National Entrance Screening Test NEST Entrance Exam 2021 for NISER 10 Full-length Mock tests Solved Edugorilla complete Practice Kit

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Inspiring Strategic Preparation

NEST 2021 Complete Practice Kit Latest Edition Practice kit

10 Tests based on Real Exam pattern • Thoroughly Revised and Updated • Detailed Analysis of all MCQs

Introduction

EduGorilla is a bunch of youth who make exam preparation curate multi language online test series to cater to the success we let students breathe victory. We don’t lead students on competitive exam preparation. Rather we guide them along success.

easy. We create and of students. In short, the crooked path of the straight path to

EduGorilla is a pioneer in online test preparation. Our website holds easy and exclusive test series for more than 1320 competitive exams. And this book is an endeavor to offer our digitized Gorilla Series in printed format. This book is not just a compilation of mocks, it rather your guide to success! The book has been written to meet the requirements of aspirants and provide them with well-conceptualized and structured practice material. It contains multiplechoice questions modeled on the relevant subjects and topics. The volume has been prepared by a team of prolific experts after thoroughly analyzing the exam pattern, syllabus, and previous years’ papers. The Gorilla Series begins with an overview of the exam and goes on to cover practice sets along with hints and solutions. The subsequent part of the book also includes previous years’ question papers and their solutions. This gives a fair idea to the students about the structure of the question paper and the type of questions asked in the exam Students can analyze their performance and gauge their preparation level.

Why NEST Gorilla Series? ☛ 10 Full-length Mock Tests For Complete Preparation ☛ Tests modeled on the latest exam pattern ☛ Detailed explanation of questions

About NEST 2021 About National Entrance Screening Test (NEST) National Entrance Screening Test (NEST) is conducted jointly by National Institute of Science Education and Research (NISER), Bhubaneswar and Mumbai University. The candidates that are able to successfully qualify the exams will be offered admission in the above mentioned prestigious universities. The NEST Exam is conducted annually to offer admission for Integrated MSc programme in Biology, Chemistry, Mathematics, Physics.

NEST Notification 2020 The recruitment process for the NEST Exam will start soon. The exam will be conducted jointly jointly by National Institute of Science Education and Research (NISER), Bhubaneswar and Mumbai University. The application fee for SC, ST and PWD, Female is INR 600, while the same for rest of the candidates is INR 1200.

Exam Calendar 2021 Events Application process starts Last date to submit application form Admit card download NEST 2021 Exam Result declaration Counselling process starts

NEST Exam Pattern 2021 Sections 1 2

Subjects Marking General Section 30 Biology 50

Dates (Tentative) 1st week of January 2021 3rd week of March 2021 4th week of April 2021 3rd week of June 2021 3rd week of June 2021 4th week of June 2021

3 Chemistry 50 4 Mathematics 50 5 Biology 50 Total —– 230 marks Candidates must go through following details regarding NEST Exam Pattern :•

Exam Mode:- NEST 2021 examination will be organized via Online (computerbased) mode.

•

Time Duration:- 3 hours time duration will be given to complete the examination.

•

Questions Type:- The question paper will consist of the Multiple Choice Questions (MCQ).

•

Total marks:- The questions will be of 230 marks.

•

Sections:- There will be total 5 sections. Section 1 will be general section, Section 2 will consist of questions from Biology, Section 3 will consist of questions from Chemistry, Section 4 from Mathematics, and Section 5 from Physics.

•

Candidates will have to attempt any other 3 sections after attempting section 1.

TABLE OF CONTENT

Paper

Page

1. Mock Test - 1

1-23

2. Mock Test - 2

24-42

3. Mock Test - 3

43-63

4. Mock Test - 4

64-84

5. Mock Test - 5

85-105

6. Mock Test - 6

106-124

7. Mock Test - 7

125-139

8. Mock Test - 8

140-157

9. Mock Test - 9

158-175

10. Mock Test - 10

176-191

C. Decreases as we move towards its end

MOCK TEST - 1

D. Increases as we move towards its end 7. The coil in a heater is made of

General Ability 1. Which one of the following can be used to confirm whether drinking water contains a gamma emitting isotope or not? A. Microscope

B. Lead plate

C. Scintillation counter

D. Spectrophotometer

2. Which one of the following pairs is not correctly matched:

A. Nichrome

B. Tungsten

C. Copper

D. Iron

8. A body is falling freely under the action of gravity alone in vacuum. Which one of the following remains constant during the fall? A. Potential energy

B. Kinetic energy

C. Total linear momentum

D. Total mechanical energy

9. Which one of the following common devices works on the basis of the principle of mutual induction?

A. Discovery of Meson -Hideki Yukawa B. Discovery of Positron-C.D. Anderson and U.F. Hess C. Theory of energy production in the sun and stars-H.A. Bethe D. Synthesis of transuranic elements -Enrico Fermi transuranic elements

A. Tubelight

B. Transformer

C. Photodiode

D. LED

10. The pressure exerted on the ground by a man is greatest A. when he lies down in the ground

3. Which one of the following is not correct:

B. when he stands on the toes of one foot

A. Theory of evolution was propounded by Charles Darwin

C. when he stands with both foot flat on the ground D. all of the above yield the same pressure

B. The breaking apart of the nucleus of an atom is called fusion. C. ‘Dry ice’ is nothing but solid carbon dioxide. D. Telephone was invented by Graham Bell. 4. Among the following radiations, which has the highest energy? A. Visible

B. X-ray

C. Ultra-violet

D. Infra-red

5. Source of Energy from the Sun is A. Nuslear fission

B. Nuclear fusion

C. Photoelectric effect

D. Cherenkov effect

Physics 11. In amplitude modulation, signal frequency is 20 kHz and carrier frequency is 1 MHz. Modulated frequency can not have the frequency. A. 1000 KHz

B. 1020 KHz

C. 20 KHz

D. 980 KHz

12. There is a certain hypothetical one electron atom in which when electron make transition from orbit

6. The magnetic field inside a long straight solenoid carrying current A. Is the same at all points B. Is zero

Mock Test - 1

𝑛=𝑝

to 𝑛 = 1 photon of wavelength ( 𝑖𝑛𝐴̂ )

𝜆=

900𝑝3 𝑝3 +1

emits. (where

potential of the atom is

𝑝 > 1) The ionization

(ℎ𝑐 = 12420𝑒𝑉 − 𝐴)

|1

A. 26.4 Volt

B. 16.4 Volt

C. 13.8 Volt

D. 21.2 Volt

13. An electromagnetic wave has a frequency

17. One (10 W, 60 Volt) bulb is to be connected to 100 V A.C line. The required induction coil has self-inductance of value (f=50 Hz): A. 0.052 H

B. 2.42 H

𝑣 = × 1010 𝐻𝑧

C. 16.2 mH

D. 1.62 mH

It is propagating in space along positive 'x' direction. Then acceptable form of magnetic field associated with it is-

18. Two identical wires 'A' and 'B' each of length ' carry the same current

6 𝜋

A.

→

𝐵 = 𝐸0 √𝜇0 𝑧0 cos[(4 × 102 )𝑥 − (12 × 1010 )𝑡]𝑘̂ →

B.

𝐵= →

C. D.

𝐸0 √𝜇0 𝜀0

𝐵=

𝐸0 √𝜇0 𝜀0

cos[(2 × 102 )𝑥 − (6 × 1010 )𝑡]𝑘̂ 2)

10 )

cos[(10 𝑥 − (10

𝑡]𝑘̂

is bent to form a square of side 'a'. If 𝐵𝐴 and 𝐵𝐵 are the values of magnetic field at the centers of the semi-circle and square respectively, then the relation between and

𝐵𝐵 is-

A.

𝐵𝐴 = 32√2 𝐵𝐵

B.

𝐵𝐴 = 8√2 𝐵𝐵

C.

𝐵𝐴 = 16√2 𝐵𝐵

D.

𝐵𝐴 =

→

𝐵 = 𝐸0 √𝜇0 𝜀0 cos[(4 × 10−2 )𝑥 − (2 × 10−10 )𝑡]𝑘̂

14. In a young’s double slit experiment when a glass plate of refractive index 1.75 is introduced in the path of one of the light beam, the fringes are displaced by a distance ‘x’. When this plate is replaced by another plate of the same thickness, the shift of fringes is 2/3 x. The refractive index of second plate is— A. 1.33

B. 1.5

C. 1.85

D. 1.25

15. A convex lens of focal length 20 cm is placed at a distance 10 cm from a glass plate (μ = 3/2) of thickness 3 cm. An object is placed at a distance 40 cm in front of lens, the image of this object from the glass plate will be formed at a distance of— A. 19 cm from glass plate

B. 30 cm from glass plate.

C. 31 cm from glass plate.

D. 28 cm from glass plate.

16. In a series ‘LCR’ circuit resistance is 60Ω and it has resonant frequency 2000 radian/second. At resonance, the voltage across resistance and inductor is 30 volt and 20 volt respectively. Capacitance of capacitor in the RLC circuit is— A.

25𝜇𝐹

B.

50𝜇𝐹

C.

10𝜇𝐹

D.

12.5𝜇𝐹

Mock Test - 1

I. Wire A is bent into a semi-circle of radius 'R' and wire B

𝐵𝐴

𝜋2

𝜋2

𝜋2

𝜋2 4√2

𝐵𝐵

19. Two bulb of power rating 200W, 200V and 400W, 200V are connected in series across a battery of emf 100 volt. Current in each bulb is– A. 0.66 Amp.

B. 1.33 Amp.

C. 0.33 Amp.

D. 1.66 Amp.

20. A battery of ‘V’ volt is connected to a capacitor of capacitance ‘C’ µF. After steady state is reached, battery is disconnected and another battery of 2V is connected to the capacitor but with reversed polarity. Amount of heat produced after the connection of second battery is– A.

𝑐𝑣 2

B.

3/2𝑐𝑣 2

C.

2𝑐𝑣 2

D.

9/2𝑐𝑣 2

21. Assume that the nuclear binding energy per nucleon (B/A) versus mass number (A) is as shown in the figure. Use this plot to choose the correct answer(s) from the choices given below.

|2

A. Fusion of two nuclei with mass numbers lying in the range of 1 < A < 50 will release energy. B. Fusion of two nuclei with mass numbers lying in the range of 51 < A < 100 will release energy. C. Fission of a nucleus lying in the mass range of 100 < A < 200 will release energy when broken into two equal fragments. D. Fission of a nucleus lying in the mass range of 200 < A < 260 will release energy when broken into two equal fragments. 22. A long current carrying wire, carrying current I1 such that I1 is flowing out form the plane of paper is placed at O. A steady state current I2 is flowing in the loop ABCD

A.

(4 ,

B.

( 32 ,

C.

(4 ,

D.

(3 ,

𝐿 √2𝐿 ) 𝜋 𝐿

√3𝐿𝐿 ) 𝜋

3𝐿 √2𝐿 ) 𝜋 2𝐿 √3𝐿 ) 𝜋

24. A charged particle having a positive charge q approaches a grounded metallic neutral sphere of radius R with constant speed v, as shown in the figure. Which of the following alternatives is/are correct?

Choose the correct option(s).

A. As the charged particle draws nearer to the surface of the sphere, a current flows into the ground. B. As the charged particle draws nearer to the surface of the sphere, a current flows out of the ground into the sphere. C. As the particle draws nearer, the magnitude of current flowing in the connector joining the sphere shell to the ground increases.

A. Net force on the loop is zero. B. Net torque on the loop is zero. C. As seen from O, the loop rotates clockwise. D. As seen from O, the loop rotates anticlockwise.

D. As the particle draws nearer, the magnitude of current flowing in the connector joining the sphere shell to the ground decreases.

23. A ray of light is incident on a reflecting surface. The ray moves in horizontal direction and is reflected vertically after striking the surface. If the surface is

25. The potential energy function of a particle moving in

denoted by

𝑦=

2𝐿 𝜋 sin ( 𝑥) , 0 𝜋 𝐿

≤ 𝑥 ≤ 𝐿, what are

the coordinates of the \ point(s) where the ray is incident?

one dimension is

𝑈 = 𝑘 [𝑥 −

𝑥3 ], where 3𝑎2

𝑎 and 𝑘

are constants. Then, A. equilibrium is stable at x = a/2 B. equilibrium is stable at x = -a C. at stable equilibrium, PE = -2ka/3 D. equilibrium is stable at x = +a

Mock Test - 1

|3

length of the square orifice the ratio of rates of effusion of gas

Chemistry 26. 10 mL of 0.2 N HCl and 30 mL of 0.1 N HCl together exactly neutralizes 40 mL of solution of NaOH, which is also exactly neutralized by a solution in water of 0.61 g of an organic acid. The equivalent weight of the organic acid is A. 61

B. 91.5

C. 122

D. 183

𝐴 to that of gas 𝐵 is:

A. π√2

B. √(π/2)

C. 2 π

D. √(2/π)

30. A dark violet colour mixture in presence of dil.HCl changes to pale yellow solution. Mixture may contain : A.

𝑀𝑛𝑂4− , 𝐹𝑒 2+

B.

𝑀𝑛𝑂4− , 𝐶2 𝑂42−

C.

𝐼 − , 𝐶𝑟2 𝑂72−

D. None of these 31.

27. A monoelectronic species in energy level with energy X was provided with excess of energy so that it jumps to higher energy level with energy Y. If it can emit six wavelengths originated from all possible transition between these group levels, then which of the following relation is correct?

A.

( ‘n’ is the principal quantum number of energy level X) A.

𝑋

3

√𝑌 = 1 + 𝑛 𝑋

B.

𝑌

C.

𝑥

D.

𝑋

𝑌 𝑌

𝑛

=6

B.

= (𝑛 − 1)2 3

=1+𝑛

28. The compressibility factor for a real gas is expressed by

𝑧 = 1+

𝐵𝑃 The value of 𝑅𝑇

𝐵 at 500𝐾 and 600 bar

0.0169𝐿/𝑚𝑜𝑙 . Molar volume of the gas at 500𝐾 and 600𝑏𝑎𝑟 is- ( bar = 1𝑎𝑡𝑚) is

C.

(𝑅 = 0.083𝐿 − 𝑎𝑡𝑚/𝑚𝑜𝑙 − 𝐾) A. 01 L B.

9 × 10−5 𝐿

C.

8.62 × 10−2 𝐿

D.

D. 1.65 L

𝐴 and 𝐵 present separately in two vessels 𝑋 and 𝑌 at the same temperature and pressure with molecular weights 𝑀 and 2𝑀 respectively are 29. Two gases

effused out. The orifice in vessel 𝑋 is circular while that in 𝑌 is a square. If the radius of the circular orifice is equal that the Mock Test - 1

6.46 × 10−5 and 𝐾𝑠𝑝 for silver benzoate is 2.5 × 10−13 . 32. The ionization constant of benzoic acid is

How many times silver benzoate is more soluble in a

|4

buffer of pH 3.19 compared to its solubility in pure water? (Antilog of

C.

0.81 = 6.46)

A. 3.32

B. 0.31

C. 6.40

D. 2.70 D.

33. The equilibrium constant

𝐾𝑝 for the reaction

𝐴(𝑔) ⇌ 𝐵(𝑔) + 𝐶(𝑔) is 1 at 27 . and 4 at 47 . For the reaction calculate enthalpy change for the

𝐵(𝑔) + 𝐶(𝑔) ⇌ 𝐴(𝑔) (Given:

𝑅 = 2𝑐𝑎𝑙/𝑚𝑜𝑙 − 𝐾)

37. Which one of the following reactions will not result in the formation of carbon-carbon bond? A. Reimer-Tieman reaction

A.

13.31𝑘𝑐𝑎𝑙/𝑚𝑜𝑙

B. Friedel Craft's acylation

B.

14.31𝑘𝑐𝑎𝑙/𝑚𝑜𝑙

C. Wurtz reaction

C.

19.2𝑘𝑐𝑎𝑙/𝑚𝑜𝑙

D. Cannizzaro reaction

D.

55.63𝑘𝑐𝑎𝑙/𝑚𝑜𝑙

34. For the water gas reaction

𝐶(𝑠) + 𝐻2 𝑂(𝑔) ⇌

38. Observe the following sequence of reactions and select the option(s) that is/are true in this context.

𝐶𝑂(𝑔) + 𝐻2 𝑂(𝑔) the standard Gibbs free energy of reaction is

(𝑎𝑡1000𝐾)

−8.1𝑘𝐽/ mol. Calculate its equilibrium constant.

A. 1

B. 10

A. The reagent 'P' is cold alkaline solution of KMnO 4.

C. 1/e

D. 2.65

B. The reagent 'P' is hot and acidic solution of KMnO 4.

35. The thermal stability of alkaline earth metal carbonates 𝑀𝑔𝐶𝑂3 , 𝐶𝑎𝐶𝑂3 decreases as

𝐵𝑎𝐶𝑂3 and 𝑆𝑟𝐶𝑂3

A.

𝐶𝑎𝐶𝑂3 > 𝑆𝑟𝐶𝑂3 > 𝑀𝑔𝐶𝑂3 > 𝐵𝑎𝐶𝑂3

B.

𝐵𝑎𝐶𝑂3 > 𝑆𝑟𝐶𝑂3 > 𝑀𝑔𝐶𝑂3 > 𝐶𝑎𝐶𝑂3

C.

𝐵𝑎𝐶𝑂3 > 𝑆𝑟𝐶𝑂3 > 𝐶𝑎𝐶𝑂3 > 𝑀𝑔𝐶𝑂3

D.

𝑀𝑔𝐶𝑂3 > 𝐶𝑎𝐶𝑂3 > 𝑆𝑟𝐶𝑂3 > 𝐵𝑎𝐶𝑂3

C. The reagent 'Q' is an oxidising agent. D. The reagent 'S' is a mixture of NaOH and CaO. 39. Consider the hypothetical reaction at equilibrium.

𝐴(𝑔) + 3𝐵(𝑠) ⇔ 2𝐶(𝑔) + 𝐷(𝑠), + Heat In light of the Le-Chatelier's principle, and characteristics of equilibrium which of the following statements are correct.

36. Which of the following substances exhibit tautomerism ?

A. If the partial pressure of 'A' is doubled, the partial pressure of 'C' reduces to half its value and the equilibrium is restored.

A.

B. Kp is greater than Kc for the forward reaction.

B.

C. Increase in temperature will shift the equilibrium in favour of the backward reaction, but both Kp as well as Kc remain constant. D. Transferring the equilibrium reaction mixture to a container, which is double the volume, will shift the equilibrium towards the forward reaction. 40. A crystalline substance composed of Ba, Ti and O crystallises in a perovskite structure. This structure may

Mock Test - 1

|5

be described as a cubic lattice, with barium ion occupying corners of the unit cell and oxide ions occupying the face centres of the unit cell. Which of the following statements is/are true? A. Empirical formula of the substance is BaTiO3. B. Titanium can be described as occupying octahedral holes in the BaO lattice. C. Titanium occupies one-fourth of the total octahedral holes. D. There is no restriction for titanium to occupy any of the octahedral holes available to it.

Harmonic mean between same numbers such that 1 𝑞

1 𝑟

1 𝑝

+

5 3

+ = then the numbers are

A. 12, 1

B. 8, 2

C. 9, 1

D. 6, 4

45. The value of 21 𝐶1 A.

221

B.

220

C.

221 − 1

D.

220 − 1

+ 21 𝐶2 + ⋯ .21𝐶10 is −

46. Find the number of four digit numbers that are less than 2000 that can be formed with the digits 1,2,3,4 such that each digit can be repeated any number of times

Mathematics

A. 24

B. 25

C. 64

41. The number of values of 𝜆 for which the given system of equations has No solution

47. If

𝑥 + 2𝑦 + 3𝑧 = 5

is false, then truth value of

4𝑥 + 6𝑦 + 2𝜆𝑧 = 21

A. T, F

B. T, T

𝑥 + 𝜆𝑦 + 3𝑧 = 3

C. F, T

D. F, F

D. 65

𝑝 → (𝑝 ∧∼ 𝑞) 𝑝 and 𝑞 are respectively

A. 2

B. 0

48. The angles of elevation of top of a pole from two

C. 1

D. Infinite

points

𝐴 and 𝐵 on the horizontal line lying on opposite 30∘ and 60∘ If 𝐴𝐵 = 100𝑚 then height of the pole is side of the pole are observed to be

3 × 3 non-singular matrix and (𝐴 − 21)(𝐴 − 𝑎𝑙) = 0, where 𝐼 = 𝐼3 , and 𝑂 = 𝑂3 42. Let 'A' is any

If

𝐴 + 𝐴−1 = (5/2)𝐼 then 𝑎 is −

A. 3/2

B. 3

C. 1/2

43. The sum of the 'n' term of the series 1+2+3 13 +23 +33

A. B.

D. 1

1+

1+2 13 +23

+

+ ⋯ …is-

𝑛 (𝑛+1)2 𝑛2 𝑛+1

C. D.

2𝑛 𝑛+1 𝑛 2(𝑛+1)

𝑎, 𝑏, 𝑐 are the Arithmetic means between two numbers such that 𝑎 + 𝑏 + 𝑐 = 15 and 𝑝, 𝑞, 𝑟 be 44. Let

Mock Test - 1

A.

20√3𝑚

B.

15√3𝑚

C.

10√3𝑚

D.

25√3𝑚

49. If the mean and standard deviation of 10 observations are 20 and 2 respectively. The sum of squares of all the observations is— A. 2020

B. 2000

C. 4040

D. 4000

50. A.

∫

𝑥 2 −1 (𝑥 2 +1)(𝑥 4 +1)1/2 1 √2

sec−1 (

𝑥+

𝑑𝑥

1 𝑥

√2

)

|6

B.

1 √2

C.

D.

sec−1 (

tan−1 ( tan−1 (

1 𝑥

𝑥−

√2

𝑥+

C. y = 0

)

1 𝑥

√2

)

1 𝑥

𝑥−

√2

Biology

)

51. The equations of two ellipses are

𝑥2 +

𝑦2 𝑝2

D. y = -30x - 50

𝑥2 𝑝2

+ 𝑦 2 = 1 and

= 1, where 𝑝 is a parameter. The locus of the

56. Scientific names higher than genus are A. Capitalised but not printed distinctively, italicised, or underlined

points of intersection of both the ellipses is a set of curves comprising

B. Italicised or underlined

A. two straight lines

B. one straight line

D. Italicised or underlined with the first word capitalised

C. one circle

D. one parabola

C. Capitalised and italicised or underlined

57. ICBN stands for 52. In a triangle ΔPQR, cos(P - R) cos(Q) + cos(2Q) = 0

A. International Code of Botanical Nomenclature

Which of the following options is/are correct?

B. International Congress of Biological Names

A. sin2 P + sin2 R = 2 sin2Q

C. Indian Code of Botanical Nomenclature

B. p2, q2 and r2 are in AP.

D. Indian Congress of Biological Names

C. p2, q2 and r2 are in AP.

58. Which of the following is set of rules and recommendations used for animal classification?

D. sin P sin R = sin2 Q 53. The internal bisector of ∠A of a triangle ABC meets side BC at D. A line drawn through D perpendicular to AD intersects the side AC at E and side AB at F. If a, b and c represent the sides of ΔABC, then-

A. ICZN

B. ICBN

C. ICVN

D. ICNCP

A. AE is HM of b and c.

The term 'species' is the least (i)______. It was coined by (ii)______.

B. C.

𝐴𝐷 =

2𝑏𝑐 𝑏+𝑐 4𝑏𝑐

cos

𝐴

59. Choose the correct option to fill in the blanks:

2

𝐴

𝐸𝐹 = 𝑏+𝑐 sin 2

A. Grade, Carl Linnaeus

B. Taxon, Charles Darwin

C. Taxon, John Ray

D. Class, Robert Whittaker

D. ΔAEF is isosceles. 60. Which of these is NOT the name of a phylum? 54. Let P(x1, y1) and Q(x2, y2), y1 < 0, y2 < 0 be the end points of the latus rectum of the ellipse x 2 + 4y2 = 4. The equation(s) of parabolas with latus rectum PQ is/are-

A. Annelida

B. Echinodermata

C. Tetrapoda

D. Arthropoda

A.

𝑥 2 + 2√3𝑦 = 3 + √3

61. A scientific name contains information about the

B.

𝑥 2 − 2√3𝑦 = 3 + √3

A. Family and Species

B. Genus and Species

C. Phylum and Order

D. Class and Family

C.

𝑥 + 2√3𝑦 = 3 − √3

D.

𝑥 2 − 2√3𝑦 = 3 − √3

2

62. A tautonym in taxonomy is

55. Find the equation(s) of common tangent(s) to y = x 2 and y = - x2 + 4x - 4.

A. Non latinised name

A. y = 4(x - 1)

C. Common name used as scientific name

Mock Test - 1

B. y = -4(x - 1)

B. Same name for genus and species

|7

D. Unscientific explanation of a phenomenon

C. Gametes

D. Flowers

63. Regarding the assertion and reason; choose the correct option. Assertion [A]: The various grades of taxa are interlinked. Reason [R]: The relations and similarities increase in the descending order (with regard to the size of the taxon). A. Assertion [A] is True and Reason [R] is False. B. Reason [R] is True and Assertion [A] is False C. Assertion [A] is True and Reason [R] is True and is a correct explanation for [A]. D. Assertion [A] is True and Reason [R] is True but is an incorrect explanation to [A]. 64. Which class do birds belong to? A. Aves

B. Amphibia

C. Reptilia

D. Felis

65. Genera having similar characters are grouped together in this taxon A. Species

B. Order

C. Phyla

D. Family

66. In lower plants gametes are produced by ________ division. A. Mitotic

B. Meiotic

C. Equal

D. Binary

67. The giant Redwood tree (Sequoia sempervirens) is a / an _______________. A. Angiosperm

B. Gymnosperm

C. Tree fern

D. Gigantic alga

68. Water is not required for fertilization inA. Pteridophytes

B. Bryophytes

C. Algae

D. Gymnosperm

69. The endosperm in gymnosperms is: A. Haploid

B. Diploid

C. Triploid

D. Polyploid

70. Angiosperms are classified into monocots and dicots on the basis of the number of _________ . A. Lifecycles

Mock Test - 1

B. Cotyledons

|8

SMART ANSWERSHEET

Correct Q.

1

Ans.

Correct Q.

Ans.

Skipped

Skipped

Skipped

21.05 %

10.53 %

C

15

C

D

D

0.0 %

A

47

26.32 %

B

C 63.15 % 36.84 %

61

B

5.27 %

78.94 %

57.9 %

57.89 %

57.9 %

15.79 %

5.26 %

15.79 %

10.53 %

21.05 %

A

20

D

A

B, D

C

0.0 % 36

A, C, D

10.53 % B, D

C

A

A, C 57.89 %

21.05 %

5.26 % 52

A, B

78.94 %

52.63 %

57.9 %

10.53 %

5.26 %

15.79 %

21.05 %

B, C

C

A, B, C

10.53 % 27

A

78.94 %

78.94 %

C

5.26 % A 63.16 % 15.79 % 67

B

57.9 %

B, C

63.16 % 26.32 % 68

B 63.15 %

15.79 % A, C

26.32 % 69

A

57.89 %

10.53 %

57.89 %

26.32 % 56

57.89 %

63.16 %

57.89 % 55

C

D

0.0 % 54

57.89 % 42

31.58 %

5.26 % 42.11 %

C

31.58 % 28

A, B, C, D

10.53 % 41

52.63 %

10.53 % B

40 57.89 %

10.53 %

53 52.63 %

10.53 % 26

78.94 % A

B, D

78.95 %

10.53 % C

39

15.79 %

66

26.32 % 25

63.16 % 65

42.11 %

C

A

10.53 % 51

B, C

15.79 % 64

10.53 % 38

57.89 %

57.89 %

78.94 % A, C

C

10.53 % 50

42.1 %

15.79 % 24

10.53 % 63

26.31 % B

57.9 %

57.89 %

10.53 %

D

B

26.32 % 49

21.05 % 37

62 57.89 %

57.89 %

78.95 % 23

D

0.0 % 35

A, C

48 42.1 %

78.95 % 22

B

D

15.79 % 21

D

34 78.95 %

78.94 %

14

33

57.89 % 60

57.9 %

26.32 %

13

C

5.26 %

C

15.79 % C

5.26 % 46

57.9 %

19

63.15 % 59

10.53 %

B

A

57.89 %

78.95 %

21.05 %

12

D

21.05 %

31.58 %

11

A

26.32 % 58

15.79 % 45

5.26 % 32

57.89 %

57.89 %

21.05 %

26.31 %

10

C

26.31 %

0.0 % A

26.32 %

9

B

A

0.0 % 44

10.53 % 31

57 57.9 %

57.89 %

78.94 %

26.32 %

8

A

10.53 %

18

C

26.32 % 30

A

43 47.37 %

78.94 % 17

B

A

10.53 % 16

B

29 78.94 %

68.42 %

7

Ans.

21.05 %

21.05 %

6

Correct Q.

Skipped

47.37 %

5

Ans.

10.53 %

26.31 %

4

Correct Q.

Skipped

63.16 %

3

Ans.

21.05 % 26.32 % 2

Correct Q.

A

26.32 % 70

63.15 %

B 57.89 %

Performance Analysis Avg. Score (%) Toppers Score (%) Your Score

Mock Test - 1

9.57% 39.13%

|9

HINTS AND SOLUTIONS 1. Scintillation counter is an instrument for detecting ionizing radiation by using the excitation effect of incident radiation. Radiation on a scintillator material and detecting the resultant light pulses. Hence, The correct answer is C. 2. In 1935 Meson particles were discovered by Japanese physicist Hideki Yukawa. Positron, the antiparticle of the antimatter counterpart of the electron was discovered by C.D. Anderson and U.F. Hess in 1932. In 1939, Hans Bethe described the nuclear reactions that power the sun and other starts. In synthesis of transuranic elements, Glenn T. Scaborg played an important role instead of Enrico Fermi. The fact is that he attempted to prepare a transuranium element in 1934 in Rome but failed to do so.

Hence, The correct answer is A. 7. Heating coils are commonly made up of metal alloys which are a combination of two or more elements. The most commonly used metal alloy is “Nichrome”. Nichrome is an alloy of nickel (80%) and chromium (20%). Hence, The correct answer is A. 8. Mechanical energy is the ability of an object to do work. This energy is equal to the sum of kinetic and potential energy, it is always constant. [As the body is falling freely under gravity, the potential energy decreases continuously and kinetic energy increases continuously as all the conservative forces are doing work. So, total mechanical energy (PE+KE) of the body will be constant.] Hence, The correct answer is D.

Hence, The correct answer is D.

9. Transformer is based on the principle of mutual induction.

3. Option (b) is not correct. It is because the breaking apart of nucleus of an atom is called fission not fusion. Fission is a radioactive decay process in which the nucleus of an atom splits into smaller parts.

[A transformer consists of two electrically isolated coils and operates on Faraday's principal of “mutual induction”, in which an EMF is induced in the transformers secondary coil by the magnetic flux generated by the voltages and currents flowing in the primary coil winding.]

4. The correct order of the following different categories of radiations are -x-rays > ultraviolet > visible light > infrared. The electromagnetic spectrum of radio waves has the lowest energy while Gama rays consist of highest energy.

Hence, The correct answer is B.

Hence, The correct answer is B.

Hence, The correct answer is B.

5. The Sun produces energy by the nuclear fusion of hydrogen into helium in its core. Since there is a huge amount of hydrogen in the core, these atoms stick together and fuse into a helium atom. This energy is then radiated out from the core and moves across the solar system. This is the main source of energy for the sun and stars. Besides that the gravitational contraction in stars is also the source of their energy.

11. In amplitude modulation, if signal frequency is

Hence, The correct answer is B.

10. Pressure is normal force per unit area, therefore, for lesser value of area pressure is greatest.

𝜔𝑆 and 𝜔𝐶 then modulated frequency can be 𝜔𝐶 + 𝜔𝑆 , 𝜔𝐶 𝜔𝐶 − 𝜔𝑆 carrier frequency is

So, here modulated frequency can be

𝜔𝐶 + 𝜔𝑆 = 1020𝐾𝐻𝑧 𝜔𝐶 = 1000𝐾𝐻𝑧 and

𝜔𝐶 − 𝜔𝑆 = 980𝐾𝐻𝑧

6. The magnetic field inside a long straight solenoid carrying current is uniform at all points.

Hence, The correct answer is C.

[The magnetic field inside a long straight solenoid-carrying current is the same at all points. It is because the magnetic field in the solenoid is constant because the lines are completely parallel to each other. Then the remains stay same throughout. It however reduces at an end when the lines move apart.]

12. lonization energy

Mock Test - 1

𝛥𝐸 is energy require to send electron from 𝑛 = 1 to 𝑛→∞ So,

𝛥𝐸 = 𝐸𝑤 − 𝐸1 | 10

𝐸𝑝 − 𝐸1 =

Now,

900𝑝3 𝑝3 +1

𝜆𝑝 =

=

ℎ𝐶 𝜆𝑝

14. Shift in fringes

900

𝐷 𝑑

1

𝑥 = (𝜇1 − 1)𝑡 … …. (i)

(1+ 3) 𝑝

𝐷 𝑑

𝑝→∞

So, when

𝐸𝑤 − 𝐸1 =

ℎ𝑐 𝜆∞

2/3𝑥 = (𝜇2 − 1)𝑡 … …. (ii) ℎ𝑐

= lim

𝑝→∞ (

=

𝐷 𝑑

= (𝜇 − 1)𝑡

⇒ from (𝑖)/(𝑖𝑖)

900 1) 1+ 3 𝑝

3/2 = (

ℎ𝑐 900

𝜇1 −1

)

𝜇2 −1

⇒ 𝜇2 − 1 = 2/3(𝜇1 − 1)

𝐸𝑤 − 𝐸1 =

12420 𝑒𝑉1 900

⇒ 𝜇2 = 2/3(𝜇1 − 1) + 1 ⇒ 𝜇2 = 2/3(1.75 − 1) + 1

[𝛥𝐸 = 13.8𝑒𝑉] lonization potential

=

𝛥𝐸 charge

=

13.8𝑒𝑉 𝑒

⇒ 𝜇2 = 1.5 Hence, The correct answer is B.

= 13.8 Volt Hence, The correct answer is C.

15.

