AND REFRACTION (TRANSMISSION) OF WAVES

t

At

•

æ v 2 - v1 ö Amplitude of reflected wave® A r = ç v + v ÷ A i è 1 2ø

æ 2v 2 ö Amplitude of transmitted wave ® A t = çè v + v ÷ø A i 1 2

A •

• •

•

56

0

io n

The frequency of the wave remain unchanged.

Se

•

•

When two waves of same frequency and amplitude travel in opposite direction at same speed, their superstition gives rise to a new type of wave, called stationary waves or standing waves. Formation of standing wave is possible only in bounded medium. • Let two waves are y1=Asin(wt–kx); y2=Asin(wt+kx) by principle of superposition y=y1+y2 =2Acoskxsinwt ¬ Equation of stationary wave • As this equation satisfies the wave equation

ss

Reflected wave

•

STATIONARY WAVES OR STANDING WAVES

If v2 > v1 i.e. medium-2 is rarer Ar > 0 Þ no phase change in reflected wave If v2 < v1 i.e. medium-1 is rarer Ar < 0 Þ There is a phase change of p in reflected wave As At is always positive whatever be v1 & v2 the phase of transmitted wave always remains unchanged. In case of reflection from a denser medium or rigid support or fixed end, there is inversion of reflected wave i.e. phase difference of p between reflected and incident wave. The transmitted wave is never inverted.

•

¶2 y 1 ¶2 y = , it represent a wave. ¶x 2 v 2 ¶t 2 Its amplitude is not constant but varies periodically with position. Nodes®amplitude is minimum :

•

l 3l 5l , , ,...... 4 4 4 Antinodes ® amplitude is maximum :

•

cos kx = 0 Þ x =

•

•

3l l cos kx = 1 Þ x = 0, , l, ,...... 2 2 The nodes divide the medium into segments (loops). All the particles in a segment vibrate in same phase but in opposite phase with the particles in the adjacent segment. As nodes are permanently at rest, so no energy can be transmitted across them, i.e. energy of one region (segment) is confined in that region.

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\10_Wave motion & Doppler's effect.p65

r Ar

19

transmitted wave

i

20

Ai

-2

v2

LL

Medium-1

yi=Aisin(w t-k1x)

v1

When two sound waves of nearly equal (but not exactly equal) frequencies travel in same direction, at a given point due to their super position, intensity alternatively increases and decreases periodically. This periodic waxing and waning of sound at a given position is called beats. Beat frequency = difference of frequencies of two interfering waves Beat frequency = |f1 – f2|

Medium-2 Ði=Ðr

Incident wave

BEATS

E

Physics HandBook

CH APTER

Transverse stationary waves in stretched string [Fixed at both ends]

Sound Waves Velocity of sound in a medium of elasticity E and density r is

[fixed end ®Node & free end®Antinode]

v= E r

Fundamental or l l= 2

f=

second harmonic

2v f= 2l

first overtone

l=l

v 2l

first harmonic or zero overtone

3v f= 2l

third harmonic

Fluids

(Bulk Modulus)

v= Y r

v= B r

• Newton's formula : Sound propagation is isothermal B = P Þ v =

3l l= 2

• Laplace correction : Sound propagation is adiabatic B = gP Þ v =

Fixed at one end

nv 2l

KEY POINTS • With rise in temperature, velocity of sound

-2

fl =

(n-1) th overtone

in a gas increases as v =

• With rise in humidity velocity of sound increases due to presence of water in air. • Pressure has no effect on velocity of sound in a gas as long as temperature remains constant.

third harmonic

5l 4

harmonic

overtone

f=

5v 4l

f=

(2n + 1) v 4l

A

N \\\\\\\\\\\\\\\\\\\\\\\\\

N

\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

fixed

\\\\\\\\\\\\\\\\\\

Sonometer

\

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\10_Wave motion & Doppler's effect.p65

n

th

th

\\ \\ \ \\\

E

(2n+1)

plucking (freeend)

[ p : number of loops]

Displacement and pressure wave

A sound wave can be described either in terms of the longitudinal displacement suffered by the particles of the medium (called displacement wave) or in terms of the excess pressure generated due to compression and rarefaction (called pressure wave). Displacement wave y=Asin(wt–kx) Pressure wave p = p0cos(wt–kx) where p0 = ABk = rAvw Note : As sound-sensors (e.g., ear or mike) detect pressure changes, description of sound as pressure wave is preferred over displacement wave. KEYPOINTS • The pressure wave is 90° out of phase w.r.t. displacement wave, i.e. displacement will be maximum when pressure is minimum and vice-versa. • Intensity in terms of pressure amplitude

ss

fifth harmonic second overtone

A

l=

3l 4

3v 4l

Se

l=

f=

first overtone

io n

LL

l 4

v 4l

gRT MW

20

f=

Fundamental

gP r

19

n th harmonic

l=

P r

EN

second overtone

Solids (Young's Modulus)

0

ALLEN

fn =

p T 2l m

I=

p20 2rv

57

Physics HandBook

C HAP TE R

Vibrations of organ pipes

Intensity of sound in decibels æ Iö Sound level, SL = 10 log10 ç ÷ è I0 ø Where I0 = threshold of human ear = 10–12 W/m2 Characteristics of sound • Loudness ® Sensation received by the ear due to intensity of sound. • Pitch ® Sensation received by the ear due to frequency of sound. • Quality (or Timbre)® Sensation received by the ear due to waveform of sound. Doppler's effect in sound : A stationary source emits wave fronts that propagate with constant velocity with constant separation between them and a stationary observer encounters them at regular constant intervals at which they were emitted by the source. A moving observer will encounter more or lesser number of wavefronts depending on whether he is approaching or receding the source. A source in motion will emit different wave front at different places and therefore alter wavelength i.e. separation between the wavefronts. The apparent change in frequency or pitch due to relative motion of source and observer along the line of sight is called Doppler Effect.

B

a b

c

S

-2

19

ss

a b

A

l/4 N

speed of sound wave w.r.t. observer observed wavelength v + v0 æ v + v0 ö n¢ = n = æ v - v s ö çè v - v s ÷ø èç n ø÷

n¢ =

v0 + vs ö æ If v0, vs as, then MV is more than la (ii) if a0 < aS, then MV is less than la

67

Physics HandBook

C HAP TE R

Application of Thermal expansion in solids I.

Thermal expansion in liquids (Only volume expansion)

Bi-metallic strip (used as thermostat or auto-cut

in electric heating circuits) R =

d ( a1 - a 2 ) DT

a 1>a 2

ALLEN

ga =

Apparent increase in volume Initial volume ´ Temperature rise

gr =

real increse in volume initial volume ´ temperature rise

R

g r = g a + g vessel Change in volume of liquid w.r.t. vessel

l DT 1 Dl Þ T µ l1/ 2 Þ = g T 2 l

V 00

DT 1 = aDq Fractional change in time period = T 2 III. Scale reading : Due to linear expansion / contraction, scale reading will be lesser / more than actual value. If temperature ­ then actual value = scale reading (1+aDq)

Initially

Finally

19

[a® vessel; g L® liquid ]

LL

Increase in height of liquid level in tube when bulb was initially completely filled. apparent change in volume of liquid V0 ( g L - 3a ) DT h= = A 0 (1 + 2a ) DT area of tube

io n

IV. Thermal Stress

Anomalous expansion of water : In the range 0°C to 4 °C water contract on heating and expands on cooling. At 4°C ® density is maximum. Aquatic life is able to survive in very cold countries as the lake bottom remains unfrozen at the temperature around 4°C.

Dl = aDq Thermal strain = l

As Young's modulus Y =

•

F/A ; Dl / l

• •

68

1 DV Coefficient of volume expansion : g V = V DT = T 0

[PV = nRT at constant pressure VµT Þ

YAaDq

So thermal stress = YAaDq = 1 + aDq ( )

• •

Thermal expansion of gases :

•

DV DT = ] V T

1 DP Coefficient of pressure expansion g P = P DT = T 0

KEY POINTS : Liquids usually expand more than solids because the intermolecular forces in liquids are weaker than in solids. Rubber contract on heating because in rubber as temperature increases, the amplitude of transverse vibrations increases more than the amplitude of longitudinal vibrations. Water expands both when heated or cooled from 4°C because volume of water at 4°C is minimum. In cold countries, water pipes sometimes burst, because water expands on freezing.

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\12_Thermal physics.p65

A

Heating [Compressive Stress]

Se

ss

Cooling [Tensile Stress]

0

h

-2

II. Simple pendulum :

T = 2p

Expansion in enclosed volume

20

Room Temperature

DV = V0 ( g r - 3a ) DT Higher Temperature

EN

d d

E

•

•

Q DT Amount of heat required to raise the temperature of a given body by 1°C (or 1K).

•

Q (m = mass) mDT Amount of heat required to raise the temperature of unit mass of a body through 1°C (or 1K)

•

Thermal capacity of a body =

Specific heat capacity =

Q (n=number of moles) nDT Water equivalent : If thermal capacity of a body is expressed in terms of mass of water, it is called water equivalent. Water equivalent of a body is the mass of water which when given same amount of heat as to the body, changes the temperature of water through same range as that of the body. Therefore water equivalent of a body is the quantity of water, whose heat capacity is the same as the heat capacity of the body. Water equivalent of the body,

Molar heat capacity=

the body. Remember that phase transformation is an isothermal (i.e. temperature = constant) change. Principle of calorimetry : Heat lost = heat gained For temperature change Q = msDT, For phase change Q = mL Heating curve : If to a given mass (m) of a solid, heat is supplied at constant rate (Q) and a graph is plotted between temperature and time, the graph is called heating curve. Temperature

The steam at 100°C causes more severe burn to human body than the water at 100°C because steam has greater internal energy than water due to latent heat of vaporization.

19

t2

t3

time

t4

1 slope of curve (or thermal capacity) Latent heat µ length of horizontal line.

Specific heat µ

KEY POINTS

ss

the temperature of one gram of water through 1°C (more precisely from 14.5°C to 15.5°C).

•

Clausins & clapeyron equation (effect of pressure on boiling point of liquids & melting point of solids

Se

A node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\12_Thermal physics.p65

E

t1

20

•

B melting

io n

Specific heat of a body may be greater than its thermal capacity as mass of the body may be less than unity.

D boiling

-2

liq ui d

O

LL

•

A

so lid

M.P.

Unit of water equivalent is g or kg.

Latent Heat (Hidden heat) : The amount of heat that has to supplied to (or removed from) a body for its complete change of state (from solid to liquid, liquid to gas etc) is called latent heat of

C

B.P.

æ specific heat of body ö W=mass of body × çè specific heat of water ÷ø

•

E

ga s

EN

•

1 cal = 4.186 J ; 4.2 J

0

CALORIMETRY •

Physics HandBook

CH APTER

ALLEN

•

Heat is energy in transit which is transferred from hot body to cold body.

•

One calorie is the amount of heat required to raise

related with latent heat)

dP L = dT T(V2 - V1 )

69

Physics HandBook

C HAP TE R

ALLEN

THERMAL CONDUCTION – q °C

Heat Transfer

Radiation

Heat transfer due to density difference

Heat transfer w ith out any medium

Due to free electron or vibration motion of molecules

Actual motion of particles

Electromagnetic radiation

Heat transfer in solid body (in mercury also)

Heat transfer in fluids (Liquid+gas)

All

Slow process

Slow process

Fast process (3 × 10 8 m/sec)

Irregular path

Irregular path

Straight line (like light)

S pectral, e mi ssiv e, absorptiv e and transmittive power of a given body surface: Due to incident radiations on the surface of a body following phenomena occur by which the radiation is divided into three parts. (a) Reflection (b) Absorption (c) Transmission

EN

Heat Transfer due toTemperature difference

l

T2

amount of incident radiation Q

19

From energy conservation

Q r Qa Q t + + =1 Q Q Q Þ r+a+t=1

Q = Qr + Q a + Q t Þ

RH

Reflective Coefficient : r = Q

•

Absorptive Coefficient : a =

dQ dt

A

•

B T2

A

dQ dT Q KA ( T1 - T2 ) = -KA = or dt dx l t l Thermal resistance R H = KA Rods in series

A

K2

K1

A1

1

2

K2

A2

l

Sl SKA K eq = ; K eq = Sl / K SA Growth of Ice on Ponds Time taken by ice to grow a thickness from x1 rL 2 ( x - x12 ) to x2: t = 2Kq 2 [K=thermal conductivity of ice, r=density of ice]

70

r = 1 and a = 0 , t = 0 a = 1 and r = 0, t = 0 (ideal black body) t = 1 and a = 0, r = 0 (daithermanous)

Þ Perfect reflector Þ Ideal absorber Þ Perfect transmitter

é Qr ù ´ 100ú % Reflection power (r) = ê Q ë û

é Qa ù ´ 100ú % Absorption power (a) = ê Q ë û

Rods in parallel

K1

l

Qa Q Qt Transmittive Coefficient : t = Q

Se

Rate of heat flow

Qr

•

ss

A T1

-2

amount of transmitted radiation Q t

io n

dQ dt

amount of reflected radiation Qr

amount of absorbed radiation Qa

LL

T1

•

0

Convection

RADIATION

20

Conduction

0°C

Radiation

•

•

é Qt ù Transmission power (t) = ê Q ´ 100ú % ë û Stefan's Boltzmann law : Radiated energy emitted by a perfect black body per unit area/sec E= sT4 For a general body E=serT4 [where 0 £ er £ 1]

Prevost's theory of heat exchange : A body is simultaneously emitting radiations to its surrounding and absorbing radiations from the surroundings. If surrounding has temperature T0 then Enet = ers(T4–T04)

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\12_Thermal physics.p65

Conduction Convection

In conduction, heat is transferred from one point to another without the actual motion of heated particles. In the process of convection, the heated particles of matter actually move. In radiation, intervening medium is not affected and heat is transferred without any material medium.

