KVPY 12 Years Solved Papers 2020-2009 Stream SB/SX 9789325298231

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5 Practice Sets

Kishore Vaigyanik Protsahan Yoiana

12 YEARS' SOLVED PAPERS 2020-2009

STREAM SB/SX Authors

Lakshman Prasad (Mathematics) Deepak Paliwal, Mansi Garg (Physics) Neha Minglani Sachdeva (Chemistry) Sanubia Saleem (Biology)

,:carihant ARIHANT PRAKASHAN (Series), MEERUT

,:carihant ARIHANT PRAKASHAN (Series), MEERUT

All Rights Reserved

© Publishers No part of this publication may be re-produced, stored in a retrieval system or distributed in any form or by any means, electronic, mechanical, photocopying, recording, scanning, web or otherwise without the written permission of the publisher. Arihant has obtained all the information in this book from the sources believed to be reliable and true. However, Arihant or its editors or authors or illustrators don't take any responsibility for the absolute accuracy of any information published and the damages or loss suffered there upon.

All disputes subject to Meerut (UP) jurisdiction only.

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O @

a�

Kishore Vaigyanik Protsahan Yojana

ABOUTTHEEXAM

KVPY i.e. Kishore Vaigyanik Protsahan Yojana is a National Level Fellowship (scholarship) Program in Basic Science (Physics, Chemistry, Mathematics & Biology) upto Pre-Phd Level, run by Department of Science & Technology, Government of India and Conducted by I/SC (Indian Institute of Science) Bangalore, Karnataka Annually. It Was Started in 1999 to Encourage Basic Sciences Students to take up Research Career in Sciences. The Objective of the Exam is to Encourage Talented Students for Research Career in Sciences.

ELIGIBILITY CRITERIA KVPY scholarships are given only to Indian Nationals to study in India. There are three streams in KVPY; SA, SB & SX. Eligibility criteria for different streams is discussed below; •

For SA Class 11 Students who passed class 10 with minimum 75% (65% for SC/ST/PWD) marks in Mathematics & Science. The fellowship of students selected in SA will be activated only if they pursue undergraduate courses in Basic Sciences (B.Sc./B.S./B.Stat./B.Math/lntegrated M.Sc. or M.S.) and have secured a minimum of 60% (50% for SC/ST/PWD) marks in science subjects in class 12th.

•

For SX Class 12 Students aspiring to pursue undergraduate program (B.Sc. etc ) with basic sciences (Physics, Chemistry, Mathematics & Biology) who passed class 10 with minimum 75% (65% for SC/ST/PWD) marks in Mathematics & Science.

The fellowship of students selected in SX will be activated only if they pursue undergraduate courses in Basic Sciences (B.Sc./B.S./B.Stat./B.Math/lntegrated M.Sc. or M.S.) and have secured a minimum of 60% (50% for SC/ST/PWD) marks in science subjects in class 12th. •

For SB B.Sc. 1st year Students who passed class 12 with 60% marks in Maths & Sciences (PCMB) & class 10 with minimum 75% marks in Mathematics & Science.

In order to activate fellowship, in the first year of undergraduate course they should secure minimum 60% (50% for SC/ST/PWD) marks. Those students who are intending or pursuing undergraduate program under distance education scheme or correspondence course of any university are not eligible.

SYLL ABUS OF KVPY There is no prescribed syllabus for KVPY aptitude test, it aims to assess the understanding & analytical ability of the students than his/her factual knowledge. However questions are framed from syllabus upto class 10/12/lst Year of Undergraduate Courses in basic sciences, as applicable. There are two Questions Papers in KVPY; one for stream SA & Other for SB/SX (Question Paper is same for SB & SX).

QUESTION PAPERS PATTERN There are two Questions Papers in KVPY; one for stream SA & Other for SX/SB (Question Paper is same for SB & SX). •

Question Paper for SA Stream caries 80 Questions for 100 marks. There are Two Parts in

the Question Paper; Part I has 15 Questions of 1 mark each for Mathematics, Physics, Chemistry & Biology while Part II has 5 Questions of 2 marks each for Mathematics, Physics, Chemistry & Biology. •

Question Paper for SB/SX Stream caries 120 Questions for 160 marks. There areTwo Parts in the Question Paper; Part I has 20 Questions of 1 mark each for Mathematics, Physics, Chemistry & Biology while Part II has 10 Questions of 2 marks each for Mathematics, Physics, Chemistry & Biology.

MODE OF EXAM KVPY is conducted in Online Mode in English & Hindi Medium.

TIME OF EXAM •

Normally the notification or advertisement for KVPY appear in National Newspapers on May 11 (Technology Day) and Second Sunday of July every year.

•

Generally the exam is conducted in the month of November.

SELECTION PROCESS After scrutiny of application forms on the basis of eligibility criteria for various streams all eligible students are called for AptitudeTest conducted in English & Hindi Medium at different centers across the country. On the basis of performance in aptitude test shortlisted students are called for an interview, which is the final stage of selection procedure.

FELLOWSHIPS The selected students are eligible to receive KVPY fellowship after class 12th/1st Year of Undergraduate course only if they pursue Undergraduate Courses in Basic Science, upto Pre­ PhD or 5 Years whichever is earlier. Details of fellowships are listed below; Basic Science

Monthly Fellowship

Annual Contingency Grant

SA/SX/SB during 1st to lllrd year of B.Sc./B.S./BB.Stat./B.Math/ Integrated M.Sc or M.S.

Rs. 5000

Rs. 20000

SA/SX/SB during M.Sc. / IVth to Vth years of Integrated M.Sc /M.S./ M.Math/ M.Stat.

Rs. 7000

Rs. 28000

CONTINUATION/ RENEWAL OF FELLOWSHIP •

The fellow should continue to study basic science and should also maintain a minimum level of academic performance as 1st division or 60% (50% for SC/ST/PWD) marks in aggregate. Also the fellow has to pass all the subjects prescribed for that particular year.

•

In each year marks are to be certified by the Dean or Head of the Institution.

•

The fellowship will be discontinued if above marks are not obtained. However if fellow passed all the subjects & obtain marks more than 60% (50% for SC/ST/PWD) in subsequent year, the fellowship can be renewed only for that year onwards.

•

If KVPY fellow opts out of the basic science at any stage then monthly fellowship and contingency grant will be forfeited from him.

KVPYTimeline 2021 IMPORTANT DATES

Opening of Application Portal Last Date of Submission of Online Application KYPY Aptitude Test

2nd Week of July 2021 1st Week of September 2021 1st Week of November 2021

APPLICATION FEE

For General Category For SC/ST/PWD (Bank Charges Extra) For more details visit:www.kvpy.iisc.ernet.in

Rs. 1000/­ Rs. 500/-

CONTENTS KVPY SB/SX QUESTION PAPERS (2020-2009) QUESTION PAPER

2020

Pg. No. 1-22

QUESTION PAPER

QUESTION PAPER

2019

Pg. No. 1-26

QUESTION PAPER

QUESTION PAPER

2018

Pg. No. 1-26

QUESTION PAPER

2017 (19 Nov)

2017 (05 Nov)

2016

QUESTION PAPER

QUESTION PAPER

QUESTION PAPER

Pg. No. 27-50

2015

Pg. No. 100-122

QUESTION PAPER

2012

Pg. No. 169-189

Pg. No. 51-74

2014

Pg. No. 123-145

QUESTION PAPER

2011

Pg. No. 190-210

Pg. No. 75-99

2013

Pg. No. 146-168

QUESTION PAPER

2010

Pg. No. 211-233

QUESTION PAPER

2009

Pg. No. 234-254

KVPY PR ACTICE SETS (1-5)

257-362

KVPY

KISHORE VAIGYANIK PROTSAHAN YOJANA

OUESTION PAPER 2020 Stream : 58/SX

MM: 160 Instructions 1. There are 120 questions in this paper. 2. The question paper contains two parts; Part I (1 Mark Questions) and Part II (2 Marks Questions).

3. There are four sections in each part; Mathematics, Physics, Chemistry and Biology.

4. Out of the four options given with each question, only one is correct.

� PART-I

(1 Mark Questions) 4. Let R be the set of all real numbers and

MATHEMATICS

f (x) = sin10 x(cos8 x + cos4 x + cos2 x + 1)

1. Consider the following statements:

XE

2 12 + (-2f I. Jim- - n- -does not exist n➔= 2

S = {A ER} there exists a point c E (0, 2 rr) with /' (c) = A f (c)}. Then,

12

+ (-3f . 3 IL hm ---does not exist n➔= 4n

Then, (a) l is true and 11 is false (c) I and II are true

(b) l is false and 11 is true (d) Neither I nor II is true

2. Consider a regular 10-gon with its vertices on the

unit circle. With one vertex fixed, draw straight lines to the other 9 vertices. Call them L1 , L2, . . . . . . L 9 and denote their lengths by �, l2 , • • • , lg respectively. Then, the product l2 , .... , lg is

(a) 10

t

(b) 10-Js

(c)� ./3

"/2

3. The value of the integral f

w�6

��4

12

(d) 20

s. 2 x m dx is 1 + rt

w�2

-TI:

R. Let

w�2 2

(a ) S = R (b) S = {0} (c) S = [0, 211]

(d) Sis a finite set having more than one clement

5. A person standing on the top of a building of height

60.f:3 feet observed the top of a tower to lie at an elevation of 45°. That person descended to the bottom of the building and found that the top of the same tower is now at an angle of elevation of 60°. The height of the tower (in feet) is

(a) 30 (c) 90 (./3 + 1)

(b) 30(./3 + 1) (d) 150(./3 + 1)

6. Assume that 3.13 :;; rr :;; 15. The integer closest to the value ofsin- 1 (sinlcos4 + coslsin4), where 1 and 4 appearing in sin and cos are given in radians, is

(a) -1

(b) 1

(c) 3

(d) 5

IWWW.JEEBOOKS.INI

2

KVPY 7. The maximum value of the function f(x) =ex + x ln x on the interval 1 sx s2 is (a) e2+ ln2+ 1 (b)e2+2ln2 12 (c) e" + _::_1n_::_ (d) e312 + �ln � 2 2 2 2

A-1 contains only integer entries is

(c)202 (d) 101 2 9. Let A= (ai)1 ,, i, J ,, 3 be a 3 x 3 invertible matrix where each a;j is a real number. Denote the inverse of the matrix A by A-1. rn:;=laij = 1 for 1 s i s3, then (b)200

(a) sum of the diagonal entries ofA is 1 (b) sum of each row ofA- 1 is 1 (c) sum of each row and each column ofA- 1 is 1 (d) sum of the diagonal entries ofA-l is 1

10. Let x, y be real numbers such that x > 2y > 0 and 2log(x - 2y) = logx + logy.

y 11. Let � + : = 1 (b < a), be a ellipse with major axis

b

a

(d) is 8 only

AB and minor axis CD . Let F1 and F2 be its two foci, with A, F1, F2, Bin that order on the segment AB. Suppose L.F1 CB = 90 °. The eccentricity of the ellipse is (a) J3 2

1

J3

(b)�

2

-1 (c) ./5

.f5

(d) �

12. Let A denote the set of all real numbers x such that :f - [x]3 = (x- [x]3 ), where [x] is the greatest integer less than or equal to x. Then, (a) A is a discrete set of a least two points (b) A contains an interval, but is not an interval (c) A is an interval, but a proper subset of (-oo, oo) (d)A = ( oo, oo)

13. Define a sequence { sn } of real number by Sn

n 1 = � , for n � 1. 2 � k-O vn + k

Then, lim Sn n➔=

(a) Does not exist (b) Exists and lies in the interval (0, 1 ) (c) Exists and lies in the interval [1 , 2) (d) Exists and lies in the interval [2, oo)

SlnX

XE

(0, 1)

x=0

11n

a, b are integers and-50 sb s50. The number of such matrices A such that A-1, the inverse of A, exists and

Then, the possible valuc(s) of� y (a) is 1 only (b)arc 1 and 4 (c)is 4 only

-j-;-,

f(X) -

1, Consider the integral

8. Let A be a 2 x 2 matrix of the form A= [ � �], where

(a) 1 01

15 • Let'

Question Paper 2020 Stream : SB/SX

14. Let R be the set of all real number and f: R ➔ R be a continuous function. Suppose If (x)- f(y) I ;c: Ix- yl for all real number x and y. Then, (a) f is one-one, but need not be onto (b) f is onto, but need not be one-one (c) f need not be either one-one or onto (d)f is one-one and onto

In = ✓n J f (x)e~ nx dx

Then, lim/n

0

n➔=

(a)Does not exist (c)Exists and is 1

16. The value of the integral

(b) Exists and is 0 (d)Exists and is 1- e-1

3

f ((x- 2)4 sin3 (x- 2)+ (x-2)2019 + l)dx is 1

(a) 0

(b) 2

(c)4

(d) 5

17. In a regular 15-sided polygon with all its diagonals

drawn, a diagonal is chosen at random. The probability that it is neither a shortest diagonal nor a longest diagonal is 8 2 5 (c)(a) (b) 6 3

18. Let M = z3° - 21 5 + 1, and M2 be expressed in base 2.

The number of l's in this base 2 representation of M 2 lS

(a) 29

(b) 30

(c) 59

(d) 60

19. Let ABC be a triangle such that AB= 15 and AC= 9. The bisector of L.BAC meets BC in D. If LACE= 2 LABC, then ED is (a) 8 (b) 9 (c) 1 0 (d)12

20. The figure in the complex plane given by lOzz- 3(z2 + z-2) + 4i (z2 (a)a straight line (c)a parabola

-

z-2 ) = 0 is (b) a circle (d)an ellipse

PHYSICS 21. Students A, B and C measure the length of a room

using 25 m long measuring tape of least count 0.5 cm, meter-scale of least count 0.1 cm and a foot-scale of least count 0.05 cm, respectively. If the specified length of the room is 9.5 m, then which of the following students will report the lowest relative error in the measured length? (a)Student A (b) Student B (c)Student C (d)Both (b) and (c)

22. Meena applies the front brakes, while riding on her bicycle along a flat road. The force that slows her bicycle is provided by the (a) front tyre (b) road (c)rear tyre (d)brakes

23. A proton and an anti-proton come close to each other in vacuum such that the distance between them is 10 cm. Consider the potential energy to be zero at infinity. The velocity at this distance will be (a) 1.17 mis (b) 2.3 mis (c) 3.0 mis (d)23 mis

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KVPY

3

Question Paper 2020 Stream : SB/SX

24. A point particle is acted upon by a restoring force -ki'. The time-period of oscillation is T, when the

amplitude is A. The time-period for an amplitude 2A will be (c) 21' (a) 1' (b) 71/2 (d) 41'

25. The output voltage (taken across the resistance) of a L-C-R series resonant circuit falls to half its peak

31. The efficiency of the cycle shown below in the figure

(consisting of one isobar, one adiabat and one isotherm) is 50%. The ratio x, between the highest and lowest temperatures attained in this cycle obeys (the working substance is an ideal gas)

P

value at a frequency of 200 Hz and again reaches the same value at 800 Hz. The bandwidth of this circuit is (a) 200 Hz (b)200"'3 Hz (c) 400 Hz (d) 600 Hz

26. A collimated beam of

charged and uncharged particles is directed towards a hole marked P on a screen as shown below. If the electric and magnetic fields as indicated below are turned ON, (a) only particles with speed ElB will go through the hole P (b) only charged particles with speed E!B and neutral particles will go through P (c) only neutral particles will go through P (d) only positively charged particles with speed EIB and neutral particles will go through P

28. The clocktower ("ghantaghar") of Dehradun is famous for the sound of its bell, which can be heard, albeit faintly, upto the outskirts of the city 8 km away. Let the intensity of this faint sound be 30 dB. The clock is situated 80 m high. The intensity at the base of the tower is (a) 60 dB (b)70 dB (c)80 dB (d) 90 dB

29. An initially uncharged capacitor C is being charged

by a battery of emf E through a resistance R upto the instant, when the capacitor is charged to the potential E/2, then the ratio of the work done by the battery to the heat dissipated by the resistor is given by (b)3:1 (a) 2:1 (c)4:3 (d) 4:1

30. Consider a sphere of radius R with uniform charge density and total charge Q. The electrostatic 0( r J = _!]__ (a + b (r I Rf). Note that the zero of

potential distribution inside the sphere is given by 41t£0R

potential is at infinity. The values of (a, b, c) are (a) (� _'.! 2' 2' (c)

1)

G,-¾, 1)

(i!2' -�2' 2) (d)G,-¾, 2) (b)

�diabat Isotherm

V

(a) x = ex-I 2 (c)x = ex -I

(b) x2 = ex - 1

(d)x2 = ex

2

-I

32. A right-angled isosceles prism is held on the surface of a liquid composed of miscible solvents A and Bof refractive index nA = 1.5 and nB = 1.3, respectively. The refractive index of prism is nP = 1.5 and that of the liquid is given by nL =CA nA + (1- CA )nn ,

where, CA is the percentage of solvent A in the liquid.

_l_______lp--

�n=2,..----

Liquid inlet _J

27. An engine runs between a reservoir at temperature

200 K and a hot body which is initially at temperature of 600 K. If the hot body cools down to a temperature of 400 K in the process, then the maximum amount of work that the engine can do (while working in a cycle) is (the heat capacity of the hot body is 1 J/K) (a) 200(1 - ln2)J (b)200(1- ln3/2)J (c) 200(1 + ln3/2)J (d) 200 J

Isobar

Liquid Loutlet

nL ------·,-

If0c is the critical angle at pris m -liquid interface, the plot which best represents the variation of the critical angle with the percentage of solvent is

ec

ec 90

90 80

80

(b) 70

(a) 70 60

60

50

ec

0

0.5

1 CA

50

ec

0

90

90 80

80

(c) 70

( ) 70 d

50

50

60

0

0.5

60

0.5

1 c,,

0

33. Instead of angular momentum quantisation a E=

-E

o , (E0 > 0) and n is a positive integer. Which of

student gives predicts that energy is quantised as

the following options is correct? (a)The radius of the electron orbit is r oc ✓n. (b)The speed of the electron is v oc ✓n. (c)The angular speed of the electron is co oc �­ n (d) The angular momentum of the electron is L oc ✓n.

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KVPY

4 34. A monochromatic beam of light

is incident at the interface of two materials of refractive index n1 and n2 as shown. If n1 > n2 and Be is the critical angle, then which of the following statements is NOT true? (a) 01 = 03 , for all values of01 . (b) cos 02 is imaginary, for 01 > 0c . (c) COS 02 = 0, for 01 = 0c . (d)cos 03 is imaginary, for 01 = 0c -

n,

35. The intensity of light from a continuously emitting

laser source operating at 6 3 8 nm wavelength is modulated at 1 GHz. The modulation is done by momentarily cutting the intensity off with a frequency of 1 GHz. What is the farthest distance apart two detectors can be placed in the line of the laser light, so that they can see the portions of the same pulse simultaneously? (Consider the speed of light in air 3 x 10 8 m/s) (a) 30 µm (d) 30 m (b)30 cm (c) 3 m

36. A conducting rod, with a

0 0 0 0 0 resistor of resistance R, is 8 pulled with constant speed v 0 0 0 0 � on a smooth conducting rail as 0 0 @ @ @ �----------,,--- v shown in figure. A constant magnetic field Bis directed into the page. If the speed of the bar is doubled, by what factor does the rate of heat dissipation across the resistance R change? (c) 2 (a) O (b).J2 (d) 4

37. The time period of a body undergoing simple

harmonic motion is given by T = pa DbS c, where p is the pressure, D is density and S is surface tension. The values of a, b and c respectively are (a)l, l., � 2 2

(b)�,- l. , 1 2 2

(c) l,- . .!:, � 2 2

(d)- � ) , 1 2 2

38. Consider the following statements regarding the real images formed with a converging lens.

I. Real images can be seen only if the image is projected onto the screen. II. The real image can be seen only from the same side of the lens as that on which the object is positioned.

Question Paper 2020 Stream : SB/SX

potential of ball after a prolonged exposure to the ultraviolet is (a)- 0.5 V (b) O (c) 0.54 V (d)o.79 V

40. A source simultaneously emitting light at two

wavelengths 400 nm and 800 nm is used in the Young's double slit experiment. If the intensity of light at the slit for each wavelength is 10 , then the maximum intensity that can be observed at any point on the screen is (a) 10 (b) 210 (c)410 (d)810

CHEMISTRY

41 . The stability of aCMe2 I

O Me2N

follows the order (a)l > II > lll (c)II > III > I

39. A zinc ball of radius 1 cm charged to a potential

- 0.5 V. The ball is illuminated by a monochromatic ultraviolet light with a wavelength 290 nm. The photoelectric threshold for zinc is 332 nm. The

Me2

II

III (b) 11 > l > lll (d)III > II > I

42. Among the following, the biodegradable polymer is (b) polyvinyl chloride (d)teflon

(a)polylactic acid (c) bakclite

43. Among the following, the compounds which can be reduced with formaldehyde and cone. 0

�H

MeO

0

ifH II

0

�Me III

aq.

KOH, are H

M

0

H v( (r'

IV

V

(a)only II and V (b) only I and V (c) only 11 and lll (d)only I, 11 and lV 44. An organic compound that is commonly used for sanitizing surfaces is (a)acctylsalicylic acid (b)chloramphenicol (c)aspartame (d) cetyltrimethyl ammonium bromide 45. The rate of reaction of Na OH with

III. Real images produced by converging lenses are not only laterally but also longitudinally inverted as with mirrors. Which of the above statement(s) is/are incorrect? (a) Both I and III (b)Only II (c) Kone of these (d) All of these

C EB

@

II follows the order (a)II > I > III (c)I > III > II

III (b) II > III > I (d)III > II > I

46. The most suitable reagent for the conversion of

2-phenylpropanamide into 1-phenylethylamine is (a)H2, Pd/C (b) Br2, NaOH (c)LiA!H 4 , Et 20 (d)KaBH 4 , MeOH

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KVPY

47. The compound X in the following reaction scheme I

lS

5

Question Paper 2020 Stream : SB/SX

H3C

C02H

Acid [HJ hydrolysis X reduction

(a) acetonitrile (c) acetaldchydc

,,,r-,... H� NH2

(b)methyl isocyanide (d) nitromethanc

48. A nucleus X captures a �-particle and then emits a neutron and y-ray to form Y. X and Y are (b) isotopes (a) isomorphs (c) isobars (d) isotones

49. The boiling point (in °C) of 0.1 molal aqueous

solution of CuSO 4- 5H 2 O at 1 bar is closest to [Given : Ebullioscopic (molal boiling point elevation) constant of water, Kb = 0.512 K kg mol-1]

(a) 1 00.36

(b)99.64

(c) 100.1 0

(d) 99.90

50. A weak acid is titrated with a weak base. Consider the following statements regarding the pH of the solution at the equivalence point.

(i) pH depends on the concentration of acid and base (ii) pH is independent of the concentration of acid and base (iii) pH depends on the pKa of acid and pKb of base

(iv) pH is independent of the pKa of acid and pKb of base

The correct statements are (a) only (i) and (iii) (b)only (i) and (iv) (c) only (ii) and (iii) (d) only (ii) and (iv)

51. Products are favoured in a chemical reaction taking place at a constant temperature and pressure. Consider the following statements

(i) The change in Gibbs energy for the reaction is negative. (ii) The total change in Gibbs energy for the reaction and the surroundings is negative. (iii) The change in entropy for the reaction is positive. (iv) The total change in entropy for the reaction and the surroundings is positive. The statements which are always true are (a) only (i) and (iii) (b) only (i)and (iv) (c) only (ii) and (iv) (d) only (ii) and (iii)

52. A mixture of toluene and benzene forms a nearly

ideal solution. Assume Pi and Pr to be the vapour pressures of pure benzene and toluene, respectively. The slope of the line obtained by plotting the total vapour pressure to the mole fraction of benzene is (a) p'JJ - Pr (b)PT - p; (c) p'JJ + p;. (d) (p'JJ + Pi-)1 2

53. Upon dipping a copper rod, the aqueous solution of the salt that can turn blue is (b)Mg(NO 3 ) 2 (a) Ca(NO 3 ) 2 (d) AgNO 3 (c) Zn(NO3 ) 2

54. Treatment of alkaline KMnO4 solution with KI solution oxidises iodide to (a) 1 2 (b) IO� (c) l03

(d) 10 2

55. If an extra electron is added to the hypothetical molecule C 2 , this extra electron will occupy the molecular orbital. (a) 1t ;P (d)cr 2P (b) it2P (c)cr;P

56. Among the following the square planar geometry is exhibited by (a)CdCl� (c)PdCl!-

(b) Zn(CN)� (d)Cu(CN)!-

57. The correct pair of orbitals involved in it-bonding

between metal and CO in metal carbonyl complexes is (a)metal dxy and carbonyl it: (b) metal dxy and carbonyl 1t x (c)metal d 2 2 and carbonyl X: (d) metal d

X -y

X2 - y 2

and carbonyl lt x

58. The magnetic moment (in µ B) of

[Ni(dimethylglyxoimate) 2] complex is closest to (a) 5.37 (b) 0.00 (c) 1 .73 (d)2.25

59. A compound is formed by two elements M and N. Element N forms hexagonal closed pack array with 2/3 of the octahedral holes occupied by M. The formula of the compound is (c)M3 N2 (a) M4 N3 (b) M2N3

(d)M3 N4

60. If the velocity of the revolving electron of He + in the first orbit (n = 1) is v, the velocity of the electron in the second orbit is (c) 2 v (a) u (b) 0.5 u (d)0.25 u

BIOLOGY

61. Species with high fecundity, high growth rates, and small body size are typically (a)endangered species (b) keystone species (d)r-selected species (c)K-selected species

62. When RNase enzyme is denatured by adding urea, which one of the following combinations of bonds would be disrupted? (a)Ionic and disulphide bonds (b)Ionic and hydrogen bonds (c)Hydrogen and peptide bonds (d) Peptide and disulphide bonds

63. The function of Aposematic coloration is to (a)attract mates (c)scare off competitors

(b) camouflage (d)warn predators

64. Maize and rice genomes have diploid chromosome

number of 20 and 24, respectively. In the absence of crossing over and mutations, which one of the following is correct about the genetic variation among their offspring? (a)Maize < rice (b) Maize = rice > 0 (d)Maize > rice (c)Maize = rice = 0

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6 65. The exponent z of the species-area curve measured at

continental scales is (a) smaller than the value of z at regional scales (b) equal to the value of z at regional scales (c) greater than the value of z at regional scales (d) unrelated to the value of z at regional scales 66. The pH of an aqueous solution of 10-8 M HCl is (b)Between 6.9 • 7.0 (a) 6.0 (d) 8.0 (c) Between 7.0 · 7.1

67. Which one of the following cannot cause

eutrophication of lakes? (a) introduction of invasive floating plants (b) Discharge of fertiliser-rich agricultural waste (c) Natural ageing of lakes (d) Discharge of industrial waste

74. Which one of the following terms in not used, while describing human vertebra? (a) Lumbar (b) Sacral (c)Thoracic

(b)RNA Pol III (d) RNA Pol iV

69. Which one of the following statements about rennin

is correct? (a) It is secreted by adrenal glands (b) lt converts angiotensinogen to angiotensin (c) lt is secreted by peptic cells of gastric glands into the stomach (d) It is a hormone 70. When one goes from a brightly lit area to a dimly lit room our eyes adjusts slowly, thereby regaining the clarity of vision. Which one of the following explains this process? (a) Regeneration ofrhodopsin in the rod cells (b) Bleaching of rhodopsin (c) Constriction of the pupil (d) increase in the number ofrod cells

71. In a diploid population at Hardy-Weinberg equilibrium, consider a locus with two alleles. The frequencies of these two alleles are denotes by p and q, respectively. Heterozygosity in this population is maximum at (a) p = 0.25, q = 0.75 (b)p = 0. 4, q = 0.6 (c) p = 0.6, q= 0.4 (d) p = 0.5, q= 0.5

72. An enzyme with optimal activity at pH 2.0 and 37 ° C is most likely to be (a) Lysozyme from hen egg white (b) Trypsin from cattle (c) DNA polymerase from Thermus aq uaticus (d)Pepsin from humans

73. While adjusting to varying environmental temperature, plants incorporate in their plasma membrane (a) more saturated fatty acids in cold and more unsaturated fatty acids in hot environment. (b) more unsaturated fatty acids in cold and more saturated fatty acids in hot environment. (c) more saturated fatty acids in both cold and hot environment. (d)more unsaturated fatty acids in both cold and hot environment.

(d)Tarsal

75. Assume a population that has reached herd

immunity for an infectious disease. If an infected individual is introduced to this population, which of the following is most likely to occur? (a) The infection will spread exponentially across the population. (b) The infection will spread linearly across the population. (c) A few individuals may get infected, but the infection will not spread across the population. (d)No other individual will be infected by the disease.

76. Match the type of cells in Column I with the organs they are part of, listed in Column II.

68. Which one of the following polymerases transcribes 5S rRNA? (a) RNA Pol I (c) RNA Pol 11

Question Paper 2020 Stream : SB/SX

Column I

Column II

P. Chondroblast

(i) Bone

Q . Osteoclast R. Microglia

(iii) Cartilage

s.