13. Electric and magnetic field both will be perpendicular to each other and both will also be perpendicular to direction of propagation of wave. Let →

𝐄 = 𝐄0 cos(𝑘𝑥 − 𝜔𝑡)𝐣̂ →

𝐵 = 𝐵0 cos(𝑘𝑥 − 𝜔𝑡)𝑘̂

Image distance "v' from lens (After refraction from lens

Now,

only)

1 𝑣

1 𝑢

1 𝑓

− =  (𝑓 = 20𝑐𝑚, 𝑢 = −40𝑐𝑚)

1

1

1

𝑣

𝑓

𝑢

⇒ = + 𝐶=

and

,𝐸 =

2𝜋 𝐸0 cos ( 𝑥 𝜆

− 2𝜋𝑣𝑡) 𝑗̂

6

𝑣 = × 1010 ⇒ 2𝜋𝑣 = 12 × 1010 𝜋

⇒𝜆= 2𝜋 𝜆

=

→

=

1 20

−1 40

+( )

𝐯 = 40𝑐𝑚

¯

So

1 𝑣

1 √𝜇0 𝜀0

𝐶 𝑣

=

3×108 6/𝜋×1010

2𝜋×6×1010 𝜋×3×108

1

= 𝑡 (1 − ) 𝜇

= 3 (1 −

= 4 × 102

𝐸 = 𝐸0 cos((4 ×

But there is a glass plate so shift in image (I I')

1 ) 3/2

= 1𝑐𝑚 So, distance of image from lens

102 )𝑥

− (12 × 1010 )𝑡)𝑗̂

= 40 + 1

→

𝐵 = 𝐸0 √𝜇0 𝜀0 cos((4 × 102 )𝑥 − (12 × 1010 )𝑡)𝑘̂ Hence, The correct answer is A. Mock Test - 1

= 41𝑐𝑚 Distance from glass plate

= 41 − 10 = 31𝑐𝑚

Hence, The correct answer is C.

| 11

For square 16. At resonance RLC circuit become pure resistive circuit and current in the circuit is given by

𝑖=

30 60

= 0.5𝐴𝑚𝑝𝑒𝑟𝑒

Also Potential drop across inductor

𝑉𝐿 = 𝑖𝑋𝐿

𝑎 = ℓ/4

𝑉𝐿 = 𝑖𝜔𝐿 𝑉𝐿 𝑖𝜔

⇒𝐿=

=

20 0.5×2000

= 20𝑚𝐻

1 𝜔𝐶 1 𝜔2 𝐿

⇒𝐶= =

𝐵𝐵 = 4 (

𝜇0 𝑖 ) 4𝜋(𝑎/2)

[sin45∘ + sin45∘ ]

At resonance,

𝜔𝐿 =

4𝑎 = ℓ

1 (2000)2 ×(20×10−3 )

𝜋(𝐶/4)

𝐵𝐵 =

8√2𝜇0 𝑖 𝜋ℓ

⇒

= 12.5𝜇𝐹

2𝜇0 𝑖

𝐵𝐵 =

𝐵1 𝐵𝐵

=

(√2)

𝜋2 32√2

Hence, The correct answer is A.

Hence, The correct answer is D. 19. Resistance of 17.

𝑉 = √𝑉𝑅2 + 𝑉𝐿2

1002 𝑖=

𝑉 𝑃

= =

102 10 60

+

𝑉𝐿2 𝑉𝐿2

𝑉2 𝑃1

=

(200)2 200

(200)2 𝑃2

=

(200)2 400

1𝑠𝑡 bulb 𝑅1 =

𝑅1 = 200 = 99.5𝑉 Resistance of II'nd bulb

𝑅2 =

= 6𝐴 ⇒ 𝑉𝐿 = 𝑖𝑋𝐿 ⇒

= 𝑖(2𝜋𝑉𝐿)

𝑅2 = 100 When bulbs are in series with 100 V battery.

99.5 = 6 × 2 × 3.14 × 50 × 𝐿 𝐿 = 0.052𝐻

Current,

𝑖=

(𝑉batlery )

(100) 100+200

Hence, The correct answer is A.

𝑖=

18. For semi circle

= 0.33𝐴𝑚𝑝

𝑅1 +𝑅2 1 3

= 𝐴

Hence, The correct answer is C.

ℓ = 𝜋𝑟 … (𝑖) 𝐵𝐴 = (

𝜇0 𝑖 ) 4𝑅

=

𝜇0 𝑖 4⋅𝐶/𝜋

=

𝜇0 𝜋 4!

Mock Test - 1

20.

| 12

Force on AB = zero, as the magnetic lines are parallel to the arm AB of the loop. dF = IdIB sinθ . or dF = IdlB sinθ = zero. Force on CD = zero, as the magnetic lines of I1 are antiparallel to the arm CD of the loop. When another battery is connected with reverse polarity, steady state change. Now, charge on capacitor 𝑞 ′

𝛥𝑞 = −2𝑞 − 𝑞 = −3𝑞

Change in charge Heat generation

= =

9𝑞2 2𝐶 9 2

=

= 2𝐶𝑉 = 2𝑞

=

(𝛥𝑞)2 2𝐶

9𝐶 2 𝑉 2 2𝐶

𝐶𝑉 2

Hence, The correct answer is D. 21. Let us examine the four alternatives separately. (1) for 1 < A < 50, on fusion mass number of the resulting nucleus will be less than 100.

dF = IdIBsinθ = IdIB sin 180° = zero. Force on BC is perpendicular to plane of paper outwards, according to Fleming's left hand rule. Force on CD is perpendicular to plane of paper inwards, according to Fleming's left hand rule. The force on BC and CD are equal and opposite. The net force on ABCD is zero. Torque on loop ABCD: The equal and opposite forces, on BC and AD, constitute a torque which, as seen from O, will rotate the loop in clockwise direction, along OO' as axis. Hence options A and C are the correct answers. 23. Consider the diagram for reflection at general point A shown below.

(2) For 51 < A < 100, on fusion mass number the resulting nucleus is between 100 and 200. B/A increases, energy will be released. (3) On fission for 100 < A < 200, the mass number for fission nuclei will be between 50 to 100. B/A decreases, no energy will be released. (4) On fission for 200 < A < 260, the mass number for fission nuclei will be between 100 to 130, B/A will increases, energy will be released. Hence options B and D are the correct answers.

Here, we have, 2i=90 degree I =45 degree So, dy/dx at A should be tan 45 degree. Therefore,

𝑦= 𝑑𝑦 𝑑𝑥

2𝐿 𝜋 sin ( 𝑥) 𝜋 𝐿

=

2𝐿 𝜋 cos ( 𝑥) 𝜋 𝐿

×

𝜋 𝐿

𝜋 𝐿

𝐼 = 2cos ( 𝑥) 𝜋𝑥 𝐿

=

22. The current carrying wire is placed at O. It carries a current I1, which flows out from the plane of paper. Field lines are circular around the wire, due to I1. Mock Test - 1

𝑥=

𝜋 3

𝐿 3

Therefore

𝑦=

2𝐿 𝜋 sin ( 𝜋 𝐿

𝐿 3

× ) | 13

= =

2𝐿 𝜋

×

𝑑2 𝑈

√3 2

|

𝑑𝑥 2 at 𝑥=−𝑎

√3𝐿 𝜋

2

=

Potential energy at

𝑈 = 𝑘 [−𝑎 +

𝐿 √3𝐿 Therefore, the point is ( , ). 3 𝜋

Another point, just opposite on the surface will also act same and its coordinates are

(

> 0, so stable

𝑎

𝑥 = −𝑎 is

𝑎3 ] 3𝑎2

=−

2𝑘𝑎 3

Hence, this is the required solution. Hence options B and C are the correct answers.

2𝐿 √3𝐿 , ) 3 𝜋

10𝑚𝑙 of 0.2𝑁𝐻𝐶𝑙 + 30𝑚𝑙 of 0.1𝑁𝐻𝐶𝑙 = 40𝑚𝑙 of

Hence options B and D are the correct answers.

26.

𝑁𝑎𝑂𝐻(0.61𝑔 organic acid ) 𝐻𝐶𝑙 = meq of 𝑁𝑎𝑂𝐻 = meq of organic acid

meq. of

10 × 0.2 + 30 × 0.1 =

24. The potential of the grounded sphere has to be zero, as the positive charge approaches the sphere, negative charge will be induced on the surface near to the charge ( bounded charge under the influence of the approaching positive charge so it cannot move). The other face will be positively charged. So, negative charge must flow from earth to the sphere to neutralize the positive charge. This negative charge is acquired by the sphere from the ground, hence current flows into the ground. As the charged particle comes nearer to the sphere, the potential of the sphere due to the positive charge increases and hence more electrons per unit time will flow from the ground to the sphere and thus magnitude of current increases. Hence, it is the correct answer. Hence options A and C are the correct answers.

25. Force corresponding to potential energy is,

−

𝑑𝑈 𝑑𝑥

= −𝑘 [1 −

For equilibrium,

1−

𝑥2 𝑎2

𝑥2 ] 𝑎2

𝐹 = 0. Therefore

𝑥 = ±𝑎

𝑑2 𝑈 𝑑𝑥 2

=−

160 5

= 122

Hence, The correct answer is C.

27. No. of different transition from

𝑛 = 𝑛1 to 𝑛 = 𝑛2 is

given by 𝑛 1𝐶𝑛2 So for 6 different wavelength, no. of different transition

=

6 𝑛

1 𝐶𝑛2

= 6 then 𝑛1 = 𝑛2 + 3

𝑛 is the principal quantum number of energy level corresponding to energy 𝑋 , the principal quantum number of energy level 𝑌 is (𝑛 + If

3) as it emit 6 wavelengths.

𝐹=

Now Energy of an orbit is directly proportional to square of the orbit no.(principlal quantum no.) So,

𝑥 𝑌

=

(𝑛+3)2 𝑛2

𝑧=1+

𝑋 𝑌

⇒√ =1+

2𝑥 𝑎2

0.0169×600 0.083×500

𝑍𝑅𝑇 𝑃

∴   𝑉𝑚 = or 𝑉𝑚

𝑑2 𝑈 | 𝑑𝑥 2 𝑎+𝑥=+𝑎 Mock Test - 1

𝐸=

28.

For stable equilibrium,

>0

0.61×1000 𝐸

× 1000

3 𝑛

Hence, The correct answer is A.

=0

𝑑2 𝑈 𝑑𝑥 2

5=

0.61 𝐸

=

= 1.244

1.244×0.083×500 600

= 8.6 × 10−2 𝐿

Hence, The correct answer is C. 2 𝑎

= − < 0, so unstable 29.

𝑟𝑥 𝑟𝑦

2𝑀 𝑀

=√

×

𝐴𝑥 𝐴𝑦

| 14

where 𝑟𝑥 𝑟𝑦

𝐴𝑥 and 𝐴𝑦 are the area of cross-section of orifice.

= √2 ×

Let the solubility of

[𝐴𝑔+ ]

𝜋𝑟 2 𝑙2

𝑟 = radius of a circular orifice ℓ = length of side of 𝑟𝑥 square orifice = √2 × 𝜋 𝑟𝑦

=𝑥

[𝐶6 𝐻5 𝐶𝑂𝑂𝐻] + [𝐶6 𝐻5 𝐶𝑂𝑂 − ] = 𝑥 10[𝐶6 𝐻5 𝐶𝑂𝑂− ] + [𝐶6 𝐻5 𝐶𝑂𝑂− ] = 𝑥 [𝐶6 𝐻5 𝐶𝑂𝑂 − ] =

Hence, The correct answer is A.

𝐶6 𝐻5 COOAg be 𝑥 mol /𝐿 Then,

𝑥 11

𝐾𝑠𝑝 [𝐴𝐺 + ][𝐶6 𝐻5 𝐶𝑂𝑂− ] 𝑀𝑛𝑂4− acts as strong oxidizing agent and itself is reduced to 𝑀𝑛(𝐼𝐼) ions. In the acidic medium, 𝐹𝑒 2+ acts as a reducing agent and itself is oxidized to 𝐹𝑒 3+ . Hence, the colour changes from dark violet to light yellow when dil 𝐻𝐶𝑙 is added to a mixture of 𝑀𝑛𝑂4− and 𝐹𝑒 2+ . 30. In acidic medium

𝑥 11

2.5 × 10−13 = 𝑥 ( ) 𝑥 = 1.66 × 10−6 𝑚𝑜𝑙/𝐿 Thus, the solubility of silver benzoate in a pH 3.19 solution is

1.66 × 10−6 𝑚𝑜𝑙/𝐿

𝐶6 𝐻5 𝐶𝑂𝑂𝐴𝑔 be 𝑥 ′ 𝑀 Then, [𝐴𝑔+ ] = 𝑥 ′ 𝑀 and [𝐶6 𝐻5 𝐶𝑂𝑂− ] = 𝑥 ′ 𝑀 Now, let the solubility of

Hence, The correct answer is A.

𝐾𝑠𝑝 = [𝐴𝑔+ ][𝐶6 𝐻5 𝐶𝑂𝑂− ]

31.

𝐾𝑠𝑝 = (𝑥 ′ )2 𝑥 ′ = √𝐾𝑠𝑝 = √2.5 × 10−13 = 5 × 10−7 𝑚𝑜𝑙/𝐿 ∴

𝑥 𝑥′

=

1.66×10−6 5×10−7

= 3.32

Hence, 𝐶6 𝐻5 COOAg is approximately 3.32 times more soluble in low pH solution. Hence, The correct answer is A.

33. The equilibrium constant

𝐾𝑝 for the reaction

𝐴(𝑔) ⇌ 𝐵(𝑔) + 𝐶(𝑔) 𝐾𝑝1 = 1 at 300𝑘; 𝐾𝑝2 = 4 at 320𝑘 4 1

log10 = 0.6 =

𝛥𝐻 1 [ 2.303𝑅 300

𝛥𝐻 2.303×2

×

−

1 ] 320

20 300×320

𝛥𝐻 = 13.31𝐾𝑐𝑎𝑙/𝑚𝑜𝑙𝑒 for

𝐵(𝑔) + 𝐶(𝑔) ⇌ 𝐴(𝑔)

𝛥𝐻 = −13.31𝐾 cal / mole.

Hence, The correct answer is B.

Hence, The correct answer is A. 32. since

𝑝𝐻 = 3.19 [𝐻3 𝑂+ ] = 6.46 × 10−4 𝑀

𝐶6 𝐻5 𝐶𝑂𝑂𝐻 + 𝐻2 𝑂 ⇋ 𝐶6 𝐻5 𝐾𝑎 =

+ 𝐻3 𝑂

[𝐻3 𝑂† ] 𝐾𝑎

=

𝐾 = 𝑎𝑛𝑡𝑖log (

𝛥𝐺 𝑜 ) ⋯ (𝑖) 2.303𝑅𝑇

Given that,

[𝐶6 𝐻5 𝐶𝑂𝑂𝐻]

=

34. We know that

𝛥𝐺 ∘ = −81. 𝑘𝐽/𝑚𝑜𝑙, 𝑅 = 8.314 × 10−3 𝑘𝐽𝐾 −1  𝑚𝑜𝑙 −1 , 𝑇 = 1000𝐾

[𝐶6 𝐻5 𝐶𝑂𝑂 − ][𝐻3 𝑂+ ]

𝐶6 𝐻5 𝐶𝑂𝑂𝐻 [𝐶6 𝐻5 𝐶𝑂𝑂−] Mock Test - 1

𝐶𝑂𝑂−

6.46×10−4 6.46×10−5

= 10 | 15

Substituting these values in eq. (i) we get −(−8.1) 𝑎𝑛𝑡𝑖log ( ) 2.303×8.314×10−3 ×1000

𝐾=

= 2.65

The Reimer Tiemann reaction is a chemical reaction used for the ortho-formylation of phenols with the simplest example being the conversion of phenol to salicylaldehyde.

Hence, The correct answer is D.

From the above examples it is clear that in Cannizzaro reaction C-C bond formation won't takes place.

35. As we move down the group alkaline earth metal carbonates require more heating to decompose, so carbonates become more thermally stable down the group.

Hence option D is correct.

𝐵𝑎𝐶𝑂3 > 𝑆𝑟𝐶𝑂3 > 𝐶𝑎𝐶𝑂3 > 𝑀𝑔𝐶𝑂3 Hence, The correct answer is C. 36. Tautomerism is spontaneous interconversion of two isomeric forms with different functional groups. The

𝐂 = 𝐎, 𝐂 = 𝐍 or 𝐍 = 𝐎 in the usual cases and an alpha 𝐻 atom. In prerequisite for this is the presence of the

case of keto enol tautomerism, the keto form is more stable. Enols can be formed by acid or base catalysis from the ketone and are extensively used in making 𝐶 − 𝐶 single bonds in organic synthesis, Compound in ( 1 ), ( 3 ) and (4) exhibit tautomerism as shown below.

38. Option (1) is incorrect as cold alkaline solution of KMnO4 would produce vicinal diols. Option (2) is correct as KMnO4 is stronger oxidising agent and in the acidic medium, it oxidises the alkene by breaking the carbon-carbon double bond and replacing it with two carbon-oxygen double bonds. Option (3) is correct as the process is ozonolysis by ozone, which is arrested at the aldehyde stage by adding zinc water. Option (4) is incorrect as decarboxylation, with soda lime, would produce ethane. Ethene is obtained by Kolbe's electrolysis. Hence, 'S' represents the process of electrolysis and not a reagent.

𝐴(𝑔) + 3𝐵(𝑠) ⇔ 2𝐶(𝑔) + 𝐷(𝑠), + Heat

39.

𝐾𝑝 =

(𝑃𝑐)2 𝑃𝐴

∴ 𝑃𝐶 ∝ √𝑃𝐴 Hence, if

𝑃𝐴 is doubled, the 𝑃𝐶 increases by a factor of

√2 Hence, option (1) is incorrect. Note: However, if the total pressure had been increased, the change would then favour the backward reaction and

Hence options A ,C and D are the correct answers.

𝑃𝐶 would then decrease. The reaction proceeds with an increase in the number of gaseous moles (𝛥𝜂9 > 0 >)𝐾𝑝 = 𝐾𝑐 (𝑅𝑇)2𝑛0 since,

tautomerism is not possible Benzoquinone is highly stable due to conjugation and does not exhibit tautomerism. 37. Cannizzaro reaction is a chemical reaction that involves the base-induced disproportionation of an aldehyde lacking a hydrogen atom in the alpha position. The oxidation product is a salt of a carboxylic acid and the reduction product is an alcohol.

𝛥𝜂0 = 2 − 1 = 1

∴ 𝐾𝑝 = 𝐾𝑐 (𝑅𝑇) ie.

𝐾𝑝 > 𝐾𝑐

Hence, option (2) is correct. The forward process being exothermic, increase in temperature will favour the backward reaction, but both

𝐾𝑝 as well as 𝐾𝐶 will decrease. Hence, option (3) is incorrect. At a constant temperature,

1 𝑉

𝑃 ∝ Hence, increase in

volume will decrease the pressure and the equilibrium will Mock Test - 1

| 16

shift in the direction of greater number of gaseous moles. Forward reaction will take place.

So, for no solution condition will be,

𝛥 = 0  and  𝛥1 ≠ 0

Hence, option (4) is correct. 40. 1) Number of Ba ions per unit cell = 1/8 x 8 = 1 Number of Ti ions per unit cell = 1 x 1 = 1 Number of O ions per unit cell = 1/2 x 6 = 3 Hence, the empirical formula of the substance should be BaTiO3.

⇒ 1(18 − 2𝜆2 ) − 2(12 − 2𝜆) + 3(4𝜆 − 6) = 0 ⇒   𝜆2 − 8𝜆 + 12 = 0 𝜆 = 2,6

Thus, (1) is correct. 2) Titanium ions can be assumed to occupy a hole in the lattice formed by barium and oxide ions. The titanium ions occupy the body centres of the face centred cubic and this is one of the octahedral holes in the fcc lattice.

⇒ 1(18 − 21𝜆) − 2(12 − 21) + 5(4𝜆 − 6) = 0

So, (2) is also true.

⇒   −𝜆, +6 = 0

3) The octahedral holes at the centres of unit cell constitute just one-fourth of all the octahedral holes in a face centred cubic lattice.

2=6 So,

Thus, (3) is also true.

Hence, The correct answer is C.

4) An octahedral hole at the centre of a unit cell, occupied by titanium ion, has 6 nearest neighbour oxide ions. The other octahedral holes located at the centres of the edges of unit cells have 6 nearest neighbours each, as in case with any octahedral hole, but 2 of the 6 neighbours are barium ions (at unit cell corners terminating the given edge) and 4 oxide ions. The proximity of two cations Ba 2+ and Ti4+ would be electrostatically unfavourable. So, titanium cannot occupy all of the octahedral holes and restrictions as stated above exist. So, (4) is wrong.

42.

𝜆 can have only one value

(𝐴 − 2𝐼)(𝐴 − 𝛼𝐼) = 0

⇒ 𝐴 ⋅ 𝐴 − (2 + 𝛼)𝐴 ⋅ 𝐼 + 2𝛼𝐼 ⋅ 𝐼 = 0 ⇒ 𝐴2 − (2 + 𝛼)𝐴𝐼 + 2𝛼𝐼 = 0 𝐴−1 both side:

Pre multiply by

⇒ 𝐴−1 ⋅ 𝐴 ⋅ 𝐴 − (2 + 𝛼)𝐴−1 ⋅ 𝐴 ⋅ 𝐼 + 2𝛼𝐴−1 ⋅ 𝐼=0 ⇒  𝐼 ⋅ 𝐴 − (2 + 𝛼)𝐼 + 2𝛼𝐴−1 𝐼 = 0 ⇒ 𝐴 − (2 + 𝛼)𝐼 + 2𝛼𝐴−1 𝐼 = 0

41. When

⇒ 𝐴 + 2𝛼𝐴−1 = (2 + 𝛼)𝐼 Now,

there can be two possibilities (i) No solution

5 2

𝐴 + 𝐴−1 = 𝐼 Compare it with equation

(ii) Infinite solution

2𝛼 = 1

But for Infinite solution

𝛼=

𝛥1 is the determinant obtained by replacing the third column of 𝛥 by the elements of the constant matrix.

Hence, The correct answer is C.

43.

1 2

𝑆 = 1+

𝑆=

1 13

+

1+2 13 +23

1+2 13 +23

+

+

1+2+3 13 +23 +33

1+2+3 13 +23 +33

+⋯

+⋯

Should also be zero Mock Test - 1

| 17

𝑛th term of the series

21 𝐶 2

= 21 𝐶19

21 𝐶 10

= 21 𝐶11

𝑇𝑛 =

1+2+3+⋯.𝑛3 13 +23 +⋯..𝑛3

𝑇𝑛 =

𝑛(𝑛+1) 2 𝑛(𝑛+1) −2 [ 2 ]

𝑇𝑛 =

2 𝑛(𝑛+1)

𝑇𝑛 =

2 𝑛

2 − 𝑛+1

𝑇𝑛 = −

2 𝑛

2 𝑛+1

= {221 − 2}

𝑇1 = 2 −

2 2

∵ (𝑛 𝐶0 + 𝑛 𝐶1 + 𝑛 𝐶2 + ⋯ … 𝑛 𝐶𝑛−1 + 𝑛 𝐶𝑛 = 2𝑛 )

2 2

2 3

= 220 − 1

2

2

3

4

𝑇2 = − 𝑇3 = −

𝑆𝑜, 21 𝐶1 + 21 𝐶2 + 21 𝐶3 + ⋯ .21 𝐶10 1 2 ⋯ .21 𝑐20 }

2𝐶 = {21 𝐶1 + 21 𝐶2 + ⋯ 10 + 21 𝐶11 + 21 𝐶12 +

1

= {(21𝐶0 + 21 𝐶1 + 21 𝐶2 + ⋯ . .21 𝐶20 +

2 2𝑐 𝐶 ) 21 1 2

Hence, The correct answer is D. 46. Given that the numbers should be less than 2000 , so the first digit can only be filled by 1 The other three digits can take any of the four given values in 4 ways each

_ _ _ 2 2 − 𝑛−1 𝑛

𝑇𝑛−1 = 2 𝑛

𝑇𝑛 = −

− (21𝐶0 + 21 𝐶21 )}

∴ Required number of 4 -digit numbers = 1 × 43 = 64

2 𝑛+1

Hence, The correct answer is C.

1 𝑛

47. Truth table:

𝑆𝑛 = 𝑇1 + 𝑇2 + ⋯ . 𝑇𝑛−1 + 𝑇𝑛 2 𝑛+1

𝑆𝑛 = 2 − 𝑆𝑛 =

2𝑛 𝑛+1

Hence, The correct answer is C.

44. Let numbers be

𝑥, 𝑦 3 (

𝑥+𝑦 ) 2

= 𝑎 + 𝑏 + 𝑐 = 15

is false,

𝑥 + 𝑦 = 10 3 1 ( 2 𝑥

+

1

1

10

𝑦

9

𝑥

1 ) 𝑦

+ =

=

So, when

1 1 ( + 𝑝 𝑞

1 + ) 𝑟

=

5 3

𝑝 → (𝑝 ∧∼ 𝑞)

𝑝 and 𝑞 both are true 'T

Hence, The correct answer is B.

𝑃𝑄 = ℎ m. And 𝑄𝐵 = 𝑑 be the distance between 𝐵 and foot of pole. tan60∘ = 48. Let the height of pole be

𝑥 = 9, 𝑦 = 1

ℎ 𝑑

Hence, The correct answer is C.

⇒𝑑=

45. 21 𝐶1

tan30∘ =

+ 21 𝐶2 + 21 𝐶3 + ⋯ .21 𝐶10

1

= {2(21 𝐶1 + 21 𝐶2 + 21 𝐶3 + ⋯ + 21 𝐶10 )} 2

Now, 21 𝑐 1 Mock Test - 1

=

21 𝐶 20

⇒−

ℎ √3

ℎ √3 ℎ 100−𝑑

+ 100 = √3ℎ

⇒ 100 = ℎ (√3 +

1 ) √3

| 18

ℎ2

⇒ ℎ = 25√3𝑚

1−𝑘 2

Hence, The correct answer is D.

49.

𝑛

ℎ2 − 𝑘 2 − (ℎ4 − 𝑘 4 ) = 0 (ℎ2 − 𝑘 2 ) − (ℎ2 − 𝑘 2 )(ℎ2 + 𝑘 2 ) = 0

10

⇒ ∑10 𝑖=1 𝑥𝑖 = 200

(ℎ2 − 𝑘 2 )[1 − (ℎ2 + 𝑘 2 )] = 0

(𝑆 tandard deviation )2 = ⇒ (2)2 + 400 = ⇒ 4040 =

2 ∑𝑛 𝑖=1(𝑥𝑖 )

𝑛

− (𝑥)2

𝑥 = ±𝑦 or 𝑥 2 + 𝑦 2 = 1

2 ∑10 𝑖=1 𝑥𝑖

𝑥 = ±𝑦 represents two straight lines 𝑥 2 + 𝑦 2 = 1 represents a circle

Hence, The correct answer is C.

𝐼=∫

𝐼=∫

𝑥 2 −1 (𝑥 2 +1)(𝑥 4 +1)1/2

1

Hence options A and C are the correct answers.

𝑑𝑥

52. Consider the expression.

(𝑥+ )(𝑥 2 + 2 ) 𝑥 𝑥

1/2 𝑑𝑥

cos(𝑃 − 𝑅)cos⌈180∘ − (𝑃 + 𝑅)⌉ + 1 − 2sin2 𝑄 = 0

1

1− 2 𝑥

𝐼=∫

𝑑𝑥

1 2

1

−cos(𝑃 − 𝑅)cos(𝑃 + 𝑅) + 1 − 2sin2 𝑄 = 0

(𝑥+𝑥)√(𝑥+𝑥) −2

Let

−cos2 𝑃 + sin2 𝑅 + 1 − 2sin2 𝑄 = 0

1 𝑥

𝑥+ =𝑡

−(1 − sin2 𝑃) + sin2 𝑅 + 1 − 2sin2 𝑄 = 0

1 𝑥

 = 𝑑 (𝑥 + ) = 𝑑𝑡  = (1 − 𝐼=∫

1 𝑥2

sin2 𝑃 + sin2 𝑅 = 2sin2 𝑄

) 𝑑𝑥 = 𝑑𝑡

𝑑𝑡 𝑡√𝑡 2 −2

We know that sin𝑃 𝑝

𝑑𝑡

=∫

=

sin𝑄 𝑞

𝑡 √𝑡 2 −(√2)2

Therefore

= =

1 √2 1 √2

𝑡

𝑑𝑥

√2

𝑥√𝑥 2 −𝑎2

sec−1 ( ) {∵ ∫

1

−1 𝑥

2

2

= sec

)

𝑥+𝑥 √2

)

=𝑘

sin𝑃 = 𝑝𝑘,sin𝑄 = 𝑞𝑘,sin𝑅 = 𝑟𝑘

𝑝 2 𝑘 2 + 𝑟 2 𝑘 2 = 2𝑞 2 𝑘 2

𝑝 2 , 𝑞 2 , 𝑟 2 are in 𝐴𝑃

51. Let the point of intersections is

(ℎ, 𝑘). Therefore

ℎ2 𝑝2

ℎ2 𝑝2

+ 𝑘 2 = 1 and ℎ2 +

Mock Test - 1

sin𝑅 𝑟

Therefore

Hence, The correct answer is A.

ℎ2 1−𝑘 2

=

𝑝 2 + 𝑟 2 = 2𝑞 2

1

sec−1 (

1 − ℎ2 =

cos(𝐏 − 𝐑)cos(𝑄) +

cos(2𝑄) = 0

1−1/𝑥 2 1

ℎ2 = 𝑘 2 or ℎ2 + 𝑘 2 = 1 𝑥 2 = 𝑦 2 or 𝑥 2 + 𝑦 2 = 1

1 2 [∑10 𝑖=1(𝑥𝑖 ) ] 10

2 ⇒ ∑10 𝑖=1 𝑥𝑖 = 4040

50.

1−ℎ2

ℎ2 − ℎ4 = 𝑘 2 − 𝑘 4

∑10 𝑖=1 𝑥1

⇒ 20 =

𝑘2

ℎ2 (1 − ℎ2 ) = 𝑘 2 (1 − 𝑘 2 )

∑𝑛 𝑖=1 𝑥𝑖

𝑀𝑒𝑎𝑛𝑥 =

=

𝑘2 𝑝2

=1

Hence options A and B are the correct answers.

= 1 − 𝑘 2 and

𝑘2 𝑝2

= 𝑝 2 and 𝑝 2 =

𝑘2 1−ℎ2

| 19

Co-ordinates of mid-point of

1

𝑃𝑄 are 𝐑 ≡ (0, − ) 2

𝑃𝑄 = 2√3 = length of latus rectum. ⇒ Two parabolas are possible whose vertices are (0, −

√3 2

1 2

− ) and (0,

√3 2

1 2

− )

Hence, the equations of the parabolas are

𝑥 2 − 2√3𝑦 =

3 + √3 and 𝑥 2 + 2√3𝑦 = 3 − √3

53. Given, internal angle bisector of A meets BC at D. we know

𝐴𝐷 =

length of AD is given by,

2𝑏𝑐 𝐴 cos Given, 𝑏+𝑐 2

△ ADE is right angled triangle with right angle at 𝐷 . ∠𝐷 = 90∘ ,  ∠𝐴𝐸𝐷 = 90 − Using sine rule in 𝐴𝐷 sin∠𝐴𝐸𝐷 𝐴𝐷 𝐴

cos 2

=

2𝑘

=

𝑚𝑥 −

𝐷𝐸

𝑚2 4

= −𝑥 2 + 4𝑥 − 4

⇒ 𝑥 2 + 𝑥(𝑚 − 4) + 4 −

𝐴

sin 2

𝐴

cos 2

=

4

=0

Now,

(𝑚 − 4)2 − (16 − 𝑚2 ) = 0

⇒ 2𝑚(𝑚 − 4) = 0

2𝑏𝑐 𝑏+𝑐

⇒ 𝐦 = 0,4

∠𝐷𝐴𝐸 = ∠𝐷𝐴𝐹 = ∠

𝐴 2

∴ ∠𝐴𝐸𝐷 = ∠𝐴𝐹𝐷 = 90 − ∠

𝐸𝐹 = 2𝐷𝐸 = 2𝐴𝐸sin =

𝑦 = 0 and 𝑦 = 4(𝑥 − 1) are the required tangents. 𝑦 = 4(𝑥 − 1), 𝑦 = 0

Correct answers:

Hence options A and C are the correct answers. 𝐴 2

△ AEF is isosceles triangle.

2𝑏𝑐 𝐴 sin 𝑏+𝑐 2

𝑚2

𝐷=0

𝑆𝑂, 𝐴𝐸 is harmonic mean of 𝑏 and 𝑐 . Consider △ ADE and △ ADF , we know that ∠𝐴𝐷𝐸 = ∠𝐴𝐷𝐹 = 90∘

𝐸𝐹 = 2 ×

𝑦 = −𝑥 2 + 4𝑥 − 4

𝐴

sin∠ 2

𝑦 = 𝑥2

𝑚2 4

𝐷𝐸

𝐴 cos 2

Hence

𝑦 = 𝑚𝑥 − Putting in

Substitute the value of AD in above equation, we get

𝐴𝐸 = 𝑏+𝑐

55. The equation of tangent to

△ ADE, we get

𝐴𝐸 sin∠𝐴𝐷𝐸

= 𝐴𝐸 =

𝐴 2

Hence options B and C are the correct answers.

𝐴 2 4𝑏𝑐 𝐴 sin 𝑏+𝑐 2

56. The scientific names of the organisms are the binomial nomenclature with genus and species names written in italics. The genus name starts with a capital and later all the letters are in small. The other names of the ranks in the hierarchical classification start with capitals but are not written in italics, or underlined or printed differently. Thus, the correct answer is option A.

Hence options A, B, C and D are the correct answers. 𝑥2 54. 4

𝑦2 + 1

=1

𝑏 2 = 𝑎2 (1 − 𝑒 2 ) ⇒𝑒=

√3 2 1 2

1 2

⇒ 𝑃 (√3, − ) and 𝑄 (−√3, − ) (given 𝑦1 and 𝑦2 less than

Mock Test - 1

0)

57. The International Code of Botanical Nomenclature (ICBN) is a set of rules regarding the botanical names given to plants. These rules are not the same as those for zoological nomenclature. The order of taxa (in descending order) for plants is thus: kingdom, division, class, series, order, family, tribe, genus, species. There are also specific endings for the names of certain taxa. For instance, the names of divisions would end with "phyta", classes with "ae", and families with "aceae". Hence, The correct answer is A.

| 20

58. The International Code of Zoological Nomenclature (ICZN) is the set of rules and recommendations used in animal classification. The expansions of the remaining options are as follows: ICBN - International Code of Botanical Nomenclature ICVN - International Code of Viral Nomenclature

64. Amphibia, Reptilia, Aves and Mammalia are all classes under superclass Tetrapoda, which further, is classified under subphylum Vertebrata. Birds are grouped under the class Aves. Examples for Amphibia, Reptilia and Mammalia are frog, snake, and bear respectively. Felis is a genus of cats.

ICNCP - International Code of Nomenclature for Cultivated Plants

Hence, The correct answer is A.

Hence, The correct answer is A.

65. A group of genera having similar characters comprise a family. In biological classification, family is one of the eight major/obligate taxonomic ranks; it is classified between order and genus. A family may be divided into subfamilies, which are intermediate ranks above the rank of genus. In vernacular usage, a family may be named after one of its common members. e.g., Blattidae - The family of common household cockroaches (Blatta is a commonly seen genus of cockroach).

59. The term species is the lowest taxonomic category or the least inclusive taxon. The term was coined by John Ray in his 1686 work titled "History of plants”. Hence, The correct answer is C. 60. Annelida, Arthropoda and Echinodermata are invertebrate phyla in the Kingdom Animalia. Tetrapoda is a superclass of jawed vertebrates (which come under phylum Chordata). Hence, The correct answer is C. 61. A scientific name contains information about its genus and species. The scientific name of an organism is used to identify it anywhere in the world. Under binomial system of nomenclature, each organism has scientific name consisting of two parts. First part is genus and second part is the species. For example- the scientific name of mango is Mangifera indica, scientific name of house fly is Musca domestica.

Hence, The correct answer is D. 66. In lower plants, the gametophyte phase is dominant. The plant body is haploid and the gametes are formed by mitotic division. Fusion of gametes forms the diploid sporophytic zygote which then undergoes meiosis to form the gametophytes. Hence option A is the correct answer. 67. Sequoia sempervirens is a gymnosperm. It is the sole living species of genus Sequoia. Its common names include coast redwood, California redwood. It is an evergreen, longliving monoecious tree. Hence option B is the correct answer.

Hence, The correct answer is B. 62. A tautonym is a binomial name in which the genus and the species are given the same name. For example, there exists a species of gorilla whose scientific name is Gorilla gorilla. Iguana iguana is another example of a tautonym.

68. All cryptogams (spore - bearing plants) i.e., thallophytes, bryophytes and pteridophytes require water for their fertilisation process. Phanerogams (seed - bearing plants) i.e. gymnosperms and angiosperms on the other hand, do not require water for their fertilisation process.