E

Physics HandBook

CH APTER

Kirchhoff's law : The ratio of emissive power to absorptive power is same for all surfaces at the same temperature and is equal to the emissive power of a perfectly black body at that temperature. e E E e = = Þ = EÞeµa a a A 1 Therefore a good absorber is a good emitter.

•

Perfectly Black Body : A body which absorbs all the radiations incident on it is called a perfectly black body.

•

Absorptive Power (a) : Absorptive power of a surface is defined as the ratio of the radiant energy absorbed by it in a given time to the total radiant energy incident on it in the same time. For ideal black body, absorptive power =1

•

Emissive power(e) : For a given surface it is defined as the radiant energy emitted per second per unit area of the surface.

•

Newton's law of cooling:

r

q1

where

q0

Wien's Displacement Law : Product of the wavelength lm of most intense radiation emitted by a black body and absolute temperature of the black body is a constant lmT=b = 2.89 × 10–3 mK = Wein's constant el

T1

T1>T2>T3

T2

T = temperature of sun » 5800 K

ss •

KEY POINTS

Stainless steel cooking pans are preferred with extra copper bottom because thermal conductivity of copper is more than steel.

•

Two layers of cloth of same thickness provide warmer covering than a single layer of cloth of double the thickness because air (which is better insulator of heat) is trapped between them.

•

Animals curl into a ball when they feel very cold to reduce the surface area of the body.

•

Water cannot be boiled inside a satellite by convection because in weightlessness conditions, natural movement of heated fluid is not possible.

•

Metals have high thermal conductivity because metals have free electrons.

T3

l

lm1 lm2 lm3 ¥

S = 2 cal cm–2min=1.4 kWm–2

Se

A

q1 - q2 é q + q2 ù = kê 1 - q0 ú [where k= constant] t ë 2 û

sun and earth.

io n

Note :-

dq µ ( q - q0 ) Þ q = q0 + ( q1 - q0 ) e - kt dt [where k = constant] when a body cools from q1 to q2 in time 't' in a surrounding of temperature q0 then

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\12_Thermal physics.p65

RS = radius of sun

20

LL

If temperature difference is small Rate of cooling

E

2

æ RS ö 4 4 pR 2S sT 4 = s çè r ÷ø T 4 pr 2

r = average distance between

t

•

-2

P

S= 4 pr 2 =

19

q

Solar constant The Sun emits radiant energy continuously in space of which an insignificant part reaches the Earth. The solar radiant energy received per unit area per unit time by a black surface held at right angles to the Sun's rays and placed at the mean distance of the Earth (in the absence of atmosphere) is called solar constant.

EN

•

0

ALLEN

Area under el - l graph=ò el dl = e = sT 4 0

71

Physics HandBook

C HAP TE R

ALLEN

KINETIC THEORY OF GASES

It related the macroscopic properties of gases to the microscopic properties of gas molecules.

•

The molecules of a gas are identical, spherical, rigid and perfectly elastic point masses.

•

The size is negligible in comparision to inter molecular distance (10–9 m)

Assumptions regarding motion : Molecules of a gas keep on moving randomly in all possible direction with all possible velocities.

•

The speed of gas molecules lie between zero and infinity (very high speed).

•

The number of molecules moving with most probable speed is maximum.

r Gay–Lussac's law For a given mass at constant volume P µ T

-2

r Avogadro's law:If P,V & T are same then no. of molecules N1=N2

Different speeds of molecules v rms =

A

Molecules constantly collide with the walls of container due to which their momentum changes. This change in momentum is transferred to the walls of the container. Consequently pressure is exerted by gas molecules on the walls of container.

2RT = MW

2kT m

v av =

8RT = pM W

8kT pm

io n

Gravitational attraction among the molecules is ineffective due to extremely small masses and very high speed of molecules.

3RT 3kT = v mp = MW m ;

Nmax

(T1)

T 1=500K

ss

No attractive or repulsive force acts between gas molecules.

Assumptions regarding pressure:

•

1 P

r Charles' law : For a given mass at constant pressure VµT

Nmax

(T2)

Nmax

(T3)

Se

•

temperature. V µ

19

The gas molecules keep colliding among themselves as well as with the walls of containing vessel. These collision are perfectly elastic. (ie., the total energy before collision = total energy after the collisions.)

Assumptions regarding force: •

r Boyle's law :For a given mass at constant

LL

•

Gas laws

20

Assumptions regarding collision:

mRT mN A kT æ N ö = = ç ÷ kT = nkT è Vø V V

EN

•

PV =µRT Þ P =

0

Every gas consists of extremely small particles known as molecules. The molecules of a given gas are all identical but are different than those another gas.

number of molecules (N)

•

Ideal gas equation

vmp=most probable speed v max =maximum speed of molecule T2=1000K

T3=2000K

v mp (T ) v mp (T )

v mp (T ) velocity of molecule v 1

2

3

Assumptions regarding density:

Kinetic Interpretation of Temperature :

•

Temperature of an ideal gas is proportional to the average KE of molecules,

The density of gas is constant at all points of the container.

Kinetic interpretation of pressure : PV =

1 mNv 2rms 3

[ m = mass of a molecule, N = no. of molecules]

72

PV =

1 1 3 2 2 mNVrms & PV = mRT Þ mv rms = kT 3 2 2

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\13_Kinetic Theory of Gases.p65

Basic postulates of Kinetic theory of gases

E

Physics HandBook

CH APTER

ALLEN Degree of Freedom (F) :

For mixture of non-reacting gases

At higher temperature, diatomic molecules have two more degree of freedom due to vibrational motion (one for KE + one for PE) At higher temperature diatomic gas has f = 7

Molecular weight : M Wmix =

m1 M W1 + m2 MW2 + .... m1 + m 2 + ....

Specific heat at constant V: CV

mix

=

Maxwell's Law of equipartition of energy: Kinetic energy associated with each degree of freedom

Specific heat at constant P: CP

mix

1 of particles of an ideal gas is equal to kT 2

m1 C P1 + m2 C P2 + ...... m1 C V1 + m2 C V2 + .....

•

Translational KE of a mole =

•

f Internal energy of an ideal gas: U = mRT 2

3 RT 2

-2

Translational KE of a molecule =

KEY POINTS

Kinetic energy per unit volume

EV =

19

•

3 1 æ mN ö 2 ç ÷v = P 2 è V ø rms 2

•

At absolute zero, the motion of all molecules of the gas stops.

•

•

dU æ dQ ö = CV = ç ÷ è mdT ø V = constant mdT

At higher temperature and low pressure or at higher temperature and low density, a real gas behaves as an ideal gas.

•

For any general process

•

æ dQ ö CP = ç = C V + R ¬ Mayer's equation è mdT ø÷ dP = 0

Total (f)

Monoatomic [He, Ar, Ne…]

3 0

3

5 =1.67 3

Diatomic [H2 , N2 … .]

3 2

5

7 =1.4 5

3 2

5

7 =1.4 5

3 3

6

4 =1.33 3

Triatomic (Linear CO2) Triatomic Non-linear-NH3 ) & Polyatomic

g =

CP CV

io n

ss (a)

Internal energy change DU = nCVdT

(b)

Heat supplied to a gas DQ = nCdT

Se

CV =

f R 2

CP=CV +R

Translational Rotational

A

LL

•

Molar specific heat of a gas C = dQ mdT

Atomicity

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\13_Kinetic Theory of Gases.p65

C Vmix

=

m1 + m2 + ......

3 kT 2

•

Specific heats (CP and CV) :

C Pmix

m1 C P1 + m2 C P2 + ......

20

f freedom= kT 2

E

g mix =

Average KE of a particle having f degree of

m1 + m 2 + ....

EN

•

=

m1 C V1 + m2 C V2 + .....

0

Number of minimum coordinates required to specify the dynamical state of a system.

3 R 2 5 R 2 5 R 2

5 R 2 7 R 2 7 R 2

3R

4R

(c)

where C for any polytropic process PVx = constant is C = CV +

R 1- x

Work done for any process DW = PDV It can be calculated as area under P-V curve

(d)

Work done = DQ – DU =

nR dT 1- x

For any polytropic process PVx = constant

73

Physics HandBook

C HAP TE R

ALLEN

THERMODYNAMICS w

æ V2 ö So, DQ = W = µRT ln ç V ÷ = µRTln è 1ø

EN

Volume Expansion coefficient =

•

1 T

For adiabatic process PVg = constant or Tg P1–g = constant or TVg–1 = constant In this process DQ = 0 and

P1 V1 - P2 V2 W = –DU = µCV (T1 – T2) = g -1

æ ¶P ö adiabatic Bulk modulus B = -v ç ÷ è ¶v ø B = gP 1

Volume Expansion coefficient = 1 - g T ( )

74

R 1- x

•

Molar heat capacity

•

Work done by gas

•

( P V - P2 V2 ) = 1 1 x -1 x -1 Slope of P-V diagram (also known as indicator

nR ( T1 - T2 )

-2

W=

C = CV +

dP P = -x ) dV V Polytropic Bulk modulus B = xP Efficiency of a cycle

diagram at any point

QH

Sink T C

W

Work done by working substance

ss h=

QC

Working substance

io n

Source TH

Heat sup plied

Q - QC Q W =1- C = H QH QH QH For carnot cycle =

Q C TC QC TC = so h = 1 = 1 – – Q H TH QH TH For refrigerator Source TH

QH

Working substance W

Coefficient of performance b=

QC QC TC = = W QH - Q C TH - TC

QC

Sink TC

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\13_Kinetic Theory of Gases.p65

For cyclic process DU = 0 Þ DQ=W For isochoric process V = constant Þ P µ T & W = 0 DQ = DU = mCVDT Isochoric DV = 0 B = not defined. Bulc modulus Volume expansion coefficient = 0 For isobaric process P = constant Þ V µ T DQ = mCPDT, DU = mCVDT W = P(V2 – V1) = mRDT Isobaric DP = 0 Bulk modulus (B) = 0

A

•

•

Se

• •

Area between P-V curve & V-axis gives work done by gas from one state to another state. Sign Convention Heat absorbed by the system ®positive Heat rejected by the system ® negative Increase in internal energy (i.e. rise in temperature)®positive Decrease in internal energy (i.e. fall in temperature)® negative Work done by the system ®positive Work done on the system ®negative

LL

•

path independent

æ -P ö B = -v ç ÷=P è v ø DT = 0 Volume expansion coefficient not defined. For any general polytropic process PVx = constant

0

dQ = dW + dU path dependent

æ P1 ö ç ÷ è P2 ø

æ ¶P ö Isothermal Bulk modulus B = - v ç ÷ è ¶v øT =const

DQ = W + DU [Here W = ò PdV ; DU = nCVDT] For differential change

For Isothermal Process T = constant or DT = 0 Þ PV = constant In this process DU = µCvDT = 0

19

•

Zeroth law of thermodynamics : If two systems are each in thermal equilibrium with a third, they are also in thermal equilibrium with each other. First law of thermodynamics : Heat supplied (DQ) to a system is equal to algebraic sum of change in internal energy (DU) of the system and mechanical work (W) done by the system

20

•

E

Physics HandBook

CH APTER

ALLEN

KEY POINTS Work done is least for monoatomic gas (adiabatic process) in shown expansion. P Isobaric

Isothermal Diatomic

Monoatomic

Adiabatic

V

W ®positive

•

-2

P

P

QS

1

W ®negative

2

4

CARNOT ENGINE

h = 1-

3

QR

T3 - 4 T1- 2

V

io n

•

V

LL

V

19

P

0

æPö isothermal ç ÷ èVø Air quickly leaking out of a balloon becomes cooler as the leaking air undergoes adiabatic expansion. First law of thermodynamics does not forbid flow of heat from lower temperature to higher temperature. First law of thermodynamics allows many processes which actually don't happen.

20

• •

EN

æ Pö At a particular pressure and volume, magnitude of slope of P-V curve is greater for adiabatic ç g ÷ then è Vø

It is a hypothetical engine with maximum possible efficiency Process 2®3 & 4®1 are adiabatic.

ss

Process 1®2 & 3®4 are isothermal

dQ T

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\13_Kinetic Theory of Gases.p65

A

ds =

E

Se

Second Law :- First law is quantitative analysis while second law is qualitative analysis of thermodynamics processes. Second law tells us in what conditions best can be converted into useful work. ds ® change in entropy, dq ® best exchanged.

75

76

0

-2

19

20

EN

C HAP TE R

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\13_Kinetic Theory of Gases.p65

io n

ss

LL

Se

A Physics HandBook ALLEN

IMPORTANT NOTES

E

Physics HandBook

CH APTER

ALLEN

ELECTROSTATICS ELECTRIC CHARGE Charge of a material body is that property due to which it interacts with other charges. There are two kinds of charges- positive and negative. S.I. unit ® Coulomb (C) Properties of charge :(a) Charge is a scalar quantity (b) Charge is quantised (c) Charge is conserved. (d) Charge is independent of frame of reference. Methods of charging :(a) Friction (b) Induction (c) Conduction

r

19

r If medium is present then F =

q2

-2

1 Nm2 9 = 9 × 10 4p Î0 C2

q1

q1q 2 1 ˆr 4p Î0Îr r2

20

where,

1 q1 q 2 ˆr 4p Î0 r 2

0

r Force between two charges F =

EN

COULOMB'S LAW

LL

NOTE :The Law is applicable only for static and point charges. Moving charges may result in magnetic interaction. And if charges are spread on bodies then induction may change the charge distribution.

r

q

ss

r kq kq r ˆr = E= r 2 r r3

io n

ELECTRIC FIELD OR ELECTRIC INTENSITY OR ELECTRIC FIELD STRENGTH Electric field intensity is defined as force on unit test charge.