Pneumocyte

(ii) Brain (iv) Lung

Codes (a)P-(iii), Q-(i), R-(ii), S-(iv) (b)P- (ii), Q -(i), R -(iii), S -(iv) (c)P-(iv), Q-(iii), R-(ii), S-{i) (d) P-(iii), Q-(ii), R-(iv), S-(i)

77. A bacterial culture was started with an inoculum of

10 cells. What will be the number of cells at the end of 10 cycles of division, assuming that every progeny cell undergoes division in each cycle? (a) 100 (b) 1024 (c) 2048 (d) 10240

78. The following family

tree traces the occurrence of a rare genetic disease. The filled symbols signify the individuals with the disease, whereas the open symbols signify healthy individuals. Based on this information, the disease is most likely to be (a)Autosomal, dominant (b) Autosomal, recessive (c)X-linked, recessive (d)X-linked, dominant

79. Which one of the following statement is correct about the mechanism of action of penicillin? (a)It inhibits transcription (b)It hydrolyses cell wall (c)It inhibits cell wall biosynthesis (d) it inhibits translation

80. Leaf extract from an infected plant was passed

through a filter with a pore size of 0.05 µm diameter. The infectious agent was detected in the filtrate. Which one of the following is the likely infectious agent ? (a) Bacteria (b) Virus (c)Nematode (d)Fungus

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� PART-II MATHEMATICS 81 . Let a = and b =

200

n = I0I

n

2"

200

2 201 - 2"

n = IOI

(c) b < d < a < c

l

I

n!

b

(c) 2

(b)� 2

82. Let a, b, c be non-zero real roots of the equation :l' + ax2 + bx + c = 0. Then,

(a) There arc infinitely many such triples a, b, c (b) There is exactly one such triple a, b, c (c) There arc exactly two such triples a, b, c (d)There are exactly three such triples a, b, c

83. Let f(x) = sin x + (l' - 3x2 + 4x - 2) cos x for x E (0, 1). Consider the following statements I. f has a zero in (0, 1) IL f is monotone in (0, 1) Then, (a) I and II arc true (c) I is false and II is true

(b)I is true and II is false (d) I and II are false

84. Let A be a set consisting of 10 elements. The number of non-empty relations from A to A that are reflexive but not symmetric is (a) 289 - 1 (b) 289- 245 (c) 245 - 1 (d)200- 245

85. In a ti. ABC, the angle bisector ED of LB intersects

AC in D. Suppose BC = 2, CD = 1 and 17 2

15 (b) 2

(c)

17 4

(d)

15 4

86. Let N be the set of natural numbers. For n E N, .

defme In =

n

(x) J sin "xsin dx . Then, for (x) + cos " (x) 0

2

2n

2

m, n E N (a) Im < IJor all m < n (b) Im > In for all m < n (c) Im = In for all m #- n (d)Im < In for some m < n and Im > In for some m < n

87. For 8 E [0, rc], let {(8) =sin(cos 8) and g(8) = cos(sin 8). Let a = max f(8), b = min f (8), c = max g(8) and

d = max g (8). The correct inequalities satisfied by o ,; 0 ,; 11

a, b , c, d are

(a) b < d < c < a

90. Let a = BC, b = CA, c = AB be the side lengths of a t. ABC and m be the length of the median through A If a = 8, b - c = 2, m = 6, then the nearest integer to b is (a) 7

(d)1 0

(c)9

(b) 8

PHYSICS 91. A camera filled with a polariser is placed on a

mountain, in a manner to record only the reflected image of the sun from the surface of a sea as shown in the figure. If the sun rises at 6.00 AM and sets at 6.00 PM during the summer, then at what time in the afternoon will the recorded image have the lowest intensity, assuming there are no clouds and intensity of the sun at the sea surface is constant throughout the day? (Refractive index of water = 1.33)

◊

Camera

West��L

BD = �- The perimeter of the t, ABC is

(a)

9, 8, 7, 10, 5 in that order. The integer nearest to the sum of the remaining two sides is (d)20 (a) 17 (b) 1 8 (c) 1 9 + l ✓2 1 x2 - 1 89. The value of the integral dxis ( J 4 x +l 1 +x i rr rr (a)� (b) __ (c)__ (d)� 1 2,/2 6J2 sJ2 4J2

J ��

Then, !!_ is (a) 1

(d)b < a < d < c

88. Six consecutive sides of an equiangular octagon are 6,

k = !Ot k !

:E

I

(2 Marks Questions)

(b) d < b < a < c

Sea (a) 1 2.32 PM (c)5.00 PM

(b) 3.32 PM (d)6.00 PM

92. Suppose a long rectangular loop of width w is moving along the x-direction with its left arm in a magnetic field perpendicular to the plane of the loop (see figure). The resistance of the loop is zero and it has an inductance L. At time, t = 0, its left arm passes the origin, 0. y

x x x x x x x x xx x x x x x x x x xx x x x x x x x x xx x x x x x x x x xx Q�-

t

.J.,

--+-->X W

V

If for t � 0, the current in the loop is I and the distance of its left arm from the origin is x; then I versus x graph will be

;-:-aw Nl � w-=-w � .J.--=Ea-=E:-= B-= oc-=oa,-;-K==sa----a. l=--a,

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95. A bottle has a thin nozzle on top. It is filled with

water, held horizontally at a height of 1 m and squeesed slowly by hands, so that the water jet coming out of the nozzle hits the ground at a distance of 2 m. If the area over which the hands squeese it is 10 cm 2 , the force applied by hand is close to (take, g = 10 m/s 2 and density of water = 1000 kg/m3 )

93. Imagine a world where free magnetic charges exist.

In this world, a circuit is made with a U-shape wire and a rod free to slide on it. A current carried by free magnetic charges can flow in the circuit. When the circuit is placed in a uniform electric field E perpendicular to the plane of the circuit and the rod is pulled to the right with a constant speed v, the magnetic emf in the current and the direction of the corresponding current, arising because of changing electric flux will be (l is the length of the rod and c is speed of light) (a) vE clockwise (b) vEL counter-clockwise l v (c) �l clockwise (d) v�l counter-clockwise

94. The box in the circuit below has two inputs marked V+ and V_ and a single output marked V0 • The output C

C

+lOV if V+ > V obeys Vo = { -lOV if V+ < VR

(c)

-10\1

v.,�

+5 V

v.,�

-5 V

t

( ) b

R

v,,fA

+l O V

t

-lO V

t

(d)

(d) 2.5 N

which the magnetic flux is increasing at a constant rate out of the plane of the page. . th 1 { w· · · ·- . � ire \

2@ ,/

Path . --- - · · ··· - ...__ ! :· a.

:

b

. ___ \ ____ _ _ ___ ../

0

i

:

. ·--- . .. .. . . __ _ _ _ .. -· -·· · / ·· '/ _ The clockwise emf around the circular loop is e0 • By definition a voltammeter measures the voltage difference between the two points given by

Vb - Va

=

-

f� · ds. We assume that a and b are

infinitesimally close to each other. The values of Vb - Va along the path 1 and Va - Vb along the path 2, respectively are

97. A beam of neutrons performs circular motion of

The output V0 of this circuit a long time after it is switched ON is best represented by

a ( )

(a) 20 N

(a)

t---o---+- V _ Va t---o--+- V+

+!OV

(c) 5 N

(b) 10 N

96. The circular wire in figure below encircles solenoid in

+!O V

v.,�

- lO V

t

-E0 , -E0

(b) -E0 ,0

(c) -E0 , E0

(d) E0 , E0

radius, r = I m. Under the influence of an inhomogeneous magnetic field with inhomogeneity extending over Ar = 0.0 1 m. The speed of the neutrons is 54 m/s. The mass and magnetic moment of the neutrons respectively are 1.67 x 10-27 kg and 9.67 xl0-27 J/T. The average variation of the magnetic field over ill' is approximately (a) 0.5 T (d) 10.0 T (b) 1.0 T (c) 5 04 T

98. A student is j ogging on a

straight path with the speed t5 . 4 km/h 5.4 km per hour. Perpendicular to the path is kept a pipe with its opening 8 m from the road (see figure). Diameter of the pipe is 0.45 m. At the other end of the pipe is a speaker emitting sound of 1280 Hz towards the opening of the pipes. As the student passes in front of the pipe, she hears the speaker sound for T seconds. T is in the range (Take, speed of sound = 320 m/s) (a) 6-12 (b) 12-18 (c) 3-6 (d) 18-22

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99. A solar cell is to be fabricated for efficient conversion of solar radiation to emf using material A. The solar cell is to be mechanically protected with the help of a coating using material B. If the band gap energy of materials A and B are EA and EB respectively, then which of the following choices is optimum for better performance of the solar cell? (a) EA = 1.5 eV, EB = 5 eV (b) EA = 1.5 eV, EB = 1.5 eV (d) EA = 0.5 eV, EB = 5 eV (c) EA = 3 eV, EB = 1.5 eV 100. The 'Kangri' is an earthen pot used to stay warm in Kashmir during the winter months. Assume that the 'Kangri' is spherical and of surface area 7 x 10-2 m 2• It contains 300 g of a mixture of coal, wood and leaves with calorific value of 30 kJ/g (and provides heat with 10% efficiency). The surface temperature of the 'Kangri' is 60° C and the room temperature is 0° C. Then, a reasonable estimate for the duration t (in h) that the 'Kangri' heat will last is (take the 'Kangri' to be a black body) (a) 8 (b) 10 (c) 12 (d) 16

CHEMISTRY

101. An organic compound X with molecular formula

C 1 1 H 14 gives an optically active compound on hydrogenation. Upon ozonolysis, X produces a mixture of compounds P and Q. Compound P gives a yellow precipitate when treated with 1 2 and NaOH but does not reduce Tollen's reagent. Compound Q does not give any yellow precipitate with 1 2 and Na OH but gives Fehling's test. The compound X is (a) Ph�

(b) Ph� (d) Ph�

(c) Ph�

102. The following transformation

o6

can be carried out in three steps. The reagents required for these three steps in their correct order, are (a) (i) NaBH 4 ; (ii) PC15 ; (iii) anhyd. AlC13 (b) (i) SOC1 2; (ii) anhyd. A1Cl3 ; (iii) Zn(Hg)/HCl (c) (i) Zn(Hg)/HCl; (ii) SOC1 2; (iii) anhyd. AIC13 (d) (i) cone. H 2S0 4 ; (ii) H 2N-NH 2 · Hp; (iii) KOH, ethylene glycol, Ll

103. In the following reaction, X

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Question Paper 2020 Stream : SB/SX

(i) 0 2, catalyst, heat (ii) aq. acid

�OH

yI

+ y (CsH 10 0)

X and Y respectively, arc

(a) 00-oH and � HO (b) (c)

(d)

oo OO

and

oo

and

C)=

C)=

o

o

104. A two-dimensional solid is made by alternating circles with radius a and b such that the sides of the circles touch. The packing fraction is defined as the ratio of the area under the circles to the area under the rectangle with sides of length x and y.

x=2a+2b

The ratio r = b /a for which the packing fraction is minimised is closest to (a) 0.41 (c) 0.50 (d) 0.32 (b) 1.0

105. Consider a reaction that is first order in both directions: Kr A�B

Initially only A is present, and its concentration is A0 . Assume At and Aeq are the concentrations of A at Kb

time t and at equilibrium, respectively. The time "t" at which At = (A0 + Aeq ) / 2 is

(a) t =

(c) t =

�(�)

2 S.ys + L'>S.urr > 0 and L'> Greaction < 0

52. (a) A mixture of toluene and benzene forms an ideal solution. p13 and p;. are the vapour pressure of pure benzene and toluene respectively. According to Raoult's law, Poot.al = XB PB + Xr P; and we know, XB + X r = 1

Ptotal = XB P; + (1 - XB )p; = p;, + (p; - p; ) x s :. The graph of Pootal us XB will be straight line with slope equal to p; - p;.

53. (d) Upon dipping a copper rod, the

aqueous solution of the salt that can turn blue is Ag(NO3 ). Ag+ get reduced to Ag and oxidises Cu(s) to Cu 2+ . Complete reactions arc as follows : 0 Cu(s)- Cu 2+ + 2e- ; E = - 0.34 V + Ag (aq) + e- - Ag(s) ; E 0 = 0.80V

54. (c) Treatment of alkaline KMnO 4 solution with KI solution oxidises iodide to iodate (103) . Complete reactions is as follows: f [MnO� + 2Hz0 + 3e-----, MnO2 + 4Of ] x 2 1 + 6OH- - 103 + 3H2O + 6e2MnO� + 1 + H2O - l03 + 2MnO 2 + 2Off 55. (d) If an extra electron is added to

the hypothetical molecule C 2, this extra electron will occupy the molecular orbital lS (J 2P . Electronic configuration of C2 = o ls2 , a* ls2 , a2s2 , a* 2s2 , n 2p; = n2p; Next electron enters in o2p,.

56. (c) Among the given options, PdCl� exhibit square planar geometry. Complex

CdC1:-

Zn(CN)!-

Electronic Hybridisation configuration

Cd 2+ (3d 1°)

sp 3d z

Zn 2+ (3d 1°)

sp 3d z

8 Pd + (3d )

PdCI� Cu(CN) ! -

dsp

1°

+

Cu (3d )

sp

3

2

dz

PdCl�- complex have dsp 2-hybridisation with square planar geometry.

57. (a) The correct pair of orbitals involved in n-bonding between metal and CO in metal carbonyl complexes is metal M- C CJ-bond is formed by the donation of lone pair of electrons of the carbonyl carbon into a vacant orbital of the metal. The M- C n-bond is formed by the donation of a pair of electron from a filled d-orbital of metal into vacant antibonding 1t • -orbital of carbon monoxide.

Synergic bonding

58. (b) The magnetic moment (in µ B ) of

[Ni (dimcthylglyxoimatc) z ] complex is closest to 0.00. Herc, Ni 2+ with strong field ligand (DMG) form low spin complex. Therefore, number of unpaired electron is zero. So, magnetic moment (µ B ) is also zero.

59. (b) Given that element N forms hexagonal closed pack (HCP) structure. :. Kumbcr of N-atoms per unit cell = _1_ + 2 + _l_ x 12 + 1 x 3 = 6 2 (j :. Number of octahedral voids per unit cell = 6 Number of M-atoms (or coins) per unit

all = � x 6 = 4 3 :. Formula of the compound is M4 N6 or MzNs 60. (b) According to Bohr's model, Velocity of electron in single electronic species is given as : 6 u = u X z (u = 2 . 1 8 X 10 m/s) 0

n

0

Velocity of He + in first orbit 2 u V = V X-= V = o

1

o

2

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Velocity of He + in second orbit 2 V' = u0 X 2

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KVPY Question Paper 2020 Stream : SB/SX 2

' V=V

v' = 0.5 v

6 1 . (d) r-selected species are

characterised by high fecundity, high growth rates and small body size. r-sclcctcd species arc the ones that arc found in the sparsely crowded ecological niche. These species produce a large number of small sized progeny with lower survival rates due to short gestation period and early attainment of maturity.

62. (b) Subsequent treatment of

RNasc with urea disrupts hydrogen bonds and ionic bond and allows the protein to be denatured. Covalent bonds such as peptide bonds and disulphide bonds are broken upon degradation of protein when the primary structure is destroyed. Thus, option (b) is correct.

63. (d) Aposematic coloration is a form of

coloration showed by an animal to potential predators that is not worth attacking or eating. So predators arc warned by this method. Visible signals may be accompanied by odours, sounds, painful bites, etc.

64. (a) Crossing over is the exchange of

chromosome segment between non-sister chromatids during the production of gametes. It occurs during meiosis. When there is no crossing over and mutation taking place, then we introduce variations among offspring by randomly orientation arrangement of homologous chromosomes present on the metaphase plate. This event is completely randoms and leads to independent assortment of homologous chromosomes. The number of variation is dependent on the number of chromosome making up a set. The number of possible alignment is 2n is a diploid cell where n is a number of chromosome per haploid set. In maize = 2n = ° ln rice = 2n = 212 Hance, Maize < Rice.

:21

65. (c) The species area relationship or

species area curve describe the relationship between the area of habitat and the number of species found within that area. l t was first studied by Alexandar von Humbolt. lt is dependent on immigration, extinction and clustering, etc. lf we analyse the species area relationship among very large area like continents, then slope of the line (Z) will be much steeper means greater Z value than at regional scales.

66. (b) The pH of an aqueous solution of 10-8 M HCl can be calculated as 10-8 M HCl ⇒ + H Cl(aq)� H (aq)+ Ci-(aq) 0-3 0 8 10-8 I

10 -

H 2O(l) � W (aq) + Olr (aq) 8 (x+l0- )

(x)

1 Kw = [W J [OWJ = (x + 1 0 -8 ) (x) = 1 0 - 4 on solving, x = 9.5 x 10-8

So, [H + ] total = X + 1 0 -8 = 9.5 X 1 0 -8 + 1 0 -8 = 1.05 X 10-7 pH is the negative log of H + ions 7 i.e. pH = - log [W] = - log (1.05 x 10- ) = 6.98 Our answer ranges between 6.9 - 7.0 Therefore option (b) is correct.

67. (a) Eutrophication is the natural

ageing of a lake by nutrient enrichment of its water due to sewage pollutants and detergents with time. Streams draining into the lake introduce nutrients such as nitrogen and phosphorus. introduction of invasive floating plants is not the cause of eutrophication, because these plants will grow well on eutropic lakes.

68. (b) All eukaryotes have three

different RNA polymerases (RNAPs) which transcribes different types of genes. RNA polymerase I transcribes rRNA genes; RNA polymerase II transcribes mRNA, miRNA, snRNA and snoRNA genes, and RNA polymerase 111 transcribes tRNA and 5 S rRNA genes. Thus, option (b) is correct. 69. (c) Chymosin, also known as rennin is a proteolytic enzyme secreted by peptic cells of gastric glands. its role in digestion is to coagulate milk in the stomach. This process is considerable important in infants.

70. (a) Rhodopsin is the visual purple pigment containing sensory protein that converts light into an electrical signal. In humans, it is required for vision in dim light and is located in the retina of the eye, specifically in the photo receptive rod cells. When a person enters brightly lit area bleaching of rhodopsin occurs to reduce the sensitivity of this pigment. But when a person enters a dimly lit area from a brightly lit area, there occurs regeneration of rhodopsin in the rod cells. This process is known as dark

adaptation and this enables rod cell to become sensitive to dim light again.

7 1 . (d) The Hardy-Weinberg equilibrium is a principle stating that genetic variation in a population will remain constant from one generation to the next in the absence of disturbing factors. For instance, mutation disrupt the equilibrium of alleles frequencies by introducing new alleles into a population. According to Hardy-Weinberg principle p 2 + q2 + 2pq = 1 where 2pq represent heterozygotes Options

(a)

(b) (c)

(d)

p

q

pq

0.25

0.75

0.375

0.5

0.5

0.50

0.4

0.6

0.6

0.4

0.48 0.48

Thus, heterozygosity in these populations is maximum at p = 0.5 and q = 0.5.

72. (d) Lysozyme activity increase with

increasing temperature up to 60° C, with pH range of 6.0 - 8.0 The optimum pH for the activity of trypsin present in cattle gut is -7.8 The DNA polymerase enzyme extracted from Thermus aq uaticus (Taq) is DNA sequencing. It works at pH 7-8, when at 80° C. Pepsin secreted from peptic cells of gastric gland in stomach works at pH LO - 2.0 maintained by HCI. When the pH of the medium increases to values greater than 30, pepsin is almost completely inactivated. Thus, option (d) is correct.

73. (b) A temperature increases, so does phospholipid bilayer fluidity. The relative fluidity of the membrane is particularly important in a cold environment. A cold environment tends to compress membranes composed of saturated fatty acids, making them less fluid and more susceptible to rupturing. So in cold environment, the proportion of unsaturated fatty acid should be larger as kinks in their tails push adjacent phospholipid molecules away and maintain fluidity in the membrane­ saturated fatty acid make the membrane dense and fairly rigid. 74. (d) The vertebrae in the human vertebral column arc divided into different regions into cervix, thoracic, lumbar, sacral and coccygeal. Tarsal is a bone of hind limb, so the correct option is (d).

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75. (c) 'Herd immunity' also known as 'population immunity' is a form of indirect protection from infectious disease that ca n occur with some diseases whe n a sufficient perce ntage of population has become immu ne to a n infection, whether through vaccination or previous infection, thereby reducing the likelihood of infection for individuals who lack immunity. 76. (a) • Clondroblast 1 s connective tissue cartilage.

a thin layer which protects

• Osteoclasts arc macrophages of bones.

• Microglia are a type of glial cells located throughout the brain and spin al cord. •

Pneumocytes arc prese nt alveoli.

in

8 1 . (a) Given , a = 2 ( � ) + z1°2 + C�11 1;21 ) 1 11 1 1 1 03 + + + z! ( ) 101! 102! 1031 1 1 1 + . . . + 2200 ( __ + __ + ...+ -- ) 200! 101! 102! l a = __ (z101 + 2102+ . . .+2 200 ) 101! l + -- (z102 + z!03 + ...+2200 ) 102 1 1 + ...+ __ (2200 ) 200! 101

⇒

⇒ ⇒

a=

a=

lung

77. (d) Bacterial culture was started with an inoculum of 10 cells. Number of cells at the end of 10 cycle of division. = 210 x 10 = 1024 X 10 = 10240. 78. (b) Autosomal recessive is one of several ways by which a trait or disease ca n be passed dow n through families. G ive n pedigree shows autosomal recessive disorder. The disease/trait in the pedigree is not dominant as is not see n in every generation. lt cannot be X linked recessive as mother is affected in I generation , but its one son is normal, which is not possible. 79. (c) Pe nicillin are a group of antibiotics used to treat a wide range of bacterial infection s. It is the first antibiotic discovered. It is effective again st cell wall containing monera ns as it inhibits cell wall biosynthcsis. He nce, the smallest organism, mycoplasma is inse n sitive to pe nicillin because it lacks cell wall. BO. (b) A bacterial filter is fine enough to preve nt the passage of bacteria (0.5 - 5 µm in diameter). It is given that infectious age nt was detected in the filtcratc. So, it should be smaller than bacteria, that's why it has come out in the filterate. Therfore, the correct option is (b) I .e. VIrUS. This is because a virus is 10 to 100 times smaller than the smallest bacteria. Other option s like nematode and fu ngus are larger than bacteria.

= ⇒

L,

200

� (2n + 2n+l + ...... + 2200 )

n=lO l n ·

1 2"(2 I, (

201

200

n=IO l n l

I

n=l01

220 1

�

n·

2

-n _ l)

2-1

For a reflexive relation (a, a) must be prese nt and others have a choice ofto be prese nt or not. So, number of reflexive relation s = 290 . For a symmetric relation if (a, b) is prese nt the n (b ,a) is also prese nt (where a c1' b). There are 45 such pairs of ordered pairs. So, number of reflexive relation s which are also symmetric = 2 45 :. Required number of relations = 290 - 245 • � + 4- 1 BS. (b) We have, cos� = 2 &J2

:. Length of angle bisector,

2ac

BD =

a+c

A

J

n

- 4J2 5

cos!!_ 2

a = b ⇒ !!:. = 1

82. (c) We have, abc = - c ab + be+ ca = b

And a + b + c = -a From Eq. (i), ab = - 1 (since c c1' 0)

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

So, from Eq. (iii), c = - 2a + _1_

a

and from Eq. (ii) we get,

-1+ 2 - �2 - 2a + 1 = _ _1_ 2 a 4 2 ⇒ 2a - 2a - a + 1 = 0 ⇒ (a - 1) (2a3 + 2a2 - 1) = 0 From here we get only two real and non-zero values of a, he nce there exists two triplets of (a, b, c). 2

83. (a) Given , f (x) = (x- 1)3 cosx+ (x - 1) cosx+ sinx ⇒ f ' (x) = 3(x - 1) 2 cosx + 2cosx - sin x [(x - 1)3 + (x- l)J 2

⇒ f ' (x) = cosx [3(x - 1) + 2] + (1- x) '-v----' positive

'-,--' positive in (0, 1)

s in x[(x - 1) 2 + lj

·: sin x > 0 and cosx > 0 V x E (O, 1)

positive

:. f ' (x) > 0 ⇒ f (x) in monotonically increasing. Also f (0) = -2 a nd f (1) = s in 1> 0 Hence, f(x) has exactly one root in (0, 1).

84. (d) Since, A x A contains 100 ordered pairs (a, b) out of which 10 ordered pairs are such that a = b .

C ⇒ c=3

⇒

AB BC

AD CD

We know that,

� - --

⇒

AD = � 2

:. Perimeter of !',.ABC = 1 + - + 3 + 2 = -

J " =J

86. (c) In = In

1t

0

X·

• sm 2n x

· 2n Sln X +

15

3

2

2n

COS

X

2

dx

· 2n x ( 1t - x) sm

. . . . (1)

dx • 2n 2n O Sln X + COS X O n adding Eqs. (i) and (ii), we get • . sm 2n x 21n = dx 2n 7t sin X + cos2" X 0

J

⇒

2In =

1t

2n

f

12

0 re / 2

in . :

Sln

2n x

2n Xdx

X + COS

. 2n sin X ... � � - 2�dX ..• (111) 2n " x x + cos s in 0 1t / 2 COS2n X ⇒ In = 1t • 2n 2n dx . . . (1v) 0 sm x + cos x O n adding Eqs. (iii) and (iv), we get

⇒

In = 1t I -

f

11 1 2

2

2

lt lt 2In = 1t f l · dx = ⇒ In = -

0

In is con sta nt for a ny

2

4

n E N.

87. (c) We have, f(B) = s in (cos8) ⇒

f' (8) = - cos(cos8) • sin 8

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19

KVPY Question Paper 2020 Stream : SB/SX We know that, cos(cos8) > 0 V 8 E [0, rej and sin 8 :2: 0 V 8 E [0, re] So, f (8) is decreasing function a = f(0) = sin l and b = /(re) = - sin l :. g' (8) = - sin(sin 0) · cos8 We know that, sin(sin8) :2: 0 V 8 E [0, re ]

%,

So, g(8) is decreasing in ( 0, increasing in (

%)

b

6

9 ✓2

7 ✓2

b

5

10

7

5 ✓2 Since, ABCD is a rectangle, then

⇒

2

C

b

✓2 a

x-2

2

4 + ___§_ ✓2

a + b = 7 + 8./2 = 18.3

(1

�))x2

tan i = µ = 1.33 = � 3 l i = tan - � = 53° 3

2 ../2

X

8

x2

dx = dt

t=✓ 2z ⇒ dt = -J2dz 2 dz = f ,j2 -12 z.Jz 2 - 1 ____!_

V_ = Ve = V0

V = V0 +

A

t = __l!_° X (90° + 53° ) 180 = ____!!_° x 143° = 9 h 32 min 180 Time will be 6 + 9 h 32 min = 3.32 PM

92. (b) The circuit can be drawn as, L

(_!!____) R+R

V_ > V+ , so

X = ut

⇒

V

dx

L

Hence, graph shown in option (b) is correct.

V+ = -10 = - 5 V

[_ , _}r 2

d d e - L i = 0 ⇒ vBl - L i = 0 dt dt di di vBI uBI = L or = . . . (i) L dt dt dx -= dt

V0 = - l0 V

1 2 1 2 P1 + - PV1 = P2 + - PV2

From the circuit,

As,

2

95. (b) Using Bernoulli's equation,

Here,

or

Vo

=

2 V_ = - lO V i.e., v+ > v_ and V0 = + 10 V As, the values are fixed, so correct graph is shown in option (a). and

V

From Eqs. (i) and (ii), we get dI Bl . . s ope - = - = positive l 7t

Initially, when the capacitor is fully charged, From the circuit,

G AM 6 PM Time taken by sun to go from A to P is

dt

! t�

Let

( 1-_!_)

lf E is assumed to be outward, then this value is clockwise and for E inward, it is counter clockwise. Hence, both options (c) and (d) can be possible. +l0V if V+ > V_ 94. (a) Given, V0 = { -l0V, if v+ < v_

For V0 = - l 0 V,

../2+ 1 + ---1:_2 ) dx x 89. (b) We have, J 1 ( x+ + :2 Let x+ I = t ⇒

dt

2

As,

� = 4+ 2 ✓ 2 2 ✓

a=

Also,

c

-

-

B

� + 6+ � 2 = ✓2 + 10+ ✓2 ✓2 ✓

:r

de)};; dx = El dt dt

X

91. (b) Lowest intensity will be observed, when the reflected light is completely polarised, which is at Brewster's angle,

⇒

d Magnetic emf = i B · di = µ 0E0 = e0 J' dt

d

For path 1, vb - Va = - Eo and for path 2, Vb - Va = 0 (as flux enclosed is zero)

97. (c) Given, r = 1 m, M = 0.01 m,

v = 54 ms- 1, M = 9.67 x 1 0-2 7 JT-1 and m = 1.67 x 1 0 -2 7 kg The magnetic force experienced by the neutron,

= 1 X 7 X 1 0 -2 X 5.735 X 108 [(333) 4 - (273) 4 ] = 26.75 W Total heat produced, H = 11m x calorific value = 0.1 X 300 X 30 X l(f = 9 x l(f J Heat produced (H) :. Time, t = Rate of emission ( �; )

9 x Hf ----- = 9.35 h "" 10 h 26.75 X 3600

⇒

1.67 X 10-27 X (54) 2 X 0.01 9.67 X 10-2 7 X 1

== 5.04 T

98. (a) Given, v = 5.4 kmh - 1 = 1.5 ms-1 D = 8 m, diameter of pipe. d = 0.45 m f = 1280 Hz, V8 = 320 ms- 1

First minima for diffraction = 1.22 !:__ D

320 _ = 1.22 X X S_ = 5.42 m 1280 0.45 Width of central maxima = 2 X 5.42 = 10.84 m Time for with sound will be heard 10·84 = = 7.23 s 1.5 It lies in the range of 6-12.

99. (a) For solar cell, we want the

d

electrons to cross the energy gap of material A, which is close to 1.5 eV. As, no emission of photon takes place in B since it is only used for covering. So, its band gap must be large, i.e. Es = 5 eV. 1 00. (b) Given, A = 7 x 10-2 m2 , m = 300 g Calorific value = 30 kJ/g Efficiency, 11 = 10% = 0.1 Tl = 0° C = 273 K T2 = 60° C = 333 K e = l (black body) From Newton's law of cooling,

dQ _ = eA Meiotic anaphasc II

1 1 7. (d) Protein structure is the three

dimensional arrangement of atoms in an amino acid chain molecule. The four levels of protein structure are primary, secondary and tertiary. Upon heating a solution of eukaryotic protein from 20° C to 95 ° C the first change

P( � } P(PA ) :

[NJ "['k-JrA + P[i l P(PA )

p PA

=

10 50 - X -100 100

__!__Q_ X � + � X � 100 100 100 100 500 = = 0. 10 5000 The probability that the person actually has influenza is 0.10.

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KVPV

KISHORE VAIGYANIK PROTSAHAN YOJANA

OUESTION PAPER 201 9 Stream : 58/SX

M M : 160

Instructions 1 . There are 120 questions in this paper. 2. The question paper contains two parts; Part I (1 Mark Questions) and Part II (2 Marks Questions) . 3. There are four sections in each part; Mathematics, Physics, Chemistry and Biology. Out of the four options given with each question, only one is correct.

4.