Hence, The correct answer is B.

Hence, The correct answer is B.

63. Taxa are arranged in a hierarchical manner from a small group of very closely related organisms to a broader group with more dissimilarities. As we go in the ascending order the number of common characteristics decreases and the grade increases and vice-versa, that is in the descending order, the number of common characteristics increases and the grade decreases. For example, any genus (lower grade of classification) has members with more similarities as compared to a family (a higher grade of classification) which has members with fewer similarities among them.

69. In gymnosperms, inside the nucellus, one cell differentiated into megaspore mother cell. It undergoes reduction division (meiosis) to form a linear tetrad of four haploid megaspores. Usually, the upper 3 megaspores towards micropyle degenerate while the lowermost functional megaspore (embryo sac cell) undergoes free nuclear division followed by wall formation to form a cellular female gametophyte or endosperm. Hence, the endosperm in gymnosperms is haploid as it is formed before fertilisation, from a single megaspore.

Hence, The correct answer is C.

Hence, The correct answer is A.

Mock Test - 1

| 21

70. Angiosperms, or flowering plants, are the most abundant and diverse plants on earth. Angiosperms are classified into two groups, namely monocots and dicots based on the number of cotyledons. Monocots are the group having single cotyledon, dicots are plants having two cotyledons. Hence, The correct answer is B.

Mock Test - 1

| 22

7. In the following question, select the odd word from the given alternatives.

MOCK TEST - 2 General Ability 1. Select the related word/letter/number from the given alternatives. FR: UI :: ? A. MS: VP

B. BH: KF

C. LR: UN

D. WD: ZG

A. Kolkata

B. Mumbai

C. Chandigarh

D. Chennai

8. Direction: Select the one which is different from the other three responses. A. Floor

B. Pillar

C. House

D. Roof

9. In the following question, select the odd number from the given alternatives.

2. Direction: In the following question, select the related word pair from the given alternatives. Ecstasy : Gloom : : ?

A. 7324

B. 9324

C. 5316

D. 3716

A. Congratulations : Occasion

10. In the following question, select the odd number from the given alternatives.

B. Diligent : Successful

A. 24

B. 12

C. 39

D. 36

C. Measure : Scale D. Humiliation : Exaltation 3. In the following question, select the related word pair from the given alternatives.

Physics

Romio & Juliet : William Shakespeare :: Gulliver Travels : ? A. Jim Corbett

B. Jonathan Swift

C. William Blake

D. Charles Dickens

4. ESTABLISHED is related to ‘RGDCROYFTUB ’ in the same way as FUNCTIONING is related to ? A. MMHFGORLODG C. MHMFLRGDOVG

B. FMHMGRLVDOG D. MHMFLRGGVOD

11. A boat capable of a speed v in still water wants to cross a river of width d. The speed of the water current increases linearly from zero at either bank to a maximum of u at the middle of the river. When the boat is rowed at right angles to the bank, its downstream drift is A.

2𝑢𝑑

B.

𝑢𝑑

C.

𝑣𝑑

D.

𝑢𝑑

𝑣 𝑣

5. In the following question, select the related number from the given alternatives.

𝑢 2𝑣

20 : 61 :: 35 : ? A. 108

B. 116

C. 106

D. 110

12. The escape velocity of an atmospheric particle which

1000𝑘𝑚 above the earth's surface, is (radius of the earth is 6400𝑘𝑚 and 𝑔 = 9.8𝑚/𝑠 2 ) is

6. In the following question, select the related number from the given alternatives. 45 : 36 : : 63 : ? A. 71

Mock Test - 2

B. 54

C. 61

D. 64

A.

6.5𝑘𝑚/𝑆

B.

8𝑘𝑚/𝑆

C.

10.4𝑘𝑚/𝑆

D.

11.2𝑘𝑚/𝑆 | 23

13. Four identical hollow cylindrical columns of mild steel support a big structure of mass 50,000 kg. The inner and outer radii of each column are 30 and 60cm respectively. Assuming the load distribution to be uniform, calculate the compressional strain of each column. A.

7.22 × 10−7

B.

7.88 × 10−7

C.

7.22 × 10−8

D.

5.22 × 10−7

perpendicular to the plane of loop and directed into the paper. The loop is connected to the network of resistances, each of value 3𝛺 . The resistance of the lead wire is negligible. The speed of the loop so as to have a steady current of

1𝑚𝐴 in the loop is

A.

2𝑚𝑠 −1

𝑆1 and 𝑆2 , emitting waves of equal wavelength 20.0𝑐𝑚, are placed with a separation of 20.0𝑐𝑚 between them. 𝐴 detector can be moved on a line parallel to 𝑆1 𝑆2 and at a distance of 20.0𝑐𝑚

B.

2𝑐𝑚𝑠 −1

C.

10𝑚𝑠 −1

D.

20𝑚𝑠 −1

from it. Initially, the detector is equidistant from the two sources. Assuming that the waves emitted by the sources are in phase, find the minimum distance through which the detector should be shifted to detect a minimum of sound.

18. The walls of a closed cubical box of edge

14. Two sources of sound,

A. 24

B. 18

C. 12

50𝑐𝑚 are

1𝑚𝑚 and coefficient of 4 × 10−4 𝑐𝑎𝑙/𝑠/𝑐𝑚/∘ 𝐶. The interior of the box is maintained at 100 ∘ 𝐶 above the made of a material of thickness thermal conductivity

outside temperature by a heater placed inside the box

D. 8

15. A reversible heat engine converts one-fourth of heat input into work. When the temperature of the sink is reduced by 150∘ , its efficiency is doubled. The

and connected across heater is

400𝑉 d.c. The resistance of the

A. 4.07 Ω

B. 5.83 Ω

C. 6.35 Ω

D. 7.54 Ω

temperature of the source is: A. 4500

B. 7500

C. 7000

D. 6000

19. In nuclear reaction, energy released per fission is 200 MeV. When uranium 235 is used as nuclear fuel in a reactor having a power level of 1 MW, the amount of fuel needed in 30 days will be

16. A long cylindrical conductor of cross-sectional area A is made of a material whose resistivity depends only on the distance r from the axis of the conductor as

𝑝=

𝛼 𝑟2

where 𝛼 is a constant. The resistance per unit length of such a conductor is A.

2𝜋𝐴2

B.

2𝜋𝑎

C.

𝜋𝑎 2

𝛼 𝐴2 𝐴

A. 48 g

B. 42 g

C. 38 g

D. 32 g

20. A nonconducting insulating ring of mass

𝑚 and

radius 𝑅 , having charge Q uniformly distributed over its circumference is hanging by an insulated thread with the help of a small smooth ring (not rigidly fixed with bigger ring). A time varying magnetic field

𝐵 = 𝐵0 sinwt is

𝑡 = 0 and the ring is released at the same time. The induced EMF in loop at time 𝑡 = 2𝜋/𝜔 switched on at is:

D.

𝜋𝐴2 𝛼

10𝑐𝑚 and resistance 1𝛺 is moved with constant velocity 𝑣0 in a uniform magnetic field of induction 𝐵 = 2𝑊𝑏𝑚−2 , as 17. A square metal wire loop of side

A. 0 B.

𝐵0 𝑇𝜔𝑅2 /2

C.

𝐵0 𝜋𝜔𝑅2

D.

−𝐵0 𝑇𝜔𝑅2 /2

shown in Figure. The magnetic field lines are Mock Test - 2

| 24

21. Directions: The following question has four choices, out of which ONE or MORE can be correct. A particle of charge `q` and mass `m` moves rectilinearly under the action of an electric field E= α - βX . Here, and are positive constants, and `x` is the distance from the point where the particle was initially at rest. Chose the correct option(s).

A. current passing through 2Ω is 2 A

A. The motion of the particle is oscillatory.

B. current passing through in 3Ω is 4 A

B. The amplitude of the particle`s motion is α/β

C. current in wire DE is zero

C. The mean position of the particle is at x = α/β

D. potential of point A is 10 V

D. The maximum acceleration of the particle is (qα)/m 22. A single velocity time graph can be drawn forA. 1 dimension only.

B. 2 dimensions.

C. 3-dimensions.

D. 3-dimensions.

23. Directions: The following question has four choices, out of which ONE or MORE can be correct. Which of the following statements is/are correct? A. If the electric field due to a point charge varies as r -25 instead of r-2, then Gauss law will still be valid. B. Gauss law can be used to calculate the field distribution around an electric dipole. C. If the electric field between two point charges is zero somewhere, then the signs of the two charges are the same. D. The work done by the external force in moving a unit positive charge from point A at potential VA to point B at potential VB is (VB - VA). 24. The network shown in figure is a part of the circuit. (The battery has negligible resistance.)At a certain instant the current I = 5 A and is decreasing at a rate of 10 ³ As⁻¹.What is the potential difference between points B and A?choose the correct option from the given options.

A. 5V

B. 10V

C. 15V

Chemistry 300𝐾 for reaction 𝐶2 𝐻4 (𝑔) + 3𝑂2 (𝑔) → 2𝐶𝑂2 (𝑔) + 2𝐻2 𝑂0; 𝛥𝐻 = −336.2𝑘𝑐𝑎𝑙. The approximate value of 𝛥𝑈 at 300𝐾 for the same reaction (𝑅 = 2 cal degree- 1𝑚𝑜𝑙 −1 ) 26. At temperature

will be A.

−320.0𝑘𝑐𝑎𝑙

B.

−335.0𝑘𝑐𝑎𝑙

C.

−337.2𝑘𝑐𝑎𝑙

D.

−353.0𝑘𝑐𝑎𝑙

𝑃𝑂43− , the formal charge on each oxygen atom and the 𝑃 − 𝑂 bond order respectively are: 27. In

A. − 0.75, 0.0

B. − 0.75, 1.0

C. − 0.75, 1.25

D. − 3, 1.25

28. Products A and B formed in the following reactions are respectively

D. 0V

25. Directions: The following question has four choices, out of which ONE or MORE can be correct. In the given circuit, the

Mock Test - 2

| 25

31. The compound which gives positive test with Tollen’s reagent but does not give a red precipitate with Fehling’s solution is

A .

A. HCHO B. CH3CHO C. fructose D. C6H5CHO

B.

32. Consider the reaction, 2𝐴 + 𝐵 → Products, When concentration of B alone was doubled, the rate did not change. When the concentration of A alone was doubled, the rate increased by two times. The unit of rate constant for this reaction is: C.

A.

𝑠 −1

B.

𝐿𝑚𝑜𝑙 −1 𝑠 −1

C.

𝑚𝑜𝑙𝐿−1 𝑠 −1

D.

𝑚𝑜𝑙 −2 𝐿5𝑠 −1

33. Which of the following order is wrong ?

D .

A.

𝑁𝐻3 < 𝑃𝐻3 < 𝐴𝑠𝐻3 : acidic nature

B.

𝐿𝑖 + < 𝑁𝑎+ < 𝐾 + < 𝐶𝑠 + − lonic radius

C.

𝐴𝑙2 𝑂3 < 𝑀𝑔𝑂 < 𝑁𝑎2 𝑂 < 𝐾2 𝑂 − Basic nature

D.

𝐿𝑖 < 𝐵𝑒 < 𝐵 < 𝐶 - lonisation energy 𝑦

34. A mixture of gaseous nitrogen and gaseous hydrogen attains equilibrium with gaseous Ammonia according to the reaction.

1 3 𝑁2(𝑔) + 𝐻2(𝑔) ⇌ 𝑁𝐻3(𝑔) 2 2

29. 0.400𝑔 of impure 𝐶𝑎𝑆𝑂4 (Molar mass = 136 ) solution when treated with excess of barium chloride

0.617𝑔 of anhydrous 𝐵𝑎𝑆𝑂4 (𝑀𝑜𝑙𝑎𝑟 mass = 233). The percentage of 𝐶𝑎𝑆𝑂4 present in solution, gave the sample is A. 80.0%

B. 90.0%

C. 95.0%

D. 75.0%

It is found that at equilibrium when temperature is

673𝐾 , atmospheric pressure is 100, moles of gaseous nitrogen and gaseous hydrogen are in ratio 1: 3 and mole percent of gaseous ammonia is 24. The equilibrium constant Kp for the reaction at the above condition isA. 0.0128 atm-1

B. 0.1876 atm-1

30. Which one of the following is wrong?

C. 0.0186 atm-1

D. 0.558 atm-1

A. Among the inert gases xenon only forms compounds with oxygen and fluorine

35. The ionization energy of the ground state of hydrogen

B. AgI is more ionic than KI C. Bi exhibit + 5 oxidation state D. Boiling point of HF is high and it is a weak acid due to hydrogen bonding

Mock Test - 2

atom is 2.18 × 10−18 𝐽. The energy of an electron in its second orbit would be: A.

−1.09 × 10−18 𝐽

B.

−2.18 × 10−18 𝐽

| 26

C.

−4.36 × 10−18 𝐽

D.

−5.45 × 10−19 𝐽

36. Directions: The following question has four choices, out of which ONE or MORE are correct. Consider the following graph:

39. Directions: The following question has four choices, out of which ONE or MORE are correct. When phenol is treated with CHCl3 and NaOH, followed by acidification, salicylaldehyde is obtained. Which of the following species is/are involved in the above mentioned reaction as intermediate(s)? A.

B.

From this graph, it is clear that A. cis-2-butene is more stable than 1-butene by 2.7 kcal

C.

B. trans-2-butene is more stable than 1-butene by 2.7 kcal C. trans-2-butene is more stable than cis-2-butene by 11 kJ D. trans-2-butene is more stable than 1-butene by 11 kJ

D.

37. Directions: The following question has four choices, out of which ONE or MORE are correct. An ionic compound PQ has the radius ratio of the cation and anion as expressed below: 𝑟+ 𝑟

3 2

=√ −1

Which of the following statements is correct about the ionic crystal? A. The coordination number of the cation is 4. B. The anions touch along the face diagonal.

40. Directions: The following question has four choices, out of which ONE or MORE are correct. Which of the following carbides can give hydrocarbon on reaction with water? A. SiC

B. CaC2

C. Be2C

D. Al4C3

C. The coordination number of the anion is 8. D. There is one cation on each body diagonal.

Mathematics

38. Directions: The following question has four choices, out of which ONE or MORE are correct. Which of the following oxides can act both as a reducing agent as well as an oxidising agent? A. SO2 B. MnO2

𝑥 3 − 3𝑥 + [𝑎] = 0, will have three real and distinct roots if − 41. The equation

(where [] denotes the greatest integer function)

C. CrO

A.

𝑎 ∈ (−∞, 2)

D. Al2O3

B.

𝑎 ∈ (0,2)

C.

𝑎 ∈ (−∞, −2) ∪ (0, ∞)

Mock Test - 2

| 27

D.

𝑎 ∈ [−1,2)

D. none of these

42. If the progressions 3, 10, 17, ....... and 63, 65, 67, ....... are such that their nth terms are equal, then n is equal to A. 13

B. 1

C. 19

−4

−4

∫−1 𝑓 (𝑥)𝑑𝑥 = 4 and ∫2 ( 3 − 𝑓(𝑥))𝑑𝑥 =

48. If

7

D. 18 then the value of

43. Ten different letters of an alphabet are given. Words with five letters are formed from these given letters. Then the number of words which have at least one letter repeated is

1

∫−2 𝑓 (−𝑥)𝑑𝑥 is

A. 2

B. -3

C. 5

D. none of these

𝑔(𝑓(𝑥)) = |sin𝑥| and 𝑓(𝑔(𝑥)) = (sin√𝑥)2 , then 49. If

A. 69760

B. 30240

C. 99784

D. none of these

𝑥 2 + 𝑦 2 = 5 at the point (1,-2) also touches the circle 𝑥 2 + 𝑦 2 − 8𝑥 + 6𝑦 + 20 = 0. Then its point of contact is 44. The tangent to the circle

A.

𝑓(𝑥) = sin2 𝑥, 𝑔(𝑥) = √𝑥

B.

𝑓(𝑥) = sin𝑥, 𝑔(𝑥) = |𝑥|

C.

𝑓(𝑥) = 𝑥 2 , 𝑔(𝑥) = sin√𝑥

D. f and g cannot be determined A. (-1, 3)

B. (3, -1)

C. (3, 1)

D. (-3, - 1)

45. Events that

50.

𝐴, 𝐵, 𝐶 are mutually exclusive events such

𝑃(𝐴) =

3𝑥+1

, 𝑃(𝐵) =

𝑥−1

and

𝑃(𝐶) =

3 4 1−2𝑥 . The set of possible values of 𝑥 are in interval 2

A.

1/4[log(𝑥 4 + 1) + 𝐶 ]

B.

−log(𝑥 4 + 1) + 𝐶

C.

log(𝑥 4 + 1) + 𝐶

D. none of these

1 2

A.

[ , ] 3 3

B.

1 13

51. There are 4 balls of different colours & 4 boxes of colours same as those of the balls. The number of ways in which the balls, one in each box, could be placed such that exactly no ball go to the box of its own colour is:

[ , ] 3 3

C. [0,1] D.

∫ 𝑒 3log𝑥 (𝑥 4 + 1)−1 𝑑𝑥 is equal to

1 1

[ , ] 3 2

→

̂, → 46. Given vectors 𝑥 = 3𝑖̂ − 6𝑗̂ − 𝑘 𝑦 = 𝑖̂ + 4𝑗̂ − ̂ 3𝑘 and →

→

→

𝑧 = 3𝑖̂ − 4𝑗̂ − 12𝑘̂ then the projection of 𝑥 × 𝑦 on

→

𝑍 is

A. 9

B. 30

C. 20

D. None

𝛥𝐴𝐵𝐶, 𝑎, 𝑏, 𝐴 are given and 𝑐1 , 𝑐2 are two values of the third side 𝑐 . The sum of the areas of the two triangles with sides 𝑎, 𝑏, 𝑐1 and 𝑎, 𝑏, 𝑐2 is 52. In

A.

1 2

A. 14

B. -14

C. 12

D. 15

B.

𝑏2 sin2𝐴

2𝑐 𝑏

47. The equation of the plane through the line of

C.

1 2

(𝑏2 + 𝑐 2 − 𝑎2 )sin𝐴

𝑎2 sin2𝐴

𝑥 + 𝑦 + 𝑧 + 3 = 0 and 2𝑥 − 𝑥 𝑦 𝑦 + 3𝑧 + 1 = 0 and parallel to the line = =

D. None of these

𝑧 𝑖𝑠 3

53. Directions: The following question has four choices,

intersection of planes

1

𝑓(𝜃) = + sin 𝜃. Then, for all the values of 𝜃

out of which ONE or MORE are correct. Let

A.

𝑥 − 5𝑦 + 3𝑧 = 7

B.

𝑥 − 5𝑦 + 3𝑧 = −7

C.

𝑥 + 5𝑦 + 3𝑧 = 7

Mock Test - 2

2

cos2 𝜃 A.

13 16

4

≤ 𝑓(𝜃) ≤ 1 | 28

B. f(θ) is a periodic function with period π

A. Gene transfer methods

C. f(θ) is a periodic function with period π/2

B. Natural pesticides

D.

3 4

C. Sterile Insect Technique

≤ 𝑓(𝜃) ≤ 1

D. All of the above

54. Directions: The following question has four choices, out of which ONE or MORE are correct. If are unit vectors satisfying A. B.

→ →

→

→

→

→ →

→

𝑎 , 𝑏 and 𝑐

𝑎 × (𝑏 × 𝑐 ) =

→

𝑏 , then √2

→

A. Acid rain

B. Global warming

C. Dust

D. Smoke

𝑎 , 𝑏 and 𝑐 are mutually orthogonal vectors →

60. Identify the natural sources of radiations.

→

𝑎 and 𝑏 orthogonal vectors

→ → C. angle between 𝑎 and 𝑐 is 45∘

D.

59. Which of the following has been causing most damage to the Taj Mahal?

→

→

angle between 𝑎 and 𝑏 is 45

𝑓(𝑥 − 𝑦), 𝑓(𝑥) ⋅ 𝑓(𝑦) and 𝑓(𝑥 + 𝑦) are in 𝐴𝑃 for all 𝑥, 𝑦 and 𝑓(0) ≠ 0, then out of which ONE or MORE are correct. If

𝑓(2) = 𝑓(−2)

B.

𝑓(3) + 𝑓(−3) = 0

C.

𝑓 ′ (2) + 𝑓 ′ (−2) = 0

D.

𝑓 ′ (3) = 𝑓 ′ (−3)

B. Gamma rays

C. Nuclear power plants

D. X-rays

∘

55. Directions: The following question has four choices,

A.

A. Carbon-40

61. The establishment of National Parks and Wildlife Sanctuaries is a strategy for A. Conservation of wildlife B. Creating awareness about wildlife C. Preventing wild animals from entering villages D. Studying wildlife biology 62. Which region among the following shows higher biological diversity? A. The Northeastern part of India B. The Western Ghats C. The Deccan Plateau D. The Himalayan Base

Biology 63. Those species whose numbers have been reduced considerably and are at the verge of extinction are called 56. Deforestation to a large extent can lead to A. Soil erosion during rainy season

A. Threatened

B. Endangered

C. Vulnerable

D. Rare

B. Flooding during rainy season C. Destruction of wildlife

64. Wildlife is conserved

D. All of these

A.

57. Reforestation is essential

In situ

B. Ex situ C. Both A and B

A. To conserve forests

D. Selective hunting of predators

B. To reduce pollution C. To conserve fossil fuels

65. What is ecological diversity?

D. none of these

A. Presence of diverse species in a huge population

58. How can plant pests in a field be controlled without polluting the environment?

B. A large landmass such as a country, having different types of ecosystems such as desert, rainforest and mangrove

Mock Test - 2

| 29

C. Presence of species from different geographical locations in an ecosystem D. None of the above 66. Deforestation to a large extent can lead to A. Soil erosion during rainy season B. Flooding during rainy season C. Destruction of wildlife D. All of these 67. Jaundice is a disorder of the _____. A. Respiratory system B. Skin and Eyes C. Circulatory system D. Digestive system 68. Most of the fat digestion occurs inA. Rectum

B. Stomach

C. Caecum

D. Small intestine

69. The intestinal villi are more numerous and larger in posterior part of small intestine because: A. Digestion is faster in posterior part B. Blood supply is poorer in posterior parterior part C. Blood supply is poorer in posterior part D. there is more digested food in posterior part 70. The duodenal wall of the small intestine secretes ––– ––. A. secretin B. Cholecystokinin (CCK) C. secretin and cholecystokinin (CCK) D. none of these

Mock Test - 2

| 30

SMART ANSWERSHEET

Correct Q.

1

Ans.

Correct Q.

Ans.

Skipped

Skipped

0.0 %

0.0 %

A

15

A

D

6

0.0 %

100.0 %

100.0 %

0.0 %

0.0 %

0.0 %

B

20

C

C

A, B, C, D

0.0 %

37 0.0 %

24

C

38

100.0 %

100.0 %

0.0 %

0.0 % 25

A, B

B

40 100.0 %

0.0 %

B, C, D

0.0 % 27

41

0.0 % 100.0 %

42 100.0 %

100.0 %

0.0 %

0.0 % 52

A 100.0 %

0.0 %

0.0 % 53

0.0 %

B, D

B 100.0 % 0.0 %

66

100.0 %

D 100.0 % 0.0 %

67

D

100.0 %

0.0 %

100.0 %

0.0 % 54

B, C

100.0 %

0.0 % 68

D

100.0 %

100.0 %

0.0 % 55

A, C

0.0 % 69

D

100.0 %

100.0 %

0.0 % 56

100.0 %

0.0 % C 100.0 % 65

100.0 %

A

0.0 % B 100.0 % 64

A

0.0 %

D

100.0 %

0.0 % 51

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Mock Test - 2

0.0% 0.0%

| 31

HINTS AND SOLUTIONS 1. The position values of the alphabets are used where A=1, B=2, C=3 and so on. There is a gap of 2 alphabets between the 1 st and 4th alphabets and similarly between 2nd and 3rd alphabets.

6. In this type of question we have to find the Relation In First one and after that this relation will be applied to Next one. we have to find the Relation in 45: 36 and the same Relation will be applied to (i) 45: 36 Relation is

63: ?

(45) − 9 = (36)

Applying same relation to

63: ? (63) − 9 = (54)

F _ _ I, R _ _ U

So the Answer is 63: 54

M _ _ P, S _ _ V

Hence, The correct answer is B.

Hence, The correct answer is A. 2. Gloom means a state of depression or despondency. Hence Ecstasy is the antonym of gloom. Similarly Humiliation means the action of making someone feel ashamed. Hence Exaltation is the antonym of Humiliation.

7. Among all the given options, only chandigarh is a union territory. Hence, option C is the correct response. 8. All the rest are parts of a house. Hence, The correct answer is C.

Hence, The correct answer is D. 9.

(7 − 1) × (3 + 1) = 24

3. Romio & Juliet is written by William Shakespeare. Gulliver Travels is written by Jonathan Swift. Hence, option B is the correct response.

(9 − 1) × (3 + 1) = 32 and not 24

Hence, The correct answer is B.

(3 − 1) × (7 + 1) = 16

4. According to question,

(5 − 1) × (3 + 1) = 16

Hence, option

𝐵 is the correct response.

10. Except the number 39, all others are even numbers. Hence, The correct answer is C. 11. Let us choose a coordinate system with its origin as the starting point of the boat, the and

+𝑥 -axis points downstream

+𝑦 -axis points at right angles to the bank of the river

Similarly, FUNCTIONING is related to MHMFLRGGVOD

X-motion of the boat is due to the water current velocity

𝑦 -motion is caused solely by the

Hence, The correct answer is D.

vcurrent while the

5. The relationship used in this question is x : (3x + 1)

velocity 𝑣 of the boat. The above two motions are independent of each other and can be treated separately.

Similarly, the second pair is 35 : ( 3 x 35 + 1) = 35 : 106

𝑡 = 0, the 𝑦 coordinate after a time 𝑡 is 𝑦 = 𝑣𝑡 … (i) The speed of water current is a function of 𝑦 and is given by

Thus, 106 is the required answer.

𝑣current =

Here, 20 : 61 may be written as 20 : (3 x 20 + 1) = 20 : 61

Assuming that the boat starts at time

2𝑢𝑣 for 𝑑

𝑑 2

𝑦 < … … .. (ii)

Hence, The correct answer is C. Mock Test - 2

| 32

Substituting for y from equation (i) By symmetry, its value at the middle of the river is

𝐷/2. The time required to

reach the middle of the river is

𝑡=

𝑑 2

𝑣

=

𝑑 Separating 2𝑣 𝐷

the variable in equation (iii) and intergrating 𝑑 2𝑦

2𝑢𝑣 ∫ 𝑑 0 𝐷 2

𝜆 = 20𝑐𝑚, 𝑆1 𝑆2 =

20𝑐𝑚, 𝐵𝐷 = 20𝑐𝑚 𝑥 for hearing the minimum sound So path difference 𝐴𝐼 = Let the detector is shifted to left for a distance

𝐵𝐶 − 𝐴𝐵 = √(20)2 + (10 + 𝑥)2 − √(20)2 + (10 − 𝑥)2

𝑡 𝑑𝑡

So the minimum distance hearing for minimum

𝑑 2

2𝑢𝑣 (2𝑣) [ 𝑑 2

=

∫02 𝑑 𝑥 =

14. According to the data

]=

𝑢𝑑 4𝑣

(2𝑛+1)𝜆 2

=

𝜆 2

=

20 2

𝐴𝐼 =

= 10𝑐𝑚

Substitute the value in equation of path difference:

Hence, The correct answer is D.

⇒ √(20)2 + (10 + 𝑥)2 − √(20)2 + (10 + 𝑥)2 = 10

2𝐺𝑀 𝑅+ℎ

12. 𝑉𝑒

=√

𝑉𝑒 =

2𝐺𝑀 √ 𝑅

Hence, The correct answer is C.

√𝑅+ℎ √𝑅

15. Efficiency of heat engine,

solving we get

𝑉𝑒 𝑉𝑒

=

6400+1000

=√

6400

𝑣′ =

11.27 1.067

=

=

11.2 115

7400 6400

=

10

2𝜂1 =

𝑇2 −(𝑇1 −150) 𝑇2

2𝜂1 = 𝜂1 +

𝑉1 = 104 𝑚/𝑠

2× =

1 6

Hence, The correct answer is C.

1 3

𝑀 = 50,000𝑘𝑔 Inner radius of the column, 𝑟 = 30𝑐𝑚 = 0.3𝑚 Outer radius of the column, 𝑅 = 60𝑐𝑚 = 0.6𝑚 Young's modulus of steel, 𝑌 = 2 × 1011 𝑃𝑎 Total force exerted, 𝐹 = 𝑀𝑔 = 50000 × 9.8𝑁 Stress = Force exerted on a 13. Mass of the big structure,

= 50000 ×

Where, Area,

𝑌=

𝐴=

9.8 4

= 122500𝑁

Stresss Strain Strain

𝜋(𝑅 2

− 𝑟2 )

=

=

𝜋((0.6)2

−

= 122500/[𝜋((0.6)2 − (0.3)2 ) × 2 × 1011 ] = 7.22 × 10−7 Strain

Hence, the compressional strain of each column is

× 10−7

Hence, The correct answer is A.

Mock Test - 2

=

1 6

1 3

= =

2𝜂1 =

𝑇2 −𝑇1 𝑇2

+

150 𝑇2

150 𝑇2

150 𝑇2

150 𝑇2 150 𝑇2

𝑇2 = 450∘ Hence, The correct answer is A. 16. Let us consider a cylindrical conductor of length 1𝑚 and radius a. Let the conductor be made up of a large

(𝐹/𝐴) 𝑌

(0.3)2 )

7.22

𝑇2 −𝑇1 𝑇2

1500 , the efficiency is doubled. So, new efficiency is

10

= 10 × 103 𝑚/𝑠

Young's modulus,

𝜂1 =

Where, 𝑇2 is source temperature and 𝑇1 is sink temperature. When the temperature of sink is reduced by

= √1.15 = 1.072

9.73 = 10.46𝑘𝑚/𝑠

single column

𝑥 = 12.0𝑐𝑚

number of thin ring shaped conductors each of length 1𝑚 Let us consider one such element of the conductor of radius

𝑟 and thickness dr. The resistance of the conductor 𝑑𝑅 = 𝑃×1 2𝜋𝑟𝑑 2

∴

1 𝑑𝑅

=

2𝜋𝑟𝑑𝑟 𝑎 𝑟2

=

2𝜋 3 𝑟 𝑑𝑟 𝛼

All ring shaped conductors are in parallel. Hence reciprocal of net resistance,

=

2𝜋 𝑎 3 ∫ 𝑟 𝛼 0

1 𝑅

=𝜀

1 𝑑𝑅

𝑎 2𝜋 3 𝑟 𝑑𝑟 𝛼

= ∫0

𝑑𝑟

| 33

1 𝑅

2𝜋 𝑟 4

=

𝛼

𝑅=

𝑎

[ ] = 4 0

2𝜋 𝑎4 𝑎

,

4

2𝛼 𝜋𝑎4

=

∴𝑅=

2𝛼 𝜋

1

⋅

𝛬

(𝜋 )

2 =

→

17. Effective resistance of circuit is

20.

4𝛺 𝐼 =

𝐸 𝑅

=

𝐵𝑙𝑣 or 𝑅

𝐼𝑅 𝐵𝑙

𝑣=

1×0.001×4 𝑚𝑠 −1 2×10×0.01

closed cubical box,

𝐻=

𝑙 𝑡

𝑣2 2

To maintain constant temperature difference between outside and inside the box, this heat escaped must be produced by the electric current in the heater. Let 𝑅 be the resistance of the coil. The heat produced per second is

∴

𝑉2 4.2𝑅

𝑅=

=

𝑅

𝐽𝑢𝑙𝑒 =

𝐽𝑅

𝑐𝑎𝑙 =

𝑎 = 0 𝛼 − 𝛽𝑥 = 0 ⇒ 𝑥 =

At mean position

𝑎=𝑣

𝑣

𝑉2

𝑚𝑎 = 𝑞𝐸 so acceleration of the particle

𝑞(𝛼−𝛽𝑥) 𝑚

𝑑𝑣 𝑑𝑥

=

𝑞

𝑥

𝛼 𝛽

𝑞(𝛼−𝛽𝑥) 𝑚

∫0 𝑣 𝑑𝑣 = 𝑚 ∫0 ( 𝛼 − 𝛽𝑥)𝑑𝑥

= 6000𝑐𝑎𝑙𝑠

𝑉2

𝑎=

Now

0.1

𝑡

𝑎𝑡𝑡 = 2𝜋/𝜔, 𝑉 = 𝜋𝑅 2 𝐵0 𝜔

is

4×10−4 ×6×50×50×100

𝐻=

= −𝐴𝐵0 𝜔cos𝜔𝑡 = 𝜋𝑅 2 𝐵0 𝜔cos(𝜔𝑡)

21. As we have,

𝑘𝐴(𝜃2 −𝜃1 ) 𝑑

𝑄

𝑑𝜙 𝑑𝑡

= 0.02𝑚𝑠 −1 = 2𝑐𝑚𝑠 −1

18. The rate of heat transmitted through the walls of the

=

→

𝜙 =𝐴⋅𝐵

Hence, The correct answer is C.

Hence, The correct answer is B.

=

]

235×864×3×109 6.02×3.2×1015

𝑉=−

or

235

Hence, The correct answer is D.

2𝜋𝛼 𝐴2

Hence, The correct answer is B.

𝑣=

6.02×1026

= 32𝑔

𝐴 𝜋

𝐴 = 𝜋𝑎2 ⇒ 𝑎2 =

But

[∵ no. of atoms in 1𝑘𝑔 of 𝑈 235 =

𝑉2 4.2𝑅

𝑐𝑎𝑙

=

𝑞 (𝛼𝛼 𝑚

−

𝛽𝑥 2 ) 2

2𝑞𝑥 (𝛼 𝑚

⇒𝑣=√

Hence velocity of the particle is zero at

𝑥 = 0 and 𝑥 =

We can conclude that particle oscillated between these two limits and have amplitude

𝑥=

400×400 4.2×6000

𝛽𝑥 ) 2

2𝛼 𝛽

𝐴=

𝛼 𝛽

Acceleration of the particle is maximum at

= 6000

−

2𝛼 , 𝑎max 𝛽

=

𝑥 = 0 and

𝑞𝛼 𝑚

Hence options A, B, C and D are the correct answers.

𝑅 = 6.35𝛺 Hence, The correct answer is C.

= 106 × 86400𝐽 Energy released per fission = 200 × 106 × 1.6 × 10−19 𝐽 19. Energy produced by the reactor in 1 day

No. of fissions required (i.e..) no. of in a month Mass of

(

=

106 ×86,400×30 200×106 ×1.6×10−19

235𝑈 having the requisite no. of atoms =

235 ) 6.02×1025

Mock Test - 2

235𝑈 atoms fissioned

(

106 ×86400×30 ) 200×106 ×1.6×10−19

22. One two-dimensional(with x-axis and y-axis) velocitytime graph can be drawn for motion of particle in onedimension only. Hence option A is the correct answer.

23. If field due to a point charge varies are

𝑟 −25 , then

𝐸 = 𝑐𝑟 −25 ; where 𝑐 is a constant. Using Gauss's law, we →

have

→

∮ 𝐸 ⋅ 𝑑𝑠 = 𝑐𝑟 −25 × 4𝜋𝑟 2 = 4𝑐𝜋𝑟 −23 ≠

𝑞 enclosed 𝑒0

Hence, Gauss law is invalid and option 1 is wrong.

| 34

Electric field due to a dipole is not symmetric; hence, a Gaussian surface cannot be used to find the electric field due to a dipole using Gauss law. Hence, option 2 is wrong.