SI unit : Newton/coulomb (N/C)

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\14_Electrostatics.p65

(a)

E

Se

A

ELECTRIC FIELD DUE TO SPECIAL CHARGE DISTRIBUTION (b)

r kq Due to point charge E = 2 ˆr r

Due to linear change distribution :-

lC/m P r

E r

a b

Ex Ey

q

Ex =

l ( sin a + sin b ) 4p Î0 r

Ey =

l ( cos b - cos a ) 4p Î0 r

77

Physics HandBook

(g)

Due to infinite line of charge

r s ˆ EP = n 2 Î0

l ˆr 2p Î0 r

r Electric field due to uniformly changed ring

Q

P

(R

2

+ x2

r

)

3/2

(i) At centre of the ring, x = 0. So E = 0 (ii) Electric field is maximum at x = ± Due to segment of ring

EP =

2

E

2kl æaö sin ç ÷ R è2ø

(a)

A Due to charged disk

sC/m2

++ R ++++ ++ + + ++ O + ++ ++ ++ + + ++ + ++ + +

EP =

x

x s æ ç1 2 2 Î0 ç R + x2 è

P

E

r

R

r

B

C

A

R

(j)

For point inside the sphere (r < R) : EA = 0

kQ For point on the surface (r = R) : E B = 2 R

(c)

kQ For point outside the sphere : E C = 2 r

Due to uniformly charged non-conducting sphere B C

E Kq/R2

ö ÷ ÷ ø

R (a)

78

s /Î0

(b)

Se

Direction of electric field is along the direction of angle bisector of the arc.

(f)

E

P

R

R

P

^ n

Due to hollow non-conducting sphere

Kq/R2

a

E

io n

l

(i)

LL

(e)

Due to infinite charged conducting plate

++ + ++ + + + + + + +++ +++ ++ + + + + + ++ + ++ + + + + + + s C/m2 + + + +

kQx

EP =

r

EN

+ + ++

R/Ö2

s /2Î0

ss

+

+

–R/Ö2

E x

++

R O

E

P

r s EP = nˆ Î0

+

+ + + ++ +

(h)

E

++

^ n

0

E

E

-2

P

r

++ + ++ + + + + + + +++ +++ ++ + + + + + ++ + + +++ ++ + + + 2 + s C/m +

r

++ A + ++ ++ + + ++ + ++ R + ++ + ++

For point inside the sphere (r < R)

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\14_Electrostatics.p65

lC/m

(d)

Due to infinite plane sheet of charge

19

r EP =

ALLEN

20

(c)

C HAP TE R

E

kQr R

3

=

ELECTRIC FLUX

rr 3 Î0

For point on the surface (r = R) : EB =

kQ

kQ For point outside the sphere (r > R) : E C = 2 r Due to solid or Hollow conducting sphere E

B

(a) (b)

R2

(c)

(i)

(ii)

C

r r For uniform electric field f = E.A = EA cos q r r where, q = Angle between E and area vector ( A ). r uuur For non-uniform field f = ò E.dA

A

(a)

For point inside the sphere (r < R) : EA = 0

(b)

For point on the surface (r = R): EB =

(c)

kQ For point outside the sphere (r > R) : E C = 2 r

where qin = net charge enclosed by the closed surface. (i) Flux through Gaussian surface is independent of its shape. (ii) Flux depends only on charges present inside the closed surface. (iii) Flux through a closed surface is independent of position of charges inside it. (iv) Electric field intensity at the Gaussian surface is due to all charges present (inside as well as outside).

kQ R2

-2

r

r uuur q in For a closed surface, total flux f = Ñò E.dA = Î 0

R

19

R

Gauss’s Law

EN

Kq/R2

ELECTRIC FIELD LINES

LL

ELECTROSTATIC POTENTIAL ENERGY It is the amount of energy required to bring any charge from ¥ to any particular point without any charge in K.E. Interection energy of a system of two charged particles r

+

qA>qB

q1

U=

Se

Electric field lines have the following properties :(a) Imaginary curves (b) Never intersect each other (c) Never form closed loops (d) Start from (+ve) charge and ends on (–ve) charge. (e) If there is no electric field then there will no field lines (f) Number of electric field lines per unit area normal to the area at a point represents magnitude of electric field intensity. Crowded lines represent strong field while distant lines weak field. (g) Number of lines originating f rom or terminating on a charge is proportional to magnitude of charge. (h) Field lines start or end normally at the surface of a conductor. (i) Tangent to the lines of force at a point in an electric field gives direction of intensity of electric field.

A node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\14_Electrostatics.p65

E

Fixed point charge near infinite metal plate

io n

B

ss

A

Scalar quantity SI unit :- Nm2/C or V–m

20

(k)

r uuur f = ò E.dA

0

EA =

(b)

Physics HandBook

CH APTER

ALLEN

q2

kq1q 2 r

{ Assuming potential energy at ¥ to be zero} ELECTRIC POTENTIAL It is the work done against the field to take a unit positive charge from infinity (reference point) to the given point P without gaining any kinetic energy. VP =

(i) (ii) (iii)

( W¥-p )ext q

=

U q

Electric potential is a scalar quantity SI unit :- Volt (V) or J/C In presence of dielectric medium, potential decreases 1 and becomes e times of its free space value. r

79

Physics HandBook

C HAP TE R

ELECTRIC POTENTIAL DUE TO SPECIAL CHARGE DISTRIBUTION :(a) Due to a point charge :P

ALLEN (b) For point on the surface (r = R) :VB =

V

(c)

r

For point outside the sphere (r > R) :VC =

q

r

(f)

kq q VP = = r 4pe0 r

+ R+ + + + + + + + + +q

2

- r2

)

R 2 + x2

A

B

3kQ/2R

C

O R

0

VP =

Q

kq

-2

P

EN

x

R

Due to segment of ring :-

r

(c)

LL

R

VB =

kQ R

x

P

A

Due to non-conducting spherical shell :(a) For point inside the sphere (r < R) :-

kQ r

A

B

C

V

kQ/R

O R

R

(a) For point inside the sphere (r < R) :VA =

kQ VA = R

kQ R

(b) For point on the surface (r = R) :V

+ + Q + C A B + + O R + + + + + + +

VC =

Due to conducting sphere or shell :-

Q

s æ 2 ö 2 ç x + R - x÷ ø 2 Î0 è

For point outside the sphere (r > R) :-

io n

(g)

kQ R

VB =

(c) R

r

kQ R

For point outside the surface (r > R) :VC =

kQ r

r node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\14_Electrostatics.p65

VP =

20

P

19

(b) For point on the surface (r = R)

due to charged disk :-

VP =

80

(3R

V

s + + + ++ ++ + + + ++ R ++ + +

(e)

2R

3

++

q + + + + + + + ++

(d)

kQ

ss

(c)

VA =

Due to a charged ring :+ + + + + + + + + +

kQ r

Due to solid non-conducting sphere :(a) For point inside the sphere (r < R) :-

Se

(b)

kQ R

E

Physics HandBook

CH APTER

POTENTIAL DIFFERENCE The potential difference between two points A & B is work done by external agent against electric field in taking a unit positive charge from B to A keeping kinetic energy constant :

VA - VB =

( WBA )ext q

Relation between electric field & electric potential :r r dV ˆr E = -DV = -gradV = dr

ELECTRIC FIELD DUE TO DIPOLE (a) At an axial point :r r 2kp E= r3 (b) On the equitorial line :r -kpr E= r3 (c)

EN

r -¶v r uur ˆi - ¶v ˆj - ¶v kˆ ; V = - E.dr E= ò ¶x ¶y ¶z r (a) Direction of E is from high potential to low potential. (b) If V = constant over a region, then E = 0 (in that region)

At any general point :r r kp E= 1 + 3cos2 q r3

+q

A node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\14_Electrostatics.p65

E

DIPOLE PLACED IN UNIFORM ELECTRIC FIELD r r (a) Torque rt = p ´ E

(d) (e)

io n

At a general point :-

V=

kpcos q r2

CONDUCTORS AND ITS PROPERTIES

Direction of dipole moment is from negative to positive charge :

(b) (c)

(c)

20

(b) At equitorial point :V=0

(a) (b)

Se

d P r r Dipole Moment :- p = qd –q

r2

ss

ELECTRIC DIPOLE A system of two equal and opposite charges separated by a small distance is called electric dipole :

kp

19

V=

-2

(a) At an axial point :-

LL

EQUIPOTENTIAL SURFACE. The locus of all points having same potential is called (a) Equipotential surfaces can never cross each other. (b) Equipotential surfaces are always perpendicular to the direction of electric field. (c) No work is done in moving a charge from one point to other over an equipotential surface.

ELECTRIC POTENTIAL DUE TO DIPOLE

0

ALLEN

Net force = 0 Work done in rotation of dipole from q1 to q2 angle in external electric field W = pE(cosq1 – cos q2) rr Electrostatic potential energy = -p.E = - pE cos q

(c)

(d) (e)

Conductors are always equipotential surfaces. Charge always reside on the outer surface of a conductor. Electric field is always perpendicular to conducting surface. Electric field lines never exist within conducting materials. When a conductor is grounded, its potential becomes zero.

In non-uniform electric field, force on electric dipole r r r dE F = -p. dr

81

Physics HandBook

C HAP TE R

ALLEN

GRAVITATION Newton's law of gravitation r

m1

m2

Outside the shell

Eg =

r

On the surface

Eg =

Inside the shell

Eg = 0, where r R

GM r2

, where r=R

GMr , where r< R R3

Acceleration due to gravity

Se

G l (cosb–cosa) (Eg)y= —– x

GM

ss

2

Eg =

io n

R

Eg =

r

At height h : gh =

g=

GM R2

GM (R+h)2 æ

2h ö

If h > R

Þ i=

(ii) If nr QR, VC < VD & PS < QR, VC > VD or

PS = QR Þ products of opposite arms are equal.

Potential difference between C & D at null point is zero . The null point is not affected by resistance of G & E. It is not affected even if the positions of G

C Q

P A

B

G R

S D

E

R

& E are interchanged.

87

Physics HandBook

C HAP TE R

Electrical Instruments

ALLEN

Applications of potentiometer :

Metre Bridge : Works on principle of wheet stone bridge

E1 l1 = E2 l2

(i) Comparision of emfs of two cells

At balance condition :

(100 - l ) R P R R l = Þ = ÞS= l Q S (100 - l) S

J

A

EN

0

19

-2

æ l - l2 ö r =ç 1 R è l 2 ÷ø

C

J'

A

B

G

ss

K

(iii) Comparision of two resistances l R1 = 1 R1 + R 2 l 2

Rh(0 –R)

primary circuit

L

A

secondary circuit E'

G

wire

B

E' iP

Se

ss

Step up VS > VP NS > N P iP > iS

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\19_Electromagnetic Induction.p65

A

Transmissions are done at high voltage and low current by using step up transformer.

E

Losses in tranformers

Coil (Cu losses)

Heat loss Can be minimize by using thick wires

Flux loss Can zero by making coupling factor1

Core (Iron losses)

Hysteresis Can be controlled by using substance of high m r (soft iron)

Induced electric field : Produced due to change in magnetic field and is non-conservative in nature

Eddy current Can be controlled by laminating the core

r

®

Ñò E.dl = -

dfB dt

107

Physics HandBook Growth Circuit

of

a

C HAP TE R

Current

in

an

L- R

ALLEN Decay of Current

Initial current through the inductor = I0 ; Current at any instant i = I0e-Rt/L

L L

I

I0

I

R t=0

I

R

S

t=0

S E I= (1 - e-Rt/L) . [If initial current = 0] R I = I0 (1 - e-t/t)

KEY POINTS

E . R (i) L behaves as open circuit at t = 0 [ as if I = 0 ] (ii) L behaves as short circuit at t = ¥ always. I0 =

non-zero.

•

(1)

LL

t

L L Large Curve (2) ¾® Small R R

because mutual inductance M µ L 1 L 2 .

ss

Curve (1) ¾®

The mutual inductance of two coils is doubled if the self inductance of the primary and secondary coil is doubled

io n

•

20

(2)

Acceleration of a magnet falling through a long solenoid decrease because the induced current produced in a circuit always flows in such direction that it opposes the change or the cause that produces it.