::'.>

PART-I

MATHEMATICS

1. The number of four-letter words that can be formed with letters a, b, c: such that all three letters occur is (a) 30

(b)36

(c)81

(d) 256

2

2. Let A ={e E R: (.!sine + � cos0) =.! sin 20 + � cos2 e}. 3 3 3 Then (a) A n [O, 7t] is an empty set (b) A n [O, 1tJ has exactly one point (c) A n [O, 1t] has exactly two points (d)A n [O, rr] has more than two points.

3

3. The area of the region bounded by the lines x = 1, x = 2, and the curves x(y- e" ) = sin x and 2xy =2 sin x + ;i is 1 7 (a) e2 - e - (b ) e2 - e - 6 6 (c) e2 - e + � (d) e2 - e + '!... 6 6

(1 Mark Questions) 4. Let AB be a line segment with mid-point C and D be the mid-point of AC. Let C1 be the circle with diameter AB and C2 be the circle with diameter AC. Let E be a point on C1 such that EC is perpendicular to AB. Let F be a point on C2 such that DF is perpendicular to AB and E and F lie on opposite sides of AB. Then, the value of sin L FEC is 1 (a)-

2 (b) -

1 (c)-

(d)-

✓IO

✓1 0

✓1 3

2

✓13

5. The number of integers x satisfying - 3x4 + det[� (a) 1 (c)5

;

1 xi

� i =O is equal to

x6

(b) 2 (d) 8

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2

KVPY Question Paper 2019 Stream : SB/SX 6. Let P be a non-zero polynomial such that

P(l + x) = P(l- x) for all real x and P(l) = 0. Let m be the largest integer such that (x- It' divides P(x) for all such P(x). Then, m equals (a) 1 (b) 2 (d) 4 (c) 3

7. Let _

J xsin(¼) when xat

x / ( ) - 11 when x = 0

01

and A ={x E R : / (x) =l}. Then, A has (a) exactly one element (b) exactly two clements (c) exactly three clements (d) infinitely many elements

8. Let S be a subset of the plane defined by

S = {(x, y ) : Ix i + 21 yl = l}. Then, the radius of the

smallest circle with centre at the origin and having non-empty intersection with S is 1 1 (a) (b) 5 5 J (c) I 2

(d) 2 J5

9. The number of solutions of the equation

sin (9x) + sin (3x) = 0 in the closed interval [0 ,2rr] is (a) 7 (b) 1 3 (c)1 9 (d) 25

10. Among all the parallelograms whose diagonals are 10 and 4, the one having maximum area has its perimeter lying in the interval (a) (1 9, 20] (b) (20, 21 ] (c) (21 , 22] (d) (22, 23]

11. The number of ordered pairs (a, b) of positive integers 2a- 1 2b - 1 such that - and are both integers is b a (a) 1 (b) 2 (c) 3 (d) more than 3

12. Let z = x + iy and w = u + iv be complex numbers on the unit circle such that z2 + w2 = 1. Then the number of ordered pairs (z, w) is (b)4 (a) 0 (c) 8 (d) infinite

13. Let E denote the set of letters of the English

alphabet, V = {a, e, i, o, u} and C be the complement of V in E. Then, the number of four-letter words (where repetitions of letters are allowed) having at least one letter from V and at least one letter from C is (a) 261 870 (b) 31 41 60 (c) 425880 (d) 851 760

14. Let O"l ' o-2 , o-3 be planes passing through the origin.

Assume that o-1 is perpendicular to the vector (1, 1, 1), o-2 is perpendicular to a vector (a , b, c) and o-3 is perpendicular to the vector (a2 , b 2 , c2 ). What are all the positive values of a , b and c so that o-1 n o- 2 n o-3 is a single point? (a) Any positive value of a, b, and c other than 1 (b) Any positive values of a,b and c where either a ;t b,b;tc or a i'- c (c)Any three distinct positive values of a , b and c (d)There exist no such positive real numbers a, b, and c.

15. Ravi and Rashmi are each holding 2 red cards and

2 black cards (all four red and all four black cards are identical). Ravi picks a card at random from Rashmi and then Rashmi picks a card to random from Ravi. This process is repeated a second time. Let p be the probability that both have all 4 cards of the same colour. Then, p satisfies (a)p ::; 5% (b) 5 %< p ::; 1 0% (c)1 0% < p ::; 1 5% (d)1 5% < p

16. Let A1 , A2 and As be the regions on R2 defined by A1 ={(x, y): x 2:0, y 2: 0,2x+ 2y- x2- y 2 > l > x + y}

A2 = {(x, y ) : x 2: 0, y 2:0,x+ y > 1 >f + y 2 }, As ={(x, y ) : x 2: 0, y 2:0,x+ y > l > x' + y3 }

Denote by I A1 1 , IA2 1 and IAs I the areas of the regions A1 , A2 and As respectively, Then, (a)IA1 l>IA2l>IA;i I (b) IA1 l>IA;i l>IA2I (c)IA1 l = IA21IA21

17. Let f : R➔ R be a continuous function such that

/(x2 ) = /(x' ) for all x E R. Consider the following statements. I. f is an odd function. II. / is an even function. III. / is differentiable everywhere. Then, (a)l is true and 111 is false (b) 11 is true and 111 is false (c)Both l and Ill are true (d)Both 11 and 111 are true

18. Suppose a continuous function /: (O, =) ➔ R satisfies /(x) =2 J; tf(t)dt + 1, V x 2: 0. Then, /(1) equals (a) e (b) e2

(c) e4

(d)e6

19. Let a > 0, a ;t 1. Then, the set S of all positive real numbers b satisfying (1 + a 2 ) (1 + b 2 ) = 4ab is

(a)an empty set (b)a singleton set (c)a finite set containing more than one clement (d) (0,oo)

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3

KVPY Question Paper 2019 Stream : SB/SX 20. Let f : R� R be a function defined by f(x) =

!� if x ,t O) X

O if x = 0

Then, at x = 0, f is (a) not continuous (b) continuous but not differentiable (c) differentiable and the derivative is not continuous (d) differentiable and the derivative is continuous

PHYSICS 21. In a muonic atom, a muon of mass of 200 times of

that of electron and same charge is bound to the proton. The wavelengths of its Balmer series are in the range of (b)infrared rays (a) X-rays (c)y-rays (d) microwave

22. We consider the Thomson model of the hydrogen

atom in which the proton charge is distributed uniformly over a spherical volume of radius 0.25 A. Applying the Bohr condition in this model, the ground state energy (in eV) of the electron will be close to (b)- 1 3.6 (a)- 1 3.6 / 4 (c)- 1 3.6/2 (d)- 2 x 1 3.6

23. A spherical rigid ball is released from rest and starts

rolling down an inclined plane from height h = 7 m, as shown in the figure. It hits a block at rest on the horizontal plane (assume elastic collision). If the mass of both the ball and the block is m and the ball is rolling without sliding, then the speed of the block after collision is close to

(a) 6 m/s (c) 1 0 mis

e

h

Taking him to be a rigid body, the instantaneous angular velocity (in rad/s) is (a) 1 . 5 (b) 2.0 (c)2.5 (d) 3.0

26. A point mass M moving with a certain velocity collides with a stationary point mass M I 2. The

collision is elastic and in one-dimension. Let the ratio of the final velocities of M and M/2 be x. The value of

XlS

(a) 2 (c) 1 /2

(b) 3 (d) 1 /4

27. A particle of mass 2 / 3 kg with velocity v = - 15 m/s at t = - 2 s is acted upon by a force F =k -Pt 2 • Here, k = 8N and p = 2N/ s2 • The motion is one-dimensional.

Then, the speed at which the particle acceleration is zero again, is (a)1 m/s (b) 1 6 m/s (d)32 m/s (c) 17 m/s

28. A certain stellar body has radius 50 Rs and

temperature 2Ts and is at a distance of 2 x 1 010 AU from the earth. Here, AU refers to the earth-sun distance and Rs and Ts refer to the sun's radius and temperature, respectively. Take, both star and sun to be ideal black bodies. The ratio of the power received on earth from the stellar body as compared to that received from the sun is close to (a)4 x 1 0-20 (b) 2 x 1 0-6 8 (c)1 0(d)1 0-16

29. As shown in the schematic below, a rod of uniform

cross-sectional area A and length l is carrying a constant current i through it and voltage across the rod is measured using an ideal voltmeter. The rod is stretched by the application of a force F. Which of the following graphs would show the variation in the voltage across the rod as function of the strain E when the strain is small. Neglect Joule heating.

(b)8 m/s (d) 12 m/s

24. A girl drops an apple from the window of a train

which is moving on a straight track with speed increasing with a constant rate. The trajectory of the falling apple as seen by the girl is (a) parabolic and in the direction of the moving train (b) parabolic and opposite to the direction of the moving train (c) an inclined straight line pointing in the direction of the moving train (d) an inclined straight line pointing opposite to the direction of the moving train

25. A train is moving slowly at 2m/s next to a railway

platform. A man, 1.5 m tall, alights from the train such that his feet are fixed on the ground.

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4

KVPY Question Paper 2019 Stream : SB/SX

30. Two identical coherent sound sources R and S with

frequency f are 5 m apart. An observer standing equidistant from the source and at a perpendicular distance of 12 m from the line RS hears maximum sound intensity. When he moves parallel to RS, the sound intensity varies and is a minimum when he comes directly in front of one of the two sources. Then, a possible value of f is close to (the speed of sound is 3 3 0 m/s) (b) 275 Hz (a) 495 Hz (c) 660 Hz (d) 330 Hz

Then, the graph of the total charge on the particle versus the applied voltage would look like

31. A photon falls through a height of 1 km through the

earth's gravitational field. To calculate the change in its frequency, take its mass to be hv I c2 • The fractional change in frequency v is close to (a) 1 0 2 -0 (b)1 0 - 17 � l�w W l� �

32. 0.02 moles of an ideal diatomic gas with initial

temperature 20 °C is compressed from 1500 cm3 to 500 cm3 • The thermodynamic process is such that p V 2 = �' where � is a constant. Then, the value of � is close to (the gas constant, R =8.31 J/K/mol). (b)1.5 x 1 02Pa - m6 (a) 7.5 x 1 0 2 - Pa - m6 6 2 (d) 2.0 x la1Pa - m6 (c) 3 x 1 0- Pa - m

33. A heater supplying constant power P watts is

switched ON at time t = 0 min to raise the temperature of a liquid kept in a calorimeter of negligible heat capacity. A student records the temperature of the liquid T(t) at equal time intervals. A graph is plotted with T(t) on the Y-axis versus t on the X-axis. Assume that there is no heat loss to the surroundings during heating. Then, (a) the graph is a straight line parallel to the time axis (b) the heat capacity of the liquid is inversely proportional to the slope of the graph (c) if some heat were lost at a constant rate to the surroundings during heating, the graph would be a straight line but with a larger slope (d) the internal energy of the liquid increases quadratically with time

34. Unpolarised red light is incident on the surface of a

lake at incident angle eR . An observer seeing the light reflected from the water surface through a polariser notices that on rotating the polariser, the intensity of light drops to zero at a certain orientation. The red light is replaced by unpolarised blue light. The observer sees the same effect with reflected blue light at incident angle 0n-

Then, (a) 9B < 9R < 45° (c) eB > aR > 45°

(b)9B = 9R (d) 9R > 9B > 45° 35. A neutral spherical copper particle has a radius of 10 nm (1 nm = 10-9 m). It gets charged by applying the voltage slowly adding one electron at a time.

36. A charge +q is distributed over a thin ring of radius r with line charge density A. = q sin2 0 /(n:r). Note that the ring is in the XY- plane and 0 is the angle made by r with the X-axis. The work done by the electric force in displacing a point charge + Q from the centre of the ring to infinity is (a) equal to qQ I 2 JtE0 r (b)equal to qQ I 4 JtE0 r (c)equal to zero only, if the path is a straight line perpendicular to the plane of the ring (d) equal to qQ I 8 nE0 r

37. Originally the radioactive beta decay was thought as

a decay of a nucleus with the emission of electrons only (Case I) . However, in addition to the electron, another (nearly) massless and electrically neutral particle is also emitted (Case II). Based on the figure below, which of the following is correct?

Energy of electrons (a) (a) in both cases I and II (b) (a) in case I and (b) in case II (c)(a) in case 11 and (b) in case I (d) (b) in both cases I and 11

38. One gram mole of an ideal gas A with the ratio of

constant pressure and constant volume specific heats YA = 5 I 3 is mixed with n gram moles of another ideal gas Bwith Yn = 7 / 5. If the y for the mixture is 19/13, then what will be the value of n? (d) 3 (c) 1 (a) 0.75 (b) 2

39. How will the voltage (V) between the two plates of a parallel plate capacitor depend on the distance (d) between the plates, if the charge on the capacitor remains the same?

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5

KVPY Question Paper 2019 Stream : SB/SX 0 (a)

H

H�

0

� 0

(b)

40. Three large identical plates are kept parallel to each other. The outer two plates are maintained at temperatures T and 2T, respectively. The temperature of the middle plate in steady state will be close to (b) l . 3 T (a) l . l T (d) l.9 T (c) l . 7 T

CHEMISTRY 41 . The major products of the following reaction. 0

(aq) NaO H

Ph A CBr3

arc

C H Br3

MeO Ill

44. Permanent hardness of water can be removed by (a)heating (b)treating with sodium acetate (CH3 COONa) (c)treating with calcium hydrogen carbonate Ca(HC03 ) 2 (d) treatment with sodium hexametaphosphate (Na 6 P6 018 )

(b) S and S (d)S and R (b) 02 (d)

ot

48. Among the following transformations, the r

ch

Br

IV

43. The major product of the following reaction.

IS

0

(a)0� (c)0 2

the compounds which can undergo an SN 1 reaction in an aqueous solution, are (a) l and lV only (b)11 and lV only (c) 11 and lll only (d) 11, Ill and IV only

EtO � CN

H

47. The diamagnetic species among the following is

0'8

0

H O�

respectively, are (a)R and R (c)R and S

CBr3C02 Na

42. Among the following,

(d)

compounds CH 2SH H YCH 0H 2 H 3C

(c) Ph A CHBr an d NaBr 2 and

H (c) EtO� 0 0

46. The absolute configurations of the following

0

(d) PhH

0

(a)MNH2 (b)MH (c)[ M(NH3 )x t + [ e (Nlfs )y ]­ (d) M3 N

AH (a) Br3C-OH and Ph

0

CN

45. Alkali metals (M) dissolve in liquid NI-Is to give

0

(b) Ph A ONa an d

H

Excess DIBAL-H Toluene, -78,C

hybridisation of the central atom remains unchanged

Ill

(a)CO 2 - HCOOH (c)Nlfs - N�

(b) BF; - BF; (d)PC13 - PC15

49. For an octahedral complex MX4Y2 (M =a transition

metal, X and Y are monodentate achiral ligands), the correct statement among the following is (a)MX4Y2 has 2 geometrical isomers, one of which is chiral (b)MX4 Y2 has 2 geometrical isomers both of which arc achiral (c)MX4Y2 has 4 geometrical isomers, all of which are achiral (d) MX4Y2 has 4 geometrical isomers, two of which are chiral

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KVPY Question Paper 2019 Stream : SB/SX

50. The values of the Henry's law constant of

Ar,CO2 , CH4 and O 2 in water at 25 °C are 40.30, 1.67, 0.4 1 and 34.86 kbar, respectively. The order of their solubility in water at the same temperature and pressure is (a) Ar> 0 2 > CO 2 > CH4 (b) CH4 > CO2 > Ar> 0 2 (c) CH4 > CO2 > 0 2 > Ar (d)Ar> CH4 > 0 2 > CO 2

57. The curve that best describes the adsorption of a gas (x)g on 1.0 g of a solid substrate as a function of pressure (p) at a fixed temperature 3 2

X

51. Thermal decomposition of N2 O5 occurs as per the equation below:

2N2 O5 � 4NO2 + 02 The correct statement is (a) 0 2 production rate is four times the NO 2 production rate (b) 02 production rate is the same as the rate of disappearance of N2O5 (c) rate of disappearance of N2O5 is one-fourth of NO2 production rate (d)rate of disappearance of N2O5 is twice the 0 2 production rate

52. For a first order chemical reaction, (a) the product formation rate is independent of reactant concentration. (b) the time taken for the completion of half of the reaction (t112) is 69.3% of the rate constant (k) (c) the dimension of Arrhenius pre-exponential factor is reciprocal of time. (d) the concentration vs time plot for the reactant should be linear with a negative slope

IS

(a) 1

p

(b) 2

(c) 3

(d)4

58. The octahedral complex [Co(NH3 )s804] Cl exists in two isomeric forms X and Y. Isomer X reacts with

AgNO 3 to give a white precipitate, but does not react with BaC12 • Isomer Y gives white precipitate with BaC1 2 but does not react with AgNO3 . Isomers X and Y are (a)ionisation isomers (b) linkage isomers (c)coordination isomers (d)solvate isomers

59. The correct order of basicity of the following amines

U ----:::: ,&

NH2

53. The boiling point of 0.001 M aqueous solutions of

NaCl, Na 28O4 , �PO4 and CH3 COOH should follows the order. (a) CH3 COOH < NaCl < Na 2SO 4 < K:J PO 4 (b) NaCl < Na 2SO 4 < K:J PO 4 < Cll:i COOH (c) Cl-ls COOH < K:i PO 4 < Na 2SO4 < NaCl (d) CH:i COOH < K3 PO 4 < NaCl< Na2SO 4

54. An allotrope of carbon which exhibits only two types of C - C bond distance of 143.5 pm and 138.3 pm, is (a) charcoal (b)graphite (c) diamond (d) fullerene

55. Nylon-2-nylon-6 is a co-polymer of 6-aminohexanoic acid and (a) glycine (c) alanine

(b)valine (d) leucine

56. A solid is hard and brittle. It is an insulator in solid state, but conducts electricity in molten state. The solid is a (a) molecular solid (b) ionic solid (c) metallic solid (d)covalent solid

IS

(a)I > II> III> IV (c)Ill > II> I> IV

(b) I > III> II> IV (d)IV > Ill> II> I

60. Electrolysis of concentrated aqueous solution of NaCl results in (a)increase in pH of the solution (b)decrease in pH of the solution (c)02 liberation at the cathode (d) H2 liberation at the anode

BIOLOGY

61. Ethanol is used to treat methanol toxicity because ethanol (a)is a competitive inhibitor of alcohol dehydrogenase (b)is a non-competitive inhibitor of alcohol dehydrogenase (c)activates enzymes involved in methanol metabolism (d) inhibits methanol uptake by cells

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KVPY Question Paper 2019 Stream : SB/SX 62. Given below is a diagram of the stomata! apparatus.

Match the labels with the corresponding names of the compounds.

68. According to the logistic population growth model,

the growth rate is independent of (a)per capita birth rate (b) per capita death rate (c)resource availability (d) environmental fluctuations

69. A violent volcanic eruption wiped out most of the life

forms in an island. Over different forms of simple organisms colonised this region, followed by the emergence of other organisms such as shrubs, woody plants, invertebrates and mammals. This ecological process is referred to as (a)generation (b) replacement (c)succession (d) turnover

70. Which one of the following microbial product is called

Choose the correct combination. (a) 1. Stomata! 2. Guard pore; cell; (b) 1. Guard cell; 2. Stomata! pore; (c) 1. Subsidiary 2. Guard cell; cell; (d) l.Guard cell; 2. Stomata! pore;

3. Epidermal cell; 3. Subsidiary cell; 3. Stomata! pore; 3. Epidermal cell;

4. Subsidiary cell 4. Epidermal cell 4. Epidermal cell 4. Subsidiary cell

63. Which one of the following pairs was excluded from Whittaker's five kingdom classification? (a) Viruses and lichens (b) Algae and Euglena (c) Lichens and algae (d)Euglena and viruses

64. When a plant species is grown in shade tends to

produce thinner leaves with more surface area, and when grown under abundant sunlight starts producing thicker leaves with reduced surface area. This phenomenon is an example of (a) character displacement (b) phenotypic plasticity (c) natural selection (d) genotypic variation

65. Sacred groves found in several regions in India are an example of (a) in situ conservation (c) reintroduction

(b)exsitu conservation (d) restoration

66. Which one of the following immune processes is most effectively controlled by anti-histamines? (a) Cell-mediated auto-immunity (b) IgE-mediated exaggerated immune response (c) IgG-mediatcd humoral immune response (d)IgM-mediated humoral immune response

67. Which one of the following is explained by the endosymbiotic theory? (a) The interaction between bacteria and viruses (b) The symbiosis between plants and animals (c) The origin of mitochondria and chloroplast (d)The evolution of multicellular organisms from unicellular ones

"clot buster'? (a)Cyclosporin-A (c)Statins

(b) Paracetamol (d)Streptokinase

71. Which one of the following elements is not directly involved in transcription? (a)Promoter (b) Terminator (d) Ori C (c)Enhancer

72. Which one of the following phyla is a pseudocoelomate? (a)Cnidaria (c)Mollusca

(b) Nematoda (d) Chordates

73. Which one of the following glands does not secrete saliva? (a)Submaxillary gland (c)Parotid gland

(b) Lacrimal gland (d)Sublingual gland

74. Which one of the following options correctly represents the tissue arrangement in roots? (a)Cortex, pericycle, Casparian strip, vascular bundle (b)Pericycle, cortex, Casparian strip, vascular bundle (c)Cortex, Casparian strip, pericycle, vascular bundle (d) Casparian strip, pericycle, cortex, vascular bundle

75. During fermentation of glucose to ethanol, glucose is (a)first reduced and then oxidised (b)only oxidised (c)neither oxidised nor reduced (d) only reduced

76. Which of the following is/are the product(s) of cyclic photophosphorylation? (a)Both NADPH and W (c)ATP

(b) NADPH (d) Both ATP and NADPH

77. Which one of the following amino acids is least likely to be in the core of a protein? (a)Phenylalanine (b) Valine (d)Arginine (c)Isoleucine

78. Which one of the following statements is a general feature of global species diversity? (a) lt increases from high to low altitudes (b)It increases from low to high altitudes (c)It changes over time but not spatially (d) It changes randomly across space and time

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KVPY Question Paper 2019 Stream : SB/SX

79. Which one of the following conditions is not responsible for the presence of deoxygenated blood in the arteries of a newborn? (a) Pneumonia (b) Atrial scptal defect (c) Shunt between pulmonary artery and aorta (d) Phcnylkctonuria

80. Rhizobium forms symbiotic association with roots in legumes and fixes atmospheric nitrogen. Which one

� PART-II

(2 Marks Questions)

MATHEMATICS

81 . The points C and D on a semicircle with AB as diameter are such that AC = l , CD = 2 and DB = 3. Then, the length of AB lies in the interval. (a) [4, 4. 1) (b) [4. 1, 4.2) (c) (4. 2, 4. 3) (d) [4.3, oo)

82. Let ABC be a triangle and let D be the mid-point of BC. Suppose cot (L CAD) : cot (L BAD) = 2 : 1. If G is the centroid of MBC, then the measure of LBGA is (a) 90° (b) 105 ° ° (c) 120 (d) 135 °

83. Let f (x) = x6 - 2.:f + i3 + x2 - x - l and g(x) = x4 - l' - x2 - 1 be two polynomials. Let a, b, c and d be the roots of g(x) = 0. Then, the value of /(a) + /(b) + /(c) + f(d) is (a) -5 (b) O (c) 4 (d) 5

84. Let a = i + j + k, b = 2i + 2j + k and c = 5i + j - k be three vectors. The area of the region formed by the set of points whose position vectors r satisfy the equations r · a = 5 and I r - b I + I r - c 1 = 4 is closest to the integer. (a) 4 (b) 9 (c) 14 (d) 19 A

A

,._

,._

85. The number of solutions to

sin (rr sin20) + sin(n cos20) = 2cos( cos 0 ) %

satisfying O ::; e ::; 2rr is (a) 1 (c) 4

86. Let

J=

x. I0 � 1+ 1

X

(b) 2 (d) 7

S

Consider the following assertions: 1 7[

4

l. J >-

II. J r). Current I follows in the short solenoid. Choose the correct statement. (a) There is uniform magnetic field µ 0 nl in the long solenoid. (b) Mutual inductance of the solenoids is nµ 0 r2nNl. (c) Flux through outer solenoid due to current I in the inner solenoid is proportional to the ratio R I r. (d) Mutual inductance of the solenoids is nµ 0 rRnN!L I (rR)u2.

93. Consider the wall of a dam to be straight with height H and length L. It holds a lake of water of height

h(h < H) on one side. Let the density of water be P w · Denote the torque about the axis along the bottom length of the wall by T1 . Denote also a similar torque due to the water up to height h 1 2 and wall length L/2 by T 2 . Then, T1 /T 2 (ignore atmospheric pressure) is (a) 2 (b)4 (d) 1 6 (c) 8

94. Two containers C1 and C2 of volumes V and 4V

respectively, hold the same ideal gas and are connected by a thin horizontal tube of negligible volume with a valve which is initially closed. The initial pressures of the gas in C1 and C2 are p and 5p, respectively. Heat baths are employed to maintain the temperatures in the containers at 300 K and 400 K, respectively.

The valve is now opened. Select the correct statement. (a) The gas will flow from the hot container to the cold one and the process is reversible. (b) The gas will flow from one container to the other till the number of moles in two containers arc equal. (c) A long time after the valve is opened, the pressure in both the containers will be 3p. (d)A long time after the valve is opened, number of moles of gas in the hot container will be thrice that of the cold one.

95. Four electrons, each of mass m, are in a one dimensional box of size L. Assume that, the electrons are non-interacting, obey the Pauli exclusion principle and are described by standing de Broglie waves confined within the box. Define ex = h 2 l 8m,L2 and UO to be the ground state energy. Then, (a)the energy of the highest occupied state is 1 6 a (b) U0 = 30 a

(c)the total energy of the first excited state is UO + 9 a (d) the total energy of the second excited state is U0 + S a

96. A rope of length L and uniform linear density is

hanging from the ceiling. A transverse wave pulse, generated close to the free end of the rope, travels upwards through the rope. Select the correct option. (a)The speed of the pulse decreases as it moves up. (b) The time taken by the pulse to travel the length of the rope is proportional to JL. (c)The tension will be constant along the length of the rope. (d) The speed of the pulse will be constant along the length of the rope.

97. A circuit consists of a coil with inductance L and an uncharged capacitor of capacitance C. The coil is in a constant uniform magnetic field such that the flux through the coil is . At time t = 0 min, the magnetic field is abruptly switched OFF. Let m 0 = 1 / .Jrc and ignore the resistance of the circuit. Then, (a)current in the circuit is J(t) = ( / L)cosro0 t (b)magnitude of the charge on the capacitor is IQ(t) I = 2Cm0 I sinm0 tl (c)initial current in the circuit is infinite (d) initial charge on the capacitor is Cm0

98. Consider the configuration of a stationary water tank of cross-section area Ao and a small bucket as shown in figure below;

Ao

H X

What should be the speed v of the bucket, so that the water leaking out of a hole of cross-section area A (as shown) from the water tank does not fall outside the bucket? (Take, h = 5m, H = 5m, g = 10m / s2 , A = 5cm2 and A0 = 500cm2 ). (a) 1 m/s

(b) 0.5 m/s

(c) 0.1 m/s

(d) 0.05 m/s

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KVPY Question Paper 2019 Stream : SB/SX

99. The circuit below is used t o heat water kept i n a bucket.

11:�I

102. The major products X, Y and Z in the following

Water b ucket

U

Assuming heat loss only by Newton's law of cooling, the variation in the temperature of the water in the bucket as a function of time is depicted by

(a)

A O A X =::���,� y

.,N.,,

l

are

0 (d)

T

T

100. A bubble of radius R in water of density p is 0

expanding uniformly at speed v. Given that water is incompressible, the kinetic energy of water being pushed is (a) zero (b) 2n pR3 v2 2 (d) 4n pR3v2 / 3 (c) 2rr pR3 v / 3

CHEMISTRY 101. The product of which of the following reactions forms a reddish brown precipitate when subjected to Fehling's test?

(b)

NH ,

(b)

0

(a)

0

T

T

0 (c)

0

sequence of transformations

R

0

CO. HCI

anh. AICl3,CuCI

Cl

� + (CH3CH 2) 2Cd �

0

COOH

1. PC15

----➔

(c ,,,:;:,

2. H2,Pd-BaS04

103. In the following reaction, P gives two products Q and R each in 40% yield.

p (Mwt.=21 0 )

OMe

__ 1. --' o,_ Q 2.Zn/H20

+

40%

R 40%

If the reaction is carried out with 420 mg of P, the reaction yields 108.8mg of Q. The amount of R produced in the reaction is closest to (a)97.6 mg (b) 1 08.8 mg (c)84.8 mg (d) 121.6 mg

104. Solubility products of Cul and Ag2CrO4 have almost

the same value (-4 x 10- 12 ). The ratio of solubilities of the two salts (Cul: Ag 2CrO4 ) is closest to (a) 0.01 (b) 0.02 (c) 0.03 (d) 0.1 0

105. Given that, the molar combustion enthalpy of

benzene, cyclohexane and hydrogen are x, y, and z, respectively, the molar enthalpy of hydrogenation of benzene to cyclohexane is (a) x - y + z (b) x - y + 3z (c)y- x + z (d)y- x + 3z

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KVPY Question Paper 2019 Stream : SB/SX 106. Among the following, the pair of paramagnetic complexes is (a) K;i [Fc(CN)6 ] (b) K;i [Fe(CN)6 ] (c) K4 [ Fe(CN)6 ] (d) K4 [Fc(CN)6 ]

111. In a population NAA and Na.a are the numbers of

and K3 [CoF6 ] and [Co(Ntla )6 ] Cl3 and K;i [Col x + y3 } then IA:i l>IA21 = IA1 I y

------o--�---x x +y = 1

1 7. (d) Given function f :R- R be a continuous function such that f (x2) = f (:I' )Vx E R then f (x) = f (xz3 ) [on replacing x byx1'3 ] Similarly, f(x) = f(xz3 ) = f (x 419 ) = f (x ei21 ) = . . . = f(:P3 ) " ) 0 = f(x ) [as x tends to infinity] = /(1) f (x) = f (1) = constant The function f (x) = constant is even and differentiable everywhere.