Hence, The correct answer is B.

Electric fleid on the line joining the charges is opposite in direction if they are of the same sign only. Hence, option 3 is correct. By definition of potential, work done by external force to

𝐴 to point 𝐵 is given by: 𝑊 = 𝑞(𝑉𝐵 − 𝑉𝐴 ) = 𝑉𝐵 − 𝑉𝐴 Hence, option

27.

4 is correct.

Bond order

move a unit positive charge from point

𝑅 = 1𝛺, emf , 𝜉 = 15𝑉 and inductance , 𝐿 = 5𝑚𝐻, 𝑑𝑖/𝑑𝑡 = 103 𝑠𝑜, 𝑉𝐴 − 𝐼𝑅 + 𝜖̂ − 24. Resistance,

𝐿

𝑑𝑖 𝑑𝑡

− 𝑉𝐵 = 0

⇒ 𝑉𝐴 − 5 × 1 + 15 − 5 × 10

−9

9

× 10 − 𝑉𝐵 = 0

=

Number of bonds Number of Resonating structures

1.25 In a given resonance structure, the 0 atom that forms double bond has formal charge of 0 and the remaining 30 atoms have formal charge of -1 each. In the resonance hybrid, a total of -3 charge is distributed over 40 atoms. Thus the formal charge of each 0 atom is

−3 4

= −0.75

Note:

⇒ 𝑉𝐴 − 𝑉𝐵 − 5 + 15 − 5 = 0

To calculate the formal charge, the following formula is

⇒ 𝑉𝐴 − 𝑉𝐵 + 15 = 0

used. Formal Charge

⇒ 𝑉𝐵 − 𝑉𝐴 = 15𝑉

atom

= [ Number of valence electrons on ] − [ non-bonded electrons + number of bonds ] For 0 atom that forms double bond with 𝑃 atom, Formal Charge = 6] − [4 + 2] = 0

Hence option (c) is correct.

25.

5 4

= =

For 0 atom that forms single bond with

𝑉𝐷 = 𝑉𝐸 = 0𝑉

Charge

𝑉𝐵 − 𝑉𝐷 = 2𝑉 ⇒ 𝑉𝐵 = 2𝑉

𝑃 atom, Formal

= 6] − [6 + 1] = −1

Hence, The correct answer is C.

𝑉𝐶 − 𝑉𝐵 = 10𝑉 ⇒ 𝑉𝐶 = 12𝑉 𝑉𝐶 − 𝑉𝐴 = 6𝑉 ⇒ 𝑉𝐴 = 6𝑉

28.

𝑉𝐴 − 𝑉𝐵 = 6 − 2 = 4𝑉 Current through the

2𝛺 resistance =

𝑉𝐴 −𝑉𝐵 2

= = 2𝐴

4 2

Current through the

3𝛺 resistance =

𝑉𝐶 −𝑉𝐷 3

=

12 3

=

4𝐴 Hence option A and B is the correct answer. Hence, The correct answer is D. 26.

𝐶2 𝐻4 (𝑔) + 3𝑂2 (𝑔) → 2𝐶𝑂2 (𝑔) + 2𝐻2 𝑂()

𝛥𝑛9 = 2 − (1 + 3) = −2

𝐶𝑎𝑆𝑂4(29) + 𝐵𝑎𝐶𝑙2(𝑎𝑞) → 𝐵𝑎𝑆𝑂4(𝑠) + 𝐶𝑎𝐶𝑙2(𝑎)

𝑇 = 300𝐾, 𝛥𝐻 = −336.2𝑘𝑐𝑎𝑙

136𝑔 CaSO

𝑅 = 2 × 10−3 𝑘𝑐𝑎𝑙 degree

Weight of

𝐴𝑡300𝐾𝛥𝐻 = −336.2𝑘𝑐𝑎𝑙

−1 𝑚𝑜𝑙 −1

𝛥𝐻 = 𝛥𝑈 + 𝛥𝑛9 𝑅𝑇 −336.2 = 𝛥𝑈 + (−2) × 300 × 2 × 10−3 𝛥𝑈 = −336.2 + 1.2 = −335.0𝑘𝑐𝑎𝑙 Mock Test - 2

29.

Mass of

4 produces

233𝑔 𝐵𝑎𝑆𝑂4

𝐵𝑎𝑆𝑂4 = 0.617𝑔

𝐶𝑎𝑆𝑂4 reacted =

Percentage purity of

0.617×136

𝐶𝑎𝑆𝑂4 =

233

= 0.3601𝑔

0.3601×100 0.400

= 90.025

Hence, The correct answer is B.

| 35

30. Ag+1 ion has 18 valence electrons so it is more covalent due to greater polarizability on anion. Hence, The correct answer is B. 31. Aromatic aldehydes do not give the Fehling test. [Fructose, or fruit sugar, is a simple ketonic monosaccharide found in many plants, where it is often bonded to glucose to form the disaccharide sucrose. It is one of the three dietary monosaccharides, along with glucose and galactose, that are absorbed directly into blood during digestion.] Hence, The correct answer is C.

Partial pressure of

1

𝑁2 = × 76 = 19 atm 4

𝑃𝑁𝐻3 1/2 3/2 2 2

𝐾𝑝 =

𝑝𝑁 +𝑝𝐻

=

24 (19)1/2 ×(57)3/2

=

24 4.358×(7.549)3

=

24 4.358×430

= 0.0128𝑎𝑡𝑚−1

Hence, The correct answer is A.

35. The ionization energy is given to be

= 2.18 ×

10−18 𝐽, which equals 32. When the concentration of 𝐵 alone was doubled, the rate did not change. Hence the reaction is zero order in B. When the concentration of A alone was doubled, the rate increased by two times. Hence, the reaction is of first-order

𝑟=

in A.

13.6𝑒𝑉 . Therefore, the energy of electron in the first level is −13.6𝑒𝑉 . We can thus use the formula, 𝐸 = −13.6 × 𝑍 2 /𝑛 2 to calculate energy of electron in the second orbit. −2.18×10−18

𝑘[𝐴]1 [𝐵]0

22

The overall order of the reaction is

1.

𝐸2 =

= −5.45 × 10−19 𝐽

Hence, The correct answer is D.

For the first-order reaction, the unit of 𝑘 is sec−1 Hence, The correct answer is A. 33. A) Acidic nature depends on elecronegativity. In the periodic table, the trend in E.N is 𝑁 > 𝑃 > As.Thus the trend in acidic nature is reverse of that of E.N and can be written as;

𝐴𝑠𝐻3 > 𝑃𝐻3 > 𝑁𝐻3

B) lonic radii increases on moving down a group from top to bottom, therefore the trend in ionic size is;

𝐾+

>

𝑁𝑎+

>

𝐶𝑠 + >

𝐿𝑖 +

C) The trend in basicity is

37.

right in a period.But first I.E of Be is more than

𝐵 ; because

(2𝑠2 ) Therefore the trend in I.E is ; 𝐶 > 𝐵𝑒 > 𝐵 > 𝐿𝑖 Be has stable fully filled valence orbital

Hence, The correct answer is D. 1 𝑁 2 2(𝑔)

𝑟+ 𝑟

3 2

= √ − 1 = 0.225 suggests that the cation is

occupying the tetrahedral void. Therefore,the coordination number of the cation is 4 and the option (1) is correct. The anions will form the lattice (ccp or hcp) and in both the cases, they will touch along the face diagonal. Hence, option (2) is also correct.

𝑃𝑄 ⇒ 𝑍cotion = 𝑍axion indicating 𝑍𝑛𝑆 type of structure with alternate The formula for the compound is

3 2

+ 𝐻2(𝑔) ⇌ 𝑁𝐻3(𝑔)

Partial pressure of

𝑁𝐻3 =

24 100

× 100

tetrahedral voids occupied. So, the anion is touching 8 octahedral voids, but only 4 of them are occupied. Therefore, the coordination number of the anion is Hence option (3) is incorrect.

= 24.0𝑎𝑡𝑚

4.

𝑁2 + 𝐻2 = 100 − 24 = 76𝑎𝑡𝑚

Partial pressure of Mock Test - 2

or 2.7 kcal x 4.2 = 11 kJ Hence options B and D are the correct answers.

D) lonisation potiential increases on moving from left to

Pressure of

30.3 kcal - 27.6 kcal = 2.7 kcal,

; 𝐴𝑙2 𝑂3 < 𝑀𝑔𝑂 < 𝑁𝑎2 𝑂
∣q2∣

Coulomb's electrostatic force acting on 2 charged particles is equal if they are kept isolated.

Hence option A is correct.

i.e it doesn't matter which bob has more charge, the force acting between them will still be the same and thus they will make the same angle from the vertical.

20. Resistance of a wire R=ρl/A where ρ= resistivity, l= length, A= cross section of the wire.

From Coulomb's law, the force acting on both the particles will be

Thus, R2/R1=A1/A2=3/1⇒R2=3R1.

Mock Test - 6

As both have same material and length so R∝1/A

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here R1 is the resistance of thicker wire so its resistance R1=10Ω (given) so, R2=3(10)=30Ω As they are connected in series so the equivalent resistance is

The wavelength emitted by molecular energy levels, which are generally grouped into several bunches, are also grouped and each group is well separated from each other. The spectrum in this case looks like a band spectrum. Hence option B & D are correct.

Req=R1+R2=10+30=40Ω 23. Loss in gratiational potential energy of M is Mgl, as M falls down by l.

Hence option C is correct.

Elastic potential energy =(1/2) x Stress x Strain x Volume

21.

= (1/2) (Mg/A)x(I/L)x L =(1/2) Mgl Hence option A & B are correct. 24. We see that the only force doing work in the direction of motion that is along X is the force. And from the work energy theorem, total work done is equal to change in kinetic energy. Hence. KE of loaded car at any instant is equal to work done by force F till that instant. Acceleration of the point of contact, 𝑎𝑐 = 𝑎𝑐𝑛 + 𝛼𝑅 Velocity of point of contact is non-zero. Hence, rolling with slipping will occur and friction will act upward. For translational motion,

𝑚𝑔sin𝜃 − 𝑇 − 𝜇𝑚𝑔cos𝜃 =

Hence option B is correct.

25. Using Biot Savart law, magnetic field at location of due a 1

𝑚𝑎 − ⋯ − 1

𝐵=

For rotational motion, torque about CM of cylinder

paper

−𝜇𝑚𝑔cos𝜃 ⋅ 𝑅 + 𝑇𝑟 = 𝑚𝐾 2 𝛼 𝑎 𝛼 = [𝐴𝑠 there is no slip of rope ] −𝜇𝑚𝑔cos𝜃 ⋅ 𝑟

𝑎 𝑅 + 𝑇𝑟 = 𝑚𝐾 2 𝑟 𝑅 𝜇𝑚𝑔cos𝜃 + 𝑇 = 𝑟

𝑚

𝐾2 𝑎 𝑟2

− ⋯ − 11

∥, 𝑎 =

sin𝜃−𝑚cos𝜃(1+ 𝑟 ) 𝐾2 1+ 2 𝑟

𝑅 𝑟

𝑎 > 0sin𝜃 > 𝜇cos𝜃 (1 + ) 𝜇
𝑟 𝜇 < tan

𝜃 2

Hence option A,C& D is correct.

26. In the dry cell, the oxidation of zinc occurs at graphite anode. Zinc loses two electrons to form Zn2+ ion. The electrode reaction is Zn→Zn2+2e− Hence, correct option is B.

22. Line mission spectra can be obtained for both atoms and molecules, an atom or molecule in an excited state emits photons by making transition from excited state to ground state. Thus, constituting line emission spectra.

27. By Raoult's law PS=P0BXB+P0AXA But A is non-volatile solute So P0A=0

Mock Test - 6

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PS=P0BXB

∴ 2.74 = 1 + 2𝛼

Hence option C is correct.

∴𝛼=

1.74 2

= 0.87 = 87%

Hence option B is correct. 28. Haematite ore of iron contain Si as impurity where it reacts with O2 to produce SiO2 which is further used in formation of slag. But in extraction of copper SiO2 is needed to form slag. Hence option C is correct.

29. Option (C) is correct. The brown ring test for

𝑁𝑂2 and

33. Out of 7 valencies of iodine 5 is satisfied by 5 fluorine atoms. Thus, iodine has one lone pair of electrons remaining after the formation of IF 7. Thus there are 7 electrons surrounding the central iodine out of which 5 are bond pairs and 1 is a lone pair. Thus IF 7 shows sp3d2 hybridization.

𝑁𝑂3 is due to the formation of complex, [𝐹𝑒(𝐻2 𝑂)5 𝑁𝑂]2+ Brown ring test is the test to find out

Hence option B is correct.

the presence of nitrate radical. For this equal volume of the

34. 2PbS+3O2→2PbO+2SO2

𝐹𝑒𝑆𝑂4 is mixed with solution to be analysed. To this concentrated 𝐻2 𝑆𝑂4 is added along the freshly prepared

sides of the test tube. A brown ring will forms at the junction of the two liquids. This is nitroso Ferrous Sulphate

(𝐹𝑒𝑆𝑂4 ⋅ 𝑁𝑂) Nitric Acid (Nitrate) reduces Ferrous to Ferric and itself is reduced to nitrogen monoxide 𝑁𝑂 . This reacts with FeSO 4 and forms the brown ring. Hence option C is correct.

PbS+2PbO→3Pb+SO2 (self reduction) Pb,Cu and Hg are obtained by self reduction process. Hence option C is correct. 35. (A) Zinc sulphide ZnS is Zinc blende. (B) Lead carbonate (PbCO3) is Cerrusite. (C) Calamine is zinc carbonate (ZnCO3) (D) Magnesium carbonate MgCO3 is magnesite.

30. In the Arrhenius equation, k=Ae−E/RT , E represents the activation energy. It is the minimum energy that the reacting molecules should possess so that the collision is effective. The colliding molecules will not react if their energy is lower than the activation energy. Hence option B is correct. 31. During the depression of freezing point in a solution, liquid solvent and solid solvent are in equilibrium. During freezing of a solution, only the solvent freezes out and the equilibrium exists between solid and liquid forms of the solvent. Vapour pressure of the solid and liquid forms must be the same at freezing point, because otherwise, the system would not be at equilibrium, the lowering of the vapour pressure leads to the lowering of temperature at which the vapour pressures of the liquid and frozen forms of the solution will be equal. Hence option A is correct.

0.1𝑀𝐵𝑎(𝑁𝑂3 )2 solution is 2.74. The degree of dissociation is 87%𝑖 = 1 + 2𝛼[ for 𝐵𝑎(𝑁𝑂3 )3 ⇌ 𝐵𝑎 2+ + 2𝑁𝑂3 ] 32. The van't Hoffs factor for

Mock Test - 6

Hence option C is correct. 36. The higher the reduction potential, the higher the tendency to get reduced, and the higher the oxidising power. Hence option B, C & D are correct. 37. Statement (1) is correct as the vapour pressure of a solvent decreases on addition of a non-volatile solute. The colligative properties depend on the number of particles in the solution. Hence, the change will not be the same for the solution of an electrolyte and a nonelectrolyte, even if they have equal concentrations. Hence, statement (2) is incorrect. However, the solutions of urea and glucose, having equal concentrations, will have the same depression in the freezing point as the number of particles of the solute are the same. Hence, statement (4) is correct. The cryoscopic constant is specific for each solvent and is a function of temperature. Hence, (3) is incorrect. Hence option A & D are correct.

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38. Co(NO3)2 and CrCl3 have unpaired electron; hence, they are coloured. Zn(NO3)2, LiNO3 and potash alum have no unpaired electron; hence, they are colorless. Hence option C & D are correct.

If \(\frac{\pi}{2}

𝜋 3𝜋

⇒ tan𝑥 ≤ 1 ⇒ 𝑥 ∈ ( , 2

4

]

considering both cases, 𝜋 𝜋 4 2

𝜋 3𝜋 ] 2 4

𝑥 ∈ [ , ]∪( ,

Hence option C is correct. 39. Option 3 is incorrect because half-life period of first order reaction is independent from initial concentration of reactants, while options 1, 2 and 4 are correct as the concentration of reactant which is following first order kinetics always decreases exponentially and becomes zero at infinity. As the temperature increases, the rate constant increases and the half-life decreases since half-life is inversely dependent on rate constant.

𝐶𝑡 = 𝐶 ⋅ 𝑒 −𝑡𝑡 𝑡1 𝛼 1

⋅ 𝐾 increases on increasing 𝑇 After eight half-lives.

2 𝐾 𝑐0

𝑐=

23

% completion =

𝑜 𝐶0 − 03 𝑞

𝐶0

42. Given

𝐴 = [𝑎𝑖𝑗 ](3×3)

𝑎𝑖𝑗 = 𝑖 2 − 𝑗 2 ∴ 𝑎𝑖𝑗 = 0 if 𝑖 = 𝑗 Now, 𝑎12 = 12 − 22 = −3 𝑎13 = 12 − 32 = −8 𝑎21 = 22 − 12 = 3 𝑎23 = 22 − 32 = −5 𝑎31 = 32 − 12 = 8 𝑎32 = 32 − 22 = 5 where,

× 100 = 99.6%

Hence option A, B & D are correct. Here,

𝐴𝑇 = −𝐴

∴ 𝐴 is a skew-symmetric matrix. 40. (1) Both benzoic acid and phenol are soluble in NaOH, hence inseparable using this reagent.

Hence option C is correct.

However, only benzoic acid (C6H5COOH) is soluble in NaHCO3, while benzyl alcohol (C6H5CH2OH) is not. Hence, the mixture can be separated using NaHCO3.

43.

(2) Although, both benzyl alcohol (C6H5CH2OH) and phenol (C6H5OH) react with NaOH, only the latter (phenol) is soluble in NaOH. Hence, the mixture can be separated using NaOH. However, both alcohols and phenol do not react with NaHCO 3 and cannot be separated using this reagent. (3) Both benzyl alcohol (C6H5CH2OH) and phenol (C6H5OH) react with NaOH, but are not soluble in it. Both alcohols and phenol do not react with NaHCO3 and cannot be separated using this reagent.

Now,

By matrix multuplication, we get

(4) -phenyl acetic acid (C6H5CH2COOH) is soluble in NaOH and NaHCO3. Benzyl alcohol (C6H5CH2OH) is not. Hence, the mixture is separable with both the reagents.

𝐴𝐵𝐶 = [(𝑎𝑥 + ℎ𝑦)𝑥 + (ℎ𝑥 + 𝑏𝑦)𝑦] 𝐴𝐵𝐶 = [𝑎𝑥 2 + ℎ𝑥𝑦 + ℎ𝑥𝑦 + 𝑏𝑦 2 ] 𝐴𝐵𝐶 = [𝑎𝑥 2 + 2ℎ𝑥𝑦 + 𝑏𝑦 2 ]

Hence option A, C & D are correct.

Hence option B is correct.

|cos𝑥| ≤ sin𝑥, 0 ≤ 𝑥 ≤ 2𝜋 𝜋 If 0 ≤ 𝑥 ≤ ⇒ cos𝑥 ≥ 0 ⇒ |cos𝑥| = cos𝑥

44.

41.

2

𝜋 𝜋 4 2

⇒ tan𝑥 ≥ 1 ⇒ 𝑥 ∈ [ , ]

Mock Test - 6

| 118

Now.

∴ 𝐴 + 2𝐵 + 𝐶 = 0 Hence option A is correct.

Now.

(A+B)2=A+B (since AB=BA=0)

𝐴 = 𝐵 , then all the elements of 𝐴 and 𝐵 should be same. ∴ 𝑥 + 3 = 0 2𝑦 + 𝑥 = −7 𝑧−1 =3 𝑧 − 4𝑎 = −2𝑎 𝑥 = −3 𝑦 = −2 𝑧=4 𝑎=2 𝑥 + 𝑦 + 𝑧 + 𝑎 = −3 − 2 + 4 + 2 =1

Hence A+B is Idempotent

Hence option C is correct.

48. If

Hence option A is correct.

45. Given A,B are Idempotent matrices A2=A, B2=B (A+B)2=A2+B2+AB+BA (A+B)2=A+B+AB+BA

Hence option B is correct. 49. We know that,

46.

𝐴 ⋅ 𝐴−1 = 𝐼 … … … … (𝑖)

𝐴 = 𝐼𝐴 (Transposing is inter changing the rows

∴  5𝐴𝑇 + 3𝐵𝑇

1& columns)

𝑅2 → 𝑅2 − 3𝑅1 , we get,

𝑅1 → 𝑅1 − 3𝑅2 , we get,

Comparing with (i), we get

Hence option A is correct.

47.

Hence option D is correct

50. Given

∴ 𝐴∣=0. (2 − 3) − 1(1 − 9) + 2(1 + 6) Mock Test - 6

| 119

= 0 + 8 − 10 = −2 ≠ 0 If 𝐶 be the matrix of cofactors of the elements in

⇒ −𝑃(𝐴 ∪ 𝐵) > −1 ⇒ 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∪ 𝐵) > 𝑃(𝐴) + 𝑃(𝐵) − 1 ⇒ ⇒

𝑃(𝐴)+𝑃(𝐵)−𝑃(𝐴∪𝐵) 𝑃(𝐴)+𝑃(𝐵)−1 > 𝑃(𝐵) 𝑃(𝐵) 𝐴 𝑃(𝐴)+𝑃(𝐵)−1 𝑃( ) > 𝐵 𝑃(𝐵)

Option (1) is correct. Choice

𝑃(𝐴 ∩ 𝐵) = 0 Finally. 𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴) ⋅ 𝑃(𝐵), if 𝐴 and 𝐵 are independent = 1 − {1 − 𝑃(𝐴)}{1 − 𝑃(𝐵)} = 1 − 𝑃(𝐴) ⋅ 𝑃(𝐵) (2) holds only for the disjointed

Hence option A & C are correct.

53. Let A, B and C, respectively, denote the events that the student passes in Maths, Physics and Chemistry.

Hence,

It is given: P (A) = m, P(B) = p and P(C) = c Probability of at least one success: p + m + c - pm - pc - cm + pmc = 0.75 ------(i) Probability of at least two successes: mc + mp + pc - 2mcp = 0.5 ------------(ii)

⇒ |𝑎𝑏𝑐| = 10

Probability of exactly two successes:

Hence option C is correct.

mc + mp + pc - 3mcp = 0.4 ------------(iii) From eq. (ii) and (iii),

|𝑧1 | = |𝑧2 | 𝑧1 +𝑧2 𝑧 𝑁𝑜𝑤, × 1 − 𝑧2 𝑧1 −𝑧2 𝑧1 𝑧1 𝑧1 −𝑧1 𝑧2 +𝑧2 𝑧1 −𝑧2 𝑧2 = |𝑧1 −𝑧2 |2

51. Given:

= =

mcp = 0.5 - 0.4 = 0.1 = 1/10 mc + mp + pc = 0.5 + 0.2 = 0.7 = 7/10 From (i),

|𝑧1 |2 +(𝑧2 𝑧1 −𝑧1 𝑧2 )−|𝑧2 |2 |𝑧1 −𝑧2 |2 𝑧2 𝑧1 −𝑧1 𝑧2 2 1 |𝑧1 −𝑧2 |2

p + m + c = 27/20 Hence option B & C are correct.

(∵ |𝑧 | = |𝑧2 |2 )

𝑧 − 𝑧 = 2𝑖𝑙𝑚(𝑧) ∴ 𝑧2 𝑧1 − 𝑧1 𝑧2 = 2𝑖𝑙𝑚(𝑧2 𝑧1 )

As, we know 𝑧1 +𝑧2 𝑧1 −𝑧2

=

54. For

2𝑖𝑙𝑚(𝑧2 𝑧1 ) ; which is purely imaginary or zero. |𝑧1 −𝑧2 |2

Hence option A & D are correct.

𝐴 52. We know that, 𝑃 ( ) 𝐵 𝑃(𝐴)+𝑃(𝐵)−𝑃(𝐴∪𝐵) 𝑃(𝐵)

since Mock Test - 6

𝑃(𝐴 ∪ 𝐵) < 1

=

𝑃(𝐴∩𝐵) 𝑃(𝐵)

0 0 D. ΔG < 0 38. Iodine reacts with hypo to give: A. NAI B. Na2SO3 C. Na2S4O6 A. I-d, II-a, III-c, IV-b

B. I-c, II-b, III-d, IV-a

C. I-b, II-c, III-a, IV-d

D. I-a, II-d, III-b, IV-c

32. Which of the following has the maximum number of unpaired d-electrons? A.

Zn2+

C.

Ni3+

B.

Fe2+

D.

Cu+

D. Na2 SO4 39. The spectrum of helium is expected to be similar to that of: A. Li+

B. He

C. H

D. Na

33. Which property of colloids is not dependent on the charge on colloidal particles?

40. There are two samples of HCl having molarity 1M and 0.25N. Find the volume of these sample taken in order to prepare 0.75N HCl solution (Assume no water is used):

A. Coagulation

B. Electrophoresis

A. 20 mL, 10 mL

B. 100 mL, 50 mL

C. Electro-osmosis

D. Tyndall effect

C. 40 mL, 20 mL

D. 50 mL, 25 mL

34. Propane-I-ol can be prepared from propene by : A. H2O/H2SO4 B. Conc.H2SO4

Mathematics

C. B2H6 followed by H2O2/OH D. CH3CO2H/H2SO4 35. A group of atoms can function as a ligand only when : A. it is a small molecule B. it has an unshared electron pair C. it is a negatively charged ion

Mock Test - 7

41. Let I be an open interval contained in the domain of a real function f′, then f(x) is called strictly decreasing function in I ifA. x1 0 for 𝑥 ∈ (0, )

Hence option B is the correct answer.

2

(B) 42. In given interval, both cot Use

𝑥 and tan𝑥 are positive.

𝐴𝑀 ≥ 𝐺𝑀 for positive quantities. 1 2

(16cot𝑥 ⋅ 9tan𝑥) 16cot𝑥 + 9tan𝑥 ≥ 24

𝑓 ′ (𝑥) = cos𝑥

46. (A)

16cot𝑥+9tan𝑥 2

𝑓 (𝑥) = sin𝑥 ′

𝑓 ′ (𝑥) ≥

𝜋 2

< 0 for 𝑥 ∈ (0, )

𝑓 ′ (𝑥) = 𝑐𝑜𝑠𝑒𝑐 2 𝑥 𝜋

𝑓 1 (𝑥) < 0 for 𝑥 ∈ (0, ) 2

(D)

𝑓 ′ (𝑥) = −cot𝑥 ⋅ 𝑐𝑜𝑠𝑒𝑐𝑥 𝜋 2

Hence option C is the correct answer.

𝑓 ′ (𝑥) < 0 for 𝑥 ∈ (0, )

43.

Hence option A is the correct answer.

47. Given,

𝑥2 + 𝑦2 + 𝑥 − 𝑦 = 0

Differentiating w.r.t 𝑑𝑦 𝑑𝑥

𝑑𝑦 𝑑𝑥

=0⇒

=

𝑥, we get 2𝑥 + 2𝑦

𝑑𝑦 𝑑𝑥

+1−

1+2𝑥 1−2𝑦

Thus slope of the tangent at origin is,

𝑑𝑦 𝑑𝑥 (0,0)

𝑚=( )

=

1 Hence required line through (-2,3) is. (𝑦 − 3) = 1(𝑥 + 2) ⇒ 𝑥 − 𝑦 + 5 = 0 Hence option C is the correct answer. Hence option B is the correct answer.

𝑥, 𝑦, 𝑧 be the magnitudes of the displacements along the three axes 𝑋, 𝑌, 𝑍 respectively. Let 𝑉𝑥 , 𝑉𝑦 , 𝑉𝑧 be the respective velocities and 𝑎𝑥 , 𝑎𝑦 , 𝑎𝑧 be the respective accelerations along the axes 𝑋, 𝑌, 𝑍 48. Let

44.

respectively.

𝑉𝑥 =

𝑑𝑥

= 1, 𝑉𝑦 =

𝑑𝑡 𝑑𝑉 𝑎𝑥 = 𝑥 𝑑𝑡

1 2

1 3 𝑑𝑧

𝑥 = 𝑡, 𝑦 = 𝑡 2 , 𝑧 = 𝑡 3 𝑑𝑦

= 𝑡, 𝑉𝑧 =

𝑑𝑡 𝑑𝑉𝑦

= 0, 𝑎𝑦 =

𝑑𝑡

𝑑𝑡

= 1, 𝑎𝑧 =

= 𝑡2 𝑑𝑉𝑥 𝑑𝑡

= 2𝑡

∴ acceleration at 𝑡 = 1 is 𝑗 + 2𝑘 Mock Test - 7

| 135

Hence option A is the correct answer. log𝑥 𝑥 1 log𝑥 1−log𝑥 ′ Therefore, 𝑓 (𝑥) = 2 − 2 = 𝑥 𝑥 𝑥2 ′ For 𝑓(𝑥) to be increasing, 𝑓 (𝑥) > 0 ⇒

49. We have

𝐴 = |𝐴|𝐴−1 𝑎𝑑𝑗(𝐴𝐵) = |𝐴𝐵|(𝐴𝐵)−1 = |𝐴||𝐵|(𝐴𝐵)−1 = |𝐴||𝐵|(𝐵−1 𝐴−1 ) = (|𝐵|𝐵−1 )(|𝐴|𝐴−1 ) = 𝑎𝑑𝑗𝐵 adj 𝐴 53. We have, adj

𝑓(𝑥) =

1 − log𝑥 >

0 ⇒ 1 > log𝑥 ⇒𝑒>𝑥 Therefore, 𝑓(𝑥) is increasing in the interval (0, 𝑒).

Hence option A, B and D are the correct answer. 54.

Hence option A is the correct answer. 50.

Hence option D is the correct answer.

55.

sin−1 (𝑥 2 + 𝑥 + 1) + cos−1 (𝜆𝑥 + 1) =

cos

−1

𝜋

51.

2

−1

(𝜆𝑥 + 1) = − sin (𝑥 2 + 𝑥 + 1) 2

Taking cos on both sides,

Hence option D is the correct answer.

𝜋

𝜆𝑥 + 1 = 𝑥 2 + 𝑥 + 1 𝑥 2 +

(1 − 𝜆)𝑥 = 0 𝑥(𝑥 + 1 − 𝜆) = 0 𝑥 = 0 or 𝜆 = 𝑥 + 1 sin−1 (𝑥 2 + 𝑥 + 1) is defined for −1 ≤ 𝑥 2 + 𝑥 + 1≤1 𝑥 2 + 𝑥 + 2 ≥ 0 is always true. 𝑥(𝑥 + 1) ≤ 0 ⇒ 𝑥 ∈ [−1,0] ⇒ 𝑥 + 1 ∈ [0,1] 𝜆 ∈ [0,1] 𝜆 ≠ −1,2 Also when 𝜆 = 1, there is only one solution to the given equation i.e.,

𝑥 = 0 𝑆𝑜, 𝜆 ≠ 1

Hence option A and D are the correct answers

Hence option A, C and D are the correct answer.

52.

56. Shoot-tip culture refers to the culture of the terminal portion of a shoot comprising the shoot apical meristem together with primordial and developing leaves and adjacent stem tissue. As the meristematic tissues are the tissues with the rapid cell division property and hence are free from viral infection. Thus, this technique is used for the production of virus-free plants. Hence option A is the correct answer.

Hence option D is the correct answer.

Mock Test - 7

| 136

57. The polyploid organism contains more than two sets of chromosomes. There are many naturally occurring flowering species which are polyploid. Polyploidy breeding is performed to create the plants with polyploids. There are two types of polyploids which can be generated, as allopolyploid and autopolyploid. The somatic cells are fused to generate the polyploid hybrids which can be used to study the natural polyploids and to develop new polyploid species. Hence option B is the correct answer. 58. PUFA is the abbreviated form of polyunsaturated fatty acid. These are the fatty acids which contain more than one double bond in their carbon structure. This is the group of fatty acids which contains essential oils like the omega-3 fatty acids, omega-6, omega-9 and other conjugated fatty acids. These compounds are required in the diet for the growth and development of animals Hence option A is the correct answer. 59. Flavr Savr is a genetically modified tomato. This was the first commercially grown genetically engineered food. Flavr Savr tomato is more resistant to rotting. This feature is due to the introduction of an antisense gene which interferes with the production of enzyme polygalacturonase. This enzyme is responsible for the degradation of pectin which is present in the cell wall of the fruit. This results in the softening and makes the fruit susceptible to fungal infections. The modified fruit has extended shelf life due to the reduction in the ripening rate.

So, the correct answer is 'General surface of plants'. Hence option D is the correct answer. 62. Fossorial adaptation is seen in those animals which live underground and dig ground for their food. These animals are diggers and burrowers. In these animals, forelimbs are modified into broad, strong and stout limbs with long clawed digits as forelimbs are used to loosen the earth. Girdle bones are also modified to provide strength during digging. Examples of fossorial animals include moles, mole rat, hedgehogs, etc. Arboreal animals lives on land but come to trees for shelter. Cursorial animals are runner. They are adapted to run fast. Volant animals are those which are adapted in such way that they are able to fly in air for long time. Hence option C is the correct answer. 63. Genetic engineering is the direct manipulation of an organism's genome by using biotechnology. This new gene or DNA may result in the development of certain characters. The genetic monsters are developed due to genetic engineering are the organisms which disturb the balance of nature. These are evil animals which are formed during genetic experiments and have abnormal genes and chromosomes. They can destroy humanity and other animals by interbreeding. There are institutions which propose guidelines for the development of organisms by genetic engineering. This institution is against the formation of genetic monsters due to fatal consequences. Hence option B is the correct answer.

Hence option B is the correct answer. 60. Non-indigenous animals and plant species introduced from other countries and which are not otherwise found local are termed exotic. In India, large varieties of exotic animal and plant species have been introduced from other parts of the world through the ages. Some exotic plants have turned into weeds, multiplying fast and causing harm to the ecosystem, e.g., water hyacinth and Lantana camara. Water hyacinth was introduced into Indian waters to reduce pollution. It has clogged water bodies including wetlands at many places resulting in the death of aquatic organisms. Hence option C is the correct answer. 61. Submerged hydrophytes grow below the water surface and thus they have the numbers of adaptation to live in water such as they respire through general leaf surface due to the absence of stomata. They do not possess an external layer on the epidermis and thus easily absorb nutrients, water and gases.

Mock Test - 7

64. Xerophytic plants have the following adaptations. A thick cuticle on stems and leaves which protect from excessive transpiration. These plants show the presence of sunken stomata which reduce the loss of water. There are certain thick walled cells of hypodermis which do not allow the passage of water and prevents water loss. The leaves are reduced in size and may be modified to form the spines that reduce the surface area for transpiration. The root system is highly developed for water absorption. In these pants the stem also performs photosynthesis. Hence option C is the correct answer. 65. Nandankanan is famous for birth of white tiger from normal coloured tigers. It is one of the major host zoo for the white tigers. A unique white tiger safari was established in the zoological park, on 1st October 1991. Hence option C is the correct answer. 66. Darwin's finches are known for their diverse beak shapes and structures. This arises due to different feeding

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habit and habitat and is an example of adaptive radiation. Adaptive radiation refers to the diversification of organisms from their ancestral species due to change in habitat, feeding habit etc. So, option D is the right choice. 67. a) Cybrids are artificial hybrid cell produced by inserting nuclear material from one organism into an enucleated cell. b) Hybrids are the recombinant individuals, that have genes from both parents and thus, show new combination of traits. c) Clones are the population of genetically identical individuals which are produced by asexual or vegetative reproduction. Sexual reproduction does not produce clones owing to the three processes that causes recombination of genetic material and produce variations in progeny. These three processes are cross over during meiosis, independent assortment of chromosomes during meiosis and random fusion of gametes. The resultant progeny will be a hybrid. Asexual reproduction does not include meiosis and fusion of gametes; omits the variation creating processes and produce clones.