19

I

An emf is induced in a closed loop where magnetic flux is varied. The induced electric field is not conservative field because for induced electric field, the line r r integral Ñ ò E.d l around a closed path is

0

•

-2

L = time constant of the circuit. R

EN

t=

t

108

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\19_Electromagnetic Induction.p65

Se

A

IMPORTANT NOTES

E

Physics HandBook

CH APTER

ALLEN

Alternating Current & EM Waves Voltage or current is said to be alternating if periodically it changes its dir. and magnitude. i = I0 sin wt v = V0 sin (wt + f) t

ò idt Average current = 0 t ò0 dt

t 2

ò0 i dt t ò0 dt

i rms =

AC ammeter and voltmeter reads RMS value of current and voltages respectively Þ i0 > irms > iAV

Full wave rectifier

p

0

–

I0 = 0.5 I0 2

2p

A E

AC CIRCUITS

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\20_AC & EM Waves.p65

R

V = V0 sin wt i=

V0 sin wt R

VR , R

i

I0

I0 2

3

L

C

L

C

V = V0 sin wt

V = V0 sin wt

i=

Resistance = R

VR = iR

I0

Se

R

p 2p

ss

0

I0

2I0 =0.637 I 0 p

io n

–

Saw Tooth wave

I0 = 0.318 I0 (full cycle) p

I0 = 0.707 I 0 2

2p

+

Square or Rectangular

2I0 =0.637 I0(half cycle) p

= 0.707 I0

LL

p

2

2p

-2

0

p

19

+

0

Half wave rectifier

I0

Average or mean

20

Sinusoidal

RMS Value

V0 (- cos wt) wL

i=

Reactance XL = wL

0

Wave–form Value for half cycle

EN

Nature of wave form

V0 cos wt 1 / ( wC )

Reactance X C =

1 wC i

X L,V L

i

VL = iXL

X C, V C

VC = iXC

109

Physics HandBook

C HAP TE R

ALLEN

AC THROUGH LCR CIRCUIT VL

V= V 2+(V –V )2 R L C

VL–VC i

VR

f

VC

VR

L

R

XL–XC

C

f

If VL > VC Þ XL > XC then

V = max R

Se

R æ ö Power factor ç cos f = Z = 1 ÷ è ø Angle (or phase defference) Between v and i = 0° VR = VSource

A

•

Z = R = min Þ i =

• •

Resonating frequency w0 =

1

LC

19

20

V

L-C-R PARALLEL COMBINATION

1 = Z

~

1 æ 1 1 ö +ç ÷ 2 R è xL xC ø

2

CHOKE COIL

It is used to control alternating current without any power loss. It is an inductor and low resistance.

Z » X L Þ Power = 0

high L, low R

i

Z

f

ss

At resonance • XL = XC or VL = VC

i

io n

LL

V = V0sin wt i = I0 sin(wt + f) Power = Vrms irms cos f = i2 R Wattfull current = irms cosf Wattless current = irms sinf Wattless power = vrms irms sinf Where cosf = Power factor

RESONANCE IN SERIES LCR CIRCUIT

•

-2

Vö Rö æ æ 1 and ç i = Z ÷ and ç cos f = Z ÷ è ø è ø Z

Impedence= Z = R2 + (X L - X C )2 and admittance = Power in AC Circuit :

0

R

R1 > R2

LC - OSCILLATION

R2 R1

w0

w

w0

X 1 Sharpnessµ quality factor = L = R R

110

w L f0 = C B and width

+ Q=CV

C

L

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\20_AC & EM Waves.p65

v=V0sinwt

EN

Z

E

Physics HandBook

CH APTER

ALLEN

d2 Q

VC + VL =0 L

dt2

•

Q + =0 C

Q = Q0 cos wt Þ i = –i0 sin wt where i0 = Q0w where w =

1

Q value =

Re sonance frequency Band width

i

frequency of oscillation

LC

f1

f0

f2

f

E.M.W Maxwell's equations

2.

Ñò B × dA = 0

3.

Ñò E × dl = -

4.

r uur dfE ù é Ñò B × dl = m0 êëi c + e0 dt úû (Empere – maxwell law)

dfE (Faraday is law) dt

Q e0

E0 B0

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\20_AC & EM Waves.p65

A

=

m 0 e0

io n

Total energy density =

B2 2m0

B2 1 B2 1 = e0 E20 = 0 e0E 2 + 2m0 2 2m0 2

If total energy transferred to a surface in time t is U, total momentum delivered to this surface is p = U/c.

ss

Electromagnetic wave : Ex = E0 sin(kz – wt) By = B0 sin(kz – wt)

1

1 e0 E 2 ; 2

Magnetic field energy density =

df 1 dQ dfE = Þ i d = e0 dt e0 dt dt

E

Electric field energy density =

Se

f = EA =

C=

r 1 r r (E ´ B) Poynting vector S = m0

LL

DISPLACEMENT CURRENT

ˆ ´B ˆ. Direction of propagation of light E

0

r uur

(Gauss's law of magnetism)

-2

r uuur

(Gauss's law of electricity)

EN

0

19

Q

Ñò E × dA = e

20

r uuur

1.

111

C HAP TE R

20

KEY POINTS

•

An alternating current of frequency 50 Hz becomes zero, 100 times in one second because alternating current changes direction and becomes zero twice in a cycle.

•

An alternating current cannot be used to conduct electrolysis because the ions due to their inertia, cannot follow the changing electric field.

•

Average value of AC is always defined over half cycle because average value of AC over a complete cycle is always zero.

•

AC current flows on the periphery of wire instead of flowing through total volume of wire. This known as skin effect.

112

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\20_AC & EM Waves.p65

Se

A

ss

io n

LL

19

-2

0

ALLEN

EN

Physics HandBook

E

Physics HandBook

CH APTER

ALLEN

Modern Physics

E hv h = = l C c INTENSITY OF LIGHT

Intensity =

Energy time ´ Area

Point source I = P 4pr2 where P = power of source Infinite linear source

P

r

V0

LL

P 2 prl Infinite planar source I = constant (independent of distance) IµP

tanf = h/e

q V0

–f/e

A node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\21_Modern physics.p65

E

Se

N I´A Nhv , n= = = t hv At No of photons incident per second Force exerted by light on a surface

I=

V

io n

(i)

ss

I=

hv f e e Graphs of Photo Electric Effect V0 =

or

0

P=

-2

• •

E 1 ; m µ . So l C2 effective mass of violet light is more then effective mass of red light photon. Rest mass of a photon is zero. Linear momentum of photon

Effective mode of photon, m =

F I = (1 + r) cos 2 q A C Photo Electric Effect (PEE) This experiment shows particle nature of light. Electrons are emitted a metal surface when light is incident upon it only when v ³ v0 ( Threshold frequency) and P.E.E. is independant of intensity of light. This is shown as ther is no time rag in emission of electron. Einsten's Photoelectric Equation hv = ft KEmax. where f = work function which depends on metal KEmax = maximum KE that an e– can have after emission. KEmax = eV0, where V0 = stopping potential or cut of potential So, hv = f + eV0

P=

19

•

Radiation Pressure (P)

20

•

PHOTON Max plank said light is mode up of discrete packets of energy called 'Photon' according to his quantum theory. Energy of photon, E = hv, where h = Plank's constant = 6.63 ×10–34 Fs and v = frequency of light used.

EN

•

i (photocurrent)

i2

I2

i1

I1

(ii)

v(potential)

IA F= (1 + r) cos q , where r = reflectivity C

i1, i2 = Saturation currents I2 > I1 (v = same)

Incident light

v0 Metal-1

q q

Metal-2

v0

(iii) –f0/e

Reflected light

–f0'/e

v0'

v

v0'>v0

113

Physics HandBook

C HAP TE R

ALLEN

Bohr atomic model

i (Photocurrent)

v (iv)

+ Nucleus (+ze)

I (Intensity)

Quantum efficiency Quantum efficiency = Electron revolves circular orbit around nucleus

nh 2p So electron has discrete angular momentum and is allowed to be present in certain fixed orbits only (called as stationary energy levels or shells.) (iii) Electrons denot radiate energy when in shells but energy is radiated or aborbed when an electrons jumps to lower or higher orbit respectively. Mathematical Analysis of Bohr's Theory (i) v = 2.2 × 106 2/n m/s

(ii)

mvr =

19

(iii)

n2 Å z Total energy of electron in nth orbit,

r = 0.53

z2 eV n2

20

(ii)

E n = -13.6

LL

Matter waves theory Light has dual nature Experiments like reflection, refraction, interference deffraction are explained only ont the basis of wave theory of light. Experiments the PEE and crompton effect, pair production and positon axilation are explained on the basis of particle nature of light. Atomic Structure Various Models for structure of Atom (i) Dalton's Theory (a) Thomson Model (b) Rutherford model

-2

EN

IAl i =n (Saturation current) hc Refer graph (iv) Photocell : Works on PEE, Photocell light energy is converted into electrical energy.

(i)

0

No. of electron emitted per second Total no. of photon incident per second

æ n3 ö T = ç 2 ÷ 1.51 ´ 10-16 s èz ø

(v)

æ z2 ö f = ç 3 ÷ 6.6 ´ 1015 Hz èn ø

(vi)

é1 1 1ù = Rz 2 ê 2 - 2 ú , l n n 2 û ë 1

114

RH = Rydberg constant =

1.097 ´ 10-3 Å

(For stationary nucleus) RH (If nuceus is not stationary) 1+ m/M i.e. mass of nucleus and revolving particle are comparable) where m = mass of revolving particle M = mass of nucleus R' =

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\21_Modern physics.p65

A

(Number of a-particles scattered at an angle q)

Se

ss

io n

(iv)

E

Physics HandBook

CH APTER

ALLEN

S.No.

Series Observed

Value of n1

Values of n2

Position in the Spectrum

1.

Lyman Series

1

2, 3, 4 ..... ¥

Ultra Violet

2.

Balmer Series

2

3, 4, 5 ..... ¥

Visible

3.

Paschen Series

3

4, 5, 6 ..... ¥

Infra-red

4.

Brackett Series

4

5, 6, 7 ..... ¥

Infra-red

5.

Pfund Series

5

6, 7, 8 ..... ¥

Infra-red

h h = , p 2mK 2

n1 ® n2 =

12.27 V 0.286

Å

It is the energy required to remove an electron from an atom

l a- Particle =

V

0.101 V

ex I.E.of Hydrogen = 0 – ( – 13.6) = 13.6 eV.

Å

io n

0.202

It is the potential required correspongin to ionization energy in order to remove the electron fro the atom

Å

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\21_Modern physics.p65

A

nh 2p

E

Se

Bohr's quantigation mvr =

Ionization Potential

ss

l Deutron =

E n2 - E n1 electron charge

Ionization energy of hydrogen atom

LL

V

Excitation potential of atom

-2

2mqV

l Pr oton =

= n–m C2

; if particle has charger and is

accelerated by V l electron =

æ ( n - m )( n - m + 1) ö ÷ 2 ø

From n1 = n to n2 = m state is = ç è

20

h

n(n - 1) 2

19

where lD = De-Broglie wvaelength of any particle. lD =

From n1 = n to n2 = 1 state =

EN

lD =

Total emission spectral lines

0

De Broglie Hypothesis He said there wave nature of very particle just the light has dual nature.

=

- En electronic charge

X - RAYS

Produced by bombording high speed electrons on a target of high atomic weight and high melting point. The we basically highly magnetic photons. Soft X–ray Hard X–ray Wavelength 10 Å to 100 Å 0.1 Å – 10 Å

• •

Energy

12400 eV–Å l

12400 eV–Å l

Penetrating power UseRadio photography

Less Radio therapy

More

Intensity of X-ray × current flowing through filament Rentetrating power × Applied Potential difference.

115

Physics HandBook • •

C HAP TE R

ALLEN

Continous X-ray – Produce when electron while they hit the target Cutt-off wavelength – Mininum wavelength of continuous X-rays which appears when as electron loosses all its KE. in its first collision only. Hence producing photon of maximum energy and of minimum wavelength. 12400 Å , where V is applied potential difference V Characteristic X-ray – Produce when electron hitting the metal target ejects on electron from shell and that vacant space when occupied by an electron from upper shell, produces a photon. l min =

From Bohr Model

O

n1 = 2,

n2 = 3, 4, 5.......L series

M

n1 = 3,

n2 = 4, 5, 6.......M series

L

K

Transition Wave– Energy Energy length difference L®K

lKa

hnKa

(2 ®1)

–(EK–EL)

= hnKa

K aK b K g K d K series

Wavelength

lKa =

hc (E K - E L )

Characteristic X-ray

12400 eVÅ = (E K - E L )

lLa

hnLa

A

M®L

(3 ®2)

= hnKb

12400 eVÅ = (E K - E M )

–(EL–EM)

lLa =

= hnLa

io n

(3 ®1)

–(EK–EM)

Kb

1

ss

hnKb

2

hc (E L - E M )

0.02 0.04

=

MOSELEY'S LAW

é1 1ù n = RcZ2 ê 2 - 2 ú ë n1 n2 û

116

Kb Ka n

Z

La

0.06 0.08 0.10 0.12 wavelength (nm)

12400 eVÅ (E L - E M )

n µ (Z – b) where v = frequancy of characteristic x-ray Z = atomic number of target n = frequency of characteristic spectrum b = screening constant (for K– series b=1, L series b=7.4) a = proportionality constant

X-ray from a molybdenum target at 35 kV Lb

Bremsstrahlung continuum

Se

lKb

hc lKb = (E K - E M )

1

Ka

3

LL

M®K

2

L series

-2

Third line of series = g

L a L b Lg

3

M series

19

Second line of series = b

Ma MbMg

4

N series

20

First line of series = a

Na Nb

0

N

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\21_Modern physics.p65

n2 = 2, 3, 4.......K series

EN

n1 = 1,

Relative intensity

•

E

Bohr model

Moseley's correction

1.

For single electron species

1.

For many electron species

2.

é1 1ù DE = 13.6Z2 ê 2 - 2 ú eV n n ë 1 2û

2.

é1 1ù DE = 13.6 (Z–1)2 ê 2 - 2 ú eV n n ë 1 2û

3.

é1 1ù n = RcZ2 ê 2 - 2 ú ë n1 n2 û

3.

é1 1ù n = Rc(Z–1)2 ê 2 - 2 ú ë n1 n2 û

4.

é1 1ù 1 = RZ2 ê 2 - 2 ú n n l ë 1 2û

4.