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KVPY Question Paper 2019 Stream : SB/SX

21 . (a) A muon is an unstable

1 8. (a) It is given that a continuous function / : [0, oo )� R satisfies

elementary particle of mass nearly 200 m, and charge ± e. Herc, a negative muon is given bound to a proton. ·-. m

f(x) = 2J tf(t)dt + 1, Vx � 0 0

f ' (x) = '2xf (x) f' (x) = '2x f(x)

BM

On integrating both sides, we get lnl /(x) I = x2 + c

,•

2

⇒ f(x) = Kex , where K = ec 0

So, m = 200 m, and M = 1836 m, (as mass of a proton is 1836 times of mass of electron) So, reduced mass of system, mM = 200 m, x 183 6m, = 180 me m' = m + M 200 m, + 1836 me

/(0) = 2J t f(t)dt + l = 1 K=l f (1) = e

0

1 9. (a) Given relation (1 + a2 ) (1 + b 2) = 4ab ⇒ a 2 + b2 - 2ab = 2ab - l- a2b2 ⇒ (a - b) 2 = -(1- ab) 2 ·: a > 0, a l and b is a positive real number :. (a - b) 2 0 -(1- ab) 2, because (a - b) 2 and (1- ab) 2 are non-negative real numbers. :. Set S is an empty set.

As mass of muon is comparable to mass of proton, we have to take account of motion of nucleus. That's way we are calculating reduced mass. Now, as energy of an electron in nth orbit is of H-atom,

* * *

E

n

=

si 2 ) [ nt

, X '#

s

.

x2

x ---. 0

sin h 2

or ⇒ ⇒ =

1

= 1 lim h ➔O h 2 So, f (x) is differentiable at x = 0 . l sin x2 2cos(x2) , 1f x OJ f ' (x) = [ � 1 , if X = 0

·: I1m f '

Steric number = 2 hybridisation = sp

F

"

/

F

I B

F

N

H H H /I"

F

le

-

Steric number = 3 hybridisation = sp2

0 H- C -:1/' "OH Steric number = 3 hybridisation = sp2

F F F / i"

Steric number = 4 hybridisation = sp3

H

-

Steric number = 4 hybridisation = sp 3

/ I"

H H H

Steric number = 4 hybridisation = sp3

Cl I )P- Cl / I "Cl Cl I Cl Cl Cl Steric number = 4 Steric number = 5 hybridisation = sp 3 hybridisation = sp3d Thus, the hybridisation of the central atom remains unchanged in NI--1:i -, NH: .

-

p

Cl

49. (b) Two geometrical isomers (cis and trans with respect to ligand Y) arc possible. The trans isomer has both centre of

y

y

symmetry and several planes of symmetry.

�

�

(trans)

(cis)

x�x �x y

y

and, therefore it is achiral. The cis isomer has two planes of symmetry and, therefore it is also achiral.

50. (c) By Henry's law, Pgas oc Xgas (in liquid solution) and Pgas = KH X Xgas KH = Henry's law constant.

:.At a given partial pressure of gas, 1 KH oc -Xgas

:. The greater the value of KH , the less its

solubility. Here, KH of Ar > 02 > CO2 > CH4 . :. Solubility of CH4 > CO2 > 0 2 > Ar.

5 1 . (d) Rate of disappearance of N205 and rate of formation of 02 must be in

2 : 1 ratio as per the stoichiometry of the reaction, 2N205 --➔ 4N0 2 + 02

52. (c) Arrhenius equation for rate constant, k is k = Ae-Ea !RT , where A is the

pre-exponential factor. Since, e- E. IRT is dimensionless, dimension of A = k. For first order reaction, dimension of k = time -I

:. Also same for A.

As for the other options, For a first order reaction, the product formation rate is directly proportional to 0.693 . ha If- 11· fe = -reactant concentrat10n,

k

or 69.3% of ( ¾} and concentration of reactant changes with time by the equation [AJ 1 = [A]0 e- kt , which shows exponential decreases, not linear. 53. (a) Elevation in boiling point, ti.Tb = i x Kb x m. (Kb is constant for a

solvent) If molality (m) is same, ti.Ti, oc van't Hoff

factor (i) CI--1:i COOH only partially dissociates,

i III > II > IV.

60. (a) The cathode half-reaction in the process 1s 2Hz0 + 2e---➔ H2 + 20Hwhich increases [OH- ] and, therefore pH and the anode half-reaction. 2c1- --➔ Cl2 + 2e- does not produce or + consume H or OH- . Overall, the reaction can be written as 2NaCl + 2H20 --➔ 2Na0H + H2 + Cl2

6 1 . (a) Ethanol is used to treat methanol

toxicity because it is a competitive inhibitor of alcohol dehydrogenase competitive inhibitor competes with substrates for the active site of enzyme. Ethanol has high binding affinity then method for alcohol dchydrogcnasc. This enzyme converts methane to formaldehyde into formate and formate accumulation causes blindness and affect central nervous system. Ethanol ingestion slows down the metabolism of methanol, so that kidneys, got sufficient time to filter out methanol.

62. (b) 1. Guard cell Stomata! guard cells

control the pore to balance CO 2 entry into the leaf for photosynthesis with water loss through transpiration. It also regulates membrane transport, signalling and homeostasis. 2. Stomatal pore It allows CO2 to diffuse into leaf for photosynthesis. 3. Subsidiary cell These arc accessory cells to guard cells. They play an important role in ion-channel mediated opening and closing of guard cells. 4. Epidermal cell Provides protection to the plants from the external environment.

63. (a) Viruses and lichens were

excluded from Whittaker's five kingdom classification because

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KVPY Question Paper 2019 Stream : SB/SX

Whittaker's classification was based on certain character like cell structure, mode and source of nutrition, body organisation and reproduction. Viruses and lichens docs not fulfil the classification criteria as viruses are acellular organism and lichens arc sy mb iotic association of algae and fungi .

64. (b) Plant species when grown in

shade produces thin leaves with more surface area to capture more of the limited light and when grown in light produce thick leaves to m aximi se photosynthesis in light. This phenom enon is an exa mple of phenotypic plasticity. It is the ability of organis m to alter the ir phenotype in response to environmental factor without changing its geno me.

65. (a) Sacred groves found in Westren

Ghats, central India and North East India are exa mple of in situ conservation of living organis m s m ainly plants and animals. Sacred groves are diversity rich regions protected by local people for re ligious and cultural beliefs. Endangered, threatend and ende mic species arc conserved in these regions.

66. (b) lgE

mediated immunity is

most effectively controlled by anti-histamines in immune response. Anti-hista mines are used to prevent the symptom s of hay fever and allergy. lgE class antibodies along with antigen and m ast cell triggrcs the allergic reaction by releasing hista mines, which in turn cause inflammatory response. Anti-histamines prevents the effects of histamines by affecting lgE mediate response.

67. (c) The origin of mitochondria and

chloroplast arc explained by endosym biotic theory. This theory first postul ated by Lynn Margulis in 1967. Mitochondria and chloroplast are eukaryotic cell organelles with bacterial characteristics. According to this theory eukaryotic mitochondria evolved from a s m all, prokaryotic autotrophic bacterium that was engulfed by large prim itive heterotrop hie eukaryotic cell. Endosymbiosis of chloroplast occurs when a cell engulfed a photosynthetic cyanobacterium .

68. (d) Growth rate is independent of

environmental fluctuation as population exhibits logistic growth when resources arc limited. Population expansion decreases as resources become scarce. After reaching to the carrying capacity

there is little or no change in population size over tim e. Birth rate lt is the ab ility of individuals of a population to produce new individuals. Natality is the sc ientific ter m for birth rate. Death rate Scientific ter m for death rate is mortality. Mortality refers to the death of individuals in population. Resource availability A resource is an substance in environment required by an organism for growth and reproduction. lt is one of the m ain factor which deter mines the ecological dynamics of species or population. 69. (c) Natural events like volcanic eruptions, wildfires and flood, etc., wipes out most of the life form and leaves the barren land. Succession starts when organis m colonise on barren land. Algae, fungi and simple plants such as lichens and mosses grows firstly on barren land upon successful colonisation of pl ants, animals and small invertebrates starts to colonise the m selves and become stable. bluster'. It is produced by beta • he molytic streptocci . Streptokinase cleaves the Arg/val bond in pl as minogcn to for m proteolytic enzym e plas min. Plas min breaks the m ain component of blood clot 'fibrin'. Cyclosporin-A is an immunosuppressant used to prevent organ rejection after kidney, heart and liver transplant. Paracetamol is used to treat pain and fever. Statins lowers the cholesterol level in blood.

70. (d) Streptokinase in known as 'clot

71 . (d) Ori C is not involved in transcription. Replication starts from sequences found on chromosom e called origin of replication the point at which DNA opens, hclicasc unwinding the DNA double helix resulting in the formation of replication fork. Promoter is a region of DNA from where transcription of gene is initiated. Terminator is a sequence of DNA that causes RNA polym erase to terminate transcription. Enhancers arc DNA sequences that cause an increase in the level of expression of gene. 72. (b) Nematoda is a pseudocoelomate. Pseudocoelomate animals possess pseudocoelom s. It is a fluid filled body

cavity between the cndodcrm and m esoder m . 73. (b) Lacrimal gland does not secrete saliva. It secretes aqueous layer of tear film. Submaxillary, partoid and sublingual glands are associated with saliva secretion.

74. (c) Cortex, Casparian strip, pericycle, vascul ar bundle represents the tissue arrangement in roots. Cortex It conducts water and minerals across the root. Casparian strip It prevents water from entering the pericycle. Casparian strip are made up of suberin. Pericycle It gives rise to lateral roots. Vascular bundles It consists of xylem and phloe m . Xylem carries water and dissolved minerals and phloe m carries food. 75. (c) ln alcoholic fermentation, conversion of glucose into ethanol occurs. The NADH generated by the oxidation of glyccraldchydc 3-phosphatc is consumed in the reduction of acetaldehyde to ethanol. Thus, there is no net oxidation • reduction in the conversion of glucose into ethanol.

76. (c) Product of cyclic

photophosphorylation is ATP. In cyclic photophosphorylation only one photosystem (PS-1) is involved when light is absorbed by PS-1 excited electrons enters into electron transport chain to produce ATP. Both ATP and NADPH are produced in non-cyclic photophosphorylation, which includes both photosystems (PS-I and PS-11). 77. (d) Arginine is a hydrophilic a mino acid found at the surface of protein.

78. (a) Global species diversity increases

from high to low altitudes. High altitudes arc less diverse then lower altitudes because atmosphere becom es less dense and due to the lack of oxygen few species are able to sustain their life at higher altitudes.

79. (d) Phenylketonuria is not

respons ible for the presence of deoxygenated blood in the arteries of a newborn. It is a genetic disorder in which level of phenylalanine is increased in blood. This occurs because of defect in PAH gene. Phenylalanine hydroxylase, which converts phenylalanine to tyrosine. In the absence of PAH gene enzym e PAH is not synthesised and breakdown of i i m l ph · [WWW .JE E BcO O Ks .

I NI

21

KVPY Question Paper 2019 Stream : SB/SX Symptoms of PKU includes seizures tremors, stunted growth, hyperactivity, etc. Pneumonia is an infection of lungs can be caused by bacteria, virus and fungi. Air sacs present in the lungs get inflamed and filled with fluid resulting in cough, fever and breath shortness. 80. (a) Rhizobium is a Gram negative bacteria fixes atmosphoric nitrogen by infecting roots of leguminous plants and leads to the formation of root nodules. Nitrogen fixation is catalysed by enzyme nitrogenase. Nitrogenase is very sensitive to oxygen. If oxygen is present, nitrogenase is not able to perform its function. Leg haemoglobin scavanges 0 2 if present for proper functioning of enzyme. It is a ATP dependent process and required anaerobic condition. 8 1 . (b) Let the diameter AB = x

L'..ACB = � and

and

L'..ADB = �

BC = ✓x2 - l 2

AD = ✓x2 - 9

83. (b) Given, 6 2 f (x) = x - 2x° + x3 + x - x - l 2 2 4 = x (x - x3 - x - 1)

then

A

B ,,

_, , -

%

IG I I I

D

i{

2

%

According to Apollonius theorem,

AD 2 =

b2 + c2 - �}

AD = !. ✓2b2 + 2c2 - a2 2 2

AG GD

AG = �AD = !. ✓2b2 + 2c2 - a2 1

Similarly BG = !_ ✓2c2 + 2a2 - b2 3

3

3

BG 2 + AG 2 -AB 2 2(BG) (AG)

So, cose = -------, where 0 = L'..AGB 1 2 1 2 2 2 2 2 2 - (2c + 2a - b ) + - (2b + 2c - a ) - c 9 = 9 2(BG) (AG) 2

2

2

a + b - 5c 18(BG) (AG)

D

Now, as

. . . (i)

2

2

4 4 =2 ⇒ 2AD · csin81 2AD · b sin82

2x + 3 = ✓x - l ✓x - 9

AB · CD + AC · BC = AD · BC ⇒

2

2

On squaring both sides, we get 2 4 4x2 + 9 + 12x = x - 10x + 9 4 2 ⇒ x - 14x - 12x = 0 ⇒ x3 - 1 4x - 12 = 0 Let /(x) = x3 - 14.x - 12 /(4) = -4 /(4.1) = -0.479 f (4.2) = 3.288 /(4.1) · f(4.2) < 0 x E l4.1, 4.2) 82. (a) Let MBC and D is the mid- point of side BC,

⇒

2

⇒ AD + b

⇒

2_

2

2

2

AD 2 = b2 + .

I

(c) _Q_(5 + 4coscp) 9

(b)

10

sin 2 ( /2)

I (5 + Scos ) (d) _Q_ (j> 9 3

Block

ch X )

Plank

Let the centre of mass of the block be at (0, l / 2) at a given instant. If a = g I 10, then the normal reaction exerted by the plank on the block at that instant acts at (a) (0, 0) (b) ( l /20, 0) (d) ( l /1 0,0) (c) ( l /1 0,0)

29. Using the Heisenberg uncertainty principle, arrange

,LD F

(a)10 cos2 ((j> /2)

y

(d) {ST

23. For an ideal gas, the internal energy is given by U =5 p V / 2 + C, where C is a constant. The equation

p

constant acceleration ai. A uniform rough cubical block of side l rests on the plank and is at rest relative to the plank.

28. A plank is moving in a horizontal direction with a

the following particles in the order of increasing lowest energy possible. (I) An electron in H2 molecule (II) A hydrogen atom in a H2 molecule (III) A proton in the carbon nucleus (IV) A H2 molecule within a nanotube (a)I < III < II < IV (b)lV < II < l < lll (c)II < IV < III < I (d) lV < l < II < lll

30. The current in flowing along the path ABCD of a

cube (shown in the left figure) produces a magnetic field at the centre of cube of magnitude B. Dashed line depicts the non-conducting part of the cube. D--------A,

E/

,,,� � ----:

: ,,

,, C

- - - -

,,.,

,,/ :

i

'

G

F

Consider a cubical shape shown to the right which is identical in size and shape to the left. If the same current now flows in along the path DAEFGCD, then the magnitude of magnetic field at the centre will be (a) zero (b) J2B (c)-/3B (d)B

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KVPY Question Paper 2018 Stream : SB/SX

3 1 . A thin metallic disc is rotating with constant angular

velocity about a vertical axis that is perpendicular to its plane and passes through its centre. The rotation causes the free electrons in the disc to redistribute. Assume that, there is no external electric or magnetic field. Then, (a) a point on the rim of the disc is at a higher potential than its centre (b) a point on the rim of the disc is at a lower potential than its centre (c) a point on the rim of the disc is at the same potential as its centre (d) the potential in the material has an extremum between centre and the rim

35. Consider a glass cube slab of dielectric bound by the planes x = 0, x = a; y = 0, y = b; z = 0 , z = c; with b > a > c. The slab is placed in air and has a refractive index of n. The minimum value for n, such that all rays entering the dielectric at y = 0 reach y = b is

36. The graph shows the log of activity log R of a radioactive material as a function of time t in minutes.

32. One mole of a monoatomic gas and one mole of a

diatomic gas are initially in the same state. Both gases are expanded isothermally and then adiabatically, such that they acquire the same final state. Choose the correct statement. (a)Work done by diatomic gas is more than that by monoatomic gas (b)Work done by monoatomic gas is more than that by diatomic gas (c) Work done by both the gases are equal (d) Change in internal energies of both the gases are equal

33. An ideal gas is made to undergo the cyclic process shown in the figure below.

� Volume

Let A W depict the work done, AU be the change in internal energy of the gas and Q be the heat added to the gas. Sign of each of these three quantities for the whole cycle will be (0 refers to no change) (a)- , 0,(b)+ , 0, + (c)0, 0, 0 (d)+,+,+ 34. Two balls of mass M and 2 M are thrown horizontally with the same initial velocity v0 from top of a tall tower and experience a drag force of- kv (k > 0), where v is the instantaneous velocity.

Then, (a)the heavier ball will hit the ground further away than the lighter ball (b)the heavier ball will hit the ground closer than the lighter ball (c) both balls will hit the ground at the same point (d) both balls will hit the ground at the same time

(b) .J2 (d) 2

(a) 1 (c) J3

a:: 6 0)

.Q 4

2 -,..............._,r-...._,...�_...�.,........-"-'-t➔ 0 5 1 0 1 5 20 25 Time (min)

The half-life (in minute) for the decay is closest to (a) 2.1 (b) 3.0 (c) 3.9 (d) 4.4

37. The magnetic field is uniform for y > 0 and points into the plane. The magnetic field is uniform and points out of the plane for y < 0. A proton denoted by filled circle leaves y = 0 in the-y-direction with some speed as shown below.

® --- r--0--------

- - -e- - ­

-L\J - -e- - ­

Which of the following best denotes the trajectory of the proton? (a)

(c) -

�

(b) (d )

-

38. The Hitomi satellite recently observed the Lyman

alpha emission line (n = 2 to n = l) of hydrogen-like iron ion (atomic number of iron is 26) from the Perseus galaxy cluster. The wavelength of the line is closest to (a) 2 A (b) 1 A (cl 50 A (d) 1 0 A

39. Assume that, the drag force on a football depends

only on the density of the air, velocity of the ball and the cross-sectional area of the ball. Balls of different sizes but the same density are dropped in an air column. The terminal velocity reached by balls of masses 250 g and 125 g are in the ratio s (a) 211 (b) i' 3 (c) 2112 (d)2213

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KVPY Question Paper 2018 Stream : SB/SX 40. An electrostatic field line leaves at an angle IX from

point charge Qi and connects with point charge- q2 at an angle � (q1 and q9 are positive) see figure below. If 3 q2 = - q1 and IX = 30 ° , then 2 +q, °

°

45. Benzaldehyde can be converted to benzyl alcohol in concentrated aqueous NaOH solution using (a) acetone (b) acetaldehyde (c)formic acid (d)formaldehyde

46. The major product of the following reaction,

(b)� = 30° (d) 60 ° < � :-;; 90 °

(a)0 < � < 30 (c) 30° < � :-, 60 °

(a) x2 for all x ,t. 0 and p' (0) = _! is 2 (a)0 (b) 1 (c) more than 1, but finite (d) infinite n --> =

L = lim ✓n f 0

(1 + x2 f

dx

exists and is larger than _!__ Then, 2

Column II

Contaminated food and water Inhalation of aerosol Contact via skin Sexual intercourse Mosquito bite

D 4 4

B 1 3

A (b) 2 (d) 2

1 (a)- < L < 2

81 . Let R be a rectangle, C b e a circle, and T b e a triangle

1

B C 1 5 3 5

1. 2. 3. 4. 5.

C 3 4

D 5 5

(2 Marks Questions)

MATHEMATICS

86. Suppose the limit

Column I Tuberculosis Dysentery Filariasis Syphilis

A (a) 2 (c) 1

79. In photosynthetic carbon-fixation, which one of the following reacts with CO 2?

(a)Phosphoglycolate (b)3-phosphoglycerate (c)Ribulose-1 , 5-bisphosphate (d) Ribose-5-phosphate

(b) 2< L < 3

87. Consider the set An of points (x, y) such that (d)L 2 4

(c)3< L < 3

0 ::; x ::; n, Oy ::; n where n, x, y are integers. Let Sn be the set of all lines passing through at least two distinct points from An . Suppose we choose a line l at random from Sn . Let Pn be the probability that l is

(

tangent to the circle x2 + y2 = n 2 1 + (1 n --> =

Then, the limit lim Pn is

(a) 0

(b) 1

1 (c)-

- 1r}

(d)_!_ J2

88. Let f : [0, 1] ➔ R be an injective continuous function that satisifes the condition- 1 < f(0) < f (1) < 1.

Then, the number of functions g : [ 1, 1] ➔ [0, 1] such that (go/) (x) = x for all x E [0, 1] is (a) 0 (b) 1 (c)more than 1 , but finite (d) infinite

y = x2 + x + 10 and a chord of the parabola of length 1

89. The maximum possible area bounded by the parabola IS

1 (a)12

1 (c)3

90. Suppose z is any root of l lz8 + 2 l iz7 + lOiz- 22 = 0 where i = N. Then, S = I zl 2 + (a)S ::; 3 (c)7 ::; S < l3

I zl + 1 satisfies

(b) 3 < S < 7 (d)S 2 1 3

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KVPY Question Paper 2018 Stream : SB/SX 95. Two rods of copper and iron with the same

PHYSICS 91. In steady state heat conduction, the equations that determine the heat current j (r) [heat flowing per unit time per unit area] and temperature T(r) in space are exactly the same as those governing the electric field E (r) and electrostatic potential V (r) with the equivalence given in the table below. Heat flow

T (r) j (r)

Electrostatics V (r) E(r)

We exploit this equivalence to predict the rate Q of total heat flowing by conduction from the surfaces of spheres of varying radii, all maintained at the same temperature. If Q oc R" , where R is the radius, then the value of n is (b) 1 (a) 2 (c)- 1 (d) - 2

92. An arrangement of spring, strings, pulley and masses is shown in the figure below.

The pulley and the strings are massless and M > m. The spring is light with spring constant k. If the string connecting m to the ground is detached, then immediately after detachment (a)the magnitude of the acceleration of m is zero and that of M is g (b)the magnitude of the acceleration of m is (M - m)g/m and that of M is zero (c) the accelerations of both masses arc same (d) the elongation in the spring is (M - m)glk

93. The potential due to an electrostatic charge - ar

distribution is V(r) =�, where a is positive. 4rc£ 0 r

The net charge within a sphere centred at the origin and of radius 1 / a is (b)(1 - 1 / e) q (a) 2q I e (d) (1+ 1 / e) q (c)q I e

94. A wheel of radius R is trapped in a mud pit and

spinning. As the wheel is spinning, it splashes mud blobs with initial speed u from various points on its circumference. The maximum height from the centre of the wheel, to which a mud blob can reach is 2 gR2 (b).!!:.__ + (a) u 2 l 2g 21-t 2u 2 (c)O

u2

(d) R + -

2g

cross-sectional area are joined at S and a steady current I flows through the rods as shown in the figure.

s

()

Choose the most appropriate representation of charges accumulated near the junction S. (a) 4ru

(c) 4ru

s

Fe / b) I ()9u ) (

s

Fe / d) I ()9u ) (

:[E D

s

:U: 9--1 =[E s

Fe / )

96. Graphs below show the entropy versus energy U of

two systems 1 and 2 at constant volume. The initial energies of the systems are indicated by U1 , i and U2, i, respectively. Graphs are drawn to the same scale. The systems are then brought into thermal contact with each other. Assume that, at all times the combined energy of the two systems remains constant. Choose the most appropriate option indicating the energies of the two systems and the total entropy after they achieve the equilibrium.

U, , ;

U2,;

U1

U2

(a)Vi increases and U2 decreases and the total entropy remains the same (b)Vi decreases and U2 increases and the total entropy remains the same (c)Vi increases and U2 decreases and the total entropy increases (d) U1 decreases and U2 increases and the total entropy increases

97. The image of an object O due to reflection from the

surface of a lake is elongated due to the ripples on the water surface caused by a light breeze. This is because the ripples act as tilted mirrors as shown below. Consider the case, where O and the observer E are at the same height above the surface of the lake. If the maximum angle that the ripples make with the horizontal is a, then the angular extent 6 of the image will be

(a) � 2

(b) a

(c) 2a

(d) 4a

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KVPY Question Paper 2018 Stream : SB/SX

98. A spiral galaxy can be approximated a s an

infinitesimally thin disc of a uniform surface mass density (mass per unit area) located at z = 0. Two stars A and Bstart from rest from heights 2z0 and 20 (z0 , bl)

�

As, intensity 1 °" A 2 or I = kA 2, where So,

f

Forces acting on the block are

k = a constant and A is amplitude. So, I = A12 + Af + 2A1A2 cos(j> Here, A1 = A and A2 = 2A ⇒ I = A2 + (2A) 2 + 2(A)(2A) cos(j> = A 2 (5 + 4cos(j>) . . . (i) Intensity is maximum (10 ), when cos(j> = 1 ⇒ 10 = A 2 (5 + 4 x 1) = 9A 2 . . . (ii) A 2 = 10 I 9 ⇒ So, resultant intensity, I = A 2 (5 + 4cos(j>) [ From Eqs. (i) and (ii)] I (5+ 4cos(j>) I = _Q_

21 . (a) By Bernoulli's equation for streamline flow, 1 p + -pv 2 + pgh = a constant 2 In given situation h = constant 1 p + -pv 2 = constant 2 When velocity of flow in narrow section increases, pressure decreases. So, current level in vertical tubes is as shown below. ⇒

9

X

---,c-+----. ------- --

1/2

mg

In above diagram, ma = pseudo force due to acceleration of plank, f = force of static friction, mg = weight of block and N = normal reaction. Here, we must note that due to motion of plank, normal reaction force does not passes through centre of mass. If line of action of normal force at a distance x from centre, then for equilibrium of block, net torque about centre of mass must be zero. ⇒ t Normal = t Friction reaction

mg and ma does not produces any torque about centre of mass as their line of action passes through centre of mass.

N - x = f- l/2 l mgx = ma

2

[·: N = mg and f =

x = (�)i

ma]

1 10 2

or

x=-

or

X= -

20 So, coordinates of point from which 1 , 0} reaction passes are (20

29. (b) A H2 molecule in a nanotube is bound only by adhesive forces. So, its energy is least. A hydrogen atom in a H 2 molecule is bound by interatomic forces. So, its energy is less than the energy of an electron in H 2 molecule. Proton in a carbon nucleus is bound because of strong nuclear forces. So, its energy is largest. :. Correct order is IV < II < I < Ill.

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17

KVPY Question Paper 2018 Stream : SB/SX 30. (c) Given magnetic field due to a square loop DABCA at centre of cube is B.

As, slope of an adiabatic curve = Also, 'Ydiatomic( =

i)

1 ( i)

< 'Ymonoatomic ( = � )

So, adiabatic expansion will be more steep for monoatomic gas, whereas isotherms for both gases are nearly same. As, work done = area under p-V graph Now, loop DAEFGCD can be viewed as super position of three square loops as shown below.

y

:. Work done is more for diatomic gas.

33. (a) Given cyclic process is

D

y=b

x=a

x= 0 Rays incident aty=O

Due to dielectric, a light ray incident at y = 0, at some angle 0 bends towards normal and it strikes plane x = a. If angle of incidence at x = a, is more than angle of critical incidence, TlR occurs and light ray, then emerges out of plane y = b.

---

y =b f-------""'-c----�

G

So, net field at centre = resultant of fields of these three loops

Bj

-8

k

V Arca under compression process CA is more than area under expansion process AB. So, net work done is negative.

1.e.

Also, in a cyclic process, change in internal energy is zero. i.e. Now, by using first law of thermodynamics, we have

Bnet = Bi +

Bj - Bk

Magnitude of resultant field is 2 2 2 I Bnet l = ,JB + B + B = Js - B

3 1 . (b) Centrifugal action causes

electrons to accumulate around rim of disc.

'-----t� � �

w

�

So, rim of disc is at a negative potential with respect to centre of disc. So, points of rim arc at lower potential than its centre.

32. (a) Expansion curves arc as shown below.

p

w e sec that,

: 90-r 90- r r� - -+-- ,'-� x=a

Ll.W < O

Ll.Q = O+ Ll.W

Bi

�90---�r ' '

Ll.Q < 0

34. (a) Only force resisting motion of

particle in horizontal direction is drag force given by

F = - lw But where, ax = acceleration in x-dircction caused by drag force. kv max = - kv ⇒ ax = -

:. Acceleration of particle is x-dircction is

inversely proportional to mass. So, decreases in velocity in x-direction for lighter particle will be more. Hence, heavier mass will hit the ground further away from the lighter ball.

35. (b) Planes bounding the dielectric are as shown below. y

(O,b,O)

➔x

For TlR at x = a, sin (90° - r) = � or cosr = �

n

. . . (i)

n

n = -.

sine sin r Now 0 at y = 0, varies from 0° to a maximum of 90° . So, for limiting incidence, sin 90° 1 n = -. . . (ii) - ⇒n = . . sm r sin r From Eqs. (i) and (ii), we get sin r = cosr ⇒ r = 45° Substituting r = 45° in Eq. (i) or Eq. (ii), l we get nmm • = _ __ = ../2 cos45°

Also,

36. (b) Activity of a radioactive sample is given by R__

dN _ _ d _ '}.,,N . e-At No e-'At o dt

- dt -

So, log R = log(A.N0 ) + log (e-"1 ) ⇒ logR = - t..t + log(t..N0 ) This equation is form of y = mx + C So, absolute value of slope of log R versus t graph gives decay constant A. Now, from graph, 10

�

8,8

-

16, �i"16

5 8 10

15

20

/(min) �

25

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18 W e get, slope =

KVPY Question Paper 2018 Stream : SB/SX

l� -6

1

= - = ').,, 8 - 16 4

So, half-life time period of sample is log 2 0.693 = _ TV2 = 3 0 min = 1/ 4 A,

⇒

2 mg = k a uj rc R

⇒

mg

3

k o vl,rc [r ] - re p 3 [from m = � rcR3 p, 3 where p = density of footballj Vrr oc m1/ 6

37. (d) Following Fleming's left hand

rule, proton initially rotates clockwise and it enters region y > 0 with its velocity in positive y-direction. Herc in region y > 0, magnetic field is into plane of paper. So, proton again follows left hand rule and now tend to rotate anti-clockwise as shown below. X

X

X

X

X

X

0

0

0

0

0

0

' '

X

@x

:

' ' '

0

V o

⇒

v1 V2

(

=

� m2

)¼ (250)¼ =

125

=

�1 6

40. (a) Consider another symmetric field line below line joining centres of charges q1 and q2.