70. As a result of the struggle for existence, only those organisms could survive which have the favourable variations or changes to adapt to the environmental conditions. In other words, the struggle for existence results in the survival of the fittest. This is called natural selection that tends to favour those organisms that are most fit to survive and reproduce in a particular environment. The adaptations are genetically controlled and are heritable. Natural selection can alter the frequency of the heritable traits in three ways: Directional, stabilizing and disruptive selection. Under Directional selection, the frequency distribution of the trait is shifted from where it was present in the parental generation. Under Stabilizing selection, the population is kept well adapted to the environment at the genetic level and therefore, the population remains genetically constant. Under Disruptive selection, the populations which are homologous previously break into different adaptive forms. Hence option B is the correct answer.

So, the correct answer is 'Clone' Hence option C is the correct answer. 68. Besides dung, the weed that can be used in biogas production is Eichornia crassipes. An aquatic weed like water hyacinth (Eichhornia crassipes) is used as a source of biogas through harvesting, chopping and crushing. Hence, the correct answer is option C. 69. A. Sexual selection is an evolutionary model that explains the selection of mates by organisms. B. Convergent evolution is a process during which organisms with different evolutionary history evolve similar phenotypic adaptations in response to common environmental conditions. C. Genetic drift is a random change in allele frequencies over the generation. In small populations frequencies of particular alleles may change drastically by chance alone. Such changes occur randomly as if frequencies were drifting and thus, are known as genetic drift. D. Isolation is condition when an organism is in the state of being isolated. Hence the species that have slightly different diets, behaviours, and habitats within the lake, diverged from a common ancestor as a result of Convergent evolution. So, the correct answer is 'Convergent evolution'. Mock Test - 7

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MOCK TEST - 8

6. The average weight of 8 person's increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What might be the weight of the new person ?

General Ability 1. Look at this series: 1.5, 2.3, 3.1, 3.9, ... What number should come next? A. 4.2

B. 4.4

C. 4.7

D. 5.1

A. 76 kg

B. 76.5 kg

C. 85 kg

D. None of these

7. The average weight of 16 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg. Find the average weights of all the boys in the class ? A. 47.55 kg

B. 48 kg

2. Look at this series: 14, 28, 20, 40, 32, 64, ... What number should come next?

C. 48.55 kg

D. 49.25 kg

A. 52

8. The hazards of radiation belts include:

B. 56

C. 96

D. 128

3. Find the statement that must be true according to the given information. Sara lives in a large city on the East Coast. Her younger cousin Marlee lives in the Mid-west in a small town with fewer than 1,000 residents. Marlee has visited Sara several times during the past five years. In the same period of time, Sara has visited Marlee only once. A. Marlee likes Sara better than Sara likes Marlee.

A. deterioration of electronic circuits B. damage of solar cells of spacecraft C. adverse effect on living organisms D. All of the above 9. The intersecting lines drawn on maps and globes are: A. latitudes

B. longitudes

C. geographic grids

D. None of the above

B. Sara thinks small towns are boring. C. Sara is older than Marlee.

10. Under Akbar, the Mir Bakshi was required to look after:

D. Marlee wants to move to the East Coast. 4. Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5? A.

A. military affairs

B. the state treasury

C. the royal household

D. the land revenue system

1

Physics

2

B.

2 5

C.

8 15

D.

9 20

5. A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 6500? A. Rs. 4991

B. Rs. 5991

C. Rs. 6001

D. Rs. 6991

Mock Test - 8

11. An ideal choke takes a current of 10 A when connected to an ac supply of 125 V and 50 Hz. A pure resistor under the same conditions takes a current of 12.5 A. If the two are connected to an ac supply of 100 V and 40 Hz, then the current in series combination of above resistor and inductor is: A. 10/√2 A

B. 12.5 A

C. 20 A

D. 10 A

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12. An aeroplane flying horizontally at a height of 980 m with velocity 100 m/s drops a food packet. A person on the ground is 100(√2−1) m ahead horizontally from the dropping point. At what velocity approximately should he move so that he can catch the food packet. A. 50√2 ms−1

B. 50/√2 ms-1

C. 100 ms-1

D. 200 ms-1

13. Two bodies with moment of inertia I1 and I2(I1>I2) have equal angular momentum. If the KE of rotation is E 1 and E2 , then : A. E1>E2 B. E1 R) is given by (in J/m3) : A.

𝑄2 /(32𝜋 2 𝜖0 𝑅2 𝑟 2 )

B.

𝑄2 /(32𝜋 2 𝜀𝑜 𝑟4 )

C.

𝑄2 /(32𝜋 2 𝜖0 𝑅4 )

D.

𝑄2 /(32𝜋 2 𝜖0 𝑅4 )

15. An aeroplane flying at constant speed 105 m/s towards East, makes a gradual turn following a circular path to fly South. The turn takes 15 seconds to complete. The magnitude of the centripetal acceleration during the turn is A. 7π/2 m/s2

B. 46π/3 m/s2

C. 23π/6 m/s2

D. 14π m/s2

A. g/2

B. g/3

C. g

D. zero

19. A block of mass M is pulled by a uniform chain of mass M tied to it by applying a force F at the other end of the chain. The tension at a point distant quarter of the length of the chain from free end will be:

16. Electric potential in a region varies as V=(1500x 2 kVolt/m2). Total electric charge lying inside the cube of side 10 cm shown in figure is (ϵ0=8.9×10−12C2/Nm2) : A. 7F/8

B. 4F/5

C. 3F/4

D. 6F/7

20. A current of 4A produces a deflection of 30 ∘ in the galvanometer. The figure of merit is:

A. −2.67×10−8C

B. +2.67×10−8C

C. −5.24×10−8C

D. +5.24×10−8C

Mock Test - 8

A. 6.5 A/rad

B. 7.6 A/rad

C. 7.5 A/rad

D. 8.0 A/rad

21. Two straight conducting rails form a right angle where their ends are joined. A conducting bar in contract with the rails starts at the vertex at t = 0 and moves with

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constant velocity v along them as shown. A magnetic field B is directed into page. The induced emf in the circuit at any time 't' is proportional to:

A. B. C. A. to

B. t

C. v

D. v2

𝜎𝑒𝑅

√2𝑚𝜀

0

𝜎𝑒𝑅

√𝑚𝜀

0

2𝜎𝑒𝑅

√ 𝑚𝜀

0

D. None of these

22. A uniform cylinder of mass M and radius R is placed on a rough horizontal board, which in turn is placed on a smooth surface. The coefficient of friction between the board and the cylinder is μ. If the board starts accelerating with constant acceleration a, as shown in the figure, then:

24. In a dark room with ambient temperature T0, a black body is kept at a temperature T. Keeping the temperature of the black body constant (at T), Sun rays are allowed to fall on the black body through a hole in the roof of the dark room. Assuming that there is no change in the ambient temperature of the room, which of the following statements is correct? A. The quantity of radiation absorbed by the black body in unit time will increase. B. Since emissivity = Absorptivity. Hence, the quantity of radiation emitted by black body in unit time will increase.

A. for pure rolling motion of cylinder, the direction of frictional force is forward and its magnitude is Ma/3

C. Black body radiates less energy in unit time in the visible spectrum.

B. the maximum value of 'a', so that cylinder performs pure rolling, is 3µg

D. The reflected energy in unit time by the black body remains same.

C. The acceleration of the center of mass of the cylinder under pure rolling condition for the given 𝑎 is

2𝑎 3

25. In YDSE, the source S is not symmetrically placed from the slits S1 and S2 (in figure).

D. for pure rolling motion of cylinder, direction of the frictional force is backward 23. An infinite dielectric sheet having charge density o has a hole of radius

𝑅 in it. An electron is released on the

axis of the hole at a distance √3𝑅 from the center. Find the speed with which it crosses the center of the hole. If the separation between slits and screen is D (d < < D), then A. the central fringe shifts by (√2-1)D above the previous location of central fringe

Mock Test - 8

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B. the shift in the third bright band of red colour is more than that of yellow colour

31. Select the correct I.U.P.A.C name for [Cu(NH3)4] [PtCl4] :

C. all the fringes have same shift and no change in brightness

A. Tetraamminecopper (II) tetrachloridoplatinum (II)

D. the pattern becomes unsymmetrical about the original central fringe

C. Tetraamminecopper(II) tetrachloridoplatinate (II)

B. Tetraamminecopper (II) tetrachloridoplatinum (IV)

D. Tetraamminecuprate(II) tetrachloridoplatinate (II) 32. Which of the following can cause liver cancer? A. CHCl3

Chemistry

B. CCl4 C. CH2Cl2

26. Cadmium oxide (CdO) has NaCl type structure with density equal to 8.27 g cm−3. If the ionic radius of O2- is 1.24 A˚, determine the ionic radius of Cd 2+.

D. C14H9Cl5

A. 1.5

A. Nylon-6,6

B. Dacron

C. PVC

D. Bakelite

B. 1.1

C. 1.9

D. 1.6

27. Equivalent conductance at infinite dilution, λ 0 of NH4Cl, NaOH and NaCl are 128.0,217.8 and 109.9 ohm−1cm2eq−1 respectivley. The equivalent conductance of 0.01N NH4OH is 9.30 ohm−1cm2eq−1, then the degree of ionization of NH4OH at this temperature would be :

33. Which of the following has an ester linkage?

34. Match the following.

List I

List II

A) Sucrose

i) 𝛽 -D-Galactose and 𝛽 -DGlucose

B) Cellulose

ii) 𝛼 -D-Glucose and 𝛽 -DFructose

28. Which of the following gives the maximum number of isomers ?

C) Starch

iii) 𝛽 -D-Glucose

D) Latose

iv) 𝛼 − 𝐷 − Glucose

A. [Co(NH3)4Cl2]

A. A-i; B-ii; C-iii; D-iv

B. A-ii; B-i; C-iii; D-iv

C. A-ii; B-iii; C-iv; D-i

D. A-iv; B-iii; C-i; D-ii

A. 0.04

B. 0.1

C. 0.39

D. 0.62

B.

[Ni(en)(NH3)4]2+

C. [Ni(C2O4)(en)2] D. [Cr(SCN)2(NH3)4]+

35. Which of the following is useful as a food preservative ?

29. How many electrons are involved in the following redox reaction?

A. Sucrolose

B. Salts of sorbic acid

C. Ascorbic acid

D. Citric acid

𝐶𝑟2 𝑂72−

+ 𝐹𝑒 2+

+ 𝐶2 𝑂42−

→

𝐶𝑟 3+

+ 𝐹𝑒 3+

+ 36. Directions: The following question has four choices, out of which one or more is/are correct.

𝐶𝑂2 (Unbalanced) A. 0

B. 2

C. 4

D. 6

30. The heat of neutralisation of a strong acid and a

57.0𝑘𝐽𝑚𝑜𝑙 −1 . The heat released when 0.5 mole of 𝐻𝑁𝑂3 solution is mixed with 0.2 mole of 𝐾𝑂𝐻 is :

Consider the following equilibrium. 1 2

𝑆𝑂3 (𝑔) ⇌ 𝑆𝑂2 (𝑔) + + 𝑂2 (𝑔)

strong alkali is

A. 57.0kJ

B. 11.4 kJ

C. 28.5 kJ

D. 34.9 kJ

Mock Test - 8

8 g of SO3 is kept in a container at 527oC. The equilibrium pressure and density are 1.6 atm and 1.6 gL -1 , respectively. (R = 0.08 L. atm/mol. K) With reference to the given data, which of the following facts about the reaction is/are true?

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A. The value of Kr is 0.5 atm1/2. B. The value of Kr is 0.253

atm1/20.

C. Average molar mass of the equilibrium mixture is 64 g/mol.

C. A secondary alcohol, on oxidation, gives a carboxylic acid containing the lesser number of carbon atoms. D. A primary alcohol, on oxidation, gives a carboxylic acid containing the same number of carbon atoms.

D. At equilibrium, Q = Keq and ΔG = (-) ve. 37. Directions: The following question has four choices, out of which one or more is/are correct.

Mathematics

Boron hydride is an electron deficient species with an incomplete octet. It dimerises through banana bonding and exists as diborone (B 2H6). Borax, Na2[B4O5(OH)4]·8H2O and borzine, B3H6N3 are other compounds of boron.

41. The vertex of the parabola x2=8y−1 is: A. (−1/8,0)

B. (1/8,0)

Which of the following statements is/are correct?

C. (0,1/8)

D. (0,−1/8)

A. In B2H6, number of 2 centre-2 electron bonds is 4 and at the most, 4 hydrogen atoms are present in the same plane.

42. The equation of the circle of radius 3 that lies in the fourth quadrant and touching the lines x=0 and y=0 is:

B. In B2H6, bond angle Ht B Ht > HbBHb (t: terminal b: bridging) C. In the structure of anion of borax, the number of BO-B units is 5.

A. x2+y2−6x+6y+9=0 B. x2+y2−6x−6y+9=0 C. x2+y2+6x−6y+9=0 D. x2+y2+6x+6y+9=0

D. Borazine is a non-planar molecule. 38. Directions: The following question has four choices, out of which one or more is/are correct. The cation precipitated by ammonium chloride and aqueous ammonia isA.

Bi+3

C.

Mg+2

B.

Pb+2

D.

Fe+3

39. Aqueous solutions of HNO3, KOH, CH3COOH and CH3COONa of identical concentrations are provided. The pair of solutions which forms a buffer upon mixing is

43. If t1 and t2 are the ends of a focal chord of the parabola y2=2x then: A. t21+t22=2 B. t1+t2=1 C. t1t2=−1 D. none of these 44. The nature of the intercepts made on the axes by the tangent at the point (16/5,9/5) to the ellipse 9x2+16y2=144 are: A. equal

A. HNO3 and CH3COOH

B. unequal

B. KOH and CH3COONa

C. equal in magnitude but opposite in sign

C. HNO3 and CH3COONa

D. in the ratio 1:2

D. CH3COOH and CH3COONa 40. Directions: The following question has four choices, out of which one or more is/are correct. Which of the following statements is incorrect? A. The substitution of hydroxyl group by a halogen group in alcohol is an electrophilic substitution reaction. B. Alcohols are weak acids as well as weak bases.

Mock Test - 8

45. Equation of the two tangents drawn from (2,−1) to x2+3y2=3 are: A. x=2, 4x+3y−7=0

B. y=−1, 4x+y−7=0

C. x+y−1=0, 4x+y−7=0

D. x+y+1=0, 4x+y−7=0

46. A double ordinate of the parabola y2=8px is of length 16p. The angle subtended by it at the vertex of the parabola is

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A. π/4

B. π/2

D. nothing can be said in general

C. π/3

D. none of these

47. The distance of a point P on the ellipse the centre is 2 the eccentric angle of P is:

x 2+3y2=6

A. π/2

D. π/3

B. π/6

C. π/4

from

54. Tangent is drawn at any point P of a curve which passes through (1, 1) cutting x-axis and y-axis at A and B, respectively. If BP : AP = 3 : 1, then A. differential equation of the curve is 3x dy/dx + y = 0 B. differential equation of the curve is 3x dy/dx - y = 0

48. The parabola

y2−2x−6y+5=0

has :

C. curve is passing through (1/8 , 2)

A. Focus ≡(−3/2,3)

B. Vertex ≡(−2,3)

C. Directrix x=−5/2

D. all of these

49. Find the latus rectum of the parabola x 2+2y−3x+5=0 ? A. 1

B. 2

C. 3

D. 4

50. The equation of the directrix of the parabola y2+4y+4x+2=0 is: A. x = −1

B. x = 1

C. x = −3/2

D. x = 3/2

D. normal at (1, 1) is x + 3y = 4. 55. Let PQ be a chord of the parabola y2 = 4x. A circle drawn with PQ as diameter passes through the vertex V of the parabola. If area of ΔPVQ = 20 square units then coordinates of P are: A. (-16, -8)

B. (-16, 8)

C. (16, -8)

D. (16, 8)

51. Directions: The following question has four choices, out of which ONE or MORE can be correct.

Biology

Three lines px + qy + r = 0, qx + ry + p = 0 and rx + py + q = 0 are concurrent, if:

56. Sustentacular cells are found in:

A. p + q + r = 0

A. Testis of mammal

B. Ovary of mammal

B. p2 + q2 + r2 = pr + rq

C. Testis of Ascaris

D. Pancreas of frog

C. p3 + q3 + r3 = 3 pqr D. None of these

57. Gametes of similar morphology fuses in: A. Oogamy

B. Isogamy

52. Directions: The following question has four choices, out of which ONE or MORE can be correct.

C. Anisogamy

D. None of the above

The equation(s) of the tangents drawn from the origin to the circle x2 + y2 + 2rx + 2hy + h2 = 0 is/are:

58. Which one of the following statements about human sperm is correct?

A. x = 0

A. Acrosome serves no particular function.

B. y = 0

B. Acrosome has a conical pointed structure used for piercing and penetrating the egg resulting in fertilization.

C. (h2 - r2) x - 2rhy = 0 D. (h2 - r2) x + 2rhy = 0 53. Directions: The following question has four choices, out of which ONE or MORE can be correct.

C. The sperm lysins in the acrosome dissolve the egg envelope facilitating fertilization. D. Acrosome serves as a sensory structure leading the sperm towards the ovum.

Let h(x) = f(x) - (f(x))2 + (f(x))3 for every real number x. Then,

59. In a population, stabilizing selection is the result of :

A. h is increasing whenever f is increasing

A. Founder effect

B. Heterozygote advantage

B. h is increasing whenever f is decreasing

C. Population bottleneck

D. Random mating

C. h is decreasing whenever f is decreasing Mock Test - 8

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60. Allelic sequence variation where more than one variant allele at a locus in a human population with a frequency greater than 0.01 is referred to as: A. DNA polymorphism

B. Multiple allelism

C. SNP

D. EST

61. "The genetic variations appearing among the members of a population bring about evolution". This statement defines:

B. Removal of stigma C. Removal of entire organisms D. Removal of petals and sepals 68. Plants obtained through tissues culture are genetically identical and they are obtained by somatic cells. What do you call them? A. Somaclones

B. Monoclones

C. Somatic hybrids

D. Cross hybrids

A. Mutation theory B. Lamarckism

69. The domestic sewage in large cities:

C. Darwinism

A. Is processed by aerobic and then anaerobic bacteria in the secondary treatment in sewage treatment plant (STPs).

D. Neo-Darwinism 62. A - DNA is A. Left-handed helix with 12 nucleotide pairs per turn B. Right-handed helix with 11 nucleotide pairs per turn C. Right-handed helix with 12 nucleotide pairs per turn D. Left-handed helix with 11 nucleotide pairs per turn

B. When treated in STPs does not really require the aeration step as the sewage contains adequate oxygen. C. Has very high amounts of suspended solids and dissolved salts. D. Has a high BOD as it contains both aerobic and anaerobic bacteria.

63. The hormone that prevents ovulation and maintainance of corpus luteum is:

70. Cultigens are:

A. Progesterone

B. Estrogen

A. Plants that require human protection

C. LH

D. FSH

B. That are protected by wild life

64. Loss of a X chromosome in a particular cell during its development, results into: A. Diploid individual

B. Triploid individual

C. Gynandromorphs

D. Both (a) and (b)

C. Both A and B D. Transposomes

65. A set of genes will be in a complete linkage when the progeny phenotypes for parental (P) and recombinant (R) types are: A. P = 0%, R = 100%

B. P = 50%, R = 50%

C. P < 50%, R > 50%

D. P = 100%, R = 0%

66. A process that uses micro-organisms to convert harmful industrial wastes to less toxic or non-toxic compounds is known as: A. Complement fixation

B. Precipitation

C. Bioremediation

D. Bioconversion

67. Emasculation is achieved byA. Removal of anther

Mock Test - 8

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SMART ANSWERSHEET

Correct Q.

1

Ans.

Correct Q.

Ans.

Skipped

Skipped

0.0 %

0.0 %

C

15

A

B

6

0.0 %

100.0 %

100.0 %

0.0 %

0.0 %

0.0 %

C

20

B

C

0.0 % 35 0.0 %

B 100.0 % 0.0 %

A

25 100.0 %

A, B, D A, C, D

0.0 % C

0.0 % 39

0.0 %

41

0.0 %

0.0 % 52

A, C 100.0 %

0.0 %

0.0 % 53

42

A, C

100.0 %

100.0 % 0.0 % C 100.0 % 0.0 % 67

A 100.0 %

0.0 % A, C

0.0 % 68

A

100.0 %

100.0 %

0.0 % 55

C, D

0.0 % 69

A

100.0 %

100.0 %

0.0 % 56

100.0 %

D

100.0 % 54

A

0.0 %

66

100.0 %

0.0 %

D

100.0 %

100.0 %

0.0 %

100.0 %

0.0 % 28

0.0 % C 100.0 % 65

100.0 %

C

100.0 %

100.0 % 64

A, C

0.0 %

A

0.0 % A

0.0 % 51

100.0 %

0.0 % 27

100.0 %

0.0 %

0.0 %

A

0.0 % B

100.0 %

0.0 % 40

100.0 %

63

D

100.0 %

100.0 %

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Performance Analysis Avg. Score (%) Toppers Score (%) Your Score

Mock Test - 8

0.0% 0.0%

| 146

HINTS AND SOLUTIONS 1. In this simple addition series, each number increases by 0.8. Hence option C is the correct answer. 2. This is an alternating multiplication and subtracting series: First, multiply by 2 and then subtract 8. Hence option B is the correct answer. 3. Since the paragraph states that Marlee is the younger cousin, Sara must be older than Marlee. There is no information to support the other choices. Hence option C is the correct answer.

4. Here,

𝑆 = {1,2,3,4, … . ,19,20}

𝐸 = event of getting a multiple of 3 or 5 = {3,6,9,12,15,18,5,10,20} Let

∴ 𝑃(𝐸) =

𝑛(𝐸) 𝑛(𝑆)

=

9 20

Hence option D is the correct answer.

8. During periods of intense space weather, the density and energy of radiation belt particles can increase and pose a danger to astronauts, spacecraft, and even technologies on the ground. Some particles in the radiation belts move at nearly the speed of light, which is about 186,000 miles per second. Hence option D is the correct answer. 9. The geographic grid is a system of imaginary arcs drawn in a grid pattern on Earth's surface. The location of any place on Earth's surface can be described by these humancreated arcs, known as meridians and parallels. The geographic grid plays an important role in telling time. Hence option C is the correct answer. 10. The head of the military was called the Mir Bakshi, appointed from among the leading nobles of the court. The Mir Bakshi was in charge of intelligence gathering, and also made recommendations to the emperor for military appointments and promotions. Hence option A is the correct answer.

11.

𝑋𝐿 = 𝑉rms /𝐼гто = 125/10 = 12.5𝛺

𝑅 = 125/12.5 = 10𝛺 𝑎𝑡40𝐻𝑧, 𝑋𝐿′′ = 12.5

5. Total sale for 5 months = Rs. (6435 + 6927 + 6855 + 7230 + 6562) = Rs. 34009. ∴ Required sale = Rs. [ (6500 x 6) - 34009 ] = Rs. (39000 - 34009)

40 50

= 10𝛺

When connected in series, imepdance

√𝑅 2 + 𝑋𝐿𝑛2 =

√102 + 102 = 10√2𝛺 =

100

= Rs. 4991.

Current in series combination

Hence option A is the correct answer.

Hence option A is the correct answer.

6. Total weight increased = (8 x 2.5) kg = 20 kg.

12. For food packet

Weight of new person = (65 + 20) kg = 85 kg.

𝑠𝑣 = −980𝑚

Hence option C is the correct answer.

𝑔 = −10𝑚/𝑠2

7.

𝑠 = 𝑢𝑣 𝑡 + 𝑎𝑡 2

10√2

=

10 √2

𝐴

𝑢ℎ = 100𝑚/𝑠 𝑢𝑣 = 0

1 2

1 2

(−980) = 0 + (−10)𝑡 2 ⇒ 𝑡 = 14𝑠 Distance travelled horizontally in 14 s by food packet

Hence option C is the correct answer.

Mock Test - 8

𝑠ℎ =

𝑢ℎ × 𝑡 = 100 × 14 = 1400𝑚 Person position is 100(√2 − 1) ahead of food packet at time of dropping = 41.42𝑚 distance person has to run = 1400 − | 147

41.42 = 1358.57𝑚 in time = 14𝑠 so his speed will be

=

1358.57 14

= 97𝑚/𝑠

17. Torque applied

=

∼ 100𝑚/𝑠

𝑀𝑔 5𝐿 2 6

𝜏 = Force × perpendicular distance

5𝑀𝑔𝐿 12

=

= 𝐼𝛼 … … (1)

where 𝐼 is the moment of inertia about the hinge and the angular acceleration. Moment of inertia of the rod

Hence option C is the correct answer.

about an axis passing through an end of the 13. 𝐼1 Let

> 𝐼2

Substituting

𝜔1 and 𝜔2 be angular speed. 𝐼1 𝜔1 = 𝐼2 𝜔2

𝐸2 2 𝜔 2 2

𝐼 𝜔

𝐼 𝜔

2

2

=

𝑀𝐿2 𝛼 3

⇒𝛼=

Hence option B is the correct answer.

1 2

∴ 𝐾𝐸 ) Rotational = 𝐼𝜔2 1 2

5𝑀𝑔𝐿 12

𝑀𝐿2 3

5𝑔 4𝐿

∴ 𝜔2 > 𝜔1

𝐸1 = 𝐼1 𝜔12 , 𝐸2 =

𝐼 in eqn(1) we get

𝑟𝑜𝑑𝐼 =

𝛼 is

18. Elongation in the spring will be Finally, balancing forces,

= ( 1 1) 𝜔1 , 𝐸2 = ( 2 2) 𝜔2

𝑥=

2𝑚𝑔 𝑘

3𝑚𝑔 − 𝑘𝑥 = 3𝑚𝑎 𝑎 =

𝑔 3

Hence option B is the correct answer.

∴ 𝜔2 > 𝜔1 ∴ 𝐸2 > 𝐸1

19. If a be the acceleration of the whole system of mass

Hence option B is the correct answer.

𝑀𝑇 = (𝑀 + 𝑀) = 2𝑀, then 𝑀𝑇 𝑎 = 𝐹, 𝑎 =

𝐹 2𝑀

acceleration a should be same where we will found tension 14. The energy density is given by,

𝐸𝐷 = 1/2 × 𝜖0 ×

𝐸𝑟2

(𝑇) Now tension due to rest of mass, 𝑚𝑟 = 𝑀 +

Therefore,

3𝑀

4

4

𝑇 = 𝑚𝑟 𝑎 =

𝐸𝐷 = 1/2 × 𝑒0 × 𝑘 2 𝑄 2 /𝑟 4 = 𝜖0 /2 × 𝑄 2 / (16𝜋 2 𝜖02 𝑟 4 ) = 𝑄 2 /(32𝜋 2 𝜖0 𝑟 4 )

20. Here

Hence option B is the correct answer.

Now, 15. For

𝑀

(𝑀 − ) = 𝑀 +

𝑇 = 60𝑠 We have, 𝑇 =

2𝜋 𝜔

=

2𝜋𝑟 or 𝑉

𝑉=

Thus, we get the centripetal acceleration as

𝑉2 𝑟

2𝜋𝑟 𝑇

=

2𝜋𝑉 𝑇

7𝜋/2𝑚/𝑠2

= =

84 11

=

7𝑀 4

7𝐹 8

𝐼 = 4𝐴𝜃 = 30∘ = (

𝑘=

1/4 cycle it takes 15 sec. Therefore, time period

7𝑀 𝑃 4 2𝑀

=

𝐼 𝜃

=

4 𝜋 6

( )

=

4×6×7 22

30×𝜋 𝑐 ) 180

=

=

𝜋𝑐 6

2×6×7 11

= 7.6𝐴/𝑟𝑎𝑑

Hence option B is the correct answer. 21. At time 't', distance travelled by rod = ED = vt

Hence option A is the correct answer.

16. Given,

𝑉 = 1500𝑥 2

Electric field,

𝐸=− →

𝜕𝑉 𝑖̂ 𝜕𝑋

𝑘𝑣𝑜𝑙𝑡 𝑚2

= 3000𝑥

𝑘 zolt 𝑚

→

, 𝜙𝐸 = ∮ 𝐸 ⋅ 𝑑𝑠 = −3000 × 0.1 × 0.12 × 3 10 𝑉𝑚( as 𝑥 = 0.1𝑚 and 𝑑𝑠 = 0.12 ) Flux

⇒ 𝑞𝑒𝑛𝑑 = 𝜙𝐸 ⋅ 𝜖0 = −2.67 × 10−8 𝐶 where 𝜖0 = 8.9 × 10−12 Using Gauss's law,

Hence option A is the correct answer.

Mock Test - 8

tan𝜃 =

𝐴𝐷 𝐸𝐷

| 148

𝐴𝐷 = 𝐸𝐷tan𝜃 = 𝑣𝑡tan𝜃 tan(90 − 𝜃) =

Force on the electron,

𝐷𝐶 𝐸𝐷

𝑚𝑦

𝐷𝐶 = 𝑣𝑡cot𝜃

=−

𝜎𝑒𝑥

𝑚 ∫0 𝑣 𝑑𝑣 = − 𝑚

𝑒𝑚𝑓 = 𝐵 ∣ 𝑣 = 𝐵𝑣(𝐴𝐶)

𝑣2 2

=

−𝜎𝑒𝑥 2𝜀0 √𝑥 2 +𝑅2

2𝜀0 √𝑥 2 +𝑅2

𝑣

𝐴𝐶 = 𝐴𝐷 + 𝐷𝐶 = 𝑣𝑡tan𝜃 + 𝑣𝑡cot𝜃 = 𝑣𝑡(tan𝜃 + cot𝜃) Induced

𝑑𝑣 𝑑𝑥

𝐹=

𝜎𝑒 0 𝑥 𝑑𝑥 ∫ 2𝜀0 √𝑅 √𝑥 2 +𝑅2

𝜎𝑒 [√𝑥 2 2𝜀0

+ 𝑅2]

0 √3𝑅

𝜎𝑒𝑅 𝑚𝜀0

= 𝐵𝑣[𝑣𝑡(tan𝜃 + cot𝜃)]

𝑣=√

𝑒 = 𝐵𝑣 2 𝑡(tan𝜃 + cos𝜃)

Hence option B is the correct answer.

Hence, induced emf

∝ 𝑣2 ∝ 𝑡

Hence options B and D are the correct answers. 22. From the frame of reference, attached to the horizontal board, there is a pseudo force backward direction.

𝑀𝑎 on the cylinder in the

24. With the incident radiation, the temperature of black body tries to increase, so body emits more energy per unit time. To keep the temperature constant, it must absorb incident radiations with increased rate. The reflection depends on the nature of surface, not on the temperature. Hence options A, B and D are the correct answers.

So to prevent slipping, the frictional force acts in the forward direction. Let this force be

𝑓

25.

So the linear acceleration of the cylinder is,

𝐴 = (𝑀𝑎 −

𝑓)/𝑀 in the backward direction. 𝑓𝑅 and corresponding angular acceleration is given as, 𝛼 = 𝑓𝑅/𝐼 = 𝑓𝑅/0.5𝑀𝑅 2 = 2𝑓/𝑀𝑅 Torque due to this frictional force is

For pure rolling.

𝐴 = 𝛼𝑅 ⇒ (𝑀𝑎 − 𝑓)/𝑀 = 2𝑓/𝑀 ⇒ 𝑓 = 𝑀𝑎/3 Suppose, the position of zero order maxima is at Now the maximum value of frictional force is,

𝜇𝑀𝑔

So the maximum value of

distance 𝑦 from waves at

𝑀𝑎max /3 ⇒ 𝑎max

𝑃 is 𝛥𝑥 = (𝑆𝑆2 + 𝑆2 𝑃) − (𝑆𝑆1 + 𝑆1 𝑃)

𝑎 is given by, 𝜇𝑀𝑔 = = 3𝜇𝑔, for pure rolling to happen.

Now acceleration of the

𝐶𝑜𝑀 is, 𝐴 = (𝑀𝑎 −

𝑓)/𝑀 = 2𝑎/3

𝑈. The path difference between two

= −(𝑆𝑆1 − 𝑆𝑆2 ) + (𝑆2 𝑃 − 𝑆1 𝑃) 𝛥𝑥 = −(√2𝑑 − 𝑑) + 𝑑sin𝜃

Hence options A, B and C are the correct answers. For small 23. Potential function is not defined for infinite nonconducting sheet and hence to solve this, either calculate potential difference or use force equations. Electric field due to infinite dielectric sheet,

𝐸1 =

𝜎 [1 2𝜀0

−

𝑥 √𝑥 2 +𝑅2

]

Resultant electric field,

𝐸 = 𝐸1 − 𝐸2 = Mock Test - 8

𝜎 Electric field at 2𝜀0

𝜎 𝑥 2𝜀0 √𝑥 2 +𝑅2

𝜃 sin𝜃 ≅ tan𝜃 =

∴ 𝛥𝑥 = −(√2𝑑 − 𝑑) + For zero order maxima, or

the axis of a disc of radius

𝐸2 =

𝑃 at a

𝑦 𝐷

𝑑⋅𝑦 𝐷

𝐷𝑥 = 0

0 = −(√2𝑑 − 𝑑) +

𝑑⋅𝑦 𝐷

∴  𝑦 = (√2 − 1)𝐷 The shift of all the colours is same as there is no expression for wavelength. The pattern does not change at all, only shift of fringes takes place.

| 149

Hence options A, C and D are the correct answers.

𝐻𝑁𝑂3 will be neutralized with 0.2 moles of KOH. 0.3 moles of 𝐻𝑁𝑂3 will remain unneutralized. 30. 0.2 moles of

26. The expression for density 𝑛𝑀 𝑉𝑁𝐴

⇒ 8.27 =

4×128 𝑎3 𝑁𝐴

(𝑑) is given below 𝑑 =

The heat of neutralisation of a strong acid and a strong

˙

⇒ 𝑎 = 4.68𝐴

𝑛 is the number of atoms per unit cell, 𝑀 is the molecular mass, 𝑎 is the edge length and 𝑁𝑎 is Here,

Avogadro's number. So, the edge length of unit cell is 4.68 A However, for NaCl type structure 𝑟 +

𝑟+ =

4.68 2

57.0𝑘𝐽/𝑚𝑜𝑙 . When 0.2 moles of 𝐻𝑁𝑂3 is 𝐾𝑂𝐻 , the heat of neutralisation is 0.2𝑚𝑜𝑙 × 57.0𝑘𝐽/𝑚𝑜𝑙 = 11.4𝑘𝐽 alkali is

+ 𝑟−

𝑎 2

=

neutralized with 0.2 moles of

Hence option B is the correct answer.

⇒ 31. Option (C) is correct. I.U.P.A.C name for

˙

− 1.24 = 1.1𝐴

[𝐶𝑢(𝑁𝐻3 )4 ][𝑃𝑡𝐶𝑙4 ] complex is tetraamminecopper(II) tetrachloridoplatinate (II). 𝑁𝐻3 is neutral, making the first

Hence option B is the correct answer.

complex positively charged overall, so cobalt is in +2

27. According to kohlrausch law of independent migration of ions, the molar conductivity of an electrolyte at infinite dilution is equal to the sum of the contributions of the molar conductivites of its ions. Hence, ∞ ∞ 𝛬∞ 𝑒𝑞 (𝑁𝐻4 𝑂𝐻) = 𝛬𝑒𝑞 (𝑁𝐻4 𝐶𝑙) + 𝛬𝑒𝑞 (𝑁𝑎𝑂𝐻) − 𝛬∞ 𝑒𝑞 (𝑁𝑎𝐶𝑙)

Substitute values in the above equation

oxidation state. 𝐶𝑙 has a - 1 charge, making the second complex the anion. As the charge on platinum is - 2 so the name ends with -ate followed by the charge in roman letters. 32. Carbon tetrachloride can cause liver cancer and kidney damage. Hence option B is the correct answer. 33.