é1 1ù 1 = R (Z – 1)2 ê 2 - 2 ú n n l ë 1 2û

EN

1 1 1 n12 n22

I = I0e–mx where

I0 = Intensity of incident X–ray

LL

I = Intensity of X–ray after passing through x distance

I0

• Intensity of X–ray decrease exponentially. • Maximum absorption of X–ray ® Lead

Se

Half thickness (x1/2)

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\21_Modern physics.p65

A

It is the distance travelled by X–ray when intensity become half the original value x1/2 =

E

x

ss

• Minimum absorption of X–ray ® Air

I

io n

m = absorption coefficient of material

DIFFRACTION OF X–RAY

1 (Z - b)2

-2

ABSORPTION OF X–RAY

When transition is same l µ

19

When target is same l µ

0

For X–ray production, Moseley formulae are used because heavy metal are used.

20

•

Physics HandBook

CH APTER

ALLEN

l n2 m

2d sinq = nl

where, d = spacing of crystal plane or lattice constant or distance between adjacent atomic plane

q = Bragg's angle or glancing angle

f q

q d

f = Diffracting angle n = 1, 2, 3 ....... For Maximum Wavelength sin q = 1, n = 1 Þ lmax = 2d so if l > 2d diffraction is not possible i.e. solution of Bragg's equation is not possible.

117

Physics HandBook

C HAP TE R

ALLEN

NUCLEAR PHYSICS BINDING ENERGY • Binding energy of a nucleus = Dmc2 where Dm = mass defect = [Zmp + (A–Z)mn] – mnuc = Totall mass of nucleus mass of nucleus. •

Binding energy per nucleus =

Stability µ

B.E. A

B.E. is maximum for A =62 (Ni), It is 8.79460 ± 0.00003 MeV/nucleon A

EN

•

B.E. B.E. = A mass number

-2

small mass numbers

0

NUCLEAR FISSION When heavy nuclei achieve stability by braking into two smaller nuclei, then it is called as nuclear fission reaction.

19

large mass number

20

small mass numbers

io n

LL

NUCLEAR FUSSION When lighter nuclei achieve stability by combining and resulting into heavy nucleus and this reaction is called fusion reaction.

large mass number

NUCLEAR FISSION OF U235 U235 + 0n1 ® Ba + Kr + 30n1 + 200 MeV or U235 + 0n1 ® Xe + Sr + 20n1 + 200 MeV and many other reactions are possible. • The average number of secondary neutrons is 2.5. • Nuclear fission can be explained by using "liquid drop model" also. • Dm is about 0.1% of mass of fissioned nucleus • About 93% of released energy (Q) is appear in the form of kinetic energies of products and about 7% part in the form of g – rays. NATURAL URANIUM : It is mixture of U235 (0.7%) and U238 (99.3%). U235 is easily fissionable, by slow neutron (or thermal neutrons) having K.E. of the order of 0.03 eV. But U 238 is fissionable with fast neutrons.

118

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\21_Modern physics.p65

Se

A

ss

Q VALUE It is teh theory released after trhe nuclei avhieve stability through nuclear fissiion or fussion reactions It is always positive when the reaction occurs towards achieving more stability. It Q value of a reaction is negative, it means taht reaction cannot occur as the products in that case will be more constable than the reaction.

E

Physics HandBook

CH APTER

ALLEN

ENRICHED URANIUM It contains 97% U238 and 3% U235. CRITICAL SIZE (OR MASS) : It is minimum mass of uranium required to sustain a chain reacter. REPRODUCTION FACTOR : rate of production of neutrons rate of loss of neutrons

(K) =

0

20

-

2b ® 94Pu239 (best fuel of fission) U238 + 0n1 ¾¾® 92U239 ¾¾¾

92

19

-2

EN

(i) If K = 1; chain reaction is steady or sustained (As in nuclear reaction) (ii) If K > 1; chain reaction will accelerate resulting in a explosion (As in atom bomb) (iii) If K < 1; chain reaction will retard and will stop. NUCLEAR REACTOR (K = 1) : • Nuclear Fuel : U235 , Pu239 .Pu239 is the best. Its critical size is less than that of U235. But Pu239 is not naturally available. So U235 is used in most of the reactors. • Moderator : They are used to slow down the fast secondary neutrons D2O, Graphite etc. • Control rods : They are used absorbs slow neutrons e.g. Boron and Cadmium. • Coolant : It is used to absorb heat and transfers it to water for further use. FAST BREADER REACTORS It is the atomic reactor in which fresh fissionable fuel (Pu239) is produced along with energy. The amount of produced fuel (Pu239) is more than consumed fuel (U235) • Fuel : Used in this reactor Natural uranium. • Process :

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\21_Modern physics.p65

E

Se

A

ss

io n

LL

• Moderator : Are not used in these reactors. • Coolant : Liquid sodium REQUIRED CONDITION FOR NUCLEAR FUSION • High temperature • High Pressure (or density) HYDROGEN BOMB Based on nuclear fusion and produces more energy than an atom bomb (based on nuclear fission). PAIR PRODUCTION e– and e+ pair is produced when in g-photon having energy >, 1.02 MeV strikes a nucleus. PAIR ANMIHILATION When electron and positron combines, 2 g-photon are formed, reach photon having energy >, 0.5 MeV.

RADIOACTIVITY

Radioactive Decays • a decay : Occurs in nucleus having A>210 • b decay : • A type (N/Z)A > (N/Z) stable To achieve stability, it increases Z by conversion of neutron into proton n1 ® 1p1 + e–1 + n , ZXA ®

0

YA +

Z+1

e -1 + n

(b particle )

This decay is called b–1 decay. Kinetic energy available for e–1 and n is, Q = Kb + K n

119

Physics HandBook •

C HAP TE R

ALLEN

B type (N/Z)B < (N/Z) stable To achieve stability it decreases Z by the conversion of a proton into neutron. That is, p ® n + e+ + (positron )

n

( neutrino)

, Z X A ® Z -1 Y A +

e+

(b particle) +

+n

g decay : When daughter nucleus in higher energy state comes to ground state. by emitting a photon or photons. a r

•

photon is released. e.g.

67 27

Co ®

67 28

Ni *

( higher energy state )

+ b- + n ,

67 28

67 Ni + g photon Ni* ®28

Properties of a, b and g rays

Identity

Helium nucleus or doubly

Fast moving electrons

Electromagnetic wave

( –b or b )

(photons) Neutral

0

–

Charge

Twice of proton (+2e) » 4mp

Electronic (– e)

Mass

(rest mass of b)

rest mass = 0

2.2 ×10 m/s. 7

(Only certain value

g–photons come out with

between this range)

same speed from all

b–particles come out with

So, can not be a

same type of nucleus.

characteristic speed.

So that it is a

So that it can not

characteristic speed.

be a characteristic speed.

» MeV

» MeV

Line and discrete

Continuous

(or linear)

(or linear)

power (a>b>g)

of g–rays

Penetration

1 times of g–rays 10000

A

10,000 times

io n

different speeds from the

nature of the nucleus.

Ionization

» MeV

Line and discrete

100 times of g–rays (or

1 times of a) 100

1 times of g–rays 100

1 (or

1 times of b) 100

1(100 times of b)

(100 times of a)

power (g>b>a) Effect of electric

types of nucleus.

Their speed depends on

ss

Energy spectrum

(All possible values

Se

K.E.

Only c = 3 × 108 m/s

LL

between this range).

1% of c to 99% of c

19

1.4 ×107 m/s. to

= (rest mass of electron)

20

Speed

-2

ionised helium atom ( 2He ) 4

mp–mass of proton

g–rays

EN

b–particles

0

a–particles

Deflection

Deflection (More than a)

No deflection

or magnetic field Explanation

By Tunnel effect

By weak nuclear

With the help of energy

of emission

(or quantum mechanics)

interactions

levels in nucleus

120

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\21_Modern physics.p65

Features

E

Physics HandBook

CH APTER

ALLEN Laws of Radioactive Decay Rate of decay µ number of nuclei. dN = -lN dt

dN = -ldt . N

where l is called the decay constant. This equation may be expressed in the form N = N0e–lt

N0 – lt

N'=N0(1–e )

N 0.37N0= e 0

(0,0)

time

Ta

l n (2 ) l

Mean or Average Life (Ta) : It is the average of age of all active nuclei i.e.

sum of times of existance of all nuclei in a sample 1 = l initial number of active nuclei in that sample

io n

LL Ta =

dN = Nl or R = R0e–lt dt

Þ

Se

SI UNIT of R : 1 becquerel (1 Bq)= 1 decay/sec Other Unit is curie : 1 Ci = 3.70 ×1010 decays/sec 1 Rutherford : (1 Rd) =106 decays/s Specific activity : Activity of 1 gm sample of radioactive substance. Its unit is Ci/gm e.g. specific activity of radium (226) is 1 Ci/gm. Parallel radioactive disintegration

A node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\21_Modern physics.p65

E

•

ss

ACTIVITY OF A SAMPLE (A OR R) (OR DECAY RATE) R =-

19

Half life (Th) : Time during which number of active nuclei reduce to half of initial value. Th =

•

Th

20

•

– lt

N=N0e

0

N0 2

-2

N

EN

0.63N0

dN A = - (l1 + l 2 ) N A dt

a l1

Þ l eff = l1 + l 2

B

A

t1 t 2 Þ t eff = t + t 1 2

b

l2 C

Radioactive Disintegration with Successive Production a l ¾¾¾¾¾¾¾ ( a=rate of production) ® A ¾¾® B

121

Physics HandBook

C HAP TE R

ALLEN

dN A = a = lN A ....(i) dt

when NA in maximum t

0

t

dN A

ò a - lN

A

a

a dNA = 0 = a - lNA = 0 , NA or max = l dt

= ò dt, Number of nuclei is N A = 0

a 1 - e -lt l

(

R

)

t

Soddy and Fajan's Group Displacement Laws : (i) a–decay : After emission of one a–particle reduces the mass number by 4 units and atomic number by 2 units. ZXA ® Z–2YA–4 + 2He4(a)

EN

(ii) b–decay : Mass number remains same and atomic bumber increases by 1. ZXA ® Z+1YA + b + n

g

(electromagnetic radiation)

-ve potential

-2

a

b

19

+ve potential

122

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\21_Modern physics.p65

Se

A

ss

io n

20

Lead

LL

0

(iii) g–decay : Both mass number and atomic number remains same, only energy is released in the form of g–photons.

E

Physics HandBook

CH APTER

ALLEN

Ray Optics & Optical Instruments REFLECTION LAWS OF REFLECTION The incident ray (AB), the reflected ray (BC) and normal (NB) to the surface (SS') of reflection at the point of incidence (B) lie in the same plane. This plane is called the plane of incidence (also plane of reflection).

Find

y ra

B

VELOCITY OF IMAGE OF MOVING OBJECT (PLANE MIRROR)

S'

VO,Y

ˆ ˆ)n ˆ In vector form rˆ = eˆ - 2 ( e.n

•

Real : Point from which rays actually diverge.

•

Virtual: Point towards which rays appear to converge

IMAGE :

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\22_Ray optics & Optical Instruments.p65

E

•

X

M

Velocity component along X-axis r r VI,M = - VO,M

ss

(i)

I

VO,X

r r r VI,X = 2VM,X - VO,X

Se

Image is decided by reflected or refracted rays only. The point image for a mirror is that point towards which the rays reflected from the mirror, actually converge (real image). OR From which the reflected rays appear to diverge (virtual image) .

A

•

O

Y

io n

LL

OBJECT :

0

-2

r

re fle ct ed

t en ci d in

S

r

i

r If m even, then n = m – 1, for all positions of object. r If m odd , then n = m, If object not on bisector and n = m – 1, If object at bisector

C

n

ra y

e

19

normal N

A

360 =m q

20

Ði = Ð r

EN

The angle of incidence (the angle between normal and the incident ray) and the angle of reflection (the angle between the reflected ray and the normal) are equal

CHARACTERISTICS OF REFLECTION BY A PLANE MIRROR : • The size of the image is the same as that of the object. • For a real object the image is virtual and for a virtual object the image is real. • For a fixed incident light ray, if the mirror be rotated through an angle q the reflected ray turns through an angle 2q in the same sense. Number of images (n) in inclined mirror

Þ

(ii) Along Y-axis r r VI,Y = VO,Y

123

Physics HandBook

C HAP TE R

ALLEN

SPHERICAL MIRRORS

M'

spherical mirror

M spherical surface

convex mirror

LONGITUDINAL MAGNIFICATION

EN

length of image v 2 - v1 = length of object u 2 - u1

For small objects only : m L = -

Linear magnification m =

io n

Se

TRANSVERSE MAGNIFICATION : h2 v =h1 u

0

wo

area of image (ma) ´ (mb) = = m2 area of object (a ´ b)

VELOCITY OF IMAGE OF MOVING OBJECT

(SPHERICAL MIRROR) Velocity component along axis (Longitudinal velocity) M

///////// ///////// // /////// ///////

O

//// // / / / // /

Image enlarged diminished inverted erect

ho

mw o

// // ///

h2 = y co-ordinate of image h1 = y co-ordinate of the object (both perpendicular to the principal axis of mirror)

ms =

w i=

/ // //

A

f = x- coordinate of focus u = x-coordinate of object v = x-coordinate of image Note : Valid only for paraxial rays.

ss

1 1 1 = + . f v u

h i wi = ho wo

M'

When an object is coming from infinite towards the focus of concave mirror

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\22_Ray optics & Optical Instruments.p65

MIRROR FORMULA :

dv = m2 du

SUPERFICIAL MAGNIFICATION

LL

Note : According to above convention radius of curvature and focus of concave mirror is negative and of convex mirror is positive.