X

' '

⇒

=

0

0

38. (a) Using Balmer's formula,

wavelength "A of emission from ion is --1:_ = RZ 2 (_-1:__ - ---1:_ "A n( n;2 J

Here, Rydberg constant, R = 109 x 107 m- 1 Atomic number of iron, Z = 26 n, = 2, nr = 1

(_!_ - _l._)

⇒ or

--1:_ = 109 X 107 X (26) 2 X A, i2 ').,, = 18 x 1 0 - 10 A A, =

ln given situation, flux (field lines) leaving charge q1 at a solid angle 2a = flux terminating over charge q2 at a solid angle 2 p. Clearly, in given situation

39. (a) Given, drag force Fa depends on

air density a, velocity of ball v and area of cross-section of ball A. Fa = a a iA C Substituting dimensions of different physical quantities, we have 1 [MLr2] = [ML-3 j° [Lr f [L2 ]c Equating dimensions of fundamental quantities, we have . . . (i) a=1 . . . (ii) - 3a + b + 2c = 1 . . . (iii) - b = - 2⇒b= 2 Putting the values of a and b in Eq (ii), WC get ⇒ -3 X 1 + 2 + 2c = 1 ⇒ C = 1 ·: a = 1, b = 2, c = 1 So, drag force on football is 2 Fa = k • a • v A • When football reaches terminal speed, its weight is balanced by drag force. ⇒ mg = Fa ⇒ mg = k a v}A where, ur = terminal speed of football.

:� •

Total flux through base of cone ¢ = J:

2rrR2 sin o. do.

= k Q 2 rrJ: sin a dcx = _9._ (1 - cosa) 2 Eo

So, equating flux of both cones, we get ...!!.L (1 - cosa) = ...!!L (1 - cos p) 2 E0 2 E0 ⇒

2

3 q1 (1- cosa) = - q1 (1 - cosP)

Substituting a = 30° in above equation, we get 2 ⇒ - (1 - cos30° ) = 1 - cosP 3 ⇒

¾ ( 1 - � ) = l - cosP

⇒

cos p = 1 -

¾(

1 - �)

⇒ cos P = 0.9 So, angle p is not more than 30° . 4 1 . (a)

22

2A

Flux through elemental ring (E l l dA) k d,1,. = Q2 • 2rcR 2 sin a da 'l' R

II

CH:,-C-CH3 + 3Br2 + 4NaOH

We can say that, flux leaving charge q1 through a cone of semi-vertical angle a = flux terminating on charge q2 through a cone of semi-vertical angle p.

-� � _ _ _ _ _ _ h_ q,

Flux leaves q1 through here

(r2

\ Flu� enters through here

To calculate flux, we first find flux through an elemental ring of base of cone and then we integrate to get total flux.

2rrr =2rrR sina

Arca of elemental ring, dA = 2rcrds = 2rcRsin a - Rda

L +

0

Ace?one

� CH3CONa + 3NaBr + 3H2 O + CHBr3 3 moles of bromine produces 1 mole of CHBr 3 . :. 1 mole of bromine produces _! mole of 3 CHB1iJ . 42. (b) Williamson ether synthesis is the reaction in which an alkyl halide is allowed to react with alkoxide to give ether. The reaction involves SN 2 attack of an alkoxidc ion on primary alkyl halide. In option (a) and (b) there are alkoxide and alkyl halide, but option (b) is preferred because 1 - is better leaving group than c1-. OH

o-

GJ�GJ VI

o:1)

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KVPY Question Paper 2018 Stream : SB/SX 43. (d)

46. (a)

H

0

V

�C02 H

H,�¥-C�l CH 3

X

Both X and Y are conformers of each other, where X is eclipsed and Y is staggered form. Conformers are the spatial arrangements of atoms, which can be converted into one another by rotation around a C-C single bond. The conformation in which the substitutents are as closed together as possible is called eclipsed conformation, whereas in staggered form the substitucnts arc far apart as possible.

44. (d) Tert-butyl cation is more stable

than isopropyl cation, because of better hyper conjugation between a-vacant and p-orbital.

NaBH

OH

4 �CO2H ----+ I

,,.,:::.

NaBH 4 is a selective reducing agent which reduces carboxyl group of aldehyde and ketone to alcohol, but does not reduces COOH group.

47. (*) As the electronegativity of the central atom increases the bond angle also increases. So, NH3 will have high bond angle than PH3 . The bond angle of NH! is highest because of its symmetrical geometry, also bond pair and lone pair on N ofNH 3 suffer repulsion, which decreases its bond angle in comparison to NH!. Thus, the correct order of bond angle is PH 3 < NH 3 < NH! But in the given question none of the option is correct because there is no H - X - H bond angle in BF3 . 48. (c) The given species arc

(Hyper-conjugation in tert-buty] carbocation)

whereas, trans-2 - butcne is more stable than propene because of hyper-conjugation a-it* -orbitals.

iI�•. oc-----o"7H c .• I

H

H-C,,,,- ,.

, 0. The light beam on the other side of the medium will emerge (a) parallel to the X-axis (b) bending downward (c) bending upward (d) split into two or more beams

29. Let the electrostatic field E at distance r from a point charge q not be an inverse square but instead an

inverse cubic, e.g. E = k ; r, here k is a constant. r Consider the following two statements: (I) Flux through a spherical surface enclosing the charge is = qenclosed I fo .

(II) A charge placed inside uniformly charged shell will experience a force. Which of the above statements are valid? (a) Only statement I is valid (b) Only statement 11 is valid (c) Both statements I and II arc invalid (d) Both statements I and II are valid

30. A star of mass M and radius R is made up of gases.

The average gravitational pressure compressing the star due to gravitational pull of the gases making up the star depends on R as (c) 1/R 2 (b) 1/R (a) 1/R 4 (d) 1/R 6

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KVPY Question Paper 2017 Stream : SB/SX

31. The black shapes in the figure below are closed

surfaces. The electric field lines are in red. For which case, the net flux through the surfaces is non-zero?

circumference of a large bowl as shown below in the figure. The ring is moving down at f\. , comes down to the lower most point P2 and is climbing up at ]1 . Let vcM denote the velocity of the centre of mass of the ring. Choose the correct statement regarding the frictional force on the ring.

(III)

)w

(a) In all cases net flux is non-zero (b) For III and IV cases (c)For l and 11 cases (d) For II, III and IV cases

P2

32. A particle of charge q and mass m enters a region of a transverse electric field of E0 j with initial velocity v0i.

!

(a) always has same width as of the p-side (b) has no bound charges (c) is negatively charged (d)is positively charged

36. A small ring is rolling without slipping on the

(I)

The time taken for the change in the de-Broglie wavelength of the charge from the initial value of A o to 1., 0 / 3 is proportional to (a)

35. The n-side of the depletion layer of a p-n junction

(c)H;

(b) :

t

(d)

33. Consider the following nuclear reactions: I. }4 N + ;He � fo + x IL � Be + i H � 2He + Y

i

Then, (a) X and Y are both protons (b) X and Y are both neutrons (c)X is a proton and Y is a neutron (d)X is neutron and Y is a proton

(a) It is opposite to vcM at the points Pi , P2 and Pa (b) lt is opposite to vcM at Pi_ and in the same direction as VCM at f'a (c) lt is in the same direction as vcM at P1 and opposite to VCM at f'a (d)It is zero at the points Pi_ , P2 and Pa

37. A bomb explodes at time t = 0 in a uniform, isotropic medium of density p and releases energy E, generating a spherical blast wave. The radius R of this blast wave varies with time t as (c) t u4 (b) t215 (d)t312 (a) t

38. A closed pipe of length 300 cm contains some sand. A speaker is connected at one of its ends. The frequency of the speaker at which the sand will arrange itself in 20 equidistant piles is close to (velocity of sound is 300 mis)

��----------/:\ �I

34. Consider a plane parallel beam of light incident on a plano-cylindrical lens as shown below. Which of the following will you observe on a screen placed at the focal plane of the lens?

(a) 1 0 kHz

Sand piles (b) 5 kHz

(c) 1 kHz

(d) 1 0 0 kHz

39. A planet of radius RP is revolving around a star of

radius R*, which is at temperature r• . The distance between the star and the planet is d. If the planet's temperature is f r*, then f is proportional to (a) .jR*/d

(b) R*ld

(c) R* RP!d2 (d)(R*/d)4

40. Some of the wavelengths observed in the emission

(a) The screen will be uniformly illuminated (b) There will be a single bright spot on the screen (c) There will be a single bright line on the screen parallel to the X -axis (d) There will be a single bright line on the screen parallel to the Y-axis

spectrum of neutral hydrogen gas are 912, 1026, 1216, 3646, 6563 A. If broad band light is passing through neutral hydrogen gas at room temperature, then the wavelength that will not be absorbed strongly is (a) 1 0 26 A (b) 1 21 6 A.. (cl 91 2 A (d) 3646 A

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KVPY Question Paper 2017 Stream : SB/SX

CHEMISTRY 41 . The major product formed in the following reaction is )CHO (

HCI gas excess MeOH

50. The complete hydrolysis of XeF6 results in the formation of (a) Xc0 2F2 (b) XcOF4

(d)Xc0 2

(c) Xe03

51 . The reactivity of the following compounds toward water is in the order (a) Cl207 < Pp5 < B203 (b) B203 < Pp5 < Cl207 (c) P205 < B203 < Cl207 (d)Bp3 < Cl2O7 < P2O5

52. Among the following complexes, the one that can

42. Which among the following is a non-benzenoid aromatic compound? (a) o-xylene (c) lndolc

(b)Phenanthrene (d) Thiophcnc

43. Natural rubber is a polymer of (a) neoprene (c)isoprene

(b)chloroprene (d) styrene

44. The following tripeptide CH3 O COOH H 2N 0 � � � O "( OH Ph can be represented as (b)phc - ala - scr (a) tyr - val - thr (c)phe - leu - cys (d) lys - ala - ser

l....,

45. The sugar units present in natural DNA and RNA, respectively are (a) D-2-deoxyribose and L-ribose (b) L-2-deoxyribose and D-ribosc (c)D-2-deoxyribose and D-ribose (d)L-2-dcoxyribose and L-ribosc

exist as facial (tac) and meridional (mer) isomers is (a) [Co(N0 2)3 (NI--1:i )3 J (b) Ks [ Fe(CN)6 J (d)[ CoCl(NI--1:i ) ] Cl 2 (c) [ Co(H20)2 (NI--1:i )4 ]Cl3 5 53. An excess of Ag2Cr0/s) is added to a 5 x 10-3 M K 2Cr04 solution. The concentration of Ag+ in the solution is closest to [ Solubility product for Ag2Cr04 = 1 . 1 x 10- 12] (a) 2.2 x l 0 - 10 M (b) l 5 x l 0 -5 M (c) lOx 1 0-6 M

54. The packing efficiency in a body centered cubic (bee) structure is closest to (a) 74% (b) 63%

47. The most abundant metal ion present in the human body is (a) Zn 2+

(b)Ca 2+

(c)Na+

(d)Fe2+

48. Phosphorus reacts with chlorine gas to give a

colourless liquid, which fumes in moist air to produce HCl and (a) POCl3 (c)PI--1:i (d)I--1:i P0 4

49. The oxidising ability of the given anions follows the order (a) Tio:- < VO!- < CrO� < Mn04 (b) VO�- < Cr◊!- < Mn04 < Ti◊!(c)Cr◊�- < Mn04 < VO!- < TiO!(d)VO!- < Tio:- < Cr◊;- < Mn04

(c) 68%

(d) 52%

55. The consecutive reaction x- Y - Z takes place

in a closed container. Initially, the container has A 0 moles of X (and no Y and Z). The plot of total moles of the constituents in the container as a function of time will be

Ao

Ao

0 (/) 2 C

0 (/) 2 a, C

a, - a,

:::J (a) 0 Ez (/) .:,l C 0

- a,

i2

Ao

46. The major product formed in the following reaction is, CH3 Br+ CI-fsCH2 0Na(a) CI--1:i CH2CH20H (b) CI--1:i OCI--1:i (c)CI--1:i CHP CI--1:i (d) CH3 CH20CH2Br

(d)5.0 x 1 0-3 M

(c)

0 (/) 2 a, C

:::J (b) 0 E :a; ]j C 0

i2

Time

----------------

0 (/) 2 a, C

:::J (d) 0 E :a:, cu C

;g

E ::: (/)

C

0

Ao

Time

- a,

a, 0 :::J

i2

(.)

(.)

Time

0

(.)

Time

56. The particles emitted during the sequential radioactive decay of

(a) 5 a and 6 P (c) 8 a and 4 P

238

U 92 to

206

Pb 82 are

(b) 6 a and 8 P (d)8 ex and 6 �

57. The allowed set of quantum numbers for an electron in a hydrogen atom is (a) n = 4, l = 2, m1 = 0, m5 = 0 (b) n = 3, l = 1, m1 =- 3, ms =- 1 / 2 (c) n = 3, l = 3, mz =- 1, m.. =1 1 2 (d) n = 2, l = 1, mz =- 1, ms = 1 1 2

a/'=---=-a,----,c==-=-=-=,........,..-=,......... W.J E E B o o K S _ l� W-=-,W l...-;NI

32

KVPY Question Paper 2017 Stream : SB/SX

58. The plot that best represents the relationship between the extent of adsorption (x/m) and pressure (p) is

(a)

x/m

( d)

identify the responsible allergen. This is because (a) of the presence of mast cells under the skin (b) lymphocytes migrate rapidly from the blood to the skin (c) hair follicles can enhance the reaction (d)ncutrophils migrate rapidly from the blood to the skin

68. Which one of the following processes in E. coli does

log (x/m)

log (p )

not directly involve RNA9 (a) DNA replication (b) Transcription (c) Translation (d)DNA repair

log (p)

59. The pH of 0.1 M acetic acid solution is closest to [Dissociation constant of the acid, Ka = 1 .8 x 1 0-5 ] (a) 2.87 (b) 1 .0 0 (c) 2.07 (d)4.76

60. The limiting molar conductivities of the given electrolytes at 298 K follow the order 11.�K + ) = 73.5 , 11.�Cl -) = 76.3 , "-:c 2 + ) = 119.0 , a 2 1 "-�soi- ) = 160.0 S cm mol(a) KC! < CaCl2 < K28O 4 (c)K28O 4 < CaCl2 < KC!

(b)KC! < K�O 4 < CaCl2 (d) CaC12 < K2SO 4 < KC!

BIOLOGY

61. Resting membrane potential of a neuron is approximately (a) -70 mV (c)-0.7V

(b)+ 70 mV (d) + 0.7V

62. Amphimixis is (a) a fusion of pronuclei of male gametes (b) a fusion ofpronuclei from male and female gametes (c) a fusion of pronuclei of female gametes (d) the development of a somatic cell into an embryo

63. Activation of sympathetic nervous system (a) decreases blood pressure (b) causes pupil contraction (c)increases heart rate (d)causes bronchoconstriction

64. At physiological temperature, sterols in biological membranes (a) increase their fluidity (b) decrease their fluidity (c)increase their permeability to water (d) decrease their permeability to water

65. Which one of the following is a heteropolysaccharide? (a) Glycogen (c)Cellulose

(a) arc non-transferable to the same bacterial species (b) arc capable of independent replication (c) have RNA as genetic material (d) always require integration in the genome for their replication

67. Skin-prick test on the forearm is conducted to

(b) x/m

log (c) x/m ( )

66. Bacterial plasmids are genetic entities that

(b)Starch (d) Hyaluronic acid

69. Which one of the following statements is incorrect for translation in cytoplasm? (a) One codon codes for only one amino acid (b)One amino acid may be coded by many codons (c)More than one amino acids are coded by one specific codon (d)There are some codons that do not code for any amino acid

70. Two homozygous parents harbouring two different

alleles of a gene, exhibiting incomplete dominance for flower colour were used for a genetic experiment. Which one of the following statements is incorrect? (a) The F2-generation will consist of plants of three different flower colours (b) The genotypic and phenotypic ratios obtained in the F2-generation will be different (c) The F1 -generation will be of a different flower colour compared to both the parents (d) The genotypic ratio obtained in the F 2 -generation will be the same irrespective of whether it is complete dominance or incomplete dominance

71 . Which one of the following is an essential condition for a population to be at Hardy-Weinberg equilibrium? (a) Random mating (b) Immigration (c) Emigration (d)Geographical isolation

72. Inbreeding in a population leads to (a) decrease in recessive disorders (b) heterosis (c) increase in homozygosity (d) increase in heterozygosity

73. Which one of the following molecules serves as a

substrate for direct synthesis of ATP? (a) 1, 3-bisphosphoglyccrate (b) Glucose-6-phosphatc (d) Fructose- 1 ,6-bisphosphate (c) Pyruvate

74. If a pure chlorophyll solution is illuminated with ultraviolet light, the solution appears (a) green (b) violet (c) red (d) black

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33

KVPY Question Paper 2017 Stream : SB/SX 75. Botanical names of plants are given in Column I and the family/order name in Column II. Choose the appropriate combination from the options below. A. B. C. D. A (a) 4 (c) 1

Column I

Tamarindus indica Cocos nucifera

Colchicum autumnale Withania somnifera

B C 1 2 2 4

D 3 3

1.

2. 3. 4.

A (b) 4 (d) 4

Column II

78. In diabetic patients, the pH of blood plasma can

decrease leading to acidosis. This is because tissues catabolise (a)amino acids leading to loss of buffering capacity of the blood (b) stored glycogen leading to the accumulation of pyruvic acid (c)stored fatty acids leading to the accumulation of beta hydroxybutyric acid and acetoacetic acid (d) nucleic acid pool leading to decrease in blood pH

Arecaceae Liliaceae Solanaceae Papilionaceae B 2 1

3 3

C

D 1 2

76. Nitrogen-fixation is inhibited by oxygen. However, in aerobic nitrogen-fixing bacteria, nitrogen is fixed in the presence of oxygen. Nitrogenase in such organisms is protected by which one of the following mechanisms? (a) Channelising oxygen to form ozone (b) Removal of oxygen by metabolic activity (c) Utilising oxygen for membrane remodelling (d)Utilising oxygen for synthesis of pentapeptide chain in peptidoglycan

77. Frederick Griffith performed an experiment where

mice were killed when injected with a mixture of Heat-Killed S-type Streptococcus (HKS) and Live R-type Streptococcus (LRS) but not with HKS or LRS separately. Mice were killed because (a) lipids from HKS made LRS virulent

� PART- I I

without changing the total alveolar volume, the gas exchange capacity of the lungs will (a) increase for both 0 2 and CO 2 (b) decrease for both 0 2 and CO 2 (c) remain unaltered for both 0 2 and CO 2 (d)increase for 0 2 and decrease for CO 2

80. In an experiment, bacteria were infected with 32P labelled virus in a ratio of 5 : 1. The culture was rigorously shaken followed by centrifugation. Radioactivity was (a) lost due to metabolic activity (b)detected in supernatant as inorganic phosphate (c)detc>cted in the supernatant in association with viral capsid (d) detected in bacterial cell pellet

84. An ellipse 2 + 2 = 1, a > b and the parabola x2

81 . Let AB be the latusrectum o f the parabola y 2 = 4ax in the XY-plane. Let T be the region bounded by the finite arc AB of the parabola and the line segment AB. A rectangle PQRS of maximum possible area is inscribed in T with P, Q on line AB , and R, S on arc AB. Then, area (PQRS)/area (T) equals (b)_1_ 3

79. If the number of alveoli in an individual is doubled

(2 Marks Questions)

MATHEMATICS

(a) _1_ 2

(b) RNA from HKS transformed LRS and made it virulent (c) proteins from HKS made LRS virulent (d) DNA from HKS transformed LRS and made it virulent

J?,

(c)--1:_

J3

(d)--1:_

82. Let S be the set of all permutations a1 , a2,

...

,a6 of 1,

2, . . . , 6 such that a1 , a 2, • • • , ak is not a permutation of 1, 2, . . ., k for any k, l < k < 5. Then, the number of elements in A is (a) 1 92 (b)408 (c)3 12 (d) 528 1 83. The area bounded by the curves y = - 14- x2 1 and 4 y = 7- l xl is (a) 1 8 (b) 32 (c) 36 (d) 64

a

y2

b

x2 = 4(y + b) are such that the two foci of the ellipse and the end points of the latusrectum of parabola are the vertices of a square. The eccentricity of the ellipse IS

(b) __3__

1 (a) -

m

m

1 (c) ­ ,/TI

m

(d)__ 3__

85. A sector is removed from a metallic disc and the

remaining region is bent into the shape of a circular conical funnel with volume 2..ffirc. The least possible diameter of the disc is (c) 8 (a) 4 (b) 6 (d) 12 1 lxl 3 1 4 2 1 3 86. Let g(x) = f t sin - dt, for all real x. Then, x -; 0

lim

g(x)

(a) =

X

0

is equal to (b) - =

t

(c) 0

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34 87. Let a

KVPY Question Paper 2017 Stream : SB/SX

r"

= l x- ll cos nxdx for all natural numbers " n. Then, the sequence (an \ ;, 0 satisfies (a) lim an = oo n➔ = (b) lim an = - oo n� oo

(c) lim a,. exists and is positive n->=

=0 (d) lima n➔ = n 88. Let f(x) be a polynomial with integer coefficients satisfying f(1) = 5 and f(2) = 7. The smallest possible positive value of f (12) is (a) 5 (c)27 (d) 1 5 (b)7

89. Suppose four balls labelled 1, 2 , 3, 4 are randomly placed in boxes B1 , B2, Ba, B4 . The probability that exactly one box is empty is (a) � 256

(b)_$}_ 16

(c)I1._ 256

(d)_$}_ 64

90. Let f(x) = log(l + x2 ) and A be a constant such that

*

If(x)- f(y) I :,; A for all x, y real and x y. Then, the l x- yl least possible value of A is (a) equal to 1 (b) greater than 1 but less than 2 (c)greater than O but less than 1 (d)greater than 2

PHYSICS 91. One mole of an ideal monoatomic gas undergoes the following four reversible processes: Step 1 It is first compressed adiabatically from volume 8.0 m3 to 1.0 m3 . Step 2 Then expanded isothermally at temperature T1 to volume 10.0 m3 . Step 3 Then expanded adiabatically to volume 80.0 m3 • Step 4 Then compressed isothermally at temperature T2 to volume 8.0 m3 • Then, T;_ IT2 is �4 W2

�6

�8

92. A solid cube of wood of side 2a and mass M is resting on a horizontal surface as shown below.

horizontally at the face opposite to ABCD at a height h above the surface to impart the cube an angular speed we, so that the cube just topples over. Then, we is (Note the moment of inertia of the cube about an axis perpendicular to the face and passing through the centre of mass is 2Ma3 /3) (a) ,.j3gM I 2ma (b) ,.j3g I 4h ___--(c) ✓3 (v'2- 1)/ 2a (d).J� 3 (-J2 _ 1)/ 4a g

g

93. A uniform thin wooden plank AB of length L and

mass M is kept on a table with its Bend slightly outside the edge of the table. When an impulse J is given to the end B , the plank moves up with centre of mass rising a distance h from the surface of the table. Then, (a) h > 9J2! 8M2g (b) h = J2!2M2g (c) J 2 /2M2g < h < 9J 2 ! 8M2g (d) h < J 2 !2M2g

94. A square shaped wire loop of mass m, resistance R

and side a moving speed u0 , parallel to the X-axis, enters a region of uniform magnetic field B, which is perpendicular to the plane of the loop. The speed of the loop changes with distance x (x < a) in the field, as B 2a 2 B 2a2 (a) v0 - -- x (b) v0 - -- x Rm 2Rm B2a (c) v0 - - x2 (d) v0 Rm -1:_ = A

nl

n2

where, R is the Rydberg constant. For a transition from n 2 to n1, the relative change fiA / A in the emission wavelength, if hydrogen is replaced by deuterium (assume that, the mass of proton and neutron are the same and approximately 2000 times larger than that of electrons) is (a) 0 .025% (b) 0.005% (c) 0.0025% (d)0.05%

96. When light shines on a p-n junction diode, the current I versus voltage V is observed as in figure below. Quad 2

Quad 3

The cube is free to rotate about the fixed axis AB. A bullet of mass m(%>� W�=% =�=% � % = �> �> % �%>�>%>� 98. Back surface of a glass (refractive index n and thickness t) is polished to work as a mirror as shown below. A laser beam falls on it and is partially reflected and refracted at the air-glass interface and fully reflected at the mirror surface, respectively. A pattern of discrete spots of light is observed on the screen.

ei

Glass mirror

')

100. Consider the L-C-R circuit given below. The circuit is driven by a 50 Hz, AC source with peak voltage 220 V. If R = 400 Q, C = 200 µF and L = 6 H, the maximum current in the circuit is closest to

(a) 0.1 2 A

(b) 0.55 A

(c) 1 .2A

(d) 5.5 A

CHEMISTRY 101. In the reaction, 0

(i'c1

(i) X

(ii) y

X and Y are (a) X = H2 , Pd / BaS0 4 ; Y = NaOAc, Ac20 (b) X = LiAIH4 ; Y = KaOAc, Ac20 (c) X = H2 , Pd I C;Y = NaOH, Ac20 (d)X = LiAlH4 ; Y = NaOH, Ac20

102. In the following reaction,

C Q)

u

Cf)

The spacing between the spots on the screen will be 2t sin0 2tcos0 (b) (a) 2 2 ✓n - sin 0 ✓n 2 - sin 2 0 2ttan0 21 sine (c) (d) . ;i - sm -e sin20 1nz -J

Which of the above statements indicate that light consists of quanta (photons) with energy proportional to frequency? (a) Statements I and Ill are correct (b) Statements II and III arc correct (c) Statements II, Ill and IV are correct (d)Statements I, II and Ill are correct

X and Y are

'J

99. Consider the following statements regarding the photoelectric effect experiment:

(I) Photoelectrons are emitted as soon as the metal is exposed to light. (II) There is a minimum frequency below which no photo-current is observed. (III) The stopping potential is proportional to the frequency of light. (IV) The photo-current varies linearly with the intensity of the light.

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36

KVPY Question Paper 2017 Stream : SB/SX

103. Acetophenone (PhCOCHa ) reacts with perbenzoic acid to produce a compound X. Reaction of X with

The equilibrium constant of the reaction is closest to (a) 4 x 1d6 (b) 2 x 1d 7 (c) 2 x l d4 (d) 4 x ld7

excess CHaMgBr followed by treatment with aqueous acid predominantly produces

111. Suppose the three non-linked autosomal genes A, B

BIOLOGY

(a) Ph AoH

104. The fusion of chromite ore (FeCr2 O4 ) with Na 2CO3 in air gives a yellow solution upon addition of water. Subsequent treatment with H2SO4 produces an orange solution. The yellow and orange colours, respectively, are due to the formation of (a) Na 2CrO 4 and Na 2Cr2O7 (b) Cr(OH)3 and Na 2Crp7 (c) Cr2 (CO3 )3 and Fe 2 (8O 4 )3 (d) Cr(OH)3 and Na 2CrO 4 105. Hybridisation and geometry of [Ni(CN)4 ] 2- are (a) sp 2d and tetrahedral (b) sd3 and square planar (c) sp3 and tetrahedral (d) dsp2 and square planar

106. The total number of geometrical isomers possible for an octahedral complex of the type [.MA2B2C2 ] is (M = transition metal; A, B and C are monodentate ligands) �4 �6 �5 W3

107. The maximum work ( in kJ mor 1) that can be derived from complete combustion of 1 mole of CO at 298 K and 1 atm is [ Standard enthalpy of combustion of CO=-2 83.0kJ moi- 1 ; standard molar entropies at 298 K : S 02 = 205 . 1 J moi- 1 , S00 = 1 97.7 J moi- 1 , S002 = 2 13.7 J mol-1] (a) 257

(b) 227

(c) 57

(d) 127

108. 1 8 g of glucose (C 6 H12 O6 ) dissolved in 1 kg of water is

heated to boiling. The boiling point (in K) measured at 1 atm pressure is closest to [Ebulioscopic constant, Kb for water is 0.52 K kg moi- 1 . Consider absolute zero to be -273.15 C 0 ] (b) 373.10 (a) 373.15 (c) 373.20 (d) 373.25

109. Polonium (atomic mass = 209) crystallises in a simple cubic structure with a density of 9.32 g cm-3 Its lattice parameter (in pm) is closest to (a) 421 (b) 334 (c) 481 (d) 193

110. The following reaction takes place at 298 K in an

electrochemical cell involving two metals A and B,

Az+ (aq ) + B (s) ➔ B2+ (aq ) + A(s)

with [A2+ ] = 4 x 10-3 M and [B2+ ] = 2 x 10-3 M in the respective half-cells, the cell EMF is 1 .091 V.

and C control coat colour in an animal and the dominants alleles A, B and C are responsible for dark colour and the recessive alleles a, b and c are responsible for light colour. If a cross between a male of AABBCC genotype and a female of aabbcc genotype produces 640 offsprings in the F9 -generation, how many of them are likely to be of the parental genotype? (c) 160 (d) 640 (b) 20 (a) 10

112. In a population of families having three children

each, the percentage of population of families having both boys and girls is (b) 25 (d) 75 (a) 10 (c) 50

113. As indicated in the gel image, lanes X and Y

represent samples obtained from a circular plasmid DNA after complete digestion using restriction enzyme X or Y with different recognition sites, respectively. How many sites for X and Y are present in the plasmid (sizes of the bands in kilo base pairs (kb) is shown)? X y 6 kb 5 kb 4 kb -

(a) 1 for X, 1 for Y (c) 1 for X, 2 for Y

(b) 2 for X, 1 for Y (d) 2 for X, 2 for Y

114. Matthew Meselson and Franklin Stahl grew E.coli

(doubling time is 20 min) in medium containing NH4Cl for many generations. Then the E.coli was transferred to medium containing 1 4 NH4Cl After 40 minutes, the cells were harvested and DNA was extracted and subjected to cesium chloride density gradient centrifugation. The proportion of light and hybrid DNA densities will be (a) 50% light and 50% hybrid DNA (b) 100% light DKA (c) 100% hybrid DNA (d) 25% light and 75% hybrid DNA 15

115. In a population interaction between the species X

and the species Y, which one of the following statements is correct? (a) When X benefits and Y is disadvantaged, it is competition (b) When both X and Y benefit, it is mutualism (c) When both X and Y arc disadvantaged, it is predation (d) When both X and Y arc disadvantaged, it is parasitism

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37

KVPY Question Paper 2017 Stream : SB/SX

119. Which one of the following graphs best describes the

116. The protein P, the oligosaccharide 0, and the

Blood Pressure (BP) change when blood moves from aorta to capillaries?

oligonucleotide N are composed of 100 amino acid residues, 100 hexose residues and 100 nucleotides, respectively. Which one of the following orders of molecular weights is correct?