𝛬∞ 𝑒𝑞 (𝑁𝐻4 𝑂𝐻) = 128.0 + 217.8 − 109.9 = 235.9  oh 𝑚−1 𝑐𝑚2 𝑒𝑞 −1 The degree of dissociation is the ratio of the equivalent conductance at given concentration to the equivalent conductance at zero concentration. Hence, 9.30 235.9

𝛼=

𝛬𝑒𝑞 𝛬∞ 𝑒𝑞

=

= 0.04

Hence option A is the correct answer.

]+ gives maximum number of

28. [𝐶𝑟(𝑆𝐶𝑁)2 (𝑁𝐻3 )4 isomers. These includes cis trans isomers and linkage isomers. In cis isomer, two SCN ligands are adjacent and in trans isomers, they are opposite. Linkage isomerism is due to ambidentate nature of the ligand SCN as it can coordinate either through S or through

𝑁.

Hence option D is the correct answer.

𝐶𝑟2 𝑂72− + 2𝐹𝑒 2+ + 2𝐶2 𝑂42− → 2𝐶𝑟 3+ + 2𝐹𝑒 3+ + 4𝐶𝑂2 29.

The oxidation number of chromium in

𝐶𝑟2 𝑂72− is +6 and it

reduces to +3 in 𝐶𝑟 3+ . On balancing the equation, we will see 2 moles of chromium ion goes from +6 to +3 . Hence, there are 6 electrons involved in the above redox reaction. Hence option D is the correct answer. Mock Test - 8

Dacron has an ester linkage. Condensation of diacid with dialcohol leads to an ester linkage. Hence option B is the correct answer.

𝛼 -D-Glucose 𝛽 -D-Fructose. Cellulose is made up of 𝛽 -D-Glucose. Starch is made up of 𝛼 -D-Glucose. Lactose is made up of two monomers 𝛽 -D-Galactose and 𝛽 -D-Glucose. 34. Sucrose is made up of two monomers and

Hence option C is the correct answer. 35. Salts of sorbic acid are used as preservative in cheese, baked food, pickles and meat. Hence option B is the correct answer.

36.

1 2

𝑆𝑂3 (𝑔) ⇌ 𝑆𝑂2 (𝑔) + + 𝑂2 (𝑔) | 150

𝑛 at Eq. (0.1 − 𝑥) ×

𝑥 2

∪ 𝑠𝑖𝑛𝑔𝑃𝑀 = 𝑑𝑅𝑇 𝑀𝑎𝑣. =

1.6×0.08×800 1.6

Hence, (3) is true. Or,

= 64𝑔𝑚𝑜𝑙 −1

∴𝑀=

8 𝑥

0.1+2

= 64

𝑥 = 0.05 and 𝑛 𝑇 = 0.125

𝜒𝑆𝑂3 =

0.05 0.125

= 0.4

𝜒𝑆𝑂2 =

0.05 0.125

= 0.4

𝜒𝑂2 = 𝐾𝑃 =

0.025 0.125

However, borozine is a planar structure, hence option (4) is incorrect.

= 0.2

0.4××0.2 0.4

Hence, option (3) is correct.

1

1

(1.6)2 = 0.2529𝑎𝑡𝑚2

Hence, option (2) is correct. Statement (4) is incorrect because at equilibrium,

𝑄 = 𝐾𝑒𝑞 and 𝛥𝐺 = 0

Hence options B and C are the correct answers. 37.

𝐵𝑖 +3 + 3𝑁𝐻4 𝑂𝐻 → 𝐵𝑖(𝑂𝐻)3 ( white 𝑝𝑝𝑡) + 3𝑁𝐻4+ 38.

𝑃𝑏 +2 + 2𝑁𝐻4 𝑂𝐻 → 𝑃𝑏(𝑂𝐻)2 ( white 𝑝𝑝𝑡) + 2𝑁𝐻4+ 𝐹𝑒 3+ + 3𝑁𝐻4 𝑂𝐻 → 𝐹𝑒(𝑂𝐻)3 ( reddish brown 𝑝𝑝𝑡) + 3𝑁𝐻4+

Only four of these bonds are in the same plane.

𝑁𝐻4+ , the 𝑂𝐻 - ion concentration is not enough to precipitate 𝑀𝑔+2 as 𝑀𝑔(𝑂𝐻)2 because of its high solubility product.

Hence, option (1) is correct.

Hence options A, B and D are the correct answers.

Due to common ion effect of 2 centre-2 electron bonds are 4

39. The mixture given in option (4) contains a weak acid (CH3COOH) and its salt with strong base NaOH i.e. CH3COONa. It will form a perfect acidic buffer. Hence, only option (4) is correct.

Bond angle

𝐻𝑡𝐵𝐻𝑡 > 𝐻𝑏𝐵𝐻𝑏

Hence, option ( 2 ) is correct In borax,

Mock Test - 8

𝐵4 05 (𝑂𝐻)4 , the number of B-O-B units is 5

40. (1) Replacement of -OH by a halogen in an alcohol is nucleophilic substitution reaction. It is the protonted alcohol which acts as a substrate. (2) Alcohols are acidic enough to react with active metals to liberate hydrogen gas. They are basic enough to accept a proton from strong acids.

| 151

(3) Secondary alcohols, on oxidation, give a ketone containing the same number of carbon atoms. Further oxidation will give carboxylic acid containing lesser number of carbon atoms.

y+1=m(x−2)

(4) Primary alcohols, on oxidation, give an aldehyde containing the same number of carbon atoms. Further oxidation will give carboxylic acid containing the same number of carbon atoms.

(−1−2m)2=3m2+1

Hence, only option (A) is incorrect.

∴ equation of tangents are y=−1 and 4x+y=7

y=mx+(−1−2m) Condition of tangency is c2=a2m2+b2

1+4m2+4m=3m2+1 m2+4m=0⇒m=0,−4

Hence option B is the correct answer. 41. Given parabola can be written as x2=8(y−1/8) Comparing with standard parabola X2=4aY X=x, Y=y−1/8, a=2

46. Length of the double ordinate is parametric form is 8pt=16p t=2

Therefore Vertex is (X,Y)=(0,0)

So coordinates of extremities will be A(8p,8p) and B(8p,−8p)

⇒(x,y−1/8)=(0,0) ⇒(x,y)=(0,1/8)

The slope of the line joining vertex to point A is 1

Hence option C is the correct answer.

The slope of the line joining vertex to point B is −1

42. Given the circle lies in 4th quadrant, and touches y=0 and x=0 and has radius 3. ∴ Equation of circle is

Product of slopes =−1 Hence the lines are perpendicular Hence option B is the correct answer.

(x−3)2+(y+3)2=32 47. Given that point P lies on ellipse

⇒x2+y2−6x+6y+9=0.

it will be in the form of (6√cos θ, 2√sin θ)

Hence option A is the correct answer.

The center of ellipse is (0,0) and its distance from P is 2 43. Given parabola y2=4ax

by using distance formula

Since, P, S and Q are collinear.

⇒6cos2θ+2sin2θ=4

mPQ=mPS

⇒cos2θ=1/2

2/(t1+t2)=2t1/(t21−1)

⇒cosθ=1/√2

⇒t21−1=t21+t1t2

therefore the value of θ=π/4

⇒t1t2=−1

Hence option C is the correct answer.

Hence option C is the correct answer.

44. Tangent at

⇒ 𝑥 5

16 𝑥( 5 )

16

+

(𝑥1 , 𝑦1 ) is

9 𝑦(5)

9

𝑥𝑥1 𝑎2

+

𝑦𝑦1 𝑏2

=1

𝑦 5

+ =1

=1

48. The given equation of the parabola can be written in the form (y−3)2 = 2(x+2) ⇒ (y−3)2=4.1/2 (c+2) So, a=1/2 Coordinates of vertex are x+2=0, y−3=0 ⇒ (−2,3) Coordinates of focus are x+2=1/2, y−3=0 ⇒ (−3/2,3)

∴ intercepts are equal

Equation of directrix is x+2=−1/2 ⇒ x=−5/2

Hence option A is the correct answer.

Hence option D is the correct answer.

45. Line passing through (2,−1) is

49. x2+2y−3x+5=0

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⇒x2−3x=−2y−5

⇒ 𝑥 2 ℎ2 + ℎ2 𝑦 2 − 2ℎ2 𝑟𝑥 − 2ℎ3 𝑦 + ℎ4 =

⇒x2−2(3/2)x+9/4 = −2y−5+9/4 = −2y−11/4

𝑟 2 𝑥 2 + ℎ2 𝑦 2 + ℎ4 + 2𝑟𝑥𝑦 − 2ℎ3 𝑦 − 2𝑟𝑥ℎ2

⇒(x−3/2)2 = −2(y−11/4)

⇒ ℎ2 𝑥 2 = 𝑟 2 𝑥 2 + 2𝑟ℎ𝑥𝑦

Hence latus rectum = 4×1/2=2

⇒ ℎ2 𝑥 2 − 𝑟 2 𝑥 2 − 2𝑟ℎ𝑦 = 0

Hence option B is the correct answer.

⇒ 𝑥(ℎ2 𝑥 − 𝑟 2 𝑥 − 2𝑟ℎ𝑦) = 0 ⇒ 𝑥(ℎ2 − 𝑟 2 )𝑥 − 2𝑟ℎ𝑦) = 0

50. y2+4y+4x+2 = 0 y2+4y+4 = −4x+2 (y+2)2

tangent to the circle

= −4(x−1/2)

(ℎ2 − 𝑟 2 )𝑥 − 2𝑟ℎ𝑦 = 0, 𝑥 = 0

Hence options A and C are the correct answers.

So equation of directrix is (x−1/2) = −a = 1 ⇒ x = 3/2 Hence option D is the correct answer.

53. Given:

51. Three lines px + qy + r = 0, qx + ry + p = 0 and rx + py + q = 0 are concurrent.

ℎ(𝑥) = 𝑓(𝑥) − (𝑓(𝑥))2 + (𝑓(𝑥))3 𝑥 , we get

On differentiating w.r.t.

ℎ′ (𝑥) = 𝑓(𝑥) − 2𝑓(𝑥) ⋅ 𝑓 ′ (𝑥) + 3𝑓 2 (𝑥) ⋅ 𝑓 ′ (𝑥) = 𝑓(𝑥)[1 − 2𝑓(𝑥) + 3𝑓 2 (𝑥)]

Then,

2 3

1 3

= 3𝑓 ′ (𝑥) [(𝑓(𝑥))2 − 𝑓(𝑥) + ] 1 2 3

1 3

1 9

= 3𝑓 ′ (𝑥) [[𝑓(𝑥) − ] + − ] 𝑅1 → 𝑅1 + 𝑅2 + 𝑅3

1 2

= 3𝑓(𝑥) [[𝑓(𝑥) − ] +

3−1

3

1 2 3

9

]

2 9

= 3𝑓(𝑥) [[𝑓(𝑥) − ] + ] ⇒ (𝑝 + 𝑞 + 𝑟)(𝑝 2 + 𝑞 2 + 𝑟 2 − 𝑝𝑞 − 𝑞𝑟 − 𝑝𝑟) = 0 𝑝3

+ 𝑞3

+ 𝑟3

− 3𝑝𝑞𝑟 = 0

Note that if

ℎ′ (𝑥) < 0 if 𝑓 ′ (𝑥) < 0 and ℎ′ (𝑥) > 0

𝑓 ′ (𝑥) > 0

Therefore, (A) and (C) are the answers.

ℎ(𝑥) is an increasing function if 𝑓(𝑥) is an increasing function, and ℎ(𝑥) is a decreasing function if 𝑓(𝑥) is a decreasing function.

52. The equation of pair of tangent to the circle

𝑥2 + 𝑦 2 + 2𝑔𝑥 + 2𝑓𝑦 + 𝑐 = 0 from point (𝑥1 , 𝑦1 ) is

Therefore, options (A) and (C) are correct answers.

(𝑥 2 + 𝑦 2 + 2𝑔𝑥 + 2𝑓𝑦 + 𝑐)(𝑥12 + 𝑦12 + 2𝑔𝑥1 + 2𝑓𝑦1 + 𝑐)

54. since,

= [𝑥𝑥1 + 𝑦𝑦1 + 𝑓(𝑥 + 𝑥1 ) + 𝑓(𝑦 + 𝑦1 ) + 𝑐]2

axes are

Thus, the equation of pair of tangent to the circle

Therefore,

𝑥2 +

𝑦 2 − 2𝑥 − 2ℎ𝑦 + ℎ2 = 0 (𝑥 2 + 𝑦 2 − 2𝑟𝑥 − 2ℎ𝑦 + ℎ2 )(02 + 02 − 2𝑟0 − 2ℎ0 + ℎ2 ) =

𝐵𝑃: 𝐴𝑃 = 3: 1. Then equation of tangent is 𝑌 − 𝑦 = 𝑓 ′ (𝑥)(𝑥 − 𝑥) The intercept on the coordinate

and

⇒ (𝑥 2 + 𝑦 2 − 2𝑟𝑥 − 2ℎ𝑦 + ℎ2 )(ℎ2 ) = [−𝑟𝑥 − ℎ𝑦 + ℎ2 ]2 Mock Test - 8

𝑦 , 0) 𝑓′ (𝑥)

𝐵[0, 𝑦 − 𝑥𝑓(𝑥)]

since,

𝑃 is internally intercepts a line 𝐴𝐵

from origin (0,0) is

[𝑥0 + 𝑦0 − 𝑟(𝑥 + 0) − ℎ(𝑦 + 0) + ℎ2 ]2

𝐴 (𝑥 −

𝑦

∴𝑥= ⇒

𝑑𝑦 𝑑𝑥

3(𝑥− ′ )+1×0 𝑓 (𝑥)

=

3+1 𝑦 −3𝑥

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256

(𝑡 4 + 4𝑡 2 ) ( 4)

64 2 (𝑡 𝑡4

𝑡4

+

64 𝑡2

) = 1600𝑡 2 (𝑡 2 +

+ 4) = 1600

(𝑡 2 + 4)2 = 𝑡2 + 4 = ±

1600 2 𝑡 64

40 𝑡 8

𝑡 2 + 4 = ±5𝑡 𝑝𝑡 = ±1, ±4 ⇒

𝑑𝑦 𝑦

=−

Hence,

1 𝑑𝑥 3𝑥

𝑃(16 ± 8), (1, ±2)

Hence options C and D are the correct answers.

On integrating both sides, we get 𝑥𝑦 3

= 𝑐 since, curve passes through (1,1), then 𝑐 = 1𝑥𝑦 3 = 1 At 𝑥 = 1 8

⇒𝑦=2

Hence, (A) and (C) are correct answers.

55. Slope of Slope of

𝑉𝑃 =

𝑉𝑄 = −

2𝑡 𝑡2

=

2 𝑡

56. A sustentacular cell is a type of cell that provides structural support. They are of two types. One type of cell is found in the Sertoli cell, in the testicle of mammals. It is present in the walls of the seminiferous tubules and supplies nutrients to sperm as the sertoli cell is nurse cell and another type of cells are found in the olfactory epithelium. Hence option A is the correct answer.

𝑡 2

57. Isogamy is a form of sexual reproduction that involves gametes of similar morphology. The gametes differ in their allele expression. As they look alike they cannot be classified as male and female gametes. Hence option B is the correct answer. 58. Acrosomes contain digestive enzymes like hyaluronidase and lysins, which helps in breaking down the outer membrane of the ovum called as the zona pellucida, allowing the haploid nucleus in the sperm cell to join the haploid nucleus in the ovum, thus enabling fertilization. Equation of

Hence, the correct answer is option C.

𝑡 2

∨ 𝑄 𝑦 = − 𝑥

Solving it with

𝑦 2 = 4𝑥

𝑡 2 𝑥2 4

− 4𝑥 = 0

𝑥(𝑡 2 𝑥 − 16) = 0 𝑥 = 0, 𝑥 = 𝑄(

162 𝑡2

16 8 ,− ) 𝑡2 𝑡

Area of

1 2

𝛥𝑃 ∨ 𝑄 = 𝑃 ∨ 𝑉𝑄 = 20

𝑃𝑉 2 ⋅ 𝑉𝑄 2 = 1600

59. Stabilizing selection in a population acts against extremes and favors intermediate phenotypes. This refers often to heterozygous advantage, in which natural selection favors the heterozygote because it possesses benefits that homozygotes lack. In other words, a heterozygous advantage describes the heterozygote genotype which has a higher relative fitness than the homozygote dominant or homozygote recessive genotype. For example, people with homozygous recessive for sickle cell anemia, meaning they have the disease, are also immune to malaria, a disease common in Africa, where sickle cell originated. Those heterozygous for the trait show no symptoms of sickle cell, but they are immune to malaria, greatly benefitting the heterozygote. Hence option B is the correct answer.

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60. DNA polymorphism can be defined as a condition where more than one different normal nucleotide sequences can exist at the same locus in DNA. These two different alleles are the product of single base pair mutation, deletions, insertions etc. The polymorphic DNA contains a variable sequence which may lead to the expression of a various random gene which brings a remarkable change in the gene pool. Thus, the correct answer is option A. 61. a) Mutation theory was proposed by Hugo de Vries. He proposed that the new species arise not by the accumulation of minor and continuous variations by natural selection but by suddenly appearing saltatory variations for which he coined the term mutation. Mutation is any heritable change in the genetic makeup of an individual. b) Lamarck explained the mechanism of evolution on the basis of use and disuse of organs. His theory is also known as the theory of inheritance of acquired characters according to which the characters which are acquired during the lifetime of an organism are transferred to the offsprings. This is known as Lamarckism. c) Darwin proposed the theory of natural selection according to which the organisms with favourable variations would survive because they are fittest to face their surroundings while the organisms which are unfit for surrounding variations are destroyed naturally. This is known as Darwinism. d) The theory of evolution as proposed by Darwin has been modified in the light of modern evidence from genetics, molecular biology, palantaeology, ecology and is known as Neo-Darwinism. It may be defined as the theory of organic evolution by the natural selection of inherited characteristics. Hence option A is the correct answer. 62. In a solution with higher salt concentrations or with alcohol added, the DNA structure may change to an A form, which is right-handed helix, but every 2.3 nm makes a turn and there are 11 base pairs per turn. B-DNA is also right handed helix having 10 base pairs per turn. Z DNA is lefthanded. One turn spans 4.6 nm, comprising 12 base pairs. Hence, option B is correct. 63. After ovulation, a structure corpus luteum is formed from the wall of the follicle that ovulated. This structure then begins to secrete estradiol and progesterone. Progesterone prevents ovulation and maintainance of corpus luteum.

64. A gynandromorph is an organism that contains both male and female characteristics. It may result from the loss of X chromosome.' Hence option C is the correct answer. 65. A set of genes will be in a complete linkage when they are present in close proximity on a chromosome and are inherited together during meiosis always. Complete linkage means no crossing over or recombination, i.e. the progeny phenotypes will be totally parental. i.e., P = 100%, R = 0% Hence option D is the correct answer. 66. Bioremediation or phytoremediation is the cleaning of environment of toxic wastes with the help of microbes, plant and animal species. The organisms involved in bioremediation are of two types: hyperaccumulators and detoxifying species. Hyperaccumulators are used in bioprospecting and removal of minerals from wastes. The detoxifying agents metabolise the harmful chemicals through various biochemical reactions. Hence option C is the correct answer. 67. Emasculation involves the removal of stamens from bisexual flowers of the female parents in order to avoid self-pollination in these flowers. It is done before the anthers are mature. If in a condition the female parent bears bisexual flowers, then elimination of anthers from the flower bud before the anther dehisces by means of a pair of forceps (a large instrument with broad blades) is necessary. And this step is known as emasculation. Thus, the correct answer is option A. 68. The genetically identical plants obtained through tissues culture or micropropagation are called as somaclones because it is produced by the culture of somatic cells. Whereas monoclone is the group of cells produced by single cell asexually, the somatic hybrid is the genetic modification of plants by the fusion of somatic cells of two different species and cross hybrid is produced by the crossing between two true breeding organism. Thus, the correct answer is option (A). 69. Sewage is wastewater having food residue, animal and human excreta and waste material from industrial establishments. The domestic sewage in large cities is processed by aerobic and then anaerobic bacteria in the secondary treatment in sewage treatment plant (STP). Treament of water:-

Therefore, the correct answer is option A. Mock Test - 8

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1. Primary treatment - The treatment utilizes settling tanks of upward, horizontal tanks. The water flow is induced to allow 15-45 min retention time during which 70% of suspended solids either settle to bottom or flow as scum on the top. 2. Secondary treatment - Digestion of organic matter dissolved in wastewater by aerobic/anaerobic processes. 3. Tertiary structure - The aim of this step is to remove suspended matter and to reduced BOD still further. The process is used are either physical or chemical. 4. Sludge treatment - Sludge is the settled solid component of water recovered during the treatment processes. It is treated mainly anaerobically to make it safe before disposal. So correct answer is A. 70. Cultigens are the plants which are cultivated for the commercial value. These are the plants which include crops, ornamentals, fruit trees, etc. These plants require human care and protection because they are cultivated by the breeding techniques and artificial selection method. There are many varieties of cultigens which are produced by genetic modification. Thus, the correct answer is option A.

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B. The nature of the surface

MOCK TEST - 9

C. The smoothness of the surface D. All the above are correct

General Ability 1. Who was the first Muslim president of the INC? A. Womesh Chandra Banerji B. Abul kalam C. Rahimtulla M. Sayani

D. Badruddin Tayabji

2. Gandhiji was the president of the Congress only on one occasion and the session was held inA. Mumbai

B. Lahore

C. Belgaum

D. Lucknow

9. Name the Indian football team head coach who has resigned from the post. A. Harendar Singh

B. Gurpreet Singh Sandhu

C. Bhaichung Bhutia

D. Stephen Constantine

10. Lenin Rajendran has passed away at 67. He was a renowned director and screenwriter of _______ film industry. A. Kannada

B. Telugu

C. Tamil

D. Malayalam

3. In which year Mahatma Gandhi was first arrested during ‘Satyagrah’ — A. 1906

B. 1908

C. 1913

D. 1917

4. Who among the following is the first Englishman to become the president of INC? A. William Wedderburn

B. George Yule

C. Alfred Webb

D. Henry Cotton

5. A ray of light travelling obliquely from denser to rarer medium A. Bends towards the normal B. Bends away from the normal C. Does deviate from its path D. None of these

11. If error in measuring diameter of a circle is 4%, the error in circumference of the circle would be A. 2%

A.

C.

B. The concave mirror only

C. The convex mirror only D. All reflecting surfaces 7. The focal length of a plane mirror is A. Positive

B. Negative

C. Zero

D. Infinity

B. 8%

D.

𝑣1 > 𝑣2 and 𝑣1 > 𝑣2 and 𝑣1 > 𝑣2 and 𝑣1 > 𝑣2 and

D. 1%

(𝑣1 −𝑣2 )2 2𝑓 (𝑣1 +𝑣2 )2 2𝑓 (𝑣1 −𝑣2 )2 2𝑓 (𝑣12 −𝑡22 ) 2𝑓

𝑑 >𝑑

>𝑑

13. Two bodies of different masses ma and mb are dropped from two different heights, a and b respectively. The ratio of times taken by the two bodies to drop through this distance is

8. The amount of light reflected depends upon

A. a2 : b2

A. The nature of material of the object

B. a/mb : b/ma

Mock Test - 9

C. 4%

12. The driver of a train moving with a constant speed v1 along a straight track sights another train at a distance d ahead of him on the same track moving in the same direction with a constant speed v2 . He at once applies the brakes and gives his train a constant retardation f. There will be a collision of the trains if:

B.

6. The laws of reflection are true for A. The plane mirror only

Physics

| 157

C.

√𝑎: √𝑏

D. a : b 14. A projectile is projected with velocity v such that its range is twice the greatest height attained. Then its range is A. 1v2/5g

B. 2v2/g

C. 4v2/5g

D. 2v/g

15. A body crosses the topmost point of a vertical circle with critical speed. What will be its centripetal acceleration when the string is horizontal: A. 6g

B. g

C. 2g

D. 3g

16. A block of mass M is at rest on a rough horizontal surface. The coefficient of friction between the block and the surface isμ. A force F=Mg acting at an angle θ with the vertical side of the block pulls it. In which of the following cases, the block can be pulled along the surface?

19. A constant force F is applied at the top of a ring as in figure. M is the mass and R is the radius of the ring. The change in angular momentum of particle about point of contact in a time t is.

A. Fr

B. a constant

C. FR/t

D. 2FRt

20. If the radius of the earth were to shrink by one per cent, its mass remaining the same, the value of g on the earth's surface would. A. increase by 0.5%

B. increase by 2%

C. decrease by 0.5%

D. decrease by 2%

21. Find the height above the surface of the earth where weight of a body becomes half.

A.

tan𝜃 ≥ 𝜇

B.

cot𝜃 ≥ 𝜇

A. R/2

B. (√2 -1)R

C.

tan𝜃/2 ≥ 𝜇

C. R/(√2 + 1)

D. R/√2

D.

cot𝜃/2 ≥ 𝜇

22. Two stones are thrown vertically upwards simultaneously from the same point on the ground with initial speed u1=30 m/sec and u2=50 m/sec. Which of the curves represent correct variation (for the time interval in which both reach the ground) of relative position (x2−x1) and relative velocity (v2−v1) of second stone with respect to respect to first with time (t). Assume that stones do not rebound after hitting.

17. Figure shows a mass M is attached to the middle of a string, the two ends of which are passing over two pulleys P1 and P2. The ends of strings are attached to masses M1 and M2. If the two parts of the string attached to M are inclined at an angle 2θ and the instantaneous downward velocity of the masses M 1 and M2 be v, what is the upward velocity of M ?

A. A. 2v cos θ

B. v/ cos θ

C. v

D. v/2

18. The potential energy of a particle is given by

𝑈=

𝑎 𝑏 where a and b are positive constants, r is the distance 𝑟2 𝑟

from the centre of the field. The stable equilibrium position of the particle corresponds to the distance given by A.

𝑟0 =

2𝑎 𝑏 𝑎

B.

𝑟0 = 𝑏

C.

𝑟0 = 𝑏

D.

𝑟0 = 2ℎ

Mock Test - 9

B.

𝑎 𝑎

| 158

C.

D.

A. Acceleration of 'A' relative to ground is in negative ydirection B. Acceleration of 'A' relative to B is in positive xxdirection

23. A particle of mass m is executing uniform circular motion on a path of radius r. If v is the speed and p the magnitude of its linear momentum, then the radial force acting on the particle is: A. mv2/r

B. pm/r

C. up/r

D. p2/mr

24. Two particles are projected under gravity with speed 4 m/s and 3 m/s simultaneously from same point and at angles 53° and 37° with the horizontal surface respectively as shown in figure. Then :

C. The horizontal acceleration of 'B' relative to ground is in negative x-direction. D. The acceleration of 'B' relative to ground directed along the inclined surface of 'C' is greater than g sin θθ.

Chemistry 26. In the reaction

4𝑁𝐻3 + 5𝑂2 → 4𝑁𝑂 +

6𝐻2 𝑂 when one mole of ammonia and one mole of oxygen are made to react to completion, then A. 1.0 mol of H2O is produced B. 1.0 mol of NO will be produced C. all the oxygen will be consumed D. all the ammonia will be consumed

A. Their relative velocity is along vertical direction B. Their relative acceleration is non-zero and it is along vertical direction C. They will hit the surface simultaneously D. Their relative velocity is canstant and has magnitude 1.4 m/s. 25. In the figure shown all the surfaces are smooth. All the blocks A,B and C are movable X-axis is horizontal and y-axis vertical as shown. Just after the system is released from the position as shown:

27. The radius of hydrogen atom in the ground state is 0.53 Å .The radius of Li2+ ion (atomic number = 3) in a similar state is A. 0.17 Å

B. 0.265 Å

C. 0.53 Å

D. 1.06 Å

28. Of the given electronic configurations for the elements, which electronic configuration indicates that there will be abnormally high difference in the second and third ionization energy for the element ? A. ls2 2s2 2p6 3s2 B. Is2 2s2 2p6 3s1 C. Is2 2s2 2p6 3s2 3p1

Mock Test - 9

| 159

D. ls2 2s2 2p6 3s2 3p2

C.

29. Ionisation energy of oxygen is less than that of nitrogen because

𝛥𝐺 0 =

0.059 𝑛

log𝐾

D. ∆G0 = RT log K 35. The oxidation number of P in Ba(H2PO2)2 is

A. More penetrating effect A. +3

B. Oxygen is larger in size than nitrogen C. Oxygen is smaller than nitrogen D. Of the extra stability of the half filled p-orbitals of nitrogen

B. +1

C. +2

D. -3

36. The correct statement(s) for the following addition reactions is(are):

30. A molecule MX3 is T-shaped. The number of non bonding pairs of electrons in it is A. 0

B. 2

C. 3

D. 4

31. A molecule having 3 bond pairs and 2 lone pairs will have ? A. T-shaped geometry B. Trigonal planar geometry C. Linear geometry D. Square pyramidal geometry

A. bromination proceeds through trans-addition in both the reactions B. O and P are identical molecules C. (M and O) and (N and P) are two pairs of diastereomers

32. A bubble of air is under water at temperature 15°C and the pressure 1.5 bar. If the bubble rises to the surface where the temperature is 25°C and the pressure is 1.0 bar, what will happen to the volume of the bubble ?

D. (M and O) and (N and P) are two pairs of enantiomers

A. Volume will become greater by a factor of 1.6.

A. The number of maxima in 2s orbital are two

B. Volume will become greater by a factor of 1.1.

B. The number of spherical or radial nodes is equal to n−l−1

C. Volume will become smaller by a factor of 0.70. D. Volume will become greater by a factor of 2.5. 33. A solution of 200ml of 1M KOH is added to 200 ml of 1M HCl and themixture is well shaken. The rise in temperature T1 is noted. The experiment is repeated by using 100 ml of each solution and increase in temperature T2 is again noted. Which of the following is correct?

37. For radial probability distribution curves, which of the following is/are correct?

C. The number of angular nodes are l D. 3dz2 has two angular nodes 38. Which of the following compounds will show optical activity? A .

A. T1 is four times as large T2 B. T1 is twice as large as T2 C. T1 = T2 D. T2 is twice as large as T1 34. The relationship between standard free energy ∆G0 and equilibrium constant of the reaction is given by A. ∆G0 = -2.303 RT log K B.

Mock Test - 9

𝛥𝐺 0 =

23.303𝑅𝑇 𝑛𝐹

log𝐾 | 160

B. 42. The distance between 3x+4y=9 and 6x+8y=15 is A. 1.5

B. 0.3

C. 6

D. 3

43. The equation of the circle having centre at (0,0) and radius 4 units is C.

A. x2 + y2 = 2 B. x2 + y2 = 1 C. x2 + y2 = 4 D. x2 + y2 = 16 44. The length of latus rectum of the parabola 13y2 = 14x is

D .

A. 14/13

B. 13/14

C. 28/13

D. 6/14

45. The length of the y – intercept of the circle x2 + y2 – 6x + 8y + 7 = 0 is A. 1

B. 2

C. 3

D. 6

46. The focus of the parabola x2 = 4ay is 39. g orbital is possible if: A. n = 5, I = 4

A. (a, 0)

B. (-a, 0)

C. (0, a)

D. (0, -a)

B. It will have 18 electrons C. It will have 19 types of orbitals D. It will have 22 electrons 40. Which of the following is/are correct statement(s)? A. Probability of finding the electron in bonding molecular orbital is more than combining atomic orbitals B. Bonding molecular orbitals are formed when same sign of orbitals are overlap

47. A rectangular parallelopiped is formed by planes drawn through (5,7,9) and (2,3,7) parallel to the coordinate planes. The length of the longest edge of the parallelopiped is A. √29

B. 5

C. 4

D. 16

48. The points (5,-4,2), (4,-3,1),(7,-6,4) and (8, -7,5) are vertices of a A. Square

B. Parallelogram

C. d-d combination of atomic orbitals gives δ and δ∗ molecular orbitals

C. Rectangle

D. None of these

D. None of these

49. The vertices of a triangle are (3,2,5), (3,2,-1) and (7,2,5). The circumcentre of the triangle is

Mathematics

A. (4,3,1)

B. (5,-2,1)

C. (5,2,2)

D. (5,-2,2)

50. The coordinates of the point of intersection of the line

41. The distance from the point (x,y) to the positive x-axis is A. x Mock Test - 9

B. y

C. |x|

𝑥+1 1

=

𝑦+3 3

=

𝑧−2 2

with the plane 3x + 4y + 5z = 5

are A. (-1, 2, 3)

B. (1, 3,-2)

D. 0

| 161

C. (-1,-2, 3)

D. (1,-3, 2)

51. A quadratic equation can contain ......... A. x C.

x3

C. Cycas

D. Thuja

58. Tube feet are locomotary organs, found in

B. z2

A. Annelids

B. A1thropoda

D. z

C. Mollusca

D. Echinoderms

52. The equation of the circle, touching the axis of x at the origin and the line 3y = 4x + 24, is

59. Which plant will lose its economic value, if its fruits are produced by induced parthenocarpy?

A. x2 + y2 + 24y = 0

A. orange

B. banana

C. grape

D. pomegranate

B.

x2

C.

x2

D.

x2

+

y2

- 6y = 0

+

y2

- 24y = 0

+

y2

+ 6y = 0

60. In a dicotyledonous stem the secquence of tissues from the outside to the inside is

53. The coordinates of two points on the circle x 2 + y2 − 12x − 16y + 75 = 0, one nearest to the origin and the other farthest from it, are

A. phellem - pericycle - endodermis - phloem

A. (3, 4)

B. (3, 2)

D. pericycle- phellem - endodermis phloem

C. (9, 12)

D. (9, -12)

B. phellem - phloerm- endodermis- pericycle C. phellem - endodermis - pericycle - phloem

61. What connects muscle to bone? 54. All points lying inside the triangle formed by the points (1,3), (5,0) and (1,2) satisfy A. 3x + 2y ≥ 0

B. 2x + y - 13 ≥ 0

C. 2x - 3y - 12 ≤0

D. -2x + y ≥ 0

𝑆1 = 𝑥 2 + 𝑦 2 − 4𝑥 − 6𝑦 − 12 = 0 and 𝑆2 = 𝑥 2 + 𝑦 2 + 6𝑥 + 4𝑦 − 12 = 0 and the line 𝐿 ≡ 𝑥 + 𝑦 = 0 55. For the circles

A. ligament

B. cartilage

C. tendon

D. sarcomere

62. Which cell organelle is concerned with glycosylation of protein ? A. ribosome

B. peroxisome

C. endoplasmic reticulum

D. mitochondria

A. L is common tangent of S1 and S2

63. During cell division a cell plate is produced during

B. L is common chord of S1 and S2

A. metaphase

B. telophase

C. L is radical axis of S1 and S2

C. cytokinesis

D. anaphase

D. L is perpendicular to the line joining the centre of S 1 and S2

64. When stomata closes which of the following events does not occur? A. guard cell become flaccid B. sugar is converted to starch

Biology 56. Linnaeus evolved a system of nomenclature called A. mononomial

B. vernacular

C. binomial

D. polynomial

57. Largest sperms in the plant world are found in A. Pinus Mock Test - 9

C. O.P of the guard cell decreases D. accumulation of O2 takes place 65. Which one of the following elements is not an essential micronutrient for plant growth? A. Zn

B. Cu

C. Ca

D. Mn

66. Identify the stage of Mitosis by inspecting the image and key changes listed.

B. Banyan

| 162

69. Division of the nucleus during cell division is called __________. A. Cytokinesis

B. Metakinesis

C. Karyokinesis

D. Amphimixis

70. Which of the following is true about spindle fibres? A. They divide the genetic material in the cell Key changes: * Chromosomes are doubled. * Spindle fibres attach to the chromosomes.