-2

All distance are measured from pole.

u1

19

•

v2

hi=mho

The direction of the incident rays is considered as positive x-axis. Vertically up is positive y-axis.

v1

u2

20

•

object

image

\

We follow cartesian co-ordinate system convention according to which the pole of the mirror is the origin.

\\\\\\\\\\\\\\\\\\ \\ \\\\\\\\ \\\\\ \ \\\ \ \\\

•

mL =

\\\\\

SIGN CONVENTION :

124

P

M

concave mirror

Rays which forms very small angle with princiapl axis are called paraxial rays. All formulae are valid for paraxial ray only.

Magnification |m|> 1 |m| < 1 m0

F

C

M

PARAXIAL RAYS :

m=

P

\ \ \ \\

F

\ \\ \

C

\\\\\\\\\\\\\\\\ \\\\ \

q

\\\ \ \ \\ \ \ \\ \\ \\

\\\\ \\\\\

P C

r i

\ \ \\

\\\

principal axis

M'

\\\

\\

M'

E

1 1 1 + = v u f

\-

1 dv 1 du =0 v 2 dt u2 dt

n

e

i

v2 r r r Þ v IM = - 2 v ox = -m2 v OM u

(b)

dv = velocity of image with respect to mirror dt

(ii) The product of refractive index and sine of angle of incidence at a point in a medium is constant. m1 sin i = m2 sin r (Snell's law)

du = = velocity of object with respect to mirror.. dt

Velocity component perpendicular to axis (Transverse velocity) m=

dh I dt

hI v f =- = Þ hI = h0 u f -u =

æ f ö ç ÷ h0 èf-uø

f h 0 du æ f ö dh 0 + ç ÷ 2 è f - u ø dt (f - u) dt

é r m 2h0 r ù ˆ r v ox ú j v iy = ê mv oy + f ë û

m2 v1 l 1 Sin i Snell's law : Sin r = 1 m2 = m = v = l 1 2 2 ˆ = m2 rˆ × n ˆ In vector form m1 eˆ × n

;

Note : Frequency of light does not change during refraction . DEVIATION OF A RAY DUE TO REFRACTION

é dh I ù r ê dt = velocity of image ^ to principal-axis ú ê ú ê dh o = velocity of object ^r to principal-axis ú êë dt úû

E

2

OPTICAL POWER :

Optical power of a mirror (in Diopters) = –

20 d

io n

angle of deviation, d = i - r

ss

REFRACTION THROUGH A PARALLEL SLAB Emergent ray is parallel to the incident ray, if medium is same on both sides. A

Se

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\22_Ray optics & Optical Instruments.p65

A

Applicable to a pair of real object and real image position only . They are called conjugate positions or foci, X1, X2 are the distance along the principal axis of the real object and real image respectively from the principal focus

N i

AIR

B

GLASS

r t

1 f

N'

C i

where f = focal length (in meters) with sign .

90° x

D

REFRACTION - PLANE SURFACE LAWS OF REFRACTION (at any refracting surface) Laws of Refraction (i) Incident ray, refracted ray and normal always lie in the same plane.

denser

r

LL

Note : Here principal axis has been taken to be along x– axis. NEWTON'S FORMULA :

X1 X 2 = f

rarer

i

0

• v OM

ˆ ˆ=0 In vector form (eˆ ´ n).r

r

-2

v IM =

r

EN

•

m1 m2

19

Q

Physics HandBook

CH APTER

ALLEN

t sin(i - r) ; t = thickness of slab cos r Note : Emergent ray will not be parallel to the incident ray if the medium on both the sides are different. Lateral shift x =

125

Physics HandBook

C HAP TE R

APPARENT DEPTH OF SUBMERGED OBJECT (h¢ < h) µ 1 > µ2

ALLEN Graph between angle of deviation (d) and angle of incidence (i) as rays goes from denser to rare medium ö -1 æ m D If i < qc µ Dsini = µ R sin r; r = sin ç m sin i÷ so è R ø

•

m2

æm ö d = r - i = sin -1 ç D sin i÷ - i è mR ø

m1 h'

h

x

m2 For near normal incidence h ¢ = m h 1

m

LL

Dx

io n

t

1ö æ Dx = Apparent normal shift = t çè 1 - m ø÷

Se

A

CRITICAL ANGLE & TOTAL INTERNAL REFLECTION (TIR)

CONDITIONS • Angle of incident > critical angle [i > qc] • Light should travel from denser to rare medium Þ Glass to air, water to air, Glass to water Snell's Law at boundary xx', mD sin qC = mR sin 90° Þ sin q C =

126

mR mD

If i > qc ; d = p – 2i

•

ss

Note : Shift is always in direction of incidence ray.

SOME ILLUSTRATIONS OF TOTAL INTERNAL REFLECTION •

Sparkling of diamond : The sparkling of diamond is due to total internal reflection inside it. As refractive index for diamond is 2.5 so C= 24°. Now the cutting of diamond are such that i > C. So TIR will take place again and again inside it. The light which beams out from a few places in some specific directions makes it sparkle.

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\22_Ray optics & Optical Instruments.p65

O'

19

O

m=1

20

m=1

-2

Note : h and h' are always measured from surface.

0

EN

O

E

•

Physics HandBook

CH APTER

ALLEN

Optical Fibre : In it light through multiple total internal reflections is propagated along the axis of a glass fibre of radius of few microns in which index of refraction of core is greater than that of surroundings (cladding)

m1 m 1 > m2

light pipe

19

-2

cold air

0

Mirage and looming : Mirage is caused by total internal reflection in deserts where due to heating of the earth, refractive index of air near the surface of earth becomes lesser than above it. Light from distant objects reaches the surface of earth with i > q C so that TIR will take place and we see the image of an object along with the object as shown in figure.

EN

•

hot air

20

ss

io n

LL

hot surface

E

Se

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\22_Ray optics & Optical Instruments.p65

A

Similar to 'mirage' in deserts, in polar regions 'looming' takes place due to TIR. Here m decreases with height and so the image of an object is formed in air if (i>qC) as shown in figure.

rarer sky denser

127

Physics HandBook

C HAP TE R

ALLEN COMBINATION OF TWO PRISMS

REFRACTION THROUGH PRISM

dmin

R

i=ig

i=e

e=90°

e=ig i=90°

angle of incidence

é n v¢ + n R¢ ù é n v + nR ù - 1ú A + ê - 1ú A ¢ û 2 2 ë û

= êë

•

d = (i + i¢) – (r + r¢)

•

r + r¢ = A

•

There is one and only one angle of incidence for which the angle of deviation is minimum.

•

EN

é nv + nR ù é n v¢ + n R¢ ù - 1ú A = - ê - 1ú A ¢ ê 2 2 ë û ë û

When d = dm then i = i¢ & r = r¢ , the ray passes symetrically about the prism, & th en

For a thin prism ( A £10o) ; d=(n–1)A

•

Dispersion Of Light : The angular splitting of a ray of white light into a number of components when it is refracted in a medium other than air is called Dispersion of Light.

•

Angle of Dispersion : Angle between the rays of the extreme colours in the refracted (dispersed) light is called Angle of Dispersion.

A q

dr dn

dv

m=

d v - d R nv - nR n + nR = n= v dy n -1 ; 2

nv, nR & n are R. I. of material for violet, red & yellow colours respectively .

P

C

I

+ve

m1 v m2 u

ss 1

m

1

+ve

1 1 1 - = v u f

æ 1 1 1ö • f = ( m - 1) ç R - R ÷ è 1 2ø

•

For small angled prism ( A £10 );

m2

Lens Formula :

•

r q mean ray v

o

w=

•

Se

Dispersive power (w) of the medium of the material of prism.

m1

v, u & R are to be O kept with sign as v = PI u = –PO R = PC (Note : Radius is with sign)

•

angular dispersion w= deviation of mean ray (yellow)

128

m 2 m1 m 2 - m 1 = v u R

io n

q = dv – dr

•

•

LL

•

Net angle of dispersion = (nv – nr) A + (nv¢ – nr¢) A¢.

19

A + dm sin é 2 ù ë û n= , where n = w.r.t. surroundings R.I. A é ù sin ë 2 û

of glass.

Direct Vision Combination : It is used for producing disperion without deviation condition for this

0

Q

i'

• m=

v u

Power of Lenses Reciprocal of focal length in meter is known as power of lens. • SI unit : dioptre (D) • Power of lens : P =

1 100 dioptre = f(m) f(cm)

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\22_Ray optics & Optical Instruments.p65

r'

It is used for deviation without dispersion . Condition for this (nv - nr) A +(n¢v - n¢r) A¢=0 wd + w¢d¢ = 0 where w, w¢ are dispersive powers for the two prisms & d , d¢ are the mean deviation. Net mean deviation

dmax

-2

d

r

Achromatic Combination :

20

P A i

•

angle of deviation

E

Physics HandBook

CH APTER

ALLEN Combination of Lenses

Two thin lens are placed in contact to each other Power of combination. P = P1 + P2 Þ

f1

1 1 1 = + f f1 f2

object

I2

f2 I1

If the distance between two positions of lens is x then x = u2 – u1 =

Use sign convention when solving numericals

Newton's Formula

D + D ( D - 4f ) 2

1 [D + D (D - 4f)] = u2 Þ v1 = u 2 2

Displacement Method

19

x1 = distance of object from focus x2 = distance of image from focus

1 [D - D (D - 4f)] = u1 Þ v 2 = u1 2 Distances of object and image are interchangeable. for the two positions of the lens. Now x = u2 – u1 and D = v1 + u1 = u2 + u1 [Qv1 = u2]

=

LL

It is used for determination of focal length of convex lens in laboratory. A thin convex lens of focal length f is placed between an object and a screen fixed at a distance D apart.

io n

ss

screen

I I D+x D-x ´ Þ 1 22 = 1 Þ O = D-x D+x O Silvering of one surface of lens

Now m1 ´ m2 =

Se

A

D-x D+x ; and u2 = v1 = 2 2

(i) For D 4f :

L2

Peff = 2PL1 + 2PL2 + PM

2 1 1 2 = + feff fL1 fL2 fM

I1 I 2

\\ \ \\ \\\ \\\\\\\

D

so u1 = v 2 =

I2 v 2 D - x I1 v1 D + x = = m1 = O = u = D - x and m2 = O u2 D + x 1

v=(D- u)

object

1 [D + D (D - 4f)] 2

0

=

-2

I

1 [D - D (D - 4f)] 2

20

f = x1 x2

L1

(ii) For D = 4f : u =

D - D(D - 4f) D + D(D - 4f ) and u2 = 2 2 So there are two positions of lens for which real image will be formed on the screen.(for two distances u1 and u2 of the object from lens)

R

When plane surface is silvered f = 2(m - 1)

u1 =

R

When convex surface is silvered f = 2m

O

O

\\\\\\\\ \\\\\\\ \\\\\\\ \\\\\\\\\\\\\\\ \\\\\\\ \\\\ \\\\\\\ \\\\ \\\\\\\ \\\\\\\ \\\ \\\\ \\\\ \\\ \\\\\\\\ \\\\\\\ \ \\\\ \\ \\\ \\ \\ \\ \\\\

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\22_Ray optics & Optical Instruments.p65

= D ( D - 4f )

EN

F2

v 2 = D - u2 = D -

E

2

D2 - x2 4D Distance of image corresponds to two positions of the lens :

F1

u

D - D ( D - 4f )

Þ x2 = D2 – 4 Df Þf =

v 1 = D - u1 = D -

O

-

129

130

0

-2

19

20

EN

C HAP TE R

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\22_Ray optics & Optical Instruments.p65

io n

ss

LL

Se

A Physics HandBook ALLEN

IMPORTANT NOTES

E

Physics HandBook

CH APTER

ALLEN

Wave Nature of Light & Wave Optics HUYGEN'S WAVE THEORY

INTERFERENCE : YDSE

Huygen's in 1678 assumed that a body emits light in the form of waves.

•

•

• •

Resultant intensity for incoherent sources I=I1+I2 Intensity µ width of slit µ (amplitude)2

•

In homogeneous medium, the wave front is always perpendicular to the direction of wave propagation.

•

S1 d

q

2

æ a + a2 ö = ç 1 è a1 - a 2 ÷ø

LL secondary wave

q

io n

ss

Distance of mth dark fringe x m = Path difference=(2m+1)

COHERENT SOURCES :

Two sources will be coherent if and only if they produce waves of same frequency (and hence wavelength) and have a constant initial phase difference.

INCOHERENT SOURCES : Two sources are said to be incoherent if they have different frequency and initial phase difference is not constant w.r.t. time.

( 2m + 1) lD 2d

l where m= 0,1,2, 3,..... 2

lD

•

Fringe width b = d

•

Angular fringe width = D = d

•

min max Fringe visibility = I + I ´ 100 % max min

•

If a transparent sheets of referactive index m and thickness t is introduced in one of the paths of interfering waves, optial path will becomes 'mt' instead of 't'. Entire fringe pattern is displaced by

Se

A B'

nl D d

Path difference = nl where n =0, 1, 2, 3, .....

•

Secondary source

2

20

n dsi

Spherical wavefront

Primary source

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\23_Wave Nature of light & Wave Optics.p65

)

2

Distance of nth bright fringe x n =

S2

E

I 1 - I2

)

-2

The forward envelope of the secondary wavelets at any instant gives the new wavefront.

A'

(

I1 + I2

19

•

Plane wavefront B A

(

I max I W a2 Þ 1 = 1 = 12 Þ I = min I2 W2 a2

0

Each point on a wave front is a source of new disturbance, called secondary wavelets. These wavelets are spherical and travel with speed of light in that medium.