(b) P > N > 0

cii I

(a) P > 0 > N

(d) 0 > P > N

-,,'

I I I I

(A) l

(c) N > 0 > P

(l_ ll]

117. An octapeptide

(NH2-Asn-Glu-Tyr-Lys-Trp-Met-Glu-Gly) is subjected to complete protease and chemical digestion. Based on the results obtained, choose the incorrect option from below. (a) Trypsin generates mixtures of dimer and trimer (b) Trypsin generates tetramers only (c) Cyanogen bromide generates a hexamer and a dimer (d)Chymotrypsin generates mixture of dimer and trimers

,,

,,

I I I

'

I I I

''

'

,,

',

Aorta -t Capillaries

Aorta -t Capillaries

I ----,

� (D) � (l_ ll]

\I

I I I I I

118. Match the enzymes in Column I with their respective biochemical reactions in column II. Choose the correct combination from below Column I A. Transaminases

(a) A

Column II 1 . Removal of phosphoryl group from

A (a) 4 (c) 2

PART-I 11 21 31 41 51 61 71

(d) (c) (b) (c) (b) (b) (a) (a)

PART-I/ 81 91 101 111

(d) (b) (a) (b)

B C 2 3

3 1

D 5 5

B 1

2

C 2 3

(c) C

,---

I I I I I

-----� .,.,

Aorta -t Capillaries (d) D

genetic disorders P and Q in Family 1 and Family 2, respectively. Family 2

Family 1

Choose the correct statement from the following options. (a) Both P and Q arc dominant traits (b) P is a dominant trait and Q is a recessive trait (c) Both P and Q arc recessive traits (d)P is a recessive trait and Q is a dominant trait

Oxidation and reduction of substrates A (b) 2 (d) 5

(b) B

'

/ I

120. The following two pedigrees describe the autosomal

a specific amino acid B. Protein kinases 2. Removal of a-amino group from a specific amino acid 3. Addition ofphosphoryl group to a C . Protein phosphatases specific amino acid D. Dehydrogenases 4. Interconversion of optical isomers 5.

Aorta -t Capillaries

I I I

D 4 1

Answers 2 12 22 32 42 52 62

3 13 23 33 43 53 63 73

(d) (c) (a) (c) (c) (b) (c) (a)

4 14 24 34 44 54 64 74

(d) (c) (c) (d) (b) (c) (a) (c)

5 15 25 35 45 55 65 75

(d) (c) (c) (d) (c) (b) (d) (a)

6 16 26 36 46 56 66 76

(c) (b) (d) (b) (c) (d! (b) (b)

7 17 27 37 47 57 67

72

(b) (b) (d) (b) (d) (a) (b) (c)

82 92 1 02 1 12

(d) (d) (d) (d)

83 93 1 03 113

(b) (c) (c) (d)

84 94 1 04 114

(b) (a) (a) (a)

85 95 1 05 115

(b) (d) (d) (b)

86 96 1 06 1 16

(c) (c) (c) (c)

8 18 28 38 48 58 68 78

(c) (a) (c) (c) (b) (c) (d) (c)

9 19 29 39 49 59 69 79

(a) (a) (b) (a) (a) (a) (c) (a)

10 20 30 40

77

(b) (a) (c) (b) (b) (d! (a) (d)

60 70 80

(d) (a) (a) (d) (c) (a) (b) (d)

87 97 1 07 117

(d) (b) (a) (a)

88 98 1 08 118

(c) (a) (c) (c)

89 99 1 09 119

(b) (d) (b) (a)

90 1 00 110 1 20

(a) (a) (b) (b)

so

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Solutions 1 . (d) Given ABCD is a square A and B. A and B slide along theirs respective axes. Let AB = BC = CD = AD = a l n !).BMC, __ _ __ _ __ _ _C(h,k) N

5. (d) Given, f (x) =

Clearly n 2 1

:. It is also true.

Let

3. (d) We have,

a2 + b2 + c2 S = ----­ ab + be + ca

a

We know that, a A

0 sin0 =

BM BC

sine = Ii_ a h = a sin0 . . . (i) k = OB + NB h = a sine + acose [·:OB = a sin0, NB = acos 0] ⇒ h = h + a cos e ⇒ . . . (ii) k - h = a cos 8 On squaring and adding Eqs. (i) and (ii), WC get h 2 + (h - h) 2 = a 2 ·: Locus of CB x2 + (y - x) 2 = a2 2x2 + y2 - 2xy = a 2 ⇒ which is ellipse not a circle. 2. (b) We have, I. n!s nn nn n n n n - = - X -- X --- X . . . X ­ 1 n ! n (n - 1) (n - 2) nn Clearly - 2 1 n! :. n n 2 n 1 True 11. (n 1) 2 s n n is False 111. lO" s n ! n! n n- 1 n- 2 1 = - X -- X -- X . . . X lOn 10 10 10 10 n l· . 21 G1ven, n > 1000. Clearly,10n Hence, it is also true. lV. n n S (2n)! 2n ! 1- 2- 3 - 4. . . n(n + l)(n + 2). . . (n + n) = n X n X n . . . n times nn n l(n + l)(n + 2)(n + 3). . . (n + 1) = n x n x n x . . . n times

(a - b) 2 + (b - c)2 + (c - a) 2 2 0 ⇒ a 2 + b2 + c2 - (ab + be + ca) 2 0 ⇒ a 2 + b2 + c2 2 ab + be + ca a 2 + b2 + c2 2 ⇒ 1 [·: ab + bc + ca > 0] ab + be + ca We know that, (a + b + c) 2 2 0 2 :. a + b2 + c2 + 2(ab + be + ca) 2 0 ab + be + ca < 0 a 2 + b2 + c2 + 2(aJI + bc + ca) p2 (b)Pi = p2 = 0.6651 (c)Pi < Pz (d) p1 = p2 = 0.3349

11. For how many different values of a does the following system have at least two distinct solutions? ax + y = 0

x + (a + lO)y =O

(a) 0 (c) 2

(b) 1 (d) Infinitely many

12. Let R be the set of real numbers and f : R ➔ R be

{x} , where [x] is the greatest 1 + [x] 2 integer less than or equal to x, and {x} = x - [x]. Which of the following statements are true?

defined by f (x) =

I. The range of f is a closed interval.

IL / is continuous on R.

III. f is one-one on R. (a) I only (c)III only

(b)II only (d) None ofl, II and III

(a) lim Xn = 00

(b) lim Xn = Js

13. Let Xn = (2 n + 3 n ) 11 2 n for all natural numbers n. Then, n➔

n➔ oo

(c) limxn = Js + J2 rH =

(d) limxn = J5 n➔ oo

14. One of the solutions of the equation

8 sin3 0- 7 sin0 + ,,/3 cos0 = 0 lies in the interval (a) (0°, 1 0°l

(b) (10°, 20°l (c) (20°, 30°l (d) (30°, 40°l

15. Let a , b, c, d, e be real numbers such that a + b < c + d, b + c < d + e, c + d < e + a, d + e < a + b. Then,

(c) X'� o. ��lc-�� 1 .0

10

Y'

Y'

18. Let A = (a1 , a2 ) and B = (b1 , b2 ) be two points in the plane with integer coordinates. Which one of the following is not a possible value of the distance between A and B? (a) -/65 (b) ../74 (c) J83 (d)J97

19. Let f(x) =max {3, x2 , � } for ½ :S: x :S: 2. Then, the value 2

of the integral f f(x)dx is

(a) � 3

11 2 13 (b) 3

(c)

14 3

(d)

16 3

20. Let a1 = i + � for i = 1, 2, . . ., 20. Put !

p = - (a1 + a2 + . . . + a20 ) and

1 20

_!_ (_!_

1 + __!__ + ... + --). Then, a 20 a2 22- p 2(22- p) 22- p , (b) q E ( (a) q E ( o, ) 21 21 ) 21 p) 22 - p 22- p 4(22- p) (c) q E ( 2(22- , ) (d)q E ( ) 21 21 7 7 '

q=

20 a1

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53

KVPY Question Paper 2018 Stream : SB/SX 27. Two circularly shaped linear polarisers are placed

PHYSICS nth orbit of hydrogen atom is a8 and that of singly

21. The magnitude of acceleration of the electron in the ionised helium atom is a 8e . The ratio of a 8 : a8e is (b) 1 :4 (a) 1 :8 (c) 1 :2 (d) dependent on n

22. A carrot looks orange in colour because of the �

carotene molecule in it. This means that the � carotene molecule absorbs light of wavelengths (a) longer than 550 nm (b)shorter than 550 nm (c)longer than 700 nm (d) shorter than 700 nm

23. If some charge is given to a solid metallic sphere, the field inside remains zero and by Gauss's law all the charge resides on the surface. Now, suppose that Coulomb's force between two charges varies as 1 / r3 • Then, for a charged solid metallic sphere (a)field inside will be zero and charge density inside will be zero (b)field inside will not be zero and charge density inside will not be zero (c)field inside will not be zero and charge density inside will be zero (d)field inside will be zero and charge density inside will not be zero

24. Using dimensional analysis, the resistivity in terms of fundamental constants h, me , c, e, Ea can be expressed as h (a) --2 m,ce E0 h2 (c)m.ce2

E m.ce2 (b) 0 h (d)

coaxially. The transmission axis of the first polariser is at 30 ° from the vertical while the second one is at 60 ° , both in the clockwise sense. If an unpolarised beam of light of intensity I = 20 W/m 2 is incident on this pair of polarisers, then the intensities 11 and 12 transmitted by the first and second polarisers respectively, will be close to (a) J1 = 10.0 W/m2 and J2 = 7.5 W/m2 (b) J1 = 20 W/m 2 and J2 = 15 W/m2 (c) 11 = 1 0.0 W/m2 and 12 = 8.6 W/m2 (d)J1 = 1 5.0 W/m 2 and 12 = 0.0 W/m2

28. An electron in an electron microscope with initial

electric field E0j . The time taken for the change in

velocity v0i enters a region of a stray transverse

its de-Broglie wavelength from the initial value of 11, to A I 3 is proportional to 1 (d) (c) /Er, (a) E0 (b) J:... Eo

-JEo

29. A bird sitting on a single high tension wire does not get electrocuted because (a) the circuit is not complete (b) the bird feet has an insulating covering (c)capacitance of the bird is too small and the line frequency is too small (d) resistance of the bird is too high

30. A positive charge q is placed at the centre of a neutral hollow cylindrical conducting shell with its cross-section as shown in the figure below.

m.Eo ce 2

Vacuum

25. Consider a bowl filled with water on which some

black pepper powder have been sprinkled uniformly. Now, a drop of liquid soap is added at the centre of the surface of water. The picture of the surface immediately after this will look like

Which one of the following figures correctly indicates the induced charge distribution on the conductor? (Ignore edge effects)

( ) b

26. It was found that the refractive index of material of a

certain prism varied as 1.5 + 0.004 / 11,2 , where 11, is the wavelength of light used to measure the refractive index. The same material was then used to construct a thin prism of apex angle 10 ° . Angles of minimum deviation S m of the prism were recorded for the sources with wavelengths 11, 1and 11, 2 , respectively. Then, (a) Om (),! ) < Om (),..2) i0..1 < '-2 (b) om (),1 ) > om 0, 2) i0..1 > '-2 (c)om (A l ) > om (A2) i0 ..1 < '-2 (d)om is the same in both cases

(c)

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54

KVPY Question Paper 2018 Stream : SB/SX

31. A transverse wave of frequency 500 Hz and speed

100 m/s is travelling in the positive x-direction on a long string. At time t = 0 s, the displacements at x = 0.0 m and at x = 0.25 m are 0.0 m and 0.02 m, respectively. The displacement at x = 0.2 m at t = 5 X 10-4 S is (a) - 0.0 4 m (b)- 0.02 m (c) 0.04 m (d) 0.02 m

The pulley rotates without any friction, whereas the friction between the rope and the pulley is large enough to prevent any slipping. Which of the following plots best represents the difference between the tensions in the rope on the two sides of the pulley as a function of the mass of the pulley?

32. A thin piece of thermal conductor

of constant thermal conductivity insulated on the lateral sides connects two reservoirs which are maintained at temperatures T1 and T2 as shown in the figure alongside. Assuming that the system is in steady state, which of the following plots best represents the dependence of the

:.:W�)\� 7 :,•ntmpy on t: ,

'!F' m2 ) tied at the ends as shown in the figure below.

(a) ,-:-

I Q

(b)

M

��

(c)

,· f :-o

M

(d)

:-o

M

:- o

M

,-� ��

35. Two satellites S1 and S2 are revolving around a

planet in the opposite sense in coplanar circular concentric orbits. At time t = 0, the satellites are farthest apart. The periods of revolution of S1 and S2 are 3 h and 24 h, respectively. The radius of the orbit of S1 is 3 x 104 km. Then, the orbital speed of S2 as observed from (a)the planet is 41t x 1 0 4 km h- 1, when S2 is closest from � (b) the planet is 21t x 1 0 4 km h-1 , when S2 is farthest from � (c) � is 7t x 1 0 4 km h- 1, when S2 is closest from � (d)� is 31t x 1 04 km h- 1 , when S2 is closest to � 36. A rectangular region of dimensions co x I (co V B 2w2 will hit the screen (b) Ions with q < B w(c) All ions will hit the screen 2Vm (d)Only ions with q = 2 2 will hit the screen B

00

37. Force F applied on a body is written as F = (n · F)n + G, where n is a unit vector. The vector G is equal to (a) n X F (c) (n x F) x Fl I Fl

(b) n X (n X F) (d) (n x F)x n

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55

KVPY Question Paper 2018 Stream : SB/SX potential _!_ mo r , where r is the distance from the 2 origin. Applying the Bohr's model in this case, the radius of the particle in its nth orbit in terms of a = .Jh I (21tmco) is (a) a.fn (b)an (c)an2 (d) an .fn

38. A particle of mass m moves around the origin in a i 2

39. Two bottles A and Bhave radii RA and RB and

heights hA and hs respectively, with Rs =2RA and hB = 2hA . These are filled with hot water at 60° C. Consider that heat loss for the bottles takes place only from side surfaces. If the time, the water takes to cool down to 50 ° C is tA and ts for bottles A and B, respectively. Then, tA and t s are best related as (a) tA = tB (b)tB = 2 tA (c)tB = 4 tA (d) tB = tA /2

40. The number of gas molecules striking per second per

square metre of the top surface of a table placed in a room at 20° C and 1 atmospheric pressure is of the order of (ks = 1.4 x 10-23 JK- 1 and the average mass of an air molecule is 5 x 10-27 kg)

(b)1023 (d) 102 9

(a) 1021 (c)1025

CHEMISTRY 41 . The major product formed in the following reaction is r:Y

�

Y

H N

CHa

cone . HN03

cone. H2S04

0

N

y

(a)

N0 2

NHCOCH 3

(c)

NHCOCH 3 2

42. Among the a-amino acids - threonine, tyrosine,

methionine, arginine and tryptophan, those which contain an aromatic group in their side chain are (a) threonine and arginine (b) tyrosine and tryptophan (c)methionine and tyrosine (d) arginine and tryptophan

43. The number of stereoisomers possible for the following compound is Wl

CHs - CH

= CH-CH(OH) - CHs �2

�3

�4

chlorobenzene, the ortho/para-directing ability of chlorine is due to its (a) positive inductive effect (+J) (b) negative inductive effect ( J) (c) positive resonance effect (+R) (d)negative resonance effect (-R)

cr o

45. Among the following,

I

II

�

I

0

�N

III

IV

H

V

the antiaromatic compounds are (a) I and IV (b) Ill and V (c) II and V (d)I and III

46. Upon reaction with CH3 MgBr followed by protonation, the compound that produces ethanol is (a) CfiaCHO (b) HCOOH (c) HCHO (d)(CH0)2

47. Which of the following is not an oxidation-reduction reaction? (a) H2 + Br2 - 2HBr (b) NaCl+ AgN03 - NaN03 + AgCI (c) 2Na 2Sp3 + ½- Na 2S4 0 6 + 2Nal (d)Cl2 + H20- HCI+ HOC!

48. The thermal stability of alkaline earth metal

carbonates MgCO3 , CaCO3 , SrCO3 and BaCO3 , follows the order (a) BaC03 > SrC03 > CaC03 > MgC03 (b) CaC03 > SrC03 > BaC03 > MgC03 (c) MgC03 > CaC03 > SrC03 > BaC03 (d)SrC03 > CaC03 > MgC03 > BaC03

49. When a mixture of diborane and ammonia is heated,

� N0 r:Y

44. In electrophilic aromatic substitution reactions of

the final product is (a) BH3 (b) NH4 BH4 (c) NH2KH2 (d) B3 N3 H6

50. Among the following metals, the strongest reducing agent is (a) Ni

(b) Cu

(c) Zn

(d) Fe

51 . The molecule, which is not hydrolysed by water at 25° C is (a) AICl3

(b) SiC1 4

(c) BF3

(d)SF6

52. Among the following compounds, the one which does not produce nitrogen gas upon heating is (b) NaN3 (a) (NH4 ) 2 Cr207 (c) NH4 N02 (d)(NH4 MCP 4 )

53. Chlorine has two naturally occurring isotopes,

Cl and 3 7 CL If the atomic mass of Cl is 35.45, the ratio of natural abundance of 35 Cl and 37 Cl is closest to (a) 3.5: 1 (b) 3: 1 (c) 2.5:1 (d)4: 1 35

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56

KVPY Question Paper 2018 Stream : SB/SX

54. The reaction C 2 H6 (g) � C 2 H4 (g) + H2 (g)is at

equilibrium in a closed vessel at 1000 K. The enthalpy change (t:,.H) for the reaction is 137.0 kJ mol- 1 . Which one of the following actions would shift the equilibrium to the right? (a)Decreasing the volume of the closed reaction vessel (b)Decreasing the temperature at which the reaction is performed (c)Adding an inert gas to the closed reaction vessel (d)Increasing the volume of the closed reaction vessel

55. The enthalpy (H) of an elementary exothermic

reaction, A � Bis schematically plotted against the reaction coordinate. The plots in the presence and absence of a catalyst are shown in dashed and solid lines, respectively. Identify the correct plot for the reaction.

59. A mineral consists of a cubic close-packed structure

formed by 0 2- ions, where half the octahedral voids are occupied by Al3 + and one-eighth of the tetrahedral voids are occupied by Mn 2 + . The chemical formula of the mineral is (a) Mn3 Al2O 6 (b) MnA1 2O4 (c) MnA1 4 O7 (d)Mn zA1 2O5

60. For a 4p-orbital, the number of radial and angular nodes, respectively are (a) 3, 2 (b) 1, 2

(c) 2, 4

(d) 2, 1

BIOLOGY

61. Interferons combat viral infection by (a) inhibiting viral packaging directly (b) increasing the binding of antibodies to viruses (c) binding to the virus and agglutinating them (d)restricting viral spread to the neighbouring cells

62. Leydig cells synthesise B

Reaction coordinate

Reaction coordinate

63. Glucagon increases the blood glucose ,;'�-,\ '\ ,/

(d) H

A/'

\..,p..-

B

A

Reaction coordinate

Reaction coordinate

56. Mg(OH)2 is precipitated, when NaOH is added to a

solution of Mg2+ . If the final concentration of Mg 2+ is 10- 10 M, the concentration of OH- (M)in the solution lS

[Solubility product for Mg(OH)2 = 5.6 x 1 0-

(a) 0.056

(b)0.12

(a) insulin (b) growth hormone (c) testosterone (d)oestrogen

(c)0.24

12 ]

(d) 0.025

57. A constant current (0.5 amp) is passed for 1 hour through (i) aqueous AgNO 3 , (ii) aqueous CuSO 4 and (iii) molten AIF3 , separately. The ratio of the mass of the metals deposited on the cathode is [MAg, Mcu, MAJ are molar masses of the respective metals.] (b) MAg : Mcu : MAJ (a) MAg : 2 Mcu : 3 MAJ (c) 6MAg : 3Mcu : 2MA1 (d) 3MAg : 2Mcu : MAl

58. A reaction has an activation energy of 209 kJ mol-1 , The rate increases 10-fold when the temperature is increased from 2 7 ° C to X° C. The temperature X is closest to [Gas constant, R = 8.134 J moi- 1 K- 1 ] (a) 35 (b)40 (c)30 (d) 45

concentration by (a) promoting glycogenolysis (b) increasing the concentration of fructose 2,6-bisphosphate (c)increasing the concentration of pyruvate kinase (d) inhibiting gluconeogenesis

64. Which one of the following is not essential for Polymerase Chain Reaction (PCR)? (a) Restriction enzyme (b) Denaturation of DNA (c) Primers (d) DNA polymerase

65. CO 2 acts as a greenhouse gas because

(a) it is transparent to heat but traps sunlight (b) it is transparent to sunlight but traps heat (c) it is transparent to both sunlight and heat (d)it traps both sunlight and heat

66. A graph of species richness vs area on log-log axes is (a) linear (c) oscillatory

(b) sigmoidal (d)parabolic

67. Concentration of Na+ ions outside a nerve cell is -100 times more than inside. The concentration of K+ ions is more inside the cells. The levels of Na + ions and K+ ions are maintained by (a)free diffusion of Na+ ions and pumping of K+ ions across the membrane (b) Na+ and K + pumps in the membrane (c)free diffusion of K + ions and pumping of Na+ ions across the membrane (d)water channels formed by lipids in the membrane

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57

KVPY Question Paper 2018 Stream : SB/SX

74. Which one of the following statements is incorrect?

68. In a chemical reaction, enzymes catalyse the

reaction by (a)lowering the activation energy (b) increasing the activation energy (c)decreasing the free energy change between reactants and products (d) increasing the free energy change between reactants and products

(a) Alleles arc different forms of the same gene (b) Alleles are present at the same locus (c) Alleles code for different isoforms of a protein (d)Alleles arc non-heritable

75. Which one of the following statements is incorrect about restriction endonucleases? (a)They serve as primitive form of immune system in bacteria (b)They digest the DNA non-randomly (c)They digest the DNA at specific location (d)They digest the DKA from free ends

69. The rigidity of cellulose is due to (a) coiled structure of glucose polymer (b) �(1 ➔ 4)glycosidic linkage (c) hydrogen bonding with adjacent glucose polymer (d)cross-linking between glucose and peptides

76. The number of net ATP molecules produced from 1

(a) always result in precipitation of the complex (b) depend only on covalent interactions (c)arc irreversible (d) depend on ionic and hydrophobic interactions

72. For a particular gene that determines the coat colour

in a diploid organism, there are three different alleles that are codominant. How many different skin colours are possible in such an organism? (a) 9 (b)6 (c)4 (d) 3

linked. During the inheritance of these traits, the Mendelian laws that would be affected is/ are (a)law of dominance, law of segregation and law of independent assortment (b)law of segregation and law of independent assortment (c)only law of independent assortment (d) only law of segregation

� PART- I I

MATHEMATICS

statements are true? I. 4 divides x or 4 divides y. II. 3 divides x + y or 3 divides x - y.

(a) l and II only (c) II only

W4

the conversion of L-alanine to a racemic mixture of D and L-alanine? (a) Pyridoxal-6-phosphatc (b) Thiamine pyrophosphate (c) Coenzyme-A (d)Flavin adenine dinucleotidc

78. The cyclic electron flow during photosynthesis generates (a) KADPH alone (c) ATP alone

(b) ATP and NADPH (d)ATP, NADPH and 02

79. Match the type of cells given in Column I with

organisms given in Column II. Choose the appropriate combination from the options below. A. B. C.

Column I Flame cells Collar cells Stinging cells

(a) A-3, B-1 , C-2 (c) A- 1 , B -2, C - 3

1. 2. 3.

Column II Sponges Hydra Planaria

(b) A-3, B-2, C-1 (d)A-2, B-3, C-1

80. Compared to the atmospheric air, the alveolar air has (a) more p02 and less pC02 (b) less p02 and more pC02 (c) more p02 and more pC02 (d)less p02 and less pC02

(2 Marks Questions)

82. How many different (mutually non-congruent)

81 . Let x, y, z be positive integers such that HCF (x, y, z) = 1 and x2 + y2 = 2z2 • Which of the following

111. 5 divides z(x2 - /).

M3

77. Which one of the following coenzymes is required for

71. Which one of the following combinations of molecular masses of polypeptides are obtained from purified human lgM when analysed on Sodium Dodecyl Sulphate Polyacrylamide Gel Electrophoresis (SDS-PAGE) under reducing conditions? (a) 55 kDa, 1 5 kDa (b) 70 kDa, 25 kDa, 1 5 kDa (c) 55 kDa, 25 kDa (d) 1 55 kDa

73. Two genetic loci controlling two different traits are

�2

glucose molecule during glycolysis is

W l

70. Antigen-antibody reactions

(b) II and Ill only (d)Ill only

trapeziums can be constructed using four distinct side lengths from the set {I, 2, 3, 4, 5, 6}? (c) 1 5 (a) 5 (b) 1 1 (d) 30

83. A solid hemisphere is mounted on a solid cylinder,

both having equal radii. If the whole solid is to have a fixed surface area and the maximum possible volume, then the ratio of the height of the cylinder to the common radius is (a) 1 :1 (c) 2: 1 (b) 1 :2 (d)J2:1

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KVPY Question Paper 2018 Stream : SB/SX

84. Let ABC be an acute scalene triangle, and O and H

be its circumcentre and orthocentre respectively. Further, let N be the mid-point of OH. The value of

PHYSICS

91 . One end of a rod of length L = 1 m is fixed to a point on the circumference of a wheel of radius R = 1 1 ,Js m. The other end is sliding freely along a straight channel passing through the centre O of the wheel as shown in the figure below.

the vector sum NA + NB + NC is ------>

------>

(a) 0 (zero vector) 1-

------>

(d)

-➔

]:OH 2

(b) HO

(c) -HO

85. The quotient when 1 + x2 + x4 + x6 + . . . + x34 1s divided by 1 + x + x2 + x3 + . . . + ;(" 7 is (a) ;,f7 _ x1-s + ;,f3 _ x11 + . . . + x (b) ;,f 7 + x1s + x1 3 + x11 + . . . + x (c) x1 1 + XIS + x1-s + ;,f4 + . . . + 1 (d) x1 7 _ xis + x1 5 _ x14 + . . . _ 1

86. Let R be the region of the disc x2 + y2

::; 1 in the first quadrant. Then, the area of the largest possible circle contained in R is (a) rr (3 - 2'/2) (b) rr (4 - 3 '/2) (d) rr (W - 2)

(c) __1t_ 6

87. Let R be the set of real numbers and / : R ➔ R be given by f (x) = .Jfxj - log(l + I x i ). We now make the following assertions:

I. There exists a real number A such that f (x) ::; A for all x.

IL There exists a real number B such that f (x) � B for all x. (a) I is true and II is false (b) I is false and II is true (c) I and II both arc true (d) I and II both are false

88. Define g(x) = J� f(x- y)f(y)dy, for all real x, where 3 1, 0 ::; t ::; 1 { /(t) = 0, elsewhere

(a) 50 (c) 5 7

(b) 52 (d) 59

I, - cos- k - 1- - cos(k + 1) 44

k� 0

0

(a) O (c) 2

l xf(x)dx= -1 + -1 f0l (f(x))2 dxis

f0

3

4

(b) 1 (d) infinity

following four reversible processes: Step 1 It is first compressed adiabatically from volume vl to 1 m3 .

Step 2 Then expanded isothermally to volume 10 m3 • Step 3 Then expanded adiabatically to volume V3 . Step 4 Then compressed isothermally to volume V1 .

If the efficiency of the above cycle is 3 I 4, then V1 is (a) 2 m3 (b) 4 m3 (d) 8 m3 (c) 6 m3

93. A neutron star with magnetic moment of magnitude

m is spinning with angular velocity w about its magnetic axis. The electromagnetic power P radiated by it is given by µ�m ywz cu , where µ 0 and c are the permeability and speed of light in free space, respectively. Then, (a) x = l, y = 2, z = 4 and u = - 3 (b) x = l, y = 2, z = 4 and u = 3 (c) x = - l, y = 2, z = 4 and u = - 3 (d) x = - l, y = 2, z = 4 and u = 3 on a horizontal surface as shown in the figure below.

0

is

90. The number of continuous functions f : [O, 1] ➔ R that satisfy

92. One mole of an ideal monoatomic gas undergoes the

94. A solid cube of wood of side 2a and mass M is resting

Then, (a) g(x) is not continuous everywhere (b) g(x) is continuous everywhere but differentiable nowhere (c) g(x) is continuous everywhere and differentiable everywhere except at x = 0, l (d) g(x) is continuous everywhere and differentiable everywhere except at x = 0, l, 2

89. The integer part of the number

The wheel is rotating with a constant angular velocity w about 0. The speed of the sliding end P, when 0 = 60 ° is 2w 2w (c) (a) (d) � (b) � 3 3 J3 J3

The cube is free to rotate about a fixed axis AB. A bullet of mass m ( P2 1 1 . (c) We have, ax + y = 0

6

J

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

x + (a + l O)y = 0 From Eqs. (i) and (ii), we get 1 � - -- ⇒ a2 + 10a - 1 = 0 1 a + 10 -10 ± ✓104 a=- -

:. Two values of a for systems has at least

two distinct solution. 1 2 . (d) We have,

f(x) = �2

⇒

f(x) =

a + b< c + d b+ c< d + e c+ d c> e > b :. Largest value is a and smallest value is b. 1 6. (b) A fair coin is tossed 5 times. Total number of sample space = i5 Probability of head docs not occur two or more time in rows. Possible cases

x - [x]

1 + [x]2

8 ) = sin 30

0 E (10° , 20° ] 1 5. (a) Given,

12, p = ..!:, q = � 6 6 P(r 2 2) = 1- [P(r = 0) + P(r = l) ] 12

1-

- 8 = 38 ⇒ 48 = 60° ⇒ 8 = 15°

=

⇒ P(E2 ) = p2 = 1-

⇒

)3 cose - . .!: sine = sin 30 2 2

⇒

E2 = Twelve fair dice rolled and at least two dice show six.