B. Spindle fibres are formed both during meiosis and mitosis C. Spindle fibres are made up of microtubules D. All of the above

* Chromosomes are arranged at equatorial plate. A. Prophase

B. Metaphase

C. Anaphase

D. Telophase

67. In the given figure, certain parts are indicated by letters. Select the answer which contains the correct matching of these letters with the correct parts.

A. A = Chromosomes, B = Spindle, C = Kinetochore B. A = Spindle, B = Kinetochore, C = Chromosomes C. A = Kinetochore, B = Spindle, C = Chromosomes D. A = Kinetochore, B = Chromosomes, C = Spindle 68. A cell has a division time of 1 minute. If in 20 minutes it could fill a culture flask by 1/8th of its volume, then by what time would the total flask be filled? A. 160 mins

B. 21 mins

C. 80 mins

D. 23 mins

Mock Test - 9

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SMART ANSWERSHEET

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Performance Analysis Avg. Score (%) Toppers Score (%) Your Score

Mock Test - 9

0.0% 0.0%

| 164

HINTS AND SOLUTIONS

9. Indian football team head coach Stephen Constantine has stepped down after the team’s shock 0-1 defeat to Bahrain which led to its exit from the Asian Cup. Hence option D is the correct answer.

1. The first Muslim president of the INC was Badruddin Tayabji. ( 1887 at Madras session ) Hence option D is the correct answer. 2. The INDIAN NATIONAL CONGRESS held its thirty-ninth session at Belgaum on the 26th & 27th Dec. 1924. Gandhiji was the president of the Congress only on one occasion and the session was held in Belgaum.

10. Malayalam director and screenwriter Lenin Rajendran has passed away at 67. He was a five-time recipient of the Kerala State Film Awards. Hence option D is the correct answer. 11.

Hence option C is the correct answer. 3. On 10 January 1908 Mahatma Gandhi was arrested for the first time in South Africa for refusing to carry an obligatory identity document card commonly known as the ‘pass’.

Hence option C is the correct answer.

Hence option B is the correct answer.

𝑣12 = 𝑣1 − 𝑣2 Relative separation = 𝑑 Relative acceleration 𝑎12 = 𝑎1 − 𝑎2 = −𝑓

12. Relative velocity 4. George Yule was a Scottish businessman who became the fourth President of INC in 1888(at Allahabad session ),the first non-indian to hold that office. Hence option B is the correct answer.

𝑠12

Let at time

1

𝑡, trains collides. Using 𝑠 = 𝑢𝑡 + 𝑎𝑡 2 ⇒ 2

1

𝑑 = (𝑣1 − 𝑣2 )𝑡 − 𝑓𝑡 2 2

⇒ 𝑓𝑡 2 − 2(𝑣1 − 𝑣2 )𝑡 + 2𝑑 = 0 For real solution of

Hence option B is the correct answer.

∴ For collision 𝑣1 > 𝑣2 and

6. The law of reflection states that the incident ray, the reflected ray, and the normal to the surface of the mirror all lie in the same plane. The laws of reflection are true for all reflecting surfaces. Hence option D is the correct answer. 7. The radius of Curvature of plane mirror is infinite, hence the focal length of plane mirror will also be infinite. Hence option D is the correct answer. 8. The amount of light reflected depends upon the nature of material of the object ,the nature of the surface and the smoothness of the surface. All the above are correct

2

𝑡, (2(𝑣1 − 𝑣2 )) −

5. When a light ray travels from rarer medium to denser medium, it bends towards the normal and vice-versa. Since air is rearer medium and water is denser medium, so the light ray bends towards he normal.

4 × 𝑓 × 2𝑑 > 0 ⇒

(𝑣1 −𝑣2 )2 2𝑓

>𝑑 (𝑣1 −𝑣2 )2 2𝑓

>𝑑

Hence option C is the correct answer. 13. Here, we have to calculate the time taken by two bodies to reach the ground which are dropped from different heights Let g be the acceleration due to gravity, the the time taken and h be the height of a body We have the time taken by a freely falling body to reach ground 2ℎ 𝑔

𝑡=√

Then, for the first particle, 2𝑎

𝑡𝑎 = √

𝑔

And for the second particle,

Mock Test - 9

| 165

15.

2𝑏

𝑡𝑏 = √

𝑔

Now, if we take the ratio

𝑡𝑎 : 𝑡𝑏 = √𝑎: √𝑏 That is, the ratio of the time taken by two bodies to reach the ground which are dropped from different heights is equal to the ratio of the square roots of the heights, a and b Hence option D is the correct answer.

ta : tb = Va: vb Hence option C is the correct answer.

16. As shown in the figure the Normal reaction

𝑁 on the

14. Here, we have to calculate the range of the projectile projected with velocity v such that its range is twice the greatest height attained.

block and Vertical component of the force 𝐹 are opposite to weight. We need only horizontal motion no vertical motion so vertical force should be balanced that is net force in vertical

Let,

direction should be zero, i.e.,

𝑁 + 𝐹cos𝜃 − 𝑀𝑔 = 0 or 𝑁 = 𝑀𝑔 − 𝐹 cos𝜃, put 𝐹 = 𝑀𝑔 as given in the question. so by putting so we get 𝑁 = 𝑀𝑔(1 − cos𝜃) Now putting another condition to make 𝐹 able to pull the block horizontally the 𝐹sin𝜃 should be greater than or

v - initial velocity θ - angle of projection g - acceleration due to gravity Then we have,

eual to Static friction and we know that maximum value of

The horizontal range, R and the maximum height reached are,

static friction is

𝑅=

𝑣 2 sin2𝜃 𝑔

𝐻=

𝑣 2 sin2 𝜃 2𝑔

That is

put

𝑣 2 sin2𝜃 𝑔

=

=

𝜃 2

𝜃 2

𝜃 2

≥ 𝜇 Option 𝐷 is correct

17. Here, we have to find out the upward velocity of the centre mass M

2×𝑣 2 sin2 𝜃 2𝑔 2

The given problem is described in the following figure.

2sin𝜃cos𝜃 = sin 𝜃 cos𝜃 =

𝜃 2

cos𝜃 = 1 − 2sin2 and sin𝜃 = 2sin cos

we will get cot

𝑅 = 2𝐻

It is given,

𝜇𝑁 so we have 𝐹sin𝜃 ≥ 𝜇𝑁 putting value of 𝑁 we get sin𝜃 = 𝜇(1 − cos𝜃)

sin2 𝜃 2sin𝜃

sin𝜃

sin𝜃 cos𝜃

2

=2

⇒ tan𝜃 = 2 𝜃 = tan−1 (2) = 63.43∘ Now the range is

= = =

𝑅=

𝑣 2 sin2𝜃 𝑔

𝑣 2 sin(2×63−43) 𝑔 𝑣2 0.8 𝑔 4𝑣 2 5𝑞

That is, the range of the projectile is given by Hence option C is the correct answer.

From figure, it is clear that velocity of the masses

𝑀2 is 𝑣 =

And the velocity of the mass

4𝑣 2 /5𝑔.

𝑀 is 𝑣 ′ =

𝑑𝑥 𝑑𝑡

Also 𝑥 2 + 𝑎 2 = 𝑦 2 Where, a is a constant Now differentiating with respect to

𝑡, we get 2𝑥 ie,

Mock Test - 9

𝑀1 and

𝑑𝑦 𝑑𝑡

𝑑𝑥 𝑑𝑡

+ 0 = 2𝑦

𝑑𝑦 𝑑𝑡

𝑥𝑣 1 = 𝑦𝑣 | 166

𝑦

20.

𝑣1 = ( ) 𝑣 = =

𝑣

𝑥

𝑣 𝑣

𝑣 cos𝜃

That is, the upward velocity of the mass

𝑀 is 𝑣/cos𝜃

Hence option B is the correct answer. 18. Hence option B is the correct answer. 21. Given, weight of the body becomes half. Let that height be h 𝑚𝑔 2

𝐺𝑀𝑚

= (𝑅+ℎ)2 𝐺𝑀

1

1

2

(1+𝑅)

𝑔 = (𝑅)2 ⇒ =

and

Hence,

ℎ 2

ℎ = (√2 − 1)𝑅

Multiplying numerator and denominator by Hence option A is the correct answer. 19. Here, we have to find out the change in angular momentum of particle about point of contact in a time, t. We have, Angular momentum, L = τt = r × F t Where, τ - torque t - time

√2 + 1ℎ =

𝑅 √2+1

Hence option B and C are the correct answer.

22. While both the stones are in flight,

𝑎1 = 𝑔 and 𝑎2 =

𝑔 So 𝑎𝑟𝑒𝑙 = 0 ⇒ 𝑉𝑟𝑒𝑙 = constant ⇒ 𝑋rel = ( const )𝑡 ⇒ Curve of 𝑥rel 𝑣/𝑠 t will be straight line. After the first particle drops on ground, the seperation (𝑥rel ) ) will decrease parabolically (due to gravitational acceleration), and finally becomes zero. and

𝑉rel = slope of 𝑥rel 𝑣/𝑠𝑡

Hence option A and D are the correct answer.

F - Force r - perpendicular distance

23. Momentum is given as

Here,

p = mv

r = 2R and F = F

the radial force is the centripetal force, so

L = 2RFt That is, the change in angular momentum of particle about point of contact in a time, t is 2RFt.

also or

𝐹=

𝐹=

(𝑚𝑣)2 𝑚𝑟

𝑚𝑣 2 𝑅

=

=

𝐹=

𝑚𝑣 2 𝑟

𝑝2 𝑚𝑟

𝑣(𝑚𝑣) 𝑟

=

𝑣𝑝 𝑟

Hence option A, C and D are the correct answer.

24.

𝑣1 = 4cos530 𝑖 + 4sin530 𝑗

𝑣2 = 3cos370 𝑖̂ + 3sin370 𝑗̂ 7 5

𝑣12 = 𝑗̂ = 1.4𝑗̂ Mock Test - 9

| 167

Relative velocity in horizontal direction is zero. Hence option A and D are correct answer. 25. Answers are A, B, C and D There is no horizontal force on block A, therefore it does not move in x-direction. There is net downward force (mg N) is acting on it, making its acceleration along negative ydirection. Hence option A is correct. Block B moves downward as well as in negative x-direction. Downward acceleration of A and B will be equal due to constrain, thus w.r.t. B, A moves in positive xdirection. Hence option B is correct. Due to the component of normal exerted by C on B, it moves in negative x-direction. Therefore the direction of acceleration is in negative x-axis. Hence option C is correct. Acceleration of block A and B is due to the force MgsinθMgsinθ, where M is the combined mass of A and B. Hence option D is correct.

26.

4𝑁𝐻3 + 5𝑂2 → 4𝑁𝑂 + 6𝐻2 𝑂

1𝑁𝐻3 + 1.25𝑂2 → 1𝑁𝑂 + 1.5𝐻2 𝑂 Hence 1 mole of NH3 reacts with 125 125 moles of O2 to produce 1 mole of NO and 1.5 moles of H2O. Hence when one mole of ammonia and one mole of oxygen are made to react to completion, then all the oxygen are made to react to completion, then all the oxygen is consume. Hence option C is the correct answer. 27. State of hydrogen atom (n) = 1 (due to ground state) Radius of hydrogen atom (r) = 0.53 A Atomic number of Li (Z) = 3 Radius of Li2+ ion

=𝑛

𝑛2 2

= 0.53 ×

(1)2 3

= 0.17

Hence option A is the correct answer.

28. Mg = ls2 2s2 2p6 3s2

electron is to the nucleus, the more difficult it will be to remove, and the higher its ionization energy will be. Ionization energies increase on moving from left to right across a period (decreasing atomic radius). As we move along a period, the nuclear charge increases and the electrons are added into the same shell. Thereby the effective nuclear charge increases and size decreases. Therefore, the energy required to remove an electron increases. Ionization energy decreases down a group in the periodic table due to the fact that the outermost electrons are further away from the nucleus. The increase in distance between the positive nucleus and the negative electrons creates a weaker attraction between the two. As a result, less energy is needed to remove an electron from elements located lower on the periodic table. Ionisation energy depends on: Atomic radius- Larger the atom lesser the attractive force felt by the valence electrons. Hence lesser energy is required to pull out an electron . Thus ionisation energy is small. For smaller atoms reverse is true. Smaller atoms have higher ionisation energy. Nuclear charge- If the nuclear charge is more the attractive force on the valence electron is more. The energy required to pull out the valence electron will be high and ionisation energy will be high. Stability of half filled and completely filled orbitals- The atoms with fully filled and half filled orbitals have greater stability than the others. Therefore they require greater energy for removing an electron. Here the electronic configuration of atomic number of nitrogen is 15 and its outer electronic configuration is 2s 2 2p3. Oxygen has atomic number 16 and its outer electronic configuration is 2s2 2p4. 2s2 2p3 is more stable than 2s2 2p4 due to half filled p sublevel. Hence nitrogen atom has greater ionisation energy than oxygen. Hence option D is the correct answer. 30. According to VSEPR theory , in a T – shaped molecule of the type MX3 , there will be two lone pairs oriented at equatorial positions . eg : ClF . The central atom (M) undergoes sp3d hybridisation

After removing of 2 electron, the magnesium acquired noble gas configuration hence removing of 3rd electron will require large amount of energy.

Hence option B is the correct answer.

Hence option A is the correct answer.

Hence, trigonal bipyramidal geometry.

29. The energy required to remove an electron from an isolated gaseous atom in its ground state is known as ionisation energy. The closer and more tightly bound an Mock Test - 9

31. For 3 bond pairs and 2 lone pairs, hybridization is sp3d.

The geometry is also called as T-shaped geometry. Example: ClF3 Hence option A is the correct answer.

| 168

36. 32. Volume will become greater by a factor of 1.6 Given

𝑃1 = 1.5 bar 𝑇1 = 273 + 15 = 288𝐾𝑉1 = 𝑉 𝑃2 = 1.0 bar 𝑇2 = 273 + 25 = 298𝐾𝑉2 =? 𝑃1 𝑉1 𝑃 𝑓𝑉 = 2 2 𝑇1 𝑇2 1.5×𝑉 1×02 = 288 298

𝑉2 = 1.55𝑉 i.e. volume of bubble will be almost 1.6 time to initial volwne of bubble Hence option A is the correct answer. 33. Given that, 200ml of 1M KOH is added to 200 ml of 1M HCl and themixture is well shaken.

Addition of halogens is stereospecific reaction. Trans-zbutene gives meso product whereas the cis-isomer gives racemic mixture. The following statements are correct statements (A) Bromination proceeds through trans-addition in both the reactions. It involces formation of cyclic bromonium ion.

Rise in temperature = T1 100 ml of each solution is again mixed. T2=? As when quantity is halved, heat released will also be halved. So, T1 = T2 Hence option C is the correct answer. 34. ∆G0 = -2.303 RT log K ∆G0 = -2.303 RT log K For a reversible reaction at constant temperature and pressure,∆G = ∆G0 + RT ln Q

(C)(Mand O) are two pairs of diastereomers. They have same configuration at one chiral carbon atom and different configuration at other chiral carbon atom. The following are incorrect statements (B) "O and P are identical molecules The correct statement is "O and P are pair of enantiomers". They are non superimposable mirror images of each other. (D) "(M and O) and (N and P) are two pairs of enantiomers". The correct statement is "(M and O) and (N and P) are two pairs of diastereomers". Hence option A and C are the correct answer. 37.

At equilibrium, ∆G = 0, Q = K Therefore 0 = ∆G0 + RT ln K Therefore ∆G0 = - RT ln K or ∆G0 = -2.303 RT log K Hence option A is the correct answer.

𝐵𝑎(𝐻2 𝑃𝑂2 )2 is 𝐵𝑎(𝐻2 𝑃𝑂2 )2 ⇒ (𝐻2 𝑃𝑂2 )2− 2 (+1) × 2 + (𝑥) − 2 × 2 = −1 +2 + 𝑥 − 4 = −1 𝑥 = −1 + 4 − 2 𝑥 = +1 So oxidation number of 𝑃 = +1 35.

Hence option B is the correct answer.

The number of maxima in 2x orbital is two. (picture attached) also as we know, Total no of nodes are n - 1 The number of spherical or radical nodes is equal to n - l - 1. The number of angular nodes are l. Hence all options are the correct answer. 38. Compounds A and B are optically inactive, as one isomer can rapidly interconvert into another isomer (mirror image). In compounds C and D, the interconversion of one isomer into another isomer (mirror image) is not possible. Also, different groups are attached to the central N atom in both

Mock Test - 9

| 169

of these compounds. So, compounds C and D are optically active. Hence option C and D are the correct answer.

The slope of 3x + 4y = 9 is −6 is 8

=

−3 4

and that of 6x + 8y = 15

−3 4

since the slopes are equal the lines are parallel. 39. (A) g orbital will have n = 5, l = 4

The distance between two parallel lines is

(B) It has 18 electrons because m = -4, -3, -2, -1, 0, +1, +2, +3, +4 i.e., 9 value 9 orbitals therefore 9 × 2 = 18e (C) Orbital will be of 9 types because m, has 9 value given in (B) (D) Is incorrect and explained in (B).

|

𝑐1 −𝑐2 √𝑎2 +𝑏

=|

Basic idea of MOT is that atomic orbitals of individual atoms combine to form molecular orbitals. Electrons in molecule are present in the molecular orbitals which are associated with several nuclei.

|

√32 +4

−18+15 2

√25

−9+ 2

|

−3

=|2| 5

Hence option A, B and C are the correct answer. 40. Molecular Orbital Theory (MOT):

15

|=| 2

=|

−3 | 2×5 3 10

= | | = 0.3 Hence option B is the correct answer.

The molecular orbital formed by the addition of atomic orbitals is called the bonding molecular orbitals.

43. Equation of the circle with centre origin and having radius r

The molecular orbital formed by the subtraction of atomic orbital is called anti-bonding molecular orbitals.

is x2 + y2 = r2 , ie x2 + y2 = 16

The sigma (s) molecular orbitals are symmetrical around the bond-axis while pi (p) molecular orbitals are not symmetrical.Bond order(b.o.) is defined as one half the difference between the number of electrons present in the bonding and theanti-bonding orbitals.So all A,B & C are basic principles in MO theory.

Hence option D is the correct answer.

44.

𝑦2 =

14 𝑥 13

length of the latus rectum = 4a

45. Given equation of the circle is Hence option A, B and C are the correct answer. 41. From the figure it is clear that the distance from P(x,y) to the x – axis is ‘y’

𝑥 2 + 𝑦 2 − 6𝑥 +

8𝑦 + 7 = 0 We know the equation of the Length of the 𝑦 - intercept is 2√𝑓 2 − 𝑐 Now comparing the given equation with the general

𝑥 2 + 𝑦 2 + 2𝑔𝑥 + 2𝑓𝑦 + 𝑐 = 0, we get 𝑔 = −3 and 𝑓 = 4 and 𝑐 = 7 The length of equation of the circle

the

𝑦 -intercept = 2√𝑓 2 − 𝑐 = 2√42 − 7 =

2√16 − 7 = 2√9 = 2 × 3 = 6 Hence option D is the correct answer. 46. x2 = 4ay

Hence option B is the correct answer. 42. The given lines are 3x + 4y = 9 and 6x + 8y = 15

Mock Test - 9

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𝐷(8, −7,5). Then 𝐴𝐵 = √(5 − 4)2 + (−4 − −3)2 + (2 − 1)2 = √12 + (−1)2 + 12 = √3 𝐵𝐶 = √(4 − 7)2 + (−3 − −6)2 + (1 − 4)2 = √(−3)2 + (3)2 + (−3)2 = √9 + 9 + 9 = √27 𝐶𝐷 = √(7 − 8)2 + (−6 − −7)2 + (4 − 5)2 = √(−1)2 + (1)2 + (−1)2 = √1 + 1 + 1 = √3 𝐴𝐷 = √(5 − 8)2 + (−4 − −1)2 + (2 − 5)2 = √(−3)2 + (−3)2 + (−3)2 = √9 + 9 + 9 = Hence option C is the correct answer.

√27 ie, opposite sides are equal. Now

√(5 −

47. Given : Points on planes which formed rectangular parallelepiped are (5,7,9) and (2,3,7). And this is parallel to the coordinate planes. Now we have to find out the length of the longest edge of the parallelopiped. We can draw the rectangular parallelepiped as follows :

7)2

+ (−4 −

−6)2

𝐴𝐶 =

+ (2 − 4)2 =

√(−2)2 + (2)2 + (−2)2 = √4 + 4 + 4 = √12 = 2√3 𝐵𝐷 = √(4 − 8)2 + (−3 − −7)2 + (1 − 5)2 = √(−4)2 + (4)2 + (−4)2 = √16 + 16 + 16 = √48 = √4 × 4 × 3 = 4√3 ie, both diagonals are not equal. Therefore the given points are vertices of a parallelogram.

𝐴(3,2,5), 𝐵(3,2, −1) and 𝐶(7,2,5), which are vertices of △ 𝐴𝐵𝐶 . Then 𝐴𝐵 = 49. Let the given points be

√(3 − 3)2 + (2 − 2)2 + (5 − −1)2 = √62 = 6 𝐵𝐶 = √(3 − 7)2 + (2 − 2)2 + (−1 − −5)2 = √(−4)2 + (−6)2 = √16 + 36 = √52 𝐴𝐶 = √(3 − 7)2 + (2 − 2)2 + (5 − 5)2 = √−42 = 4 ie, 𝐵𝐶 2 = 𝐴𝐵2 + 𝐴𝐶 2 △ 𝐴𝐵𝐶 is a right angled triangle and angle 𝐴 = 90∘ Circumcentre of 𝛥𝐴𝐵𝐶 is the midpoint of 𝐵𝐶 . Midpoint of Lengths of the edges are

|𝑥1 − 𝑥2 ||𝑦1 − 𝑦2 | and

|𝑧1 − 𝑧2 | Here 𝑥1 = 5, 𝑦1 = 7, 𝑧1 = 9 and 𝑥2 = 2, 𝑦2 = 3, 𝑧3 = 7 −|𝑥1 − 𝑥2 | = |5 − 2| = 3 |𝑦1 − 𝑦2 | = |7 − 3| = 4 |𝑧1 − 𝑧2 | = |9 − 7| = 2 The length of the longest edge is 4 Hence option C is the correct answer. 48. Let the points be

𝐴(5, −4,2), 𝐵(4, −3,1), 𝐶(7, −6,4) and Mock Test - 9

𝐵𝐶 = (

3+7 2+2 −1+5 , , ) 2 2 2

= (5,2,2)

Hence option C is the correct answer. 50. since the point is intersect with both line and plane, it must be in both line and plane Equation of line is

𝑥+1 1

=

𝑦+3 3

=

𝑧−2 2

and equation of plane is 3x + 4y + 5z = 5 From the given option the point lie on both line and plane is (1,3,-2) If we apply this values in equation of plane 3 × 1 + 4 × 3 + 5 × (- 2) = 5

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Which is true

Hence option A and C are the correct answer.

∴ The point is (1, 3, -2) 54. All points lying inside the triangle formed by the points given, then this implies that all the three vertices of triangle lies on or same side of the given conditions:

Hence option D is the correct answer. 51. An equation in which at least on term is squared and none of the terms has a degree more than 2 is know as quadratic equation ⇒ A quadratic equation can contain any terms of degree o or 1 or 2. The term in option C is of degree 3 and hence the correct options are A, B and D. Hence options A, B and D is the correct answer.

52. Let the equation of the circle be

A

3𝑥 + 2𝑦 ≥ 0 where (1,3),(5,0) and (1,2) satisfies the above condition. Therefore A is True. B

2𝑥 + 𝑦 − 13 ≥ 0

𝑥 2 + 𝑦 2 + 2𝑔𝑥 +

2𝑓𝑦 + 𝑐 = 0

(1, 3), does not satisfy it. Therefore, B is false C

Equation of tangent to (1) at origin ( 0,0 ) is

𝑥. 0 + 𝑦. 0 =

𝑔(𝑥 + 0) + 𝑓(𝑦 + 0) + 𝑐 = 0 or, 𝑔𝑥 + 𝑓𝑦 + 𝑐 = 0 But it is given to be axis of 𝑥, l.e., 𝑦 = 0 ∴ 𝑔 = 0, 𝑐 = 0 ∴ Equation (1) becomes 𝑥 2 + 𝑦 2 + 2𝑓𝑦 = 0  … (2) If it touches the line 3𝑦 = 4𝑥 + 24 i.e., 4𝑥 − 3𝑦 + 24 = 0 then the length of ⊥ from centre (0, −𝑓) on the line is numerically = radius

𝑓∴

4(0)−3(−𝑓)+24 √16+4

=

2𝑥 − 3𝑦 − 12 ≤ 0 It is satisfied for all points. Therefore C is true. D

−2𝑥 + 𝑦 ≥ 0 is not satisfied by (5,0) Therefore, D is false Thus (a), (c) are correct answers. 55. Clearly line L is radical axis of of the given pair of circle.

±𝑓 ⇒ 3𝑓 + 24 = ±5𝑓 ⇒ 𝑓 = 12, −3 putting these values of 𝑓 in (2) , the equation of circle are 𝑥 2 + 𝑦 2 + 24𝑦 = 0 and 𝑥 2 + 𝑦 2 − 6𝑦 = 0

And we know radical axis is perpendicular to the line joining the centre of the circles.

Hence option A and B are the correct answer.

Thus line is also common chord.

53. The equation of the circle is

𝑥 2 + 𝑦 2 − 12𝑥 −

16𝑦 + 75 = 0 Join its centre 𝐶 with O Let 𝑂𝐶 meets the circle in 𝐴 and 𝐵 . Centre of the circle is 𝐶 ≡ (6,8) and radius 𝐴𝐶 = √36 + 64 − 75 = 5 Also, 𝑂𝐶 = √36 + 64 = 10 ∴ 𝑂𝐴 = 𝑂𝐶 − 𝐴𝐶 = 5 Now, the point on the circle nearest to the origin is 𝐴 and farthest from it is 𝐵 . ∵ 𝑂𝐴: 𝐴𝐶 = 5: 5 = 1: 1 ⇒ 𝐴 is the mid-point of 𝑂𝐶 the coordinates of 𝐴 are 0+6 0+8 ( , ) i.e. (3,4) Let the coordinate of 𝐵 be 2 2 3+ℎ 4+𝑘

since

𝐶 is the mid point of 𝐴𝐵

2

ℎ = 9, 𝑘 = 12 the coordinates are of

Mock Test - 9

𝐵 are (9,12)

= 6,

2

Also 𝐶1 𝐶2 two point.

< 𝑟1 + 𝑟2 . So both the circles intersect at

56. Linnaeus evolved a system of nomenclature called the binomial system of nomenclature. In this system, the organism is named with two words, the first word provides the name of the genus and the second word provides the name of the species the organism belongs to. Hence option C is the correct answer. 57. The sperms of Cycas are the largest in the plant kingdom reaching a size of 300 µm . Egg of Cycas and its nucleus are also the largest in the plant kingdom. Hence option C is the correct answer.

(ℎ, 𝑘)

=8⇒

58. The underside of echinoderms' body bears a mouth at the center and a groove running along each arm. The grooves contain rows of tiny, flexible appendages called tube feet. Starfish move by means of the tube feet, which are operated by a hydraulic or water-vascular system

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unique to echinoderms. Sea water, circulates through the radiating canals of this system, enters and extends the tube feet. Each tube foot can be withdrawn by its attached muscles. The tube feet are equipped with suction cups, and the animal moves in any direction by gripping with some of its tube feet and pulling itself forward. A starfish that is turned upside down can right itself by turning an arm under and walking with the tube feet.

interconversion, active K+ transport. According to the photosynthetic production theory,stomata close when CO 2 concentration is maximum and O2 concentration is less.

Hence option D is the correct answer.

Hence option C is the correct answer.

59. Seeds having fleshy testa are the edible parts in pomegranate. In case of induced parthenocarpy, the seeds will not be produced thereby making fruit seedless. Therefore they become useless and lose their economic importance.

66. Based on key changes and the given image, this stage of Mitosis is Metaphase.

So, the correct answer is 'pomegranate' 60. Phellem is outermost layer of cork cambium, which is suberised. Hence, this is the first layer to be found. Endodermis is one the layer of cortex, which is to be found under the phellem. Pericycle is the layer present between endodermis and vascular bundles.

Hence option D is the correct answer. 65. Essential micro elements are Fe, Mn, Zn, B, Cu, Mo and essential macronutrionts are C, H, N, P, S, Ca, K, Mg.

[Metaphase is a stage of mitosis in the eukaryotic cell cycle in which chromosomes are at their second-most condensed and coiled stage. These chromosomes, carrying genetic information, align in the equator of the cell before being separated into each of the two daughter cells.] 67.

Dicotyledonous stem possess conjoint, collateral vascular bundle, where xylem is internal and phloem is external. Therefore, phloem is to be encountered after the pericycle. Hence option C is the correct answer. 61. Tendons are modified cords of white fibrous connective tissues which connect the skeletal muscles with the bone. Hence option C is the correct answer. 62. The bonding of sugar residue to another organic compound is known as glycosylation. Glycoprotein are formed in the lumen of granular or rough endoplasmic reticulum(RER), but may subsequently be modified in the lumen of the Golgi apparatus while other amino acids of the protein may become glycosylated.

A = Kinetochore - It is a complex of proteins associated with the centromere of a chromosome during cell division, to which the microtubules of the spindle attach.

Hence option C is the correct answer.

B = Spindle - Spindle fibres are protein structures that form early in mitosis or cell division. They consist of microtubules that originate from the centrioles, two wheel-shaped bodies located in the centromere area of the cell.

63. During cytokinesis in animal cells, a ring of actin filaments forms at the metaphase plate. The ring contracts, forming a cleavage furrow, which divides the cell in two. In plant cells, Golgi vesicles coalesce at the former metaphase plate, forming a phragmoplast.

C = Chromosomes - Thread-like structure of nucleic acids and protein found in the nucleus of most living cells. These carry genetic information in the form of genes.

Hence option C is the correct answer.

68. Cell division is the process by which a parent cell divides into two or more daughter cells.

64. Stomata are minute specialised pores on the epidermal layers of leaves. Closing and opening mechanism of stomata depends on some factors like photosynthetic production, turgidity of guard cell , sugar-starch Mock Test - 9

Generation time is 1 minute means after every min. The cell is going to be double of its initial value. In 20 mins culture tube ->1/8 filled

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Means after 1 min it will be doubled so in 21-minute culture tube will be 1/8 *2 = 1/4 quarterly filled Again after 1 min. It will be 1/4 *2=1/2 half filled Again after 1 min, it will be1/2*2 = 1 (fully filled) means in 23 minutes vessel will be completely filled. So, the correct answer is '23 minutes'. 69. In Biology, "Karyon" refers to the nucleus of the cell. So karyokinesis is the division of the nucleus during cell division. Hence option C is the correct answer. 70. Spindle fibres are made up of microtubules. These fibres form a protein structure that divides the genetic material in a cell. The spindle is necessary to equally divide the chromosomes in a parental cell into two daughter cells during both types of nuclear divisions - mitosis and meiosis. Hence option D is the correct answer.

Mock Test - 9

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MOCK TEST - 10

A. Pasteurisation

B. Sterlisation

C. Disinfection

D. Autoclave

8. Knot is a measure of-

General Ability

A. Solar radiation B. The curvature of spherical objects

1. Who among the following Indian-origin film director has received Lifetime Achievement Award of Academy of Canadian Cinema?

C. The speed of ship

A. Meera Nair

B. Deepa Mehta

C. Knight Shyamlan

D. Arun Desai

9. Which one of the following is used to restore the colour of old oil paintings?

D. Intensity of earthquake shock

2. Which union ministry has launched Shehri Samridhi Utsav? A. Union Ministry of Housing & Urban Affairs

A. Ozone

B. Hydrogen Peroxide

C. Barium Peroxide

D. Sodium Peroxide

B. Ministry of Culture C. Ministry of Earth Sciences

10. ‘Kelp’ is .

D. Ministry of External Affairs

A. Sulphide mineral of iron

3. Central government’s Umbrella programme for the development of Scheduled Tribes has been extended to the period of? A. March 2020

B. March 2025

C. March 2021

D. March 2022

B. Partially decomposed vegetation C. Sea weed rich in iodine content D. An aluminum silicate mineral

4. Who among the following has been appointed as Director General of Punjab Police? A. Dinkar Gupta

B. Suresh Arora

C. Rakesh Madan

D. Anil Srivastava

Physics 11. Hologram is based on phenomenon of A. Diffraction

5. Who has been appointed as the new US ambassador to the UN?

B. Polarisation C. Interference

A. Nikki Haley

B. Kelly Knight Craft

C. Ivanka Trump

D. Sharmili Roger

6. Foreign materials entering the cell, such as bacteria or food, as well as old organelles end up in the _____________. A. Vacuoles

B. Mitochondria

C. Plastids

D. Lysosomes

7. Which is the process of killing diseases producing micro-organism in food items by heat-

Mock Test - 10

D. Total internal reflection 12. Process of emission of electrons from hot metal surface is calledA. electronic emission

B. electron capture

C. thermionic emission

D. none of the above

13. Which of the following relates to photons both as wave motion and as a stream of particles? A. Interference

B. E = mc2

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C. Difraction

D. E = hv

C. Does not depend on the frequency of photon but depends only on intensity of incident light

14. Light of wavelength 200nm shines on an aluminium, 4.20eV is required to eject an electron. What is the kinetic energy of the fastest ejected electrons?

D. Depends both on intensity and frequency of incident beam

A. 0.5eV

B. 1.00eV

C. 2.00eV

D. 4.00eV

21. A particle is moving along a vertical circular path of radius r. When the circular path is just completed: A. the velocity at highest position is √rg

15. In an inelastic collision an electron excites a hydrogen atom from its ground state to a M-shell state. A second electron collides instantaneously with the excited hydrogen atom in the M-shell state and ionizes it. At least how much energy the second electron transfers to the atom in the M-shell state? A. +3.4eV

B. +1.51eV

C. −3.4eV

D. −1.51eV

B. the velocity at highest position is zero C. the velocity at lowest position is √5rg D. the tension in string at the lowest position is 6mg 22. A ball moves over a fixed track as shown in the figure. From A to B the ball rolls without slipping. Surface BC is frictionless. KA, KB and Kc are kinetic energies of the ball at A, B and C, respectively. Then

16. The surface of a metal is illuminated with the light of 400 nm. The kinetic energy of the ejected photoelectrons was found to be 1.68 eV. The work function of the metal is : (hc = 1240 eV.nm) A. 3.09 eV

B. 1.41 eV

C. 1.51 eV

D. 1.68 eV

17. If λ is the wavelength of hydrogen atom from the transition n = 3 to n = 1 then what is the wavelength for doubly ionised lithium ion for same transition? A. λ ÷ 3

B. 3λ

C. λ ÷ 9

D. 9λ

18. Work function of a metal is 5.2 × 10-18 J then its threshold wavelength will be A. 736.7A˚

B. 760.7A˚

C. 301Å

D. 380.7Å

19. The formula of kinetic mass of photon is? Where h is Planck's constant, v is frequency of the photon and c is its speed. A. hv/c

B. hv/c2

C. hc/v

D. c2/h

A. hA>hC; KB>KC B. hA>hC; KC>KA C. hAKC D. hA=hC; KB=KC 23. It is found that a body on an inclined plane just starts sliding down if inclination is sin −1(3/5), then the angle of repose (friction) is A. sin−13/5

B. tan−13/4

C. sin−13/4

D. tan−13/5

24. A point mass of 1 kg collides elastically with a stationary point mass of 5 kg. After their collision, the 1 kg mass reverses its direction and moves with a speed of 2 ms−1. Which of the following statement(s) is (are) correct for the system of these two masses? A. Total momentum of the system is 3 kg ms −1

20. In photoelectric effect, the photo current. A. Increases with increase of frequency of incident photon

B. Kinetic energy of the centre of mass is 0.75 J C. Momentum of 5 kg mass after collision is 4 kg ms −1 D. Total kinetic energy of the system is 4J

B. Decreases with increase of frequency of incident photon

Mock Test - 10

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25. A thin ring of mass 2 kg and radius 0.5 m is rolling without slipping on a horizontal plane with velocity 1m/s. A small ball of mass 0.1 kg, moving with velocity 20 m/s in the opposite direction, hits the ring at a height of 0.75 m and goes vertically up with velocity 10 m/s. Immediately after the collision:

B.