I = I1 + I2 + 2 I1I 2 cos f0

EN

•

Each point source of light is a centre of disturbance from which waves spread in all directions. The locus of all the particles of the medium vibrating in the same phase at a given instant is called a wavefront.

Resultant intensity for coherent sources

b

I

D ëé( m - 1) t ûù d

=

l

-I

b ( m - 1) t towards the side in which l

the thin sheet is introduced without any change in fringe width.

131

Physics HandBook

C HAP TE R

SHIFTING OF FRINGES

•

ALLEN

Newton's Ring : When a lens of large Radius of curvature is placed on a plane glass plate, an air film is formed between lower surface of the lens and the upper surface of the plate. It this film is illuminated by sodium

P

light, due to interference concentric bright and dark rings called Newton's Rings are seen.

y S1

•

S2

If the film is wedge shaped. The fringes will be straight lines parallel to the edge of apex with minimum at the

l t where q = . 2mq x

EN

apex and fringe width b =

• If case of Newton's rings the centre is dark spacing

• Path difference produced by a slab Dx = (m–1)t

between rings goes on decreasing as we move away from

D b • Fringe shift, Dx = ( m - 1) t = ( m - 1) t l d

square root of all natural number while bright rings 1.0

shift t ( m-1) D / d ( m - 1) t = = fringe width lD / d l

S

Real source

0

Screen P

d

l 2

Mirror

S

(t =thickness of film, m=R.I. of the film) The position of dark and bright fringes are reversed relative to the pattern of two real sources because there is a 1800 phase change produced by reflection.

132

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\23_Wave Nature of light & Wave Optics.p65

Minima ®2mt cosr =(2n+1)

ss

Minima® 2mt cosr = nl For transmitted light : Maxima ®2mt cosr =nl

l 2

Se

•

io n

For reflected Light :

Maxima ® 2mt cosr = (2n+1)

-2

Llayd's Mirror :

LL •

A

INTEREFERENCE IN THE FILM

•

20

=

the square root of odd numbers.

19

• Number of fringes shift

centre and radius of dark rings is proportional to the

E

Physics HandBook

CH APTER

ALLEN

Screen

Fresnel's Biprism A S1

d

x

Region of intercference

d d

d

S1

d A

b

a x = ad

Seperation between coherent sources d = 2ad = 2aA(µ – 1) Seperation between slit plane and screen D = a + b l ( a + b) AD Frindge width on screen b = d = 2aA m - 1

)

EN

(

Frillet split lens as a limiting case of YDSE

x

O

from lens

DIFFRACTION

af a-f

D = a + b – |V|

A

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\23_Wave Nature of light & Wave Optics.p65

d = 2x + 2

E

Frindge width

b=

lD d

v x u

Fresnel's diffraction : In Fresnel's diffraction, the source and screen are placed close to the aperture or the obstacle and light after diffraction appears converging towards the screen and hence no lens is required to observed it. The incident wave fronts are either spherical or cylindrical. Fraunhofer's diffraction : The source and screen are placed at large distances from the aperture or the obstacle and converging lens is used to observed the diffraction pattern. The incident wavefront is planar one. r For minima : a sinqn = nl

ss

•

Se

formula v =

b D

v

io n

a

20

LL

S2

Region of interference

19

x

-2

S1

0

f

Screen

æ v ö d = 2x ç 1 + ÷ uø è

•

r For maxima : a sinqn = (2n + 1)

l 2

2lD a 2l r Angular width of central maxima Wq = a

r Linear width of central maxima : Wx =

r Intensity of maxima where I0 = Intensity of central maxima 2

é sin (b / 2 ) ù 2p I = I0 ê a sin q ú and b = l ë b/2 û

133

Physics HandBook

C HAP TE R

POLARISATION OF LIGHT If the vibrations of a wave are present in just one direction in a plane perpendicular to the direction of propagation, the wave is said to be polarised or plane polarised. The phenomenon of restricting the oscillations of a wave to just one direction in the transverse plane is called polarisation of waves.

ALLEN

POLARISATION BY REFLECTION Brewster's Law : The tangent of polarising angle of incidence at which reflected light becomes completely plane polarised is numerically equal to refractive index of the medium. m=tan ip; ip =Brewster's angle and ip+rp=90° POLARISATION BY SCATTERING If we look at the blue portion of the sky through a polaroid and rotate the polaroid, the transmitted light shows rise and fall of intensity.

EN

Incident sunlight (Unpolarised)

0

20

19

MALUS LAW The intensity of transmitted light passed through an analyser is I=I0cos2q (q=angle between transmission directions of polariser and analyser)

-2

Scattered light (polarised)

Nitrogen molecule

KEY POINTS

n (b)longer = ( n + 1) (b)shorter 1

•

No interference pattern is detected when two coherent sources are infinitely close to each other, because b µ d

•

If maximum number of maximas or minimas are asked in the question, use the fact that value of sinq or cosq can't be greater than 1.

nmax =

•

Limit of resolution for microscope =

•

Limit of resolution for telescope =

134

d Total maxima = 2nmax +1 l 1.22l 1 = 2a sin q resolving power

1.22l 1 = a resolving power

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\23_Wave Nature of light & Wave Optics.p65

ss

A

• •

The law of conservation of energy holds good in the phenomenon of interference. Fringes are neither image nor shadow of slit but locus of a point which moves such a way that its path difference from the two sources remains constant. In YDSE the interference fringes for two coherent point sources are hyperboloids with axis S1S2. If the interference experiment is repeated with bichromatic light, the fringes of two wavelengths will be coincident for the first time when

Se

• •

io n

LL

The scattered light screen in a direction perpendicular to the direction of incidence is found to be plane polarised.

E

A

E

io n

ss

Se 20

CH APTER

0

-2

19

EN

LL

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\23_Wave Nature of light & Wave Optics.p65

ALLEN Physics HandBook

IMPORTANT NOTES

135

Physics HandBook

C HAP TE R

ALLEN

Errors Significant Digits Rules for determining the number of

Error

significant digits in number with indicated decimals. • •

Absolute Error : Expressed in absolute term ex-least count

a b r A + , then also = B R A B

Power rule : When a quantity Q is raised to a power P, the relative error in the result is P times the relative error in Q.

r = R

The quotient rule is not applicable if the numerator and denominator are dependent on each other. e.g if R =

XY . We cannot apply quotient rule to X+Y

find the error in R. Instead we write the equation as follows

1 1 1 = + . Differentiating both the R X Y

-

sides, we get

Thus

136

r

R

2

dR

R =

2

=x

X

2

+

dX

X

2

y

Y2

-

dY

Y2

.

-2

A sum or difference can have no more indicated positions to the right of the decimal as the number involved in the operation with the LEAST indicated positions to the right of its decimal.

Rules for rounding off digits

There are a set of conventional rules for rounding off. • Determine according to the rule what the last reported digit should be. • Consider the digit to the right of the last reported digit. • If the digit to the right of the last reported digit is less than 5 round it and all digits to its right off. • If the digit to the right of the last reported digit is greater than 5 round it and all digits to its right off and increased the last reported digit by one. • If the digit to the right of the last reported digit is a 5 followed by either no other digits or all zeros, round it and all digits to its right off and if the last reported digit is odd round up to the next even digit. If the last reported digit is even then leave it as is.

Se

q Q

A

P´

19

substraction :

This also holds for negative powers. If R = QP,

4.

significant digits in an addition or

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\24_Error & measurements.p65

and if R = 3.

Express in ecientific notation

Rule for expressing the correct number of

r a b + = R A B

R = AB,

decimals.

LL

Thus if

R=

Determining the number of significant digits in number is not having an indicated

io n

2.

Addition and subtraction rule : The absolute random errors add. Thus if R = A + B, r = a + b and if A – B, r=a+ b Product and quotient rule : The relative random errors add.

ss

1.

Absolute error Size of measurement

20

Relative/ Fractional Error :

EN

•

All nonzero digits (1-9) are to be counted as significant. Zeros that have any nonzero digits anywhere to the LEFT of them are considered significant zeros. All other zeros not covered in rule (2) above are NOT be considered significant digits.

0

Syotematic (Can be removed) Ex- zero error, Bench error Fixed sign

Random (Cann't be removed but minimized) No fixed sign

E

Physics HandBook

CH APTER

ALLEN Vernier Callipers P

Q

Vernier Scale

S

Principle of Vernier

Main Scale

0

1

2

1

0

M

2 cm

N

15

V

1 C

0

D

5

Main scale (S)

10

Vernier scale (V)

Vernier Callipers

EN

Least count of Vernier Callipers

N -1 ö 1ms ÷ ms = , which is equal to the value of the smallest division on the N ø N

19

æ è

Vernier constant = 1 ms – 1 vs = ç1 -

N -1 ms N

10

0

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\24_Error & measurements.p65

A

Vernier scale without zero error

E

0

0

5

10

Main scale

0

io n

5

1

0

Vernier scale with positive zero error

Se

0

Main scale

1

ss

Main scale

20

LL

main scale divided by total number of divisions on the vernier scale. Zero error:

0

-2

N (VS) = (N – 1) ms Þ 1VS =

0

The least count or Vernier constant (v. c) is the minimum value of correct estimation of length without eye estimation. If N division of vernier coincides with (N-1) division of main scale, then

0

th

4 division coinciding Positive zero error (+0.04 cm) and its correction

10

(ii)

0

0

5 10

5

Vernier scale with negative zero error

(i)

1

1

1

5 10

th

6 division coinciding

Negative zero error = (- 0.04 cm) and its correction

Negative zero error = – [Total no. of vsd – vsd coinciding] ×L.C.

137

Physics HandBook

C HAP TE R

ALLEN

Screw Gauge T

P Q S

M

Spindle

R 5 0 95

Screw head

Pitch

Thimble

Sleeve

Direction of motion

Screw Gauge

M 0 1 45

S

T

M

Spindle

40 Thin sheet

R 5 0 95

Thimble

Sleeve

E

E

5 0 95 96

Circular scale

LL

0

Main scale reference line

Negative zero error

(3 division error) i.e., - 0.003 cm

R

85

Se

M

5 0

A

T

Circular scale

Line of graduation

E

Positive zero error

(2 division error) i.e., + 0.002 cm

Constants of the Screw Gauge (a) Pitch (b) Least count (c) Measurement of length by screw gauage

zero of the circular scale is above the zero of main scale

ss

E

PQ S

10 5 0 95 90

io n

Line of graduation

0 Main scale reference line

15 10 5 0 95 90

Zero of the circular scale is below the zero of main scale

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\24_Error & measurements.p65

M

20

R

T

S

19

Screw gauge with no zero error

PQ

138

Q S

0

P

EN

P Q

Principle of a micrometer

R

T

Pitch

Nut

-2

E

E

Physics HandBook

CH APTER

ALLEN

Semiconductor & Digital Electronics COMPARISON BETWEEN CONDUCTOR, SEMICONDUCTOR AND INSULATOR Conductor 10 –2 – 10 –8 Wm 10 2 – 10 8 mho/m Positive

Semiconductor 10 –5 – 10 6 Wm 10 -6 – 10 5 mho/m Negative

Insulator 10 11 – 10 19 Wm 10 –19 – 10 –11 mho/m Negative

Due to free electrons

Due to electrons and holes

No current

Semi conductor

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\25_Semiconductor.p65

³ 3eV

Ge, Si, GaAs, GaF2

Wood, plastic, Diamond, Mica

19

20

DEg 2kT

io n

-

ss

Extrinsic semiconductor (doped semicondutor)

Se

A

(pure form of Ge, Si) ne =nh =ni

E

Insulator

@ 1eV

SEMICONDUCTOR

w

Valence Band

Number of electrons reaching from valence band to conduction band : n = AT 3 / 2 e CLASSIFICATION OF SEMICONDUCTORS :

Intrinsic semiconductor

Eg ³ 3eV

Forbidden Gap

0

Valence Band

LL

w w

Pt, Al, Cu, Ag

Eg @ 1ev

-2

Valence Band

@ 0eV

Electron Energy

Forbidden Gap

EN

No gap

Conductor

Forbidden energy gap Example

Conduction Band

Conduction Band Electron Energy

Overlapping region

Conduction Band Electron Energy

Properties Resistivity Conductivity Temp. Coefficient of resistance (a) Current Energy band diagram

N-type pentavalent impurity (P, As, Sb etc.) donor impurity (N D) ne>> n h

MASS-ACTION LAW : n2i = n e ´ n h r For N-type semiconductor ne ; ND

P-type trivalent impurity (Ga, B, In, Al) acceptor impurity (N A) nh>> n e

r For P-type semiconductor nh ; NA

CONDUCTION IN SEMICONDUCTOR Intrinsic semiconductor ne = nh J = ne [ ve + vh] (Current density)

s=

1 = en [µ e + µ h] (Conductivity) r

P - type nh >> ne J @ e nh vh

s=

1 @ e nh mh r

N - type ne >> nh J @ e ne ve

s=

1 @ e ne me r

139

C HAP TE R

ALLEN

N-type (Pentavalent impurity)

Intrinsic Semiconductor

CB

CB

CB

donor

acceptor impurity level

level

VB

VB

P-type (Trivalent impurity)

VB

free electron positive donor ion

hole negative acceptor ion

Mainly due to electrons

Mainly due to holes

ne = nh= ni

nh >ne (N A nh)

I = Ie + Ih

I Ie

Entirely neutral

Entirely neutral Majority Electrons Minority Holes

I

Majority Holes Minority - Electrons

Forward Bias

+

P

n

N

+

V0

distance

A

Direction of diffusion current : P to N side and drift current : N to P side

4 5 6 7

Current flows mainly due to majority carriers. Forward characteristic curves.

to prevent the movement of electron from

1 2 3 4 5 6 7

+

Potential Barrier increases. Width of depletion layer increases. P-N jn. provide high resistance Very small current flows. Order of current is micro ampere for Ge or Nano ampere for Si. Current flows mainly due to minority carriers. Reverse characteristic curve

break down voltage

Reverse saturation current

0 V f(volt)

8

Forward Resistance : Rf =

9

D Vf @ 100W DI f

potential difference called a barrier potential.