2

13

2 �

1 + [x]

1 ,0

⇒

+

=

=

m x = n + (- 1r....::_

p

I..!:3

5, p = ..!:, q = � 4 4 2 :. Required probability = 5 C2 (p) if x p

2

p + q+ r =

=

1 0. (a) E1 = Six fair dice are rolled and at least one die shows six P(E1 ) = Pi = 1- (no die show six)

x = me + (-lt �

⇒

1 4. (b) Given, 8sin3 0 - 7sin0 + J3cos8 = 0 ⇒ 2(4sin3 0) - 7sin0 + J3cos8 = 0 2(3sin0 - sin 30) - 7sin0 + J3cos8 = O ⇒ 6sin0 - 2 sin 30 - 7sin0 + J3 cos8 = 0 ⇒ J3 cos8 - sin8 - 2sin 30 = 0

+ )= % 1 n: ( 1 %

IA1 - A2I

=

3x = nn + (-l) n �

3

A2 = Area of circle - A1

5

1 1 Case I All tails = ( - ) 2 t Case II 4 tails and one head

sc.rnr -;

Case III 3 tails and 2 head x T x T x T x

• c2 x

(i J ;

Case IV 2 tails and 3 heads x T x T x 1

+l r

1

:. Required probability =

i5

1 5 6 1 13 - + - + - + -= -

1>

2°

i5

t i'

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64

KVPY Question Paper 2 0 1 8 Stream : SB/SX p 22 'I'o prove q + - < 21 21

1 7. (a) We have, x(0) = l cos401 cos0 and y(0) = l cos 4Bl sin8 8

0

x(El)

1

✓2

✓2

✓2

1

y

✓2 ✓2

21

20 O,_i

1[ =- L

1

0

Pi,

1

.E. = _.!__ [_.!_ + _.!_ + _.!_ . . . _1_]

q+

135° 180 ° 225 ° 1 1 -1

0

1

0

y (B)

9 0°

4 5° 1

20

a2

11:J

1 + --- [Ui_ + 02 + Os , , , a2o l 20 X 2 1

i

i2 + l

l ( . 1)7 + - Li + L21

✓(q - 0,_i )

.!]

+ (/J.i - a2 )

1 9. (c) We have, 2

1,

max{ 3, x , :2 } for x E [ y =x2

f

(·h o ) (.'3. o)

f

20. (a) Given 0,_i P

i + for i

.! i

Clearly, q > 0. 22 - P Let q< 21

( c2

0

zs

⇒ �. ~ ;;,

So, ratio of magnitude of acceleration of electron in nth orbit of hydrogen atom and that of singly ionised helium atom is given by

f

], 2, 3, , , .,20

1 - (0,_i + a2 + 11:J + . . . + a20 ) 20 _.!_ _.!_ _.!_ 1 . . . + __ q= ( + _.!_ + 20 °t a20 ) a2 11:J =

a c :: c �' . (!�l: )

2]

[ x3 ]

=

n;)

where, C1 and C2 are constants. So, acceleration of electron in nth orbit is 2

2 2 . 11./3 l dx+ '" 3dx+ '" x2dx f(x) dx = 2 /3 v3 v3 1 2 112 1 1 1 X 2 11 / [- .!] .. 3 + [3xl[ Fa + 3 X 112 .. 3 / 3J3 14 8 , = - Js + 2 + 3J3 - 3J3 + - - - = 3 3 3 =

2 1 . (a) I n a hydrogen ion, velocity and radius of nth orbit are vn =C1 · ( � ) and r,, = C2 . (

y= 3

f

( o,

2

83 is not a sum of squares of two number. Hence, option (c) is correct.

j

1 , 1 1 1 2 < - + - - + - X 1 9 + - X 20] 2 20 [ 2 5 21 1 1 1 1 22 < - + - (1 + 8 + 1] < - + - < 1< 2 2 21 2 20 22 - p ) qE 21

q, �

J65 = .J64 + 1, it is possible. ffi = .J49 + 25, it is also possible. J9"i = JSl + 16, it is also possible.

f(x)

k ⇒ Electric field, E oc _.!_ ⇒E = q r3 r3

i J

.! + _.!_ .! + � __ i + _.!_ + � 2 20 [ 2 i = 2 i 2 + 1 2 1 i = 1 i 1 20 1 1 1 2 20 < - + - [- + - L l + - L l 2 1i = l ] 2 20 2 5i = 2

AB = .jsum of squares of two number

=

r

_.!_ .! 2o x 2 1 _.!_l i = + � -- + + � i = 1 21 20 [ 2 , = 2 i + l 2

Y' 1 8. (c) Given, A = (0,_i , a2 ), B = (q , /J.i), where O,_i , a2 , arc integers. Distance between AB 2

F oc _.!_3

23. (d) As Coulomb's force,

a20

=

=

Clearly, an orange object must absorbs wavelengths shorter than 550 nm.

where, Z ⇒ ⇒

ZH

::.

aH _ aHe = =

(

:;

)

�• (� )

_ -

zt zt.

atomic number. l and ZHe = 2

= (½) =¾

22. (b) Carrot appears orange in white

light because it reflects orange portion and absorbs remaining part of spectrum. Visible part of spectrum is Violet - 450 nm This part is Blue l Green absorbed Yellow - 550 nm This part makes { Orange orange colour Red - 750 nm

Gaussian surface Now, for Gaussian surface (r < R). Electric flux linked with surface is kq cjl = J 3 · 211rdr r = 2 1! kqf �;- = 211kq ( - � ) As flux is non-zero, charge density is also non-zero. Also, by symmetry of charge distribution electric field is zero.

24. (c) Let resistivity depends on given

fundamental constants. p = h a m�c"ed f.b where, k = a numeric constant. Now, substituting dimensions of different physical constants, we have [ML3 r3 A-2 ] = k [(ML2r1 ]a [Mf [Lr1 ]" [AT]d [M-1 L -3 T 4A2 ]f ] Equating dimensions, we have

l = a + b- f 3 = 2a+ c - 3( - 3 = - a - c + d + 4( - 2 = d + 2f Solving these, we get

a=2 b=-l c=-1

d = -2 f=0 So, resistivity can be expressed as p = k( ) � m,ce

25. (c) Soap solution lowers the surface

tension of water. So, water surface is punctured and is pulled back near circumference of vessel due to surface tension.

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65

KVPY Question Paper 2018 Stream : SB/SX 0.004 1 .::,- + --2"­ So, deviation is given by O. 4 0 = - :: = �� ( 15 + �f )

26 • (c) G1. ven, µ

⇒

o=

0.008

=

⇒0 =

__!_

If A1 < A 2, then o()..1 ) > o(A 2 ). om ("-1 ) > 0m ("- z ) ⇒ /1.3

3 '),,,

Deviation produced by a thin prism is o = (µ - l)A 004 = (15 + 0. /1.2 - l) A

Alternative method

⇒

o = ( 0.5 +

4

O.�f

)A

Clearly, o is less for larger A.

:. When 'A,1 < 'A, 2,

⇒

27. (a)

Unpolarised beam

....______,__ ___ i0 =20 wm-2

o(A1 ) > o(A 2) om (A1 ) > om (A 2 ) , 30'

As beam incident over first polaroid is unpolarised, its intensity is reduced to half (Malus' law is not applicable). So, intensity 11 after first polaroid is I 20 = I1 = o = 10 wm-2 2 2 As light from first polaroid is polarised, Malus' law is now applicable. So, intensity 12 obtained after second polaroid is 12 = 11 · cos2 8 where, 8 = angle between transmission axis of first and second polaroid. So, 12 = 10 x cos2 30° 2 = 10 x ( � ) = 7.5 wm -2

28. (b) de-Broglie wavelength of electron

initially, Finally,

mu

A. = !!. = .!!_ p

'}.., = ....!!_

"3

So, final velocity is v' = 3v0 • Acceleration of electron due to electric field is in y-direction, its magnitude is eE0 :J a = -m Final velocity v' of electron after time t is

v' = u + at � eE0 t : , V = v0 1 + -- ] m

⇒

Magnitude of final velocity is I V'I

_

2 Vo + ( -

-

l v'I = 3u0 As We have, ⇒

eE0 t m

9u5 = u5 + ( e!o .

-)

2

tr

t=

t = _2_

⇒

Ea

29. (*) When bird sits over a high tension wire, its both feet are 2 ! i02 x cos (02- 01 )

lS

_l:.._ h l mv0 = '},,_ / 3 h l mv'

⇒

h

mv0

mu'

Angular wave number, lOOO rc = 10 1t (m-1 ) h=�= 100 V So, wave equation is y = a sin (1000 rc · t- 10 rc · x) Given at t = 0 at x = 0.25 m and y = 0.02 m. So, 0.02 = a sin (- 10 rc x 0.25) ⇒ ⇒

30. (a) As no field exists inside the shell body or in the cavity, correct charge distribution is as shown below in the figure, + +

+

>

+

+

+ + +

> E =O

+

+

+ + + +

31 . (b) Equation for a travelling wave is y = a sin(wt - hx + 0 ) = As at t 0 at x = 0 and y = 0, ⇒ o = 0 So, equation of wave over string is y = a sin (wt - hx) Herc, w = 2nf = 2n x 500 = 1ooo rr ( r: ) d

- a sin rn rr )

=

- a sin (2 rc + %)

[:, sin(- 0) = - sine]

0.02 = - a sin (-i ) or a = - 0.02

Herc note that a is amplitude and its positive and negative values are same. When we arc getting a negative value this means particle is displaced below mean position. So, we have y = - 0.02 sin (1000 n t- l0rrx) Now, at t = 5 x 10-4 s and x = 0.2 m, value of displacement of particle is y = - 0.02 sin (1000rc X 5 X 1 0-4 - 10 TC X 0.2) =

=

( As AV � 0, both feels are placed nearby over wire)

nearly at same potential. So, no current flows through bird's body.

0.02

=

- 0.02 sin (

%-

2rr )

0.02 sin ( 2rr - % )

[:. sin(% - 2rr) = - sin( 2rc

- %}

Also, sin(21t - 8) = - sin 8] - 0.02 m

32. (b) Entropy change for a system or body is =

Now, for given system to be in steady state, heat lost by resorvoir at temperature T1 = heat gained by rcsorvoir at temperature T2 ( = Q say). So, change in entropy for heat conduction process is (+ Q ) AS = Q + Tl T2 ⇒

AS = Q

(__!_ -__!_) Tz

Ti

So, time rate of change of entropy is

dS _ d f dt dt =(

(1 _

lQ T

2

)1

1 T1 J

;2 -�} ( �; )

)•T.

T _ dS =( 1 1 dt T2

1

( dQ ) dt

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66

KVPY Question Paper 2018 Stream : SB/SX ⇒

(171i.- m2) g = a {� + (171i. + � )}

⇒

a=

So, path of charged particle will be as shown below in the figure,

- )g 171i. � -+ (rr!i_ + m2) 2 (

Screen

M

Substituting for a in Eq. (iv), we have M(171i. - � )g TI - T2 M + 2(171i. + rr½ ) T1 - T2 -

⇒

Also, graph is asymmetrical. So, correct option is (b). 33. (a) Time period of a simple pendulum depends on its amplitude as, T=

21!H -( �!) 1+

where, 00 = angular amplitude from mean position. Clearly, T increases with 0 and its relation is parabolic. Its variation is as shown below in the figure. T

As M ➔ oo, T1 - 1'2 = constant.

Also, when M increases from 0, T1 - T2 increases. So, correct graph is option (c). T 2 "' R3

35. (d) From Kepler's law, we have

r� r (�J

So, for satellites Si_ and S2 is ⇒

=

m 1g

m29 111i_g- T1 = 111i_a

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

... (iii)

and

a)

From Eq. (iii), we have MR 2 Tl- T2 = (-- 2 R

1 X R

⇒

... (iv)

So, T1- T2 increases as mass M of pulley increases. Now, substituting for T1 and T2 from Eqs. (i) and (ii)in Eq. (iii), we get 111i_ (g- a)- m (g+ a)= Ma 2

-

2

3

Radius of orbit of satellite S2 around planet is R2

0 34. (c) As pulley is massive, tensions on both sides of pulley are not same. From free body diagram, we have

(111i_ - m2) g 2 1+ - (111i_ + m2) M

x H; )3 = (Tlr,2 I

=(

24 x 24 x (3 x 1 04 )3 3x 3

)s 1

4

= 4 x 3 x 1 0 km = 1 2 x 1 04 km

When satellites are closest to each other, orbital speed of S2 as observed from S1 is Vrelative = ½. + V2 = Ri_(1)1 + R2W2 211 211 = Ri x+ R2 xT1 T2

211 211 = 3 X 1 04 X - + 1 2 X 1 04 X 3 24 = 211 X 1 04 + 11 X 1 04 = 311 x 1 04 kmh- 1

36 . (b) Trajectory of charged particle in rq,ri.on of perpendicular magnetic field to its velocity is a circle.

:::t,� Iv

Clearly, charged particle will hit the screen, if R > w. mu mv2 ⇒ - > w [ from, Bq v = --) Bq R . J 2mK ⇒ --- > OJ Bq [:. momentum, p = mv = .J2mK] .J2mq V > OJ ⇒ �-Bq 2mV or > w2 B2q

37. (d)

(n x F) x n =- n x (n x F)

[:.A x B = -B x A] F)- F =:. Vector triple product is defined as, A x (B x C) = B (A- C) - C (A B) = Fx F) x = FF) So, ⇒ F= F) + x x So, G = (n X F)X n 38. (a) Energy of particle is 1 1 -mw2r 2 = - mv2 2 2 where, v = velocity of particle around the path. ⇒ v = rw Now, angular momentum of particle will be L = mur = mr2ro

{n (n .

(n . n)}

n (n - F) (n n n (n · n (n (n F) n

By Bohr's model, we have nh L= 211 nh nh 2 ⇒ ⇒ r2 = -mr w = 211 211mw h x ✓n or r=✓ 2nmw ⇒

r·:

r = a✓n

v�

given, � =

al

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67

KVPY Question Paper 2018 Stream : SB/SX 39. (b) Rate of heat loss for two bodies of

same specific heat's and same density for which temperature difference (of bodies and surroundings) is same is given by t,.T A = k• tJ.t V where, tJ.T = temperature change (T1 - T;), M = time interval, k = a constant depending on shape of bodies, A = surface area and V = volume. As in given case, bottles are identical and they cool down by same temperatures, M=V A A.h or M = h M= So,

A

t tA hA = ⇒ A = hA tB 2hA tB hB tB = 2tA

⇒

40. (a) Using, vrm, = and We have, N

=

P_ __ 2m · vrm,

[:. hB = 2hA]

✓

p X ✓m _ (p / 2) 2m .J3kT .J3mkT

101 X 10 5 2 X )(3 X 5 X 1 0-2 7 X 14 X 10-23 X 293) = 6.4 X 1027

� octt ,

N02

ln the above given reaction acetylated aniline is nitrated to get mononitro substituted product. In case of electrophilic substitution of aniline the reactivity is very high. Substitution tends to occur at all ortho and para-position. In order to get monosubstituted product the NH2 group of aniline is protected by acctylation with acetic anhydride, then carrying out the nitration. NH2 NHCOCH 3

© ";;,�" ©J �coc:�•�::;

N0 2

HaC

OH

O

� NH2 Threonine

0

OH

iH,

�OH

HO)l)

Tyrosine

CP 0

0 (III) lOn-electrons 61t-electrons 4n-electrons It has It has sp3 All C-atoms has sp2 (4n+2) n-electrons hybridised hybridisation :. It is aromatic. atom with ( 4n+2) n-electrons with electrons :. It is C-atom :. Jt is non anti-aromatic. aromatic. (I)

L

Tryptophan

H1

ffl Grignard \.a reagent Formaldehyde H I H O+ H - C - CH.3 � CH2 -CH3 I I - OMgBr OH Ethanol

47. (b) Oxidation-reduction reactions are those reactions in which there is a change in oxidation state of any clement. 0

-2

+l -1

(a) H 2 + Br2 - 2 H Br + l -1

+l

-1

+l

-1

+l -1

(b) Na Cl + Ag NO3 ------> Na NQ 3 + Ag Cl +2

-6

(c) 2Na 2 82 03 + -2

+2 -2

½+ l -1

+2

+12

+l -1

Na 2 84 06 + 2NaI +l-2 -1

(d) Cl 2 + H2 O - HCl+ H O Cl As there is no change in oxidation state in reaction given in (b). Thus, it is not an oxidation- reduction reaction.

48. (a) Thermal stability of alkaline earth

Due to this effect, the electron density increases more at ortho and para-positions than at meta- position.

45. (b) One of the criteria to know anti-aromaticity is the molecule is that it should have 4nn:-clectrons and all the carbon atoms should have sp2 • hybridisation.

metal carbonates increases down the group. This is because, down the group the size of metal increases, thus the carbonates require more heating to decompose. Thus, the order of thermal stability is BaCO3 > SrCO3 > CaCO3 > Mg CO3

49. (d) When a mixture of diborane and

ammonia is heated, the final product is borazinc, which is also known as inorganic benzene.

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68

KVPY Question Paper 2018 Stream : SB/SX

Most of its chemical properties is similar to that of benzene.

Borazine

H

(B3N3Hs)

50. (c) The order of reducing agent can be

determined with the help of electrochemical series. l t is a series of electrodes or half-cells arranged in order of their decreasing standard reduction potential. As the reduction potential decreases the strength of reducing agent also decreases. According to this series, the order of increasing reducing agent, Fe< Cu< Ki< Zn. Thus, among the given metals, Zn is the strongest reducing agent. 5 1 . (d) The moelcule, which is not hydrolysed by water at 25°C is SF6 • This is because it is kinetically stabilised, attack at S is impossible due to steric Hindrance and also because of the rigidity of F- SF angles. The hydrolysed product of other molecules are as follows : AlC13 + Hp- Al(OH)3 + HCl SiC1 4 + 2H20 - 4HC1+ SiO 2 B.l: X

�-

Y'

g(x) =

0

cos(k + 1) 0 sin k 0 ] cosk 0 cos(k + 1)0

g(x - y)f (y)dy

⇒ ⇒

g(x) = J> 2

Now, when angle e = 60°, then A L

l 44 S = -0 I, ( tan(k + 1)° - tan k 0 ) . sm l k = O

1 ⇒ S = --[tan 1° - tan 0° + tan 2 ° sin 1° - tan 1° + . . . + tan 45° - tan 44° ] l_ ⇒ S=_ tan 45° sin l° l l ⇒ S = __0 = __ = 5 7.29 sin l 0.0174 [·: sin 1° = 0.01 7] [SJ = [57.29] = 57 90. (b) We have, 1 xf (x)dx = � + � f\f(x))2 dx o 3 4 Jo

J

�[J1

[(f (x)) 2 - 4xf(x)] dx -� = 3 4 O

= 1[J�(f(x)) 2 - 4xf(x) + (2x)2 - (2x) 2 ] dx

�[J1 ff (x)- 2x] dx - J1 4x dx] 2

0

2

µ:_::....Jj_-

⇒ -� = 3

L'.A = 90°, L = 1 m and R = � m

./3

So, x = ✓L2 + R2 = ✓1 + � = 2 m 3 ./3 Substituting values in Eq. (i), we get u=

⇒

0

2

1 J rt(x) - 2x]2 dx = 0 0

⇒

:. Only one continuous function. 91 . (a) Let rod OA makes angle e at some instant t with OP as shown below in the figure.

r- 1

Clearly, g(x) is continues at for all x 0, x ::; 0 1, O < x < l g ' (x) = -1, 1 ::; x ::; 2 0, x> 2 g(x) is not differentiable at 0, 1, 2 .

Then, by applying cosine formula to L'lOAP, we get R 2 + x2 - L2 cose = --2Rx 2 2 = R + x - L2 - 2Rx cos e Differentiating above equation with respect to time t, we get ⇒

dx

2

dx = 2R ( x !!._cos e + cos e ) dt dt dt

(

Js Js x

x sin 60° } w

(2 - � cos60 ) Js Js x °

u=[

if

w⇒u=�w 1 ] 3

2

./3 - 2-Js

92. (d) Given p-V cycle is

0

J1 lf (x)- 2x] dx- �3

0"___co,,p +----'3"'-

In t,OAP, we have

p

T2

B

C � T1 A D -+---------i v

1 1 4x3 4 ⇒-- = J [f(x) - 2x] 2 dx- -] 0 3 l 3 0

I g(x) = 0 + J/(x - y)dy + 0

⇒

1 S = -sin 1° sin(k + 1) 0 cosk 0 cosk 0 cos(k + 1)0

dx dx . de + cose⇒ 2x - = 2R ( - xsme) dt dt dt ⇒ xv = - xRw sine + R u cose de dx - = w and - = u dt dt R xw · sine ⇒ U=. . . (i) X - R eas e

I

+ J) = { 0, elsewhere

g(x) =

I, [

⇒

Clearly from graph: I is false and I I is true. 88. (d) We have,

⇒

89. (c) Let S =

71

AB:Adiabatic compression, VA = Vi , VB = 1 m3 BC : Isothermal expansion, Ve = V = 10 m3 CD : Adiabatic expansion, VD

=

V3

DA : isothermal compression, VA = V1 Cycle efficiency is given, TJ = � 4 For given Carnot's cycle, -l Tj = l- Tl = l- V2 -y ) ( T2 V1

⇒

1

⇒

vi

[:. Process AB is adiabatic, 5 y = - ,.1or monoatom1c gas] 3

= l=

5

�J

3

(

3

8m

-1

⇒

v1�3

=

1

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72

KVPY Question Paper 2018 Stream : SB/SX

93. (a) Given, power radiated P is y z u p = µ �m O) C

Substituting dimensions of different physical quantities involved, we have [ML2 r3 ] = [MLr2A-2Y [L2AY 1r 1 ]2 [Lr 1 t Equating powers of fundamental quantities, we have . . . (i) x=l . . . (ii) X + 2y + ll = 2 . . . (iii) - 2x - z - u = - 3 . . . (iv) - 2x+ y = O From Eq. (i), putting the value of x in Eq. (iv), we get ⇒ -2 X 1 + y = 0 ⇒ y = 2 . . . (v) Now, from Eqs. (i) and (v), putting the values of x and y in Eq. (ii), we get ⇒ . . . (vi) 1+ 2 x 2 + u = 2 ⇒ u = -3 Now, again from Eqs. (i) and (vi), putting the values of x and u in Eq. (iii), we get ⇒ -2 x l - z + 3 = -3 ⇒ z = 4

94. (d)

So, :x: = 1, y = 2,z = 4, u = - 3.

V

m

3

4a

--- M

(J)

A

= '!:. Ma2 + M (./2a) 2 = !}_ Ma2 3 3 Applying angular momentum conservation about A, we have w L; = Lr ⇒ mvr = ]A

mvr W = --

where, m = mass of bullet and ro = initial angular speed of block. Substituting values in Eq. (i), we have

mv (1) ⇒ w = = 2

95. (a) Process equation is 5 = C1 p V 3 - eli:; p ( Vl !}_ Ma

dV p dV 3V E0 ⇒ dp + � dV = __.!__ (pdV + Vdp) p

⇒

mu 2Ma

3 V

As,

-dV = dp

⇒

-

For n = 1,

r=

(½ - f) ( 3�)

97. (b) Opposing force on loop is F = Bla where, I = induced current in loop.

=[ ir-!o

Also, I =

_.!_ _ dV __ V dp

2 2 -B 2a2 F = - -B--a - v ⇒ mA = - ·v

R

( ¾.

-

du B 2a2v dt R du dx B 2a 2v m.- . - = --dx dt R

m - = ---

du =

⇒

¾)

main scale is as shown below in the figure. 3 MSD X 3

- -,.- ?

⇒

mv 2r = �ke2

(2r)

21t

5

11

2 . . . (i)

From Bohr's postulate, we have

nh mvr = -

- y- 4

, Main Scale I I I I I I I I � I IIIIIIII Vernier Scale II I I I 10 O 3 VSD

So, net centripetal force on a rotating electron is

k e2

d:]

d

98. (*) Given, position of Vernier and

e ________ ...,.________ e

k e2

[:. v =

where, C = constant of integration. :. v versus x is a straight line with negative slope. Also, emf is induced only when the loop is going in or emerging out of the region of magnetic field. So, correct graph is (b) .

96. (*) Given situation is

_- _

�

Rm

l As T ➔ oo, __ ➔ O RT So, k approaches a constant value at higher temperatures.

mu 2

- B 2a 2 dx

Integrating above equation, we have - B 2a2x ⇒ V = --- + C

P ; - 3;1' ) ( 0

F-

R

where, A = acceleration of loop and m = mass of loop.

3:T )

V dp

-

-Bau

So, opposing force is

(� - ¾) -

I=

⇒

3RT

P ( ;o

E

R

where, E = induced emf.

_!!__ - ap l E0

41t£o h 2 = _1_ 41t£o h 2 = _l_ aB ( ) 3 3 me2 3 me2

(No option is matching)

pV = RT

dV dp

⇒

E0

;o

. . . (i)

JA

w

(

Differentiating with respect to volume, we have _!_ . dp + � = O + __!_ P + Vdp )

But this is compressibility of gas, so compressibility, k = _ _.!_ dV

As there is no external torque on the system of bullet and block, angular momentum about A remains constant. Now, moment of inertia of block about A using parallel axes theorem is IA = IcM + Mh 2

⇒

Squaring Eq. (ii) and dividing by Eq. (i), we have

Taking log to e base, we have pV log(p) + � log(V) = log(C1 ) + 3 E0

⇒

f--2a -----► Centre of "-, C mass ',�2a

pV 5 pV3 = C1 eEo

⇒

. . . (ii)

From above figure, x = 3MSD - 3VSD = 3 x lMSD - 3 x 1 VSD

= ( 3 - 3 x � ) mm 1 = 3 - 0.3 mm = 2 . 7 mm = 0.2 7 cm (No option is matching)

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73

KVPY Question Paper 2018 Stream : SB/SX 99. (a) Due to standing waves, density of

water column is more at nodes and less at antinodes. So, optical path length of light beams is different. This causes interference in light beams when they overlaps over screen. Hence, correct graph of intensity variation on screen is of (a). 1 00. (c) From given graph, time period of rotation of planet around star is obtained 3 days. 80 60 � 40 :;; 20 ·o 0 >20 2 -40 (/) -60 -80

formed. This addition reaction is in accordance to anti- Markownikoffs rule. Cr03 in the presence of H2SO 4 is known as Jones reagent is selective oxidising agent. It converts primary alcohol to carboxylic and secondary alcohol to kctoncs.

1 02. (a) O

A

CH3

days

Now, we use T2 =

41t

2

GM

5

4

3

. r3

. . . (i)

If we take time period in years and 41t 2 radius in AU, then =1

a

CH3

0

0

H OH

ACH(iJ N•BH') A CH3

(i)NaBD4

CH

x

H

CH

CHa

OD

(iiJ o,o•

H OD

A

CH3 +

(ii) H3o

CH

CHa

D OH

ACHa

H3C

y

In case I, KaBH4 is used as a reducing agent, where H- gets transferred to kctonic group and alcohol is formed. In step (ii), H of the OH group gets substituted by D+ , when hydrolysis occurs in the presence of Pi o+ .

In case II, NaBD4 is used as a reducing agent where gets transferred to ketonic group and alcohol is formed . In step (ii), D of OD group gets substituted by H+ , when hydrolysis + occurs in the presence of Ha o .

v-

1 03. (c) The hydrogenation products of given alkenes are as follows

GM

As for earth, r = lAU, T = lyr 2

T2 = ±::...__ r3 GM

So,

⇒

�

III

( 3 ); ( 8 ); = ( 64)6

H2/Ni

�

Optically active

113

365

1000

10

= _.i._ = 0.04AU 102

Thus, alkene I and I l l generate optically active compounds upon hydrogenation.

1 0 1 . (b)

Y (i B2H6 ( �OH ) � (ii) H202/NaOf1 �

X

Cr03/H2S0,

rY �

3d

state

I D

II

□

II

4s

4p

As Cl- is a weak field ligand, pairing of electron will not occur. [NiCl4] 2- = I 1l I 1l I 1l I 1 I 1 I 3d

4s

si{- hybridisa.tion

As the hybridisation of Ni in [NiC1 4 ] 2- is sp3 • Thus, the geometry will be tetrahedral. It is paramagnetic in nature as there are 2 unpaired electrons.

1 06. (c) Optically active complexes arc those complexes in which the mirror images arc non-supcrimposablc on each other. Complexes of type [M(AA)3 1 cis [M(AA)2B2] arc optically active. The type of complexes given in the options are as follows :

f

Types of complexes [M(AAh]

Complexes [Cr(en) 8

trans-[M(AA) i Cl 2 ] cis-[M (AA )2 Cl 2 ] [MA.4B2 l

trans-[Cr (en ) i CI 2 i + cis-[Cr(en) 2 CI 2 i + [Co(NH 3 ) 4 Cl 2 t

Thus, among the given options [Cr(en)3 ] 3+ and cis [Cr(enbC1 2 are optically active.

t

active

r = T 21 3

z

III

Ni2+ground 1l 1l 1l 1 1 1 1

Optically

12 41t 2 -=-=1 GM 1 3 Now, in given question is 3 1, = 3 days = -- yr 365 So, from Eq. (i),

=

1 05. (a) The oxidation state of Ni in [NiC1 4 ]2- is +2. The electronic configuration of Ni 2+ is [Ar]3d 8 4s0 .