C.

A. the ring has pure rotation about its stationary CM. B. friction between the ring and the ground is to the left.

D.

C. the ring comes to a complete stop. D. there is no friction between the ring and the ground.

Chemistry 26. 'Z' in the following sequence of reaction is:

𝐶6 𝐻6 A.

𝑍𝑛 /𝐻𝐶 𝐻𝑁𝑂3 /𝐻2 𝑆𝑂4 𝑊 → 𝛥

𝑋

𝑁𝑎𝑁𝑂2 𝐻2 𝑂/𝐻3 𝑃𝑂2 𝑌 𝑍 𝐻𝐶𝑙

27. Identify the product formed in Friedal-Crafts reaction of acetyl chloride. A. Ketone

B. Benzene

C. Aldehyde

D. Phenol

28. Which of the following is not correct regarding terylene? A. Step-growth polymer

B. Synthetic bre

C. Condensation polymer

D. Thermosetting plastic

29. The acid which does not form an anhydride when treated with P2O5 is: A. Formic acid

B. Acetic acid

C. Propionic acid

D. Benzoic acid

30. Identify the correct statement(s) out of the following: a. Deciency of vitamin A causes xerophthalmia b. The function of vitamin C is maintenance of redox potentials of cells Mock Test - 10

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c. Vitamin B12 contain ionone ring.

A. [H+]1 / [H+]2

d. Folic acid (vitamin B9) consists of corrin ring.

B. α1 / α2

A. a only

B. a and b

C. a, b and c

D. all

31. When ethyl alcohol is passed over red hot copper at 300°C, the formula of the product formed is A. CH3CHO B. CH3COCH

C. C1α1 / C2α2 D. Ka1C1 / Ka2C2 38. 2CaSO4(s)⇌2CaO(s)+2SO2(g)+O2(g),ΔH>0 Above equilibrium is established by taking sufficient amount of CaSO4(s) in a closed container at 1600 K. Then which of the following may be correct option(s)? (Assume that solid CaSO4 is present in the container in each case)

C. C2H4 D. CH3COOH 32. Which of the following is a non-reducing sugar? A. glucose

B. sucrose

C. fructose

D. lactose

33. Which of the following reactions is called Rosenmund reaction? A. Aldehydes are reduced to alcohols B. Acids are converted to acid chlorides C. Alcohols are reduced to hydrocarbons

A. Moles of CaO(s) will increase with the increase in temperature B. If the volume of the container is halved partial pressure of O2(g) at new equilibrium will remain same C. If the volume of the container is doubled at equilibrium then partial pressure of SO 2(g) will change at new equilibrium D. If two moles of He gas is added at constant pressure then the moles of CaO(s) will increase. 39. Which of the following statements is/are correct?

D. Acid chloride are reduced to aldehydes 34. Acetic anhydride with Benzene in presence of anhydrous gives: AlCl3. A. acetyl chloride

B. acetophenone

C. benzophenone

D. none of the above

35. Identify the co-polymer from the following: A. [−CH2 − CH = CH2 − CH2 − CH(C6H5 ) − CH2−]n

A. If for a reaction, ΔH is −ve and ΔS is + ve, then the reverse reaction is always non-spontaneous at all temperatures B. If ΔH is + ve and ΔS is − ve, then the forward reaction is non-spontaneous at all temperatures C. If both ΔH and ΔS are − ve, then the reverse reaction is spontaneous at high temperatures D. If both ΔH and ΔS are + ve, then the reverse reaction is spontaneous at all temperatures

B. [−CF2 − CF2−]n C. [−CH2 − C(Cl) = CH − CH2−]n D. [−CH2 − CH(Cl)−]n

40. Vinyl bromide undergoes: A. addition reaction

B. elimination reaction

C. substitution reaction

D. rearrangement reaction

36. Which of the following statements is/are correct? A. Boric acid is a hydrogen-bonded molecule B. Boric acid combines with CuO to give meta borate in the borax bead test

Mathematics

C. Boric acid is a lewis acid D. Al2O3 is amphoteric while B2O3is acidic 37. If concentration of two weak acids are different and degree of ionization (α) is very less, then their relative strength can be compared by: Mock Test - 10

41. The area enclosed by the curve |x| + |y| = 1 in sq. units is⋯ A. 2

B. 4

C. 1

D. 1/2

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42. The area bounded by the parabola y2 = 4x and its latusrectum is: A. (8/3) sq. units

B. (3/8) sq. units

C. 12 sq. units

D. (1/3) sq. units

B. (3/8) sq. units

C. 24 sq. units

D. 25 sq. units

B. (125/3) sq. units

C. (122/3) sq. units

D. (121/3) sq. units

C. 6

D. -(2/3) sin4 𝑥 2

+

cos4 𝑥 3

1 5

= then

2

A.

tan2 𝑥 = 3

B.

sin8 𝑥 8

44. The area bounded by the parabola y = 4x 2, X - axis between the ordinates x = 2, x = 4 is: A. (224/3) sq. units

B. -4

51. If

43. The area of the region bounded by the curve y = x 2, x axis and the ordinates x = 0, x = 2 is: A. (8/3) sq. units

A. 4

+

cos8 𝑥 27

1

= 125

1

C.

tan2 𝑥 = 3

D.

sin8 𝑥 8

+

cos8 𝑥 27

2

= 125

𝑓(𝑥) = 2|𝑥| + |𝑥 + 2| − ||𝑥 + 2| − 2|𝑥 ∣ has a local minimum or a local maximum at 52. The function x=?

45. The area bounded by the curve

𝑦 = sin𝑥, 𝑥 − 𝑥 = −𝜋, 𝑥 = 𝜋 is:

A. −2

B. -2/3

axis between the ordinates

C. 2

D. 2/3

A. 2 sq. units

B. 3 sq. units

C. 4 sq. units

D. (1/2) sq. units

53. Let

𝑎, 𝑥, 𝑏 be in A.P. ; 𝑎, 𝑦, 𝑏 be in G.P. and 𝑎, 𝑧, 𝑏 𝑥 = 𝑦 + 2 and 𝑎 = 5𝑧 then

be in H.P. If

46. The area of the region y = ax − bx2 bounded by x-axis in sq. units is

A. y2=xz

B. x>y>z

C. a=9,b=1

D. a=9/4,b=1/4

A. a3/6b2 54. The equations of the circles, which touch both the axes and the line 4x+3y=12 and have centres in the first quadrant, are

B. a3/6 C. a D. b

A. x2+y2+x−y+1=0

47. The area bounded by y= x2 + 2, X-axis, x = 1 and x = 2 is:

B. x2+y2−2x−2y+1=0

A. 16/3

B. 17/3

D. x2+y2−6x−6y+36=0

C. 13/3

D. 20/3

48. If α and β are the roots of the equation x 2 - a(x + 1) - b = 0, then A. b

(𝛼 + 1)(𝛽 + 1) = B. -b

C. 1-b

D. b-1

49. If α and β are the roots of the equation x 2 + 6x +

C. x2+y2−12x−12y+36=0

55. If the equation x2+y2−10x+21=0 has real roots x=α and y=β then A. 3≤x≤7

B. −2≤y≤2

C. 3≤y≤7

D. −2≤x≤2

𝜆=

0 3𝛼 + 2𝛽 = −20 then 𝜆 A. -8

B. -16

C. 16

D. 8

50. If the sum of the roots of the equation

Biology 𝜆𝑥 2 +

2𝑘 + 3𝜆 = 0 be equal to their product, then λ =

56. Which of the following categories occupies the topmost position of the classification triangle? A. Genus

Mock Test - 10

B. Species

| 179

C. Order

D. Kingdom

C. Their mode of nutrition is both autotrophic and heterotrophic

57. Human beings belong to the category ________ because they have a jawline.

D. They have a nucleoid

A. Chordata

B. Mammalia

C. Gnathostomata

D. Agnatha

66. If we assume that the 'lock and key model' of the enzyme-substrate interaction is correct, which of these conditions is absolutely true?

58. The term "New Systematics” was introduced by A. Bentham and Hooker

B. Carl Linnaeus

C. Julian Huxley

D. A.P. de Candolle

A. The enzyme can bind to substrates of only one shape B. Substrates can change the shape of the active site C. The enzyme can bind to substrates of many different shapes and sizes

59. Which of the following is correctly matched with its particular taxonomic category ?

D. There is a change in the shape of the active site after binding to the substrate

A. Triticum aestivum - Species

67. One strand of DNA has sequence of nucleotide 3' ATTCGCTAT 5’ then other strand of DNA has

B. Fishes - Pisces - Phylum C. Man - Primate - Family

A. 3' TAAGCGATA 5’

B. 3' TAGCACGTA 5’

D. Mango - Sapindales - Class

C. 5' TAGCACGTA 3’

D. 5' TAAGCGATA 3’

60. Binomial nomenclature of plants was given by A. Engler

B. Linnaeus

68. Choose the sugars arranged in the ascending order of the number of carbon atoms they possess.

C. Prantl

D. Both (a) and (c)

A. Hexose, Pentose, Tetrose, Triose

61. Fungi causing hair loss are _______________. A. Keratophilous

B. Pyrophilous

C. Coprophilous

D. None of these

B. Tetrose, Hexose, Triose, Pentose C. Triose, Pentose, Tetrose, Hexose D. Triose, Tetrose, Pentose, Hexose

62. Fungi differ from other kingdoms in being

69. The major function of simple proteins like Keratin and Collagen is _____.

A. Unicellular decomposers

A. Catalysts

B. Structural integrity

B. Unicellular consumers

C. Constituents

D. Functional integrity

C. Multicellular decomposers D. Multicellular consumers

70. In telophase, which of the following structures are reorganized?

63. Edible part of mushroom is

A. Spindle fibres

A. Basidiocarp

B. Primary mycelium

B. Chromosomes

C. Fungal hyphae

D. Basidiospores

C. Centrioles D. Nuclear envelope and nucleolus

64. The algal component of a lichen is known as A. Mycobiont

B. Phycobiont

C. Phycocyanin

D. Phycoerythrin

65. Which of the following is not a character of Protista? A. Protists are eukaryotic. B. Some protists have cell walls.

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| 181

HINTS AND SOLUTIONS 1. The Academy of Canadian Cinema & Television has announced that it will be honouring filmmaker Deepa Mehta with the Lifetime Achievement Award. Hence option B is the correct answer. 2. Shehri Samridhi Utsav, an initiative of Ministry of Housing & Urban Affairs (MoHUA),aims to extend the outreach of Deendayal Antyodaya Mission – National Urban Livelihoods Mission (DAY-NULM), to the most vulnerable, showcase its initiatives and facilitate access of Self-Help Group (SHG) members to the other government schemes.This was launched across the country earlier this week.

8. A knot is one nautical mile per hour (1 knot = 1.15 miles per hour ). The term knot dates from the 17th century, when sailors measured the speed of their ship by using a device called a "common log." Hence option C is the correct answer. 9. The white pigment in old painting turns black due to formation of PbS. This white pigment is restored by using hydrogen peroxide. Hence option B is the correct answer. 10. Kelps are large seaweeds (algae) beloging to the brown algae in the order laminariales. Because of its high concentration of iodine, kelp has been used to treat goiter, an enlargement of the thyroid gland caused by a lack of iodine. Hence option C is the correct answer.

Hence option A is the correct answer. 11. Hologram is based on the phenomenon of interference. 3. Cabinet Committee on Economic Affairs (CCEA) has approved the continuation of sub-schemes under Umbrella programme for the development of Scheduled Tribes from the period till March 2020.

[There are two physical phenomena as the principles of the holography: interference and diffraction of light waves. Holograms are photographs of three dimensional impressions on the surface of light waves.]

Hence option A is the correct answer. Hence option C is the correct answer. 4. Senior IPS officer Dinkar Gupta took over as the new Director General of Punjab Police, succeeding Suresh Arora, who was on an extension after his retirement on September 30 last year. Hence option A is the correct answer. 5. US President Donald Trump has announced the nomination of Kelly Knight Craft as the new US ambassador to the UN. The post of the US ambassador to the UN was vacant due to the resignation of Nikki Haley. Hence option B is the correct answer.

12. Thermal energy given to the charge carriers overcome the work function of the metal and thus low of charge carriers takes place from the metal surface. These charge carriers can be electrons and ions. So, process of the emission of electrons from the hot metal surface is called as thermionic emission. Hence option C is the correct answer. 13. (a) Interference exhibits the wave nature of light (b) E = mc2 exhibits the particles nature of light (c) Difraction exhibits the wave nature of light.

6. Lysosomes are basically a class of waste disposal system of the cell that helps to maintain the cleanness of the cell by digesting any strange material as well as useless cell organelles. Hence option D is the correct answer. 7. Pasteurization is the application of heat to a food product in order to destroy pathogenic (disease-producing) microorganisms, to inactivate spoilage-causing enzymes, and to reduce or destroy spoilage microorganisms. Hence option A is the correct answer.

Mock Test - 10

(d) E = hν shows both particle nature and wave nature of light, i.e., light is having a particular wavelength and the energy is in the form of packets. Hence option D is the correct answer. 14. Using the Einstein's equation of photoelectric effect, ℎ𝑐 𝜆

= 𝐾𝐸 + 𝜙

𝐾𝐸 =

1240 200

− 4.2

𝐾𝐸 = 2𝑒𝑉 | 182

Hence option C is the correct answer.

𝜆0 =

6.6×10−3 ×3×108 5.2×10−18

= 380.7𝐴0

Hence option D is the correct answer.

15. Given that the electron is in M state. This corresponds tothe principal quantum number n = 3 From Bohr's model, energy of a state with quantum number n is given by

𝐸=−

13.6 𝑒𝑉 𝑛2

∴ mkc2 = hv

Thus, the energy of the electron in M shell is

−

13.6 32

19. Energy of photon having frequency v is given by E = hv Energy of photon can be written as E = mkc 2 where mk is the kinetic mass of photon

𝐸=

⟹ mk = hv/c2` Hence option B is the correct answer.

≈ −1.51𝑒𝑉

In order to ionise the atom, the minimum energy required is thus +1.51eV Hence option B is the correct answer. 16.

20. When intensity of incident photon increases number of electrons emitted from the surface increases due to which current increases. Frequency of incident photon only limits the maximum kinetic energy of the photoelectron emitted and so the current is independent of frequency of photon. Hence option C is the correct answer.

Hence option B is the correct answer. 17. For wavelength: 1 𝜆

= 𝑅𝑍 2 (

1 𝑛12

−

1 ) 𝑛22

Here, transition is same 𝜆𝐻 𝜆𝐿

=

(𝑍𝐿𝐴 )

𝜆𝐿𝑖 =

1 2𝐼 2

2

(3)1 = 2 (𝑍𝐻 )2 (1) 𝜆𝐻 𝜆 9

𝑆𝑜, 𝜆 ∝

=

21.

=9

Considering the upper most point as 1 and the lowest point as

9

Hence option C is the correct answer. 18. Given - work function of metal W = 5.2 function of a metal is given by,

𝑊= or

ℎ𝑐 𝜆0

𝜆0 =

where

2, also consider 𝑉𝑡 and 𝑉𝑏 as velocity at top most and

bottom most point, ×10-18

The work

𝑇1 + 𝑚𝑔 =

𝑚𝑉𝑖2 𝑅 1 2

𝑇1 = 0 at highest point 𝑉𝑡 = √𝑔𝑅 𝑚𝑉𝑏2 − 1 𝑚𝑉𝑡2 2

= 2𝑚𝑔𝑅

𝑉𝑏 = √5𝑅𝑔( at lowest point ) 𝑇2 − 𝑚𝑔 =

ℎ𝑐 𝑊

𝜆0 = threshold wavelength

ℎ = 6.6 × 10−34 𝐽𝑠

𝑇2 = 𝑚𝑔 +

𝑚𝑉𝑏2 𝑅 𝑚𝑉𝑏2

( at lowest point )

𝑅

𝑇 = 6𝑚𝑔 Hence option A, C and D are the correct answer.

8

𝑐 = 3 × 10 𝑚/𝑠 Hence, by

Mock Test - 10

𝜆0 =

ℎ𝑐 𝑊

| 183

22.

vertical direction. Thus on the ring at the point of contact there is a horizontal and a vertical impulse. These will have components along the tangent of the ring, which will provide angular impulses. Using angular impulse = change in angular momentum we get: =2cos30° 1 / 2 − 1 sin 30° 1 / 2 = 2 (1 / 4) (ω2−ω1) note that we have assumed that direction of angular velocities is same before and after and since LHS of the above equation is positive ω2>ω1 thus the ring must be slipping to right and hence the friction will be to the left as it will be opposite to the direction of motion. Thus option C is correct 26. The first reaction is nitration of benzene to give

Hence option A, B and C is the correct answer.

nitrobenzene(W)

𝐶6 𝐻6 + 𝐻𝑁𝑂3 /𝐻2 𝑆𝑂4 →

𝐶6 𝐻5 𝑁𝑂2 23. We know, angle of repose

= angle of friction 𝐴𝑙𝑠 o,

3 5

𝜇 = tan𝜃 𝜃 = sin−1 ( ) Also, by trigonometry.

The second reaction is reduction of nitrobenzene to aniline

𝐶6 𝐻5 𝑁𝑂2 + 𝑍𝑛/𝐻𝐶𝑙 → 𝐶6 𝐻5 𝑁𝐻2 The third reaction is formation of arenediazonium salt.

3 4

𝜃 = tan−1 ( )

Hence option A and B are the correct answer.

𝐶6 𝐻5 𝑁𝐻2 + 𝑁𝑎𝑁𝑂2 − 𝐻𝐶𝑙 → 𝐶6 𝐻5 𝑁2+ 𝐶𝑙 The fourth step is hydrolysis of arene diazonium salt to give benzene

𝐶6 𝐻5 𝑁2+ 𝐶𝑙 − + 𝐻2 𝑂/𝐻3 𝑃𝑂2 → 𝐶6 𝐻6

Hence option B is the correct answer. 24. 27.

By conservation of linear momentum v = 5v - 2 ..... (i) By Newton's experimental law of collision u = v + 2 ....(ii) using (i) and (ii) we have u = 1m/s and u = 3m/s Kinetic energy of the centre of mass 1 𝑚 𝑉2 2 𝑠𝑦𝑠𝑡𝑒𝑚 𝑐𝑚

=

Acyl chlorides react with aromatic hydrocarbons in Friedel Crat acylation reaction to produce aromatic ketone. Acetyl group (CH3 − CO) is introduced in benzene ring. It is electrophilic aromatic substitution reaction. Hence option A is correct answer.

= 0.75. 𝐽

Hence option A and B are the correct answer. 25. Lets assume that friction between the ground and the ring gives no impulse during the collision with the ball. Using conservation of momentum along the x-axis we get that the CM of the ring will come to rest. Thus option A is correct.

28. Terylene is a synthetic polyester bre or fabric based on terephthalic acid, charectrized by lightness and crease resistance and used for clothing, sheets, ropes, sails, etc. It is produced by polymerizing terephthalic acid and ethylene glycol using condensation techniques. Thermosetting plastics are polymer materials which are liquid or malleable at low temperatures, but which change irreversibly to become hard at high temperatures. Terylene is not a thermosetting plastic hence option D is correct.

Secondly the question tells us that the ball gets a velocity in the vertical direction hence there must be an impulse in the Mock Test - 10

| 184

29. Phosphorus pentoxide is a dehydrating agent.

34.

It abstracts a molecule of water from formic acid to form carbon dioxide.

𝑅𝑂3

𝐻𝐶𝑂𝑂𝐻 → 𝐶𝑂2 + 𝐻2 𝑂

Thus, formic acid which does not form an anhydride when treated with P2O5. Hence option A is the correct answer. 30. Option (B) is correct. A. Deciency of vitamin causes xerophthalmia. It is a condition in which an eye becomes abnormally dry because it can't maintain an adequate layer of tears to coat its surface. B. The function of vitamin C is maintenance of redox potentials of cells important for defence against infections such as common colds. It also acts as an inhibitor of histamine, a compound that is released during allergic reactions. As a powerful antioxidant it can neutralize harmful free radicals and it aids in neutralizing pollutants and toxins. C. Vitamin A contain ionone ring. All forms of vitamin A have acbeta - ionone ring to which an isoprenoid chain is attached, called a retinyl group.

It is an example of Friedel Crat acylation reaction in which acetyl group (CH3 − CO) is introduced in benzene ring. It is electrophilic aromatic substitution reaction. 35. SBR (Buna S), [−CH2 − CH = CH2 − CH2 − CH(C6H5 ) − CH2−]n is a co-polymer. It is made from 2 monomers 1,3-butadiene and styrene. It is an example of addition polymerisation. Note: Two or more types of monomers are used in the preparation of a co-polymer. Hence option A is the correct answer. 36.

D. Folic acid (vitamin B12 ) consists of corrin ring. It contains a cobalt ion as part of the porphyrin-like corrin ring. Hence option B is the correct answer. 31. When ethyl alcohol is passed over red hot copper at 300°C, the product formed is acetaldehyde, CH3CHO The reaction is as follows:

𝐶2 𝐻5 𝑂𝐻

red hot 𝐶𝑢 𝐶𝐻3 𝐶𝐻𝑂 300∘ 𝐶

Hence option A is the correct answer. 32. There is no free aldehyde group present in sucrose due to the glycosidic linkage. While in others, there is a free aldehydic/ketonic group present. If an aldehydic group is present, the molecules can reduce various compounds (reagents), which are also used as a test for the same. Even if ketonic group is present, it isomerizes to aldehydic group. So, sucrose is a nonreducing sugar. Hence option B is the correct answer. 33. in Rosenmund reaction, acid chloride are reduced to aldehydes. [The Rosenmund reduction is a hydrogenation process in which an acyl chloride is selectively reduced to an aldehyde. The reaction was named after Karl Wilhelm Rosenmund, who first reported it in 1918.]

A) Boric acid molecules are given structure.

𝐻 - bonded as shown in the

𝐻𝐵𝑂2 + 𝐵2 𝑂3 which react with 𝐶𝑢𝑂 to give 𝐶𝑢(𝐵𝑂2 )2 (blue B) Boric acid, on heating gives a mixture of

bead in borax bead test) C) A lewis acid is an electron pair acceptor. The boron atom in boric acid, 𝐵(𝑂𝐻)3 , has an empty shell, able to accept a fourth electron pair. For example, it can accept oxygen's electrons in an acid-base reaction with water:

𝐵(𝑂𝐻)3 + 𝐻2 𝑂 → 𝐵(𝑂𝐻)4 + 𝐻 + D) 𝐴𝑙2 𝑂3 reacts with both acid as well as base whereas 𝐵2 𝑂3 reacts only with base. Hence all options are correct answer.

Hence option D is the correct answer.

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| 185

37. The Relative strength (R.S) [𝐻 + ∣furnirhed by acid 𝐼 [𝐻 + ]furnishod by acid 𝐼𝐼

=

=

𝐶1 𝛼1 𝐶2 𝛼2

=

41. 𝐶1 𝐶2

𝐾𝑎1 𝐶2

×√

𝐾02 𝐶1

=

𝐾𝑎1 𝐶1

√𝐾

𝑎2 𝐶2

Relative strength of various acids can be compared using

𝐾𝑎 larger the value of 𝐾𝑎 , the stronger the acid. So their relative strength can be compared by

∣𝐻 + ]1 𝐶1 𝛼1 or [𝐻 + ]2 𝐶2 𝛼2

Hence option A and C are the correct answer. 38. For endothermic reaction (△ 𝐻 > 0) if temperature increases equlibrium shift towards right. Therefore option A is correct. For any change in the volume in the container not going to change the equilibrium pressure at a given temperature if the CaSO4 is available. Therefore Partial pressure of SO2 will not change. incorrect. Same reason as for B . Option C is correct. If any external gas is added to the equlibrium it will shift towards left. Thus option D is wrong.

|𝑥| + |𝑦| = 1 Area bounded = 4 × (1/2) × 1 × 1 = 2 Hence option A is the correct answer.

Hence option A and C are the correct answer.

𝛥𝐺 = 𝛥𝐻 − 𝑇𝛥𝑆 A) 𝛥𝐺 is always − ve; the forward reaction is 39.

spontaneous and backward reaction is always nonspontaneous.

𝛥𝐺 is always + ve; the forward reaction is nonspontaneous at all temperatures. C) 𝛥𝐺 is - ve at low temperatures and + ve at high temperatures; the forward reaction is spontaneous at low temperatures. D) 𝛥𝐺 is + ve at low temperatures and − ve at high temperatures; non spontaneous at low temperatures, spontaneous as the temperatures increases. Hence option A, B and C are the correct answer. 40. Vinyl Bromide (𝐵𝑟𝐶𝐻 = 𝐶𝐻2 ) shows both addition and elimination reactions. In addition reaction 𝜋 , bond

42. 1

Area bounded

= 2 ∫0 2 √𝑥𝑑𝑥

Area bounded=

= 2 (2 × ) = 2(4/3) = 8/3

2 3

Hence option A is the correct answer.

breaks and it forms ethyl bromide and in elimination reactions hydrogen bromide eliminates and forms allyyne. Substitution is not possible as that would mean sp 2 a carbon having a positive charge, which is unfavorable. Rearrangement reaction is also not possible because of the simple reason that there are only two carbon atoms and rearrangement to a different structure cannot happen, as no other structure exists. Hence option A and B are the correct answer.

Mock Test - 10

| 186

43. y = x2 ⇒ area is the shades region

2

∫0 𝑥 2 𝑑𝑥 =

𝑥3

2

|

3 0

Hence option A is the correct answer.

8

= sq units. 3

Hence option A is the correct answer.

44. Area of shaded area =

=

224 sq units. 3

45.

𝑦 = sin𝑥

Area of shaded region

4

∫2 4 𝑥 2 𝑑𝑥 =

4𝑥 3 3

4

|

2

𝜋

= 2 ∫0 sin 𝑥𝑑𝑥 = −2cos𝑥 |𝜋0

= −2(−1 − 1) = 4 sq units. Hence option C is the correct answer.

47.

𝑦 = 𝑥2 + 2 area of the shaded region.

46.

2

∫1 (𝑥 2 + 2) 𝑑𝑥 =

2 𝑥3 | 3 1

2

= ∫− 𝑦 𝑑𝑥 = + 2𝑥 |12

7 3 13 squnits 3

= + 2(1) =

Hence option C is the correct answer. 48.

Hence option C is the correct answer.

49. Mock Test - 10

𝛼 + 𝛽 = −6. . . . (𝑖) | 187

𝛼𝛽 = 𝜆. . . (𝑖𝑖) and given 3α + 2β = -20 ......(iii)

53. 𝑥, 𝑦 and respectively.

Solving (i) and (iii), we get β = 2, α = -8

⇒𝑥=

Substituting these values in (ii), we get λ = -16

⇒ 𝑦 = √𝑎𝑏 … (2)

Hence option B is the correct answer.

⇒𝑧=

50. Under condition

2

2

𝜆

3

− =3⇒𝜆=−

Hence option D is the correct answer. 51.

𝑧 are 𝐴𝑀, 𝐺𝑀 and 𝐻𝑀 between 𝑎, 𝑏

𝑎+𝑏 … … (1) 2 2𝑎𝑏 … … … (3) 𝑎+𝑏

𝑎 = 5𝑧 and 𝑥 = 𝑦 + 2 Substituting 𝑎 = 5𝑧 in (3) gives 𝑎 = 9𝑏 Substituting 𝑎 = 9𝑏 in (1) and (2) ⇒ 𝑥 = 5𝑏 and 𝑦 = 3𝑏 𝑥 = 𝑦+2⇒𝑏 =1 But, given

Also, we get

9 5

𝑧= 𝑏

𝑠𝑖𝑛𝑐𝑒, 𝐴𝑀 ≥ 𝐺𝑀 ≥ 𝐻𝑀 and 𝐺𝑀2 = (𝐴𝑀)(𝐻𝑀) ∴ 𝑥 ≥ 𝑦 ≥ 𝑧 and 𝑦 2 = 𝑥𝑧 and 𝑎 = 9, 𝑏 = 1 Hence, option 𝐴, 𝐵 and 𝐶 are correct.

Hence option A and B are the correct answer.

𝑦 = 2|𝑥| + |𝑥 + 2| − ||𝑥 + 2| − 2|𝑥|| case- 1: −𝑥 < −2 𝑦 = −2𝑥 − 𝑥 − 2 − | − 𝑥 − 2 + 2𝑥| 𝑦 = −2𝑥 − 4 … … … . (1) \(\operatorname{case}-2:-2 𝑦 = −𝑥 + 2 − |3𝑥 + 2| 52.

𝑆𝑜, for 𝑥
0 𝑦 = 3𝑥 + 2 − |𝑥 − 2| For \(0 𝑦 = 4𝑥 … … … … (4) And for 𝑥 > 2 𝑦 = 2𝑥 + 4 … … … (5) point of minimum and

𝑥=

(𝑟) = perpendicular distance on line 4𝑥 +

3𝑦 = 12 from centre ⇒ 𝑟 = 𝑟 =

|4𝑟+3𝑟−12| √16+9

(i) When centre is (1,1) and radius is 1 then equation of

Case

After drawing this graph, we can conclude

Radius

⇒ |7𝑟 − 12| = 5𝑟 ⇒ 7𝑟 − 12 = ±5𝑟 ∴ 2𝑟 = 12 ⇒ 𝑟 = 6 and 12𝑟 = 12 ⇒ 𝑟 = 1

3

𝑦 = 2𝑥 + 4 And for

54.

(𝑥 − 1)2 + (𝑦 − 1)2 = 1 ⇒ 𝑥 2 + 𝑦 2 − 2𝑥 − 2𝑦 + 1 = 0

circle is

(ii) When centre is (6,6) and radius is 2 then equation of circle is

𝑥 = −2,0 are

−2 is a point of maxima 3

⇒

𝑥2

(𝑥 − 6)2 + (𝑦 − 6)2 = 36 + 𝑦 2 − 12𝑥 − 12𝑦 + 36 = 0

Hence option B and C are the correct answer.

Hence option A and B are the correct answer.

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𝑥 2 − 10𝑥 + 21 + 𝑦 2 = 0

55.

𝑥= =

10±√100−4(21+𝑦 2 )

2 10±√16−4𝑦 2 2

Hence option A is the correct answer.

= 5 ± √4 − 𝑦 2 For real roots 4 − 𝑦 2 ≥ 0 or 𝑦 2 ≤ 4 or −2 ≤ 𝑦 ≤ 2 … (𝑖) 𝑥max = (5 + √4 − 𝑦 2 )

Now

𝑦=0

= 5+2 =7 And

𝑥min = (5 − √4 − 𝑦 2 )

Mango belongs to class dicotyledonae and order sapindales.

60. Carolus Linnaeus employed the binomial system of nomenclature in the first edition of his book 'Species Plantarum' in 1753. According to this, a plant name consists of two Latin words, the first of which is the generic name and the second is the specific epithet. Engler and Prantl are famous for their contributions in the field of plant taxonomy. Hence option B is the correct answer.

𝑦=0

= 5−2 =3 Hence 3 ≤ 𝑥 ≤ 7 Hence option A and B are the correct answer. 56. The taxonomy triangle is an upright triangle which is divided into various levels, each of which represents a particular taxon. The taxa are arranged from bottom to top, in the order Kingdom (bottom-most), Phylum, Class, Order, Family, Genus, Species (top-most). The largest taxon thus occupies the bottom-most position, and the smallest taxon, the top-most

61. Keratophilous fungi appear on nails, feathers and hair. Hence option A is the correct answer. 62. Fungi are the microorganisms that are multicellular i.e their body structure consists of multiple cells. There are heterotrophic in nature and they get their food from other organisms. They are found in soil and are the decomposers of organic matter in the soil. Hence option C is the correct answer. 63. Basidiocarp is the edible part of the mushroom Hence option A is the correct answer.

Hence option B is the correct answer. 57. Humans belong to Phylum Chordata because of the presence of an embryonic notochord. Since this notochord gets replaced with a vertebral column in the fully developed humans, they belong to subphylum Vertebrata. Further, since they have a jawline, they belong to Division Gnathostomata. Mammalia is a class under Gnathostomata. Hence option C is the correct answer. 58. The term "New Systematics” was coined by Julian Huxley in 1940. This classification takes into account the cytological, morphological, genetic, anatomical, paleontological and physiological characteristics of an organism. Hence option C is the correct answer. 59. Triticum aestivum is the scientific name of wheat. Triticum is the genus name while aestivum is the species name. Pisces is the class to which fishes belong. Man belongs to order primate and family hominidae. Mock Test - 10

64. Lichen is a symbiotic association between algae and fungi. Algal component is called phycobiont and fungal component is called mycobiont. Hence option B is the correct answer. 65. Protists are microscopic organisms. The cell structure is eukaryotic. It is surrounded by plasmalemma (cell membrane). There may be an outer covering of pellicle, cuticle. shell or cellulose wall. It contains organelles like mitochondria, golgi complex, endoplasmic reticulum, 80s ribosomes, etc. Protista have both modes of nutrition, autotrophic as well as heterotrophic. Nucleoid is the genetic material of prokaryotes. It is not present in a protista. Hence option D is the correct answer. 66. The specific action of an enzyme with a single substrate can be explained using a Lock and Key analogy first postulated in 1894 by Emil Fischer. In this analogy, the lock is the enzyme and the key is the substrate. Only the correctly sized key (substrate) fits into the key hole (active site) of the lock (enzyme).

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Hence option A is the correct answer. 67. In a double stranded DNA, the sequence of nucleotides is complementary to each other. That is, A pair with T and G pair with C. So, the sequence of nucleotide for 3 ‘ATTCGCTAT 5’ will be 5’TAAGCGATA 3’. Hence option D is the correct answer. 68. These are sugars which have Carbon molecules ranging from three to six. As the terms indicate, Tri-3, Tetra-4, Penta-5, and Hexa-6. Therefore, D is the answer 69. Proteins, at times, may be simple in structure, but in significance, they play a major role. They function to help in Catalysis, Structural integrity, Functional integrity and as Constituents. For example; simple proteins like Keratin and Collagen provide Structural integrity. Hence option B is the correct answer. 70. During telophase, the microtubules become even longer, and daughter nuclei begin to form at the two poles of the cell. Nuclear envelopes are formed, the nucleoli reappear, the chromatin of the chromosomes uncoil. Mitosis is now complete: one nucleus has divided into two genetically identical nuclei. Cytokinesis follows and involves the formation of a cleavage furrow, which pinches the cells into two. Chromosomes, centrioles and spindle fibres are formed during prophase as a preparation for cell division. Hence option D is the correct answer.

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