Rr = 10 3 : 1 for Ge Rf

140

V

–

knee voltage

Order of knee or cut in voltage Ge® 0.3 V Si ® 0.7 V Special point : Generally

the N region into the P region. This

N

if (mA)

= drift current. So total current is zero

to the P side. This potential difference tends

P

Vr(volt)

If there is no biasing then diffusion current

In junction N side is at high potential relative

Reverse Bias

on

3

Se

free electron

electric potential

hole

Potential Barrier reduces Width of depletion layer decreases P-N jn. provide very small resistance Forward current flows in the circuit Order of forward current is milli ampere.

ss i

1 2

V –

20 19

LL E –

0

Entirely neutral

COMPARISON BETWEEN FORWARD BIAS AND REVERSE BIAS

P-N JUNCTION

(At equilibrium condition) p

Ih

-2

Quantity of electrons and holes are equal

N

Current due to electron and hole

Ir ( mA)

8

Reverse Resistance : R r = DVr @ 10 6 W DIr

9

Breakdown voltage Ge ® 25 V Si ® 35 V Rr = 10 4 : 1 for Si Rf

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\25_Semiconductor.p65

Physics HandBook

E

Physics HandBook

CH APTER

ALLEN

BREAKDOWN ARE OF TWO TYPES Zener Break down

Avalanche Break down

Where covalent bonds of depletion layer, itself

Here covalent bonds of depletion layers are broken

break, due to high electric field

by collision of "Minorities" which aquire high kinetic energy from high electric field

This phenomena take place in

This phenomena takes place in

(i)

P – N junction having "High doping"

(i)

(ii)

P – N junction having thin depletion layer

(ii) P – N junction having thick depletion layer

P – N junction having "Low doping"

Here P – N junction does not damage permanently

Here P – N junction damages permanentaly

"In D.C voltage stabilizer zener phenomena is used".

due to abruptly increment of minorities

APPLICATION OF DIODE

EN

during repeatative collisions.

Zener diode

:

It is highly doped p-n junction diode used as a voltage regulator.

•

Photo diode

:

A p-n junction diode use to detect light signals operated in reverse bias.

•

LED

:

A p-n junction device that emits optical radiation under forward bias conditions

•

Solar cell

:

Generates emf of its own due to the effect of sun radiations.

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\25_Semiconductor.p65

E

V0

t

V0

io n

-

t

V0

ss

Se

A

CENTRE – TAP FULL WAVE RECTIFIER

+

20

v=Vmsinwt

LL

HALF WAVE RECTIFIER

-2

19

V

ideal diode

0

•

V0 2

D1 D2 D1 D2 D1 D2 D1 D2 D1 D2

Vin

FULL WAVE BRIDGE REACTIFIER D1D2 D3D4 D 1D 2 D 3D 4

I RIPPLE FACTOR : r = ac Idc

RECTIFIER EFFICIENCY: h =

r For HWR r For FWR

For HWR : h% =

r = 1.21 r = 0.48

Pdc I2 R = 2 dc L Pac Irms (R F + R L )

40.6 81.2 & FWR h% = RF R 1+ 1+ F RL RL

141

Physics HandBook

C HAP TE R

ALLEN

FOR TRANSISTOR I E = IB + IC C

C

VCB

VBC PNP

B

+

B

NPN

VBE

VEB

E

E

+

(b)

EN

(a)

COMPARATIVE STUDY OF TRANSISTOR CONFIGURATIONS

C

CB

B

B

B

CE

E

IC

IE

LL

IB

B

B

E

IE

IB

IC

C

Very High

High

Low

(AI or a)

(AI or b)

a=

IC 1 IB

RL @ 150 Ri

Po 2 R =a L Pi Ri

AV =

Vo Vi

Av= b Ap =

=

I CR L IB R i

RL @ 500 Ri

Po 2 R =b L Pi Ri

(AI or g )

g=

IE >1 IB Vo

AV =

Vi

Av = g Ap =

=

IE R L I BR i

RL 100 GHz

ss

io n

LL

(iii) Television

Remarks Normally operated below 18 GHz

19

Coaxial Cable) 2 Free space (radio waves) (i) Standard AM broadcast FM (ii)

Frequency range 750 MHz (Bandwidth)

20

Service

1 Wire (most common :

0

BANDWIDTH OF TRANSMISSION MEDIUM Different types of transmission media offer different band width of which some are listed below

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\25_Semiconductor.p65

E

Se

A

GROUND WAVE PROPAGATION (a) Th e radio waves wh ich travel t hrough Hence, ground wave propagation can be sustained atmosphere following the surface of earth are only at low frequencies (500 kHz to 1500 kHz). known as ground waves or surface waves and (b) The ground wave transmission becomes weaker with their propagation is called ground wave increase in frequency because more absorption of propagation or surface wave propagation. ground waves takes place at higher frequency during These waves are vertically polarised in order to propagation through atmosphere. prevent short-circuiting of the e lect ric (c) The ground wave propagation is suitable for low and component. The electrical field due to the wave medium frequency i.e. upto 2 MHz only. induce charges in the earth's surface. As the wave travels, the induced charges in the earth (d) The ground wave propagation is generally used for also travel along it. This constitutes a current local band broadcasting and is commonly called in the earth's surface. As the ground wave passes medium wave. over the surface of the earth, it is weakened (e) The maximum range of ground or surface wave as a result of energy absorbed by the earth. Due propagation depends on two factors : to these losses the ground waves are not suited for very long range communication. Further (i) The frequency of the radio waves and (ii) Power these losses are higher for high frequency. of the transmitter

147

Physics HandBook

C HAP TE R

(a) The sky waves are the radio waves of frequency between 2 MHz to 30 MHz.

(d) Height of transmitting Antenna : The transmitted waves, travelling in a straight line, directly reach the received end and are then picked up by the receiving antenna as shown in figure. Q

(b) The ionospheric layer acts as a reflector for a certain range of frequencies (3 to 30 MHz). Therefore it is also called has ionospheric propagation or short wave propagation. Electromagnetic waves of frequencies higher than 30 MHz penetrate the ionosphere and escape.

d

h

d

P

T

S

90° R

R

(c) The highest frequency of radio waves which when sent straight (i.e. normally) towards the layer of ionosphere gets reflected from ionosphere and returns to the earth is called critical frequency. It

O

EN

Due to finite curvature of the earth, such waves cannot be seen beyond the tangent points S and T. (R+h)2 = R2 + d2

is given by fc = 9 Nmax , where N is the number

As R>>h, So h2 + 2Rh = d2 Þ d = 2Rh Area covered for TV transmission : A = pd2 = 2pRh Population covered=population density ×area covered If height of receiving antenna is also given in the question then the maximum line of sight dM

de

hT

hR

Line of sight communication by space waves

io n

where ; R=radius of earth (approximately 6400 km) hT = height of transmitting antenna hR = height of receiving antenna

ss

(c) The range of communication of space wave propagation can be increased by increasing the heights of transmitting and receiving antenna.

2Rh T + 2Rh R

dT

LL

(b) The space waves can travel through atmosphere from transmitter antenna to receiver antenna either directly or after reflection from ground in the earth's troposphere region. That is why the space wave propagation is also called as tropospherical propagation or line of sight propagation.

dM =

-2

(a) The space waves are the radio waves of very high frequency (i.e. between 30 MHz. to 300 MHz or more).

19

SPACE WAVE PROPAGATION

20

density of electron/m . 3

0

SKY WAVE PROPAGATION

ALLEN

148

Modulation

Continuous wave Modulation

AM (Amplitude Modulation)

FM (Frequency Modulation)

PM (Phase Modulation)

Pulse Wave Modulation

PAM (Pulse Amplitude Modulation)

PTM (Pulse Time Modulation)

PWM (Pulse Width Modulation)

PCM (Pulse Code Modulation)

PPM (Pulse Position Modulation)

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\25_Semiconductor.p65

A

The phenomenon of superposition of information signal over a high frequency carrier wave is called modulation. In this process, ampitude, frequency or phase angle of a high frequency carrier wave is modified in accordance with the instantaneous value of the low frequency information.

Se

MODULATION

E

Physics HandBook

CH APTER

ALLEN

AMPLITUDE MODULATION

NEED FOR MODULATION (i) To avoid interference:

If many

Modulating Signal

AM Carrier Signal

Carrier Signal

modulating signals travel directly through the same transmission channel, they will interfere with each other and result in distortion. (ii) To design antennas of practical

Modulation factor, m =

size : The minimum height of antenna (not of antenna tower) should be l/4

If v m = Vm cos wm t and v c = Vc cos wc t then m =

where l is wavelength of modulating •

As amplitude of the carrier wave varies at signal frequency fm so the amplitude of AM wave = VC + mVC cos wm t &

EN

impracticale because the frequency of

the modulating signal can be upto 5 kHz

frequency of AM wave =

which corresponds to a wavelength of

Þ v = VC cos wc t +

height of l/4 = 15 km. This size of

mVC mVC cos ( wC + wm ) t + cos ( wC - wm ) t 2 2

A theoretical study of

•

2R

antenna in which power is dissipated.

radiation from a linear antenna (length

LL

l) shows that the power radiated is

2 Power of carrier wave : PC = VC where R = resistance of

powers and hence this also points out

to the need of using high frequency transmission.

Total power of side bands : Psidebands = 2 ´

•

Total power of AM wave = PC ç 1 +

•

Fraction of total power carried by sidebands =

io n

For a good transmission, we need high

æ è

2

1 æ mVC ö m2 PC ç ÷ = 2R è 2 ø 2

•

m2 ö ÷ 2 ø

ss

proportional to (frequency)2 i.e. (l/l)2.

20

(iii) Effective Power Radiated by an

19

POWER IN AM WAVE

-2

require an antenna of the minimum an antenna is not practical.

wc 2p

Therefore v = éë Vc (1 + m ) cos wmt ùû cos wc t

3 × 108/5 × 103 = 60 km. This will

Antenna :

Vm Vc

0

signal. This minimum size becomes

amplitude of modulating wave amplitude of normal carrier wave

m2 2 + m2

Se

FREQUENCY MODULATION (FM)

When the frequency of carries wave is changed in accordance with the instantaneous value of the modulating signal, it is called frequency modulation.

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\25_Semiconductor.p65

A

FREQUENCY SPECTRUM OF AM WAVE

E

Signal

VC

mVC 2

Carrier

(fC-fm)

fC

FM Wave

fC+fm

side band frequency

Normal Max freq. freq.

Normal freq.

Min Normal freq. freq.

Max freq.

Normal Min Normal freq. freq. freq.

149

Physics HandBook

C HAP TE R

ALLEN

MODULATION FACTOR OR INDEX AND CARRIER SWING (CS)

2 3 4 5 6 7

3

Frequency Modulation The amplitude of AM signal varies depending on modulation index. Band width* is very small (One of the biggest advantage). Relatively simple and cheap.

4

Area of reception is Large.

5

It is difficult to eliminate effect of noise.

6

Most of the power which contained in carrier is not useful. Therefore carrier power transmitted is a waste. The average power in modulated wave is greater than carrier power. Maximum m = 1, otherwise over modulation (m > 1) would result in distortion. It is not possible to operate without interference.

1 2

The average power is the same as the carrier wave. No restriction is placed on modulation index (m).

7

It is possible to operate several independent transmitter on same frequency.

9

8

MODEM

FAX is abbreviation for facsimile which means exact reproduction. The electronic reproduction of a document at a distance place is called Fax.

Se

The name modem is a contraction of the terms Modulator and Demodulator. Modem is a device which can modulate as well as demodulate the signal.

FAX ( Facsimile Telegraphy)

ss

9

io n

LL

8

Amplitude Modulation The amplitude of FM wave is constant, whatever be the modulation index. It require much wider channel (Band width) [7 to 15 times] as compared to AM. Transmitters are complex and hence expensive. Area of reception is small since it is limited to line of sight. (This limits the FM mobile communication over a wide area) Noise can be easily minimised amplitude variation can be eliminated by using limiter. Power contained in the FM wave is useful. Hence full transmitted power is useful.

-2

1

0

Df = fmax.–fc=fc–fmin. ; vFM = VCcos[wCt + mfcoswmt] Carrier Swing (CS) The total variation in frequency from the lowest to the highest is called the carrier swing Þ CS = 2xDf Side Bands FM wave consists of an infinite number of side frequency components on each side of the carrier frequency fC, fC ± fm, fC ± 2fm, fC ± 3fm, & so on.

19

•

max. frequency deviation Df = Modulating frequency fm

20

•

Modulation factor : m =

EN

•

Receiving antenna

Amplifier

AM Wave

IF Stage

Rectifier

Amplifier

Detector

Envelop Detector

Output

m(t) Output

time

time

150

time

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\25_Semiconductor.p65

A

DETECTION OF AMPLITUDE MODULATION WAVE

E

E

io n

ss

CH APTER

0

-2

19

20

EN

LL

Se

A

node06\B0AI-B0\Kota\JEE(Advanced)\Leader\Phy\Sheet\Hand book (E+L)\Eng\25_Semiconductor.p65

ALLEN Physics HandBook

IMPORTANT NOTES

151