COOH

y Whenever B2H6 adds in the presence of peroxide to an alkene, an alcohol is

1 04. (c) When heated in air, brown copper turns black. This black powder would turn brown again when heated with hydrogen (H2) . This is because here, H2 will act as air reducing agent that reduces CuO into Cu, whereas air (02 ) will act as an oxidising agent. Cu + 0..,9 � CuO

Brown

Black

Black

CuO + H2 � Cu + H20

[Cr(enl 3] 3+

1 07. (d)

22 7

k

Ac

z21 Th 223

Fr As radioactive compounds are first order reactions 0.693 _ 0.693 tu2 0f AC _ - -- Or k, - -22 k % of Th .5._ = __!_ k-i_, 100

k % of Fr = 2 = � kzc 100

hi. = 2 98 k

Also,

2

k = hi_ + k2 0.693 98 = hi_ + hi_ 2 22 R-i_ = 6.3 X 1 0 -4

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KVPY Question Paper 2018 Stream : SB/SX

1 08. (d) For process, A � B

W = -pt.V = - 1(15 - 1) = -0.5 L atm t. U = Cv t.T t. U = 0 (isothermal process) :. From first law of thermodynamics, t. u = W1 + qi 0 = -0.5+ q q1 = 0.5 L atm

For porccss, B � C W = 1 L atm (given) t. U = 0 (isothermal process) !!. U = W2 + q2 0 = 1+ q2 q2 = -lL atm For process, A- C I!. U = 0 (isothermal process) W3 = -pt. V = 0 (volume is constant) % =0 :. Heat exchanged in entire reaction, q = ql + q2 + q3 q = + 0.5 - l+ 0 = -0.5 L atm Here, - ve sign represents that heat is released.

1 09. (b) According to Raoult's law,

P = P?X 1 0 = Pr Pf X i + P X 2 = P?X 1 + Pg (l- X 1 ) P?X 1 + Pg - PgX 1 (pf - pg ) X 1 + pf

⇒

PT

= P toluene + P benzene

1.013 = (0.74 2- 18) x toluene + 18 1.01 3 = -l.058 X X toluene + 1.8 -0. 787 = -1. 058 X X toluene X toluene= 0.744

1 1 0. (a) Represents a unit cell

Number of Y- atoms in a unit cell = 8 Thus, the formula for the compound based on the unit cell from pattern is XY8 .

1 1 1 . (b) The genetic distance between genes A and B is 10 cm. It means, the recombination frequency is 10%. If the cross is made between the two parents with genotype Ab and aB, out of their progeny, 10% will be recombinant type (that is AB and ab) and 90% will be parental type (that is aB and Ab). AB allele represents half of the recombination type, that is 5%. Thus, the correct answer is option (b). 1 1 2. (a) A cell membrane is a double layer of lipids and proteins that surrounds a cell, and separates the cytoplasm from its surrounding environment. It consists of various proteins embedded in this lipid bilayer i.e. receptor proteins, integral proteins and peripheral proteins. Out of these peripheral membrane proteins are temporarily attached either to the lipid bilayer or to integral proteins by a combination hydrophobic, electrostatic and other non-covalent interactions. Thus on treating the cell membrane with high ionic strength buffer these can be easily removed. Thus we can conclude that in the experiment P and Q arc peripheral membrane proteins. 1 1 3. (d) Photosystem - 11 is the first membrane protein complex in oxygenic photosynthetic organisms in nature. lt produccss atmospheric oxygen to catalyse the photooxidation of water by using light energy. Two molecules of water into one molecule of molecular oxygen occurs as Light

2H 2 0 --- 4H+ + 4e- + 02 Chlorophyll

i

1 1 4. (a) Proteins arc among the most important chemical to all life on the planet. Each protein is made up of many of the 20 different amino acids. So, the number of proteins which can be formed these different 20 amino acids if each protein contains 100 amino acids = 20th 10° .

The whole solid is a matrix of 9 x 9, so unit cell will be matrix of 3 x 3. Thus, from above figure, it is can be seen that a unit cell have 8 white square and 1 black square. :. Number of X-atoms in a unit cell = 1

1 1 5. (d) 660 g/mol is the average molecular weight of nucleotide pair. Each nucleotide pair = 0. 34 nm of the length of DNA. The length of E.coli DNA molecule will be 0.34x 3.lx 109 660

= l.6 x 106 nm = 1.6mm

1 1 6. (d) Emigration refers to the number of individuals of the population who left the habitat and gone elsewhere during a given time period. Emigration thus leads to decrease in the number of individuals in a habitat. 1 1 7. (b) Cysteine is a sulphur containing amino acid. lt is unstable in the air. In proteins it usually exists as cystine by forming a disulphide (S-S) bond between two cysteine residue. Zinc ions have the ability to be chelated to cysteine residues within protein scaffolds. Thus, option (b) is correct. 1 1 8. (c) The minimum number of plants to be screened to obtain a plant of the genotype AabbCcDd is 32. Because AaBbCcDd xAABbCCDd AabbCcDd 1 1 1 1 1 - + - + - +- = 2 4 2 2 32 1 1 9. (d) A cross between red flower producing plant and white flower producing plant is shown below RR

rr

X

l

(red flower)

(white

flower)

Rr (pink

F1-generation

flower)

When F;_ to I > A 1 and d > > Au. Then, the A 1 / Au 1s close to (a) 0.25 (b)0.5 (c)0. 7 (d) 1 .0

The bar is given an initial velocity v0 towards the right at t = 0. X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

Rx

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

X

� X

X

X

X

X

X

X

X

X

X

X

X

X

Then, the (a) induced current in the circuit is in the clockwise direction (b) velocity of the bar decreases linearly with time (c) distance the bar travels before it comes to a complete stop is proportional to R (d) power generated across the resistance is proportional to l

27. A particle with total mechanical energy, which is

small and negative. Its under the influence of a one dimensional potential U(x) = x4/4- x2/2 J, where x is (in metre). At time t = 0 s, it is at x = - 0.5 m. Then, at a later time it can be found (a) anywhere on the X-axis (b) between x = - 1 .0 m to x = 1 .0 m (c) between x = - 1 .0 m to x = 0.0 m (d)between x = 0.0 m to x = 1 .0 m

28. A nurse measures the blood pressure of a seated

patient to be 190 mm of Hg. (a) The blood pressure at the patient's feet is less than 1 90 mm of Hg (b) The actual pressure is about 0.25 times of atmospheric pressure (c) The blood pressure at the patient's neck is more than 1 90 mm of Hg (d) The actual pressure is about 1 .25 times of atmospheric pressure

29. A particle at a distance of 1 m from the origin starts

moving, such that dr I d0 = r, where r and 0 are polar co-ordinates. Then, the angle between resultant velocity and tangential velocity is (a) 30 ° (b) 45 ° (c) 60° (d) dependent on where the particle is

30. Electrons accelerated from rest by an electrostatic

potential are collimated and sent through a Young's double slit experiment. The fringe width is m. If the accelerating potential is doubled, then the width is now close to (a) 0.5 w (b) 0.7w (c) l.0 w (d)2.0 w

31. A metallic sphere is kept in between two oppositely

charged plates. The most appropriate representation of the field lines is

26. A conducting bar of mass m and length l moves on two frictionless parallel rails in the presence of a constant uniform magnetic field of magnitude B directed into the page as shown in the figure.

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KVPY Question Paper 2018 Stream : SB/SX

32. An electron with kinetic energy E collides with a

hydrogen atom in the ground state. The collision will be elastic (a) for all values of E (b) for E < 1 0.2 eV (c)for 1 0.2 eV< E < 1 3.6 eV only (d)for 0 < E < 3.4 eV only

33. The continuous part of X-ray spectrum is a result of the (a) photoelectric effect (c)compton effect

(b)raman effect (d) inverse photoelectric effect

34. Thermal expansion of a solid is due to the

(a) symmetric characteristic of the inter atomic potential energy curve of the solid (b) asymmetric characteristic of the inter atomic potential energy curve of the solid (c) double well nature of the inter atomic potential energy curve of the solid (d)rotational motion of the atoms of the solid

35. An electron and a photon have same wavelength of

10-9 m. If E is the energy of the photon and p is the momentum of the electron, then the magnitude of E I p (in SI unit) is (a) 1 .00 X 1 0-9 (b)l.50x 1 08 8 (c)3.00x 1 0 (d) 1 .20 X 1 07

36. If one takes into account finite mass of the proton, then the correction to the binding energy of the hydrogen atom is approximately (take, mass of proton = 1.60 x 10-27 kg and mass of electron = 9.10 X 10-31 kg) (a) 0.06%

(b)0.0006% (c)0.02%

(d)0.00%

37. A monochromatic light source S of wavelength

440 nm is placed slightly above a plane mirror M as shown below. Image of S in M can be used as a virtual source to produce interference fringes on the screen. The distance of source S from O is 20.0 cm and the distance of screen from O is 100.0 cm (figure is not to scale). If the angle 8 = 0.50 x 10-3 radians, then the width of the interference fringes observed on the screen is S � --- ------ _ ___20 . 0 cm

M mmmll�;�>JT/7 0

C Q) () Cl)

(a) 2.20 mm (b) 2.64 mm (c) 1 . 1 0 mm (d) 0.55 mm

38. A nuclear fuel rod generates energy at a rate of

5 x 10 8 W/m3 . It is in the shape of a cylinder of radius 4.0 mm and length 0.20 m. A coolant of specific heat 4 x 103 Jkg-1K-1 flows past it at a rate of 0.2 kgs-1. The temperature rise in this coolant is approximately (d) 30° C (c) 1 2° C (b)6 °C (a) 2° C

39. A hearing test is conducted on an aged person. It is

found that her thresold of hearing is 20 dB at 1 kHz

and it rises linearly with frequency to 60 dB at 9 kHz. The minimum intensity of sound that the person can hear at 5 kHz is (a) 1 0 times than that at 1 kHz (b) 1 00 times than that at 1 kHz (c) 0.5 times than that at 9 kHz (d) 0.05 times than that at 9 kHz

40. Two infinitely long parallel wires carry currents of

magnitude /1 and /2 are at a distance 4 cm apart. The magnitude of the net magnetic field is found to reach a non-zero minimum value between the two wires and 1 cm away from the first wire. The ratio of the two currents and their mutual direction is 1 1 (a) 2 = 9, anti-parallel (b) 2 = 9, parallel 11 11 1 1 (d) 2 = 3, parallel (c) 2 = 3, anti-parallel 11 11

CHEMISTRY 41. The shape of SCl 4 is best described as a (a) square (c) square pyramid

(b) tetrahedron (d)see-saw

42. Among the following atomic orbitial overlaps, the non-bonding overlap is

(a) �

( ) b

(c) �

( ) d

� +

� +

43. Among the following complexes, the one that can

r

t

exhibit optical activity is (b) [Co(en)Cl4 (a) [CoC1 6 f(c) cis- [Co(en)2Cl2 (d) trans- [Co(en)2Cl2

t

44. The pKa of oxoacids of chlorine in water follows the order (a) HClO < HC1O3 < HC1O 2 < HC1O 4 (b) HC1O 4 < HC1O3 < HC1O2 < HClO (c) HC1O 4 < HC1O 2 < HC1O3 < HClO (d)HC1O 2 < HClO < HC1O3 < HC1O 4

45. The packing efficiency of the face centered cubic (fee), body centered cubic (bee) and simple/primitive cubic (pc) lattices follows the order (a) fee > bee > pc (b) bee > fee > pc (c) pc > bee > fee (d)bee > pc > fee

46. The ratio of root mean square velocity of hydrogen at 50 K to that of nitrogen at 500 K is closest to (a) 1 .1 8 (b) 0.85 (c) 0.59 (d) 1 .40

47. The molecule with the highest dipole moment among the following is (a) NHs (b) N1'3

(c) CO

(d) HF

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79

KVPY Question Paper 2018 Stream : SB/SX 48. The most stable Lewis acid-base adduct among the following is (a) H20 ➔ BC13 (c) Ha N ➔ BCl3

(b) H2S ➔ BC13 (d) Ha P ➔ BCl3

49. The reaction of D-glucose with ammoniacal AgNO3 produces

CHO

H --+-- OH (a)

HO --+-- H H-

-+-

53. The compounds containing sp-hybridised carbon

H (b)

- OH

atom are

HO --+-- H H-

-+--

OH

H --+-- OH

H --+-- OH CHO

C02 H C0 2H

H (c)

HO -

H -+-

-H

(d)

H --+-- OH

HO -

OH -+-

-H

H --+-- OH H

-+-

- OH

CHO

50. The reagent(s) used for the conversion of benzene diazonium hydrogensulphate to benzene is/are (b) Ha P02 + H20 (a) H20 (c) H2S0 4 + HzO (d) CuCI/HCI

5 1 . The major product obtained in the reaction of toluene with 1 - bromo-2- methyl propane in the presence of anhyd. AlC½ is

(I)

H3 C

"-... / r CH3

(III) H3 C - CN (a) I and II (c) 11 and lll

CH O

(II)

0

(IV) H2 C-C-CHCH3 (b) III and IV (d) l and lV

54. Upon heating with acidic KMnO4 , an organic

compound produces hexan - 1,6-dioic acid as the major product. The starting compound is (a) benzene (b) cyclohexene (c) 1-mcthylcyclohcxcnc (d) 2-methylcyclohexene

55. It takes 1 h for a first order reaction to go to 50% completion. The total time required for the same reaction to reach 87.5% completion will be (a) l . 7 5 h (6) 6.00 h (c) 3.50 h (d) 3.00 h

56. A unit cell of calcium fluoride has four calcium ions. The number of fluoride ions in the unit cell is (a) 2 (b) 4 (c) 6 (d) 8

reaction at 298 K is 3.8 x 10-3 • The cell potential E 0 0 (in V) and the free energy change D.G (in kJ mol- 1) for this equilibrium, respectively are (a) -0.071, - 1 3.8 (b) -0.071, 13.8 (c) 0.71, - 13.8 (d) 0.0 7 1, - 1 3.8

57. The equilibrium constant of a 2 electron redox

o;

58. The number of stereoisomers possible for the

52. The major product in the following reaction is H

0

OH

(CH,CO>,O

(CH,COOH

following compound is CHa -CH=CH-CH(Br)-CH2 -CHa (b) 3 (a) 2 (c) 4 (d) 8

232

208

90

82

59. In the radioactive disintegration series Th ➔ Pb, involving a and � decay, the total number of a and � particles emitted are (a) 6 a and 6 � (b) 6 a and 4 � (c) 6 a and 5 � (d) 5 a and 6 �

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KVPY Question Paper 2018 Stream : SB/SX

60. In the following compressibility factor (Z) versus pressure graph at 300 K, the compressibility of CH4 at pressure < 200 bar deviates from ideal behaviour because 1 .8

Ideal gas

0.2 0

200

600 1 000 Pressure (bar)

(a)the molar volume of CH4 is less than its molar volume in the ideal state (b)the molar volume of CH4 is same as that in its ideal state (c)inrcrmolccular interactions between CH4 molecules decreases (d)the molar volume of CH4 is more than its molar volume in the ideal state

67. Insectivorous plants such as Venus fly trap catch and

digest insects in order to supplement the deficiency of (b) nitrogen (a) sulphur (c) potassium (d)phosphorus

68. Which of the following statements about nucleosome is true? (a) lt consists of only DNA (b) It is nucleus-like structure found in prokaryotes (c) It consists of DKA and proteins (d) lt consists of only histone proteins

69. Epithelial cells in animals are held by specialised

junctions one of them being gap junction. Function of a 'gap junction' is to (a)facilitate cell-cell communication by rapid transfer of small molecules (b)cement the neighbouring cells (c)stop substances from leaking (d) provide gaps in the tissue to facilitate uptake of small molecules across tissues

70. Which of the following statements is true about glandular epithelium in salivary gland? (a) It consists of isolated single cells (b) It consists of multicellular cluster of cells (c) its secretions are endocrine (d)It consists of squamous epithelial cells

7 1 . Which one of the following ion pairs is involved in

BIOLOGY

61 . Which of the following molecules is a primary

nerve impulse? (a) Ka+ .K+ (b) Na + . c1-

(c) K + .c1-

acceptor of CO2 in photosynthesis? (b) 3-phosphoglycerate (a) Pyruvate (c) Phosphocnol pyruvate (d) Oxaloacctatc

72. Which of the following hormones that controls blood

forms a hydrogen bond between them? (a) Water and water (b)Water and glucose (c)Water and ethanol (d) Water and octane

73. Oxytocin and vasopressin are synthesised in

62. Which one of the following pairs of molecules, never

63. Lactase hydrolyses lactose into (a) glucose + glucose (c)galactosc + galactose

(b) glucose+ galactose (d) galactose + fructose

64. Which of the following statements is incorrect

regarding biological membrane ? (a) lt is composed of lipids and proteins (b) Peripheral proteins arc loosely associated with the membrane (c) Integral proteins span the lipid bilayer (d) Lipids and membrane proteins do not provide structural and functional asymmetry

65. The percentage of sunlight captured by plants is (a) 2- 10% (c)60-80%

(b) 10-20% (d) 100%

66. The hard outer layer of pollens, named exine, is made of (a) cellulose (c)sporopollcnin

(b)tapetum (d) pectin

pressure is secreted by human heart? (a) Erythropoietin (b) Atrial Natriurctic Factor (c) ACTH (d) Glucocorticoid (a) hypothalamus (c) pituitary gland

(b) adrenal gland (d)ovary

74. If you exhale multiple times into a conical flask

containing lime water through a single inlet fixed through a stop cork, lime water will (a) become cooler (b) turn milky (c) remain unchanged (d)turn yellow

75. The path of passage of stimulus when you accidentally touch a hot plate is (a) receptor ➔ brain ➔ muscles (b) muscles ➔ spinal cord ➔ receptor (c) muscles ➔ brain ➔ receptor (d)receptor ➔ spinal cord ➔ muscles

76. In the presence of glucose and lactose, Escherichia

coli utilises glucose. However, lactose also enters the cells because (a) low level of permease is always present in the cell (b) it uses the same transporter as glucose (c) it diffuses through the bacterial cell membrane (d)it is transported through porins

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KVPY Question Paper 2018 Stream : SB/SX

79. If 14C0 2 is added to a suspension of photosynthesising

77. Passive immunisation is achieved by administering (a) heat-killed vaccines (b) toxoids (c) live attenuated vaccines (d)antibodies

chloroplasts, which of the following will be the first compound to be radioactive? (a)ATP (b) NADPH (c)KADH (d)3-phosphoglycerate

78. Which of the following anions neutralises the acidic

pH of the chyme that enters into the duodenum from the stomach? (b) HSOi (a) H2P04(c) HC03 (d)CI-ls coo-

� PART- I I 1 + x2 + x4 + x6 + ... + x22 is divided by 1 + X + :C' + i3 + . . . + x1 1 is

(b) 2 (d) 2(1 + x2+ x4 + ... + x1 ° )

82. The range of the polynomial P(x) = 4i3 - 3x as x varies

(a) [ 1, lj

(b)( 1, lj

(c)( 1, 1)

(-.!) ) 2 2

(d)

83. Ten ants are on the real line. At time t = 0, the kth

ant starts at the point k 2 and travelling at uniform speed, reaches the point (1 1- h) 2 at time t = 1. The number of distinct times at which at least two ants are at the same location is (b) 1 1 (c) 1 7 (d)9 (a) 45

84. A wall is inclined to the floor at an angle of 135° . A

ladder of length l is resting on the wall. As the ladder slides down, its mid-point traces an arc of an ellipse. Then, the area of the ellipse is ', Wall

' '­

,-_' ' -­ -

(a)

rc/2 4

,

', ,

(b) rc/2

%)

(d)2 rc/2

85. Let AB be a sector of a circle with centre O and radius d. LAOB =0 (
0.

81 . The remainder when the polynomial

over the interval (- ½ , ½) is

flower in the world?

(2 Marks Questions)

MATHEMATICS

(a) 0 (c)l+ x2 + x4 + ... + x1°

80. Which of the following species makes the largest true

e)

90. Let a be a fixed non-zero complex number with I a I < 1 and w = (

1 - az 2!:

),

where z is a complex number. Then,

(a) there exists a complex number z with l zl < l such that lwl> 1 (b) I wl > lfor all z such that l zl ---+-->---+---,----->

,,�]\

a

Jd � ' VO,) ue V u ) d

a

(d)

(c)

� ----� v

(d) µ a 21tr l

94. A photon of wavelength A is absorbed by an electron

confined to a box of length .J(35 h),. I Smc). As a result, the electron makes a transition from state k = 1 to the state n. Subsequently the electron transist from the state n to the state m by emitting a photon of wavelength '},_' = 1.75 A. Then, (a) n = 4, m = 2 (b)n = 5, m = 3 (c)n = 6, m = 4 (d) n = 3, m = 1

95. Consider two masses with m1 > m2 connected by a

light inextensible string that passes over a pulley of radius R and moment of inertia I about its axis of rotation. The string does not slip on the pulley and

s

The most appropriate representation of above cycle on a internal energy U - volume V diagram is

u

'' ''

The magnetic field at the point P is l l (b)µ o (a) µ o + � ) (c)zero r 1t 4 1t r

''' d '' ''' '

a ''' '' '' ''

92. Consider a cube of uniform charge density p. The

ratio of electrostatic potential at the centre of the cube to that at one of the corners of the cube is -J2 (d) 1 (a) 2 (b)J3 /2 (c)

c

�----� v

97. The heat capacity of one mole an ideal is found to be Cv = 3 R (1 + aRT) I 2, where a is a constant. The equation obeyed by this gas during a reversible adiabatic expansion is (a) TV 312eaRT = constant (b) TV 312J aRT/ 2 = constant (c) TV 312 = constant (d)TV 312e2aRTl 3 = constant

98. If the input voltage Vi to the circuit below is given by

Vi (t) = A cos(2rr f t) and the output voltage is given by V0 (t) = B cos(2rr f t + q> ).

i

+t -le C

R

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83

KVPY Question Paper 2018 Stream : SB/SX Which one of the following four graphs best depict the variation of versus f?

102. The reagents required for the following two-step

u

transformation are

99. A glass prism has a right-triangular cross-section ABC, with LA = 90 °. A ray of light parallel to the hypotenuse BC and incident on the side AB emerges grazing the side AC. Another ray, again parallel to the hypotenuse BC, incident on the side AC suffers total internal reflection at the side AB. Which one of the following must be true about the refractive index µ of the material of the prism? (a)

R

(c)µ


2x = 3 ⇒ x = �

⇒ x2 - 2x + l + y2 = x2 + y2 - 2y + l x = y = 3l2 ⇒ (x, y) =

/(x) is always positive x E [O, 1] J; t (x)dx
0 be a real number. Then the limit x

3

2

x

. a + a - - (a + a) . hm -----� -1s a3 - x - ax/ 2

x -> 2

(a) 2 log a

4 (b)- - a 3

86. Let f(x) = ax2 - 2 +

(c)

2

a

+a 2

marked I and II are equal. If a , b are the x-coordinates of A, Brespectively, then a + b equals

a3 i3

and let A = a0- a3 + a 6 - ... + a 4026 , + ... + a 4028x B = a1- a 4 + a1 - ...- a 4021 , C = az - a5 + as - · · · + O402s· Then, (a) IA l = I B l > I C I (c)IA I = I C I > I B I

88. The figure shows a portion of the graph y = 2x- 4x'3. The line y = c is such that the areas of the regions

2 (d)- (1- a)

.!.,X where a is a real constant. The

smallest a for which f (x) ;::: 0 for all x > 0 is 24 22 'i" � (b) (c) (d) (a) j j j j

87. Let f : R ➔ R be a continuous function satisfying

J7

x➔

-oo

(c) f(x) has more than one point in common with the X-axis (d)f(x) is an odd function

(d)� J7

chosen so that every pair of elements of A differ by at least 3. (For example, if n = 5, A can be , {2} or {1, 5} among others). When n = 10, let the probability that 1 E A be p and let the probability that 2 E A be q. Then, 1 1 (b) p < q and q- p = (a) p > q and p - q = 6 6 1 1 (d) p < q and q - p = (c) p > q and p- q = 10 10

90. The remainder when the determinant 2014 2014 20152015 2016 2016 2017 2017 2018 2018 2019 2019 20202020 2021 202 1 2022 2022

is divided by 5 is (a) 1 (b) 2

(d)4

(c) 3

PHYSICS 91. A cubical vessel has opaque walls. An observer (dark

circle in the figure below) is located, such that she can see only the wall CD, but not the bottom. Nearly to what height should water be poured, so that she can see an object placed at the bottom at a distance of 10 cm from the corner C? Refractive index of water is 1.33.

•

A LJ O

0

(b) Jim f (x) =- 2

J7

89. Let Xn = {l, 2, 3, ... , n} and let a subset A of Xn be

f (x) + J tf (t) dt + x2 = o,

for all xE R. Then (a) lim f(x) = 2

4 (c) -

2 (a) -

B (a) 1 0 cm

(b) 1 6 cm

C (c) 27 cm

(d)45 cm

92. The moments of inertia of a non-uniform circular disc (of mass M and radius R) about four mutually perpendicular tangents AB, BC, CD, DA are 11 , 12 , 13 and 14, respectively (the square ABCD circumscribes the circle).

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107

KVPY Question Paper 2015 Stream : SB/SX The distance of the centre of mass of the disc from its geometrical centre is given by l .JB = __cl,__ BA

d . rot · 2rra·° - B0 sin dt

=

dt

dt

2B0 rra2 · OJ · cos rot 2 E = 2B0 rra ro . loop, I = Current m · cos rot =

R

R

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115

KVPY Question Paper 2015 Stream : SB/SX So, current oscillates with a frequency ro. Heat loss per unit time = J 2R 4BO2 1t 2a 4ro2 cos2 rot · R

R2

38. (a) ln an n-sided polygon magnetic field due to one of the side at p

4B02 1t 2a 4ro2 · cos2 rot

:. Heat loss oc a

i

4

Force on a small segment dl of ring is 2B ita2ro F = Bidl= dl x B0 sinrot · o cos rot

R

:. Force per unit length on loop is

F

di

2B02 na 2co . smrot cos rot

Centre of loop using Biot-Savart's law is . 0 + sm8 µ o l (sm . . . . (1) B1 = 2) 1

4nl

As, there arc n sides angle made by one of the side at centre is

27! ct = -

R

:. Force per unit length ocBJ

Also, net force on loop is zero. 36. (b) Given, A = G " M�c1

Substituting dimensions of A,M and c, we have

⇒ ⇒ Equating dimensions on both sides, we . . . (i) -a + p = 0 have 3a + y = 2 . . . (ii) . . . (iii) - 2a - y = O Adding Eqs. (ii) and (iii), we get ⇒ 3a + y - 2a - y = 2 + 0 ⇒a = 2 . . . (iv) Now, putting the value ofa in Eq. (i), we get -2 + P = O ⇒ P = 2 ⇒ Again, putting the value ofa in Eq. (iii), we get ⇒ -2 x 2 - y = O ⇒ y = -4 So, a = 2, P = 2 and y = -4 37. (d) Total photons emitted from source 1s

n

7t) x _1_2 = __7t_n

2 81 = 82 = ( n

So,

So, from Eq. (i), we have l • l B1 = µ o ( sin _Jt_ + sin _Jt_ ) = � o ( sin _Jt_ )

4nl

n

n

2nl

At centre field due to all n segments are added up. So, magnetic field at centre due to complete polygon is

µ 0 nl sin . B = n x B1 = --

( 1t ) n

2nl

39. (a) Sound level P is generally expressed (in decibel) as, P = 1 0 (dB) · log1 0 where,

(i)

10 = reference intensity

= 1 0- 12 W/m 2 . Now, let intensity is initially I, then sound level is P 1 = 10log10 (

l)

When intensity increases by 1 0 0 times, sound level will be P 2 = 10log10 (

1

��J )

= 1 0 ( log10 100 + log10

160 x 50000 x 1 0 - 10 8 3 he / Jc 6.6 x 10 - 4 x 3 x 1 0 So, photon flux falling over spherical surface of radius r = 18 m is

N=

Power

=

N=Nn= -

A

4nr 2

160 X 50000 X 1 0 -6 6.6 X 10-34 X 3 X 108 X 4 7t X (18) 2 = 1020 m- 2 s-1

n

(i ))

⇒ P 2 = 10log10 10 0 + P 1 ⇒ P 2 - P 1 = 1 0 X 2 = 20 dB So, sound level increases by 20 dB.

40. (a) Kinetic energy of block is

converted into potential energy of stretched wire. ⇒

_l_ mv2 = _l_ Stress x Strain xVolumc

2

2

Stress . ⇒ mu 2 = -- x (Stram) 2 x Volume Strain 2

2

L ⇒ mv2 = y x M2 x A x L ⇒ M2 = mv AY L

So, distance travelled by block in mL . the wire . 1s . t.. l = v✓-. stretc hmg AY

41. (c) Given compounds of boron are electron deficient molecule, which have tendency to accept a pair of electrons to achieve stable electronic configuration and thus behave as Lewis acids. Lewis acid strength ofBBr3 , BCl3 and BF3 increases in the order < BC13 < BB13 . BF3 has least acidity due to effective back bonding between 2p-orbital of B and 2p-orbital of F. Whereas, no such back bonding is observed in BBr3 and thus has maximum acidity.

m;

42. (b) Isoclcctronic species arc those

species, which have same number of electrons. Total no. of e-s in 02 = 8 + 2 = 1 0 Total no. of e-s in Zn 2+ = 30- 2 = 28 Total no. of e-s in Mg 2+ = 1 2 - 2= 10 Total no. of e-s in K+ = 1 9 - 1 = 1 8 Total no. of e-s in Ni2+ = 28- 2 = 26 Thus 02- is isoelectronic with Mg 2+ .

43. (b) The bond angle in a molecule gets effected by the electronegativity of the central atom. More is the electronegativity of the central atom, lesser will its bond angle. Among the given compounds, the electronegativity of central atom increases in the order. C Hp :. Bond angles of H -C-H , H - N - H and H-0-H in methane, ammonia and water respectively arc 1 09.5, 107.1, 1 0 4.5.

44. (d) ln alkaline medium, the reaction of hydrogen peroxide with potassium permanganate produces, manganese oxide, Mn0 2 . -oH KMn0 4 + H202 - Mn02 + Hp

Let the oxidation state of Mn in MnO -? be x. X + 2(-2) = 0

X= + 4 45. (c) The temperature dependance of rate of a chemical reaction is expressed by Arrhenius equation h = Ae-Ea !R1' When T is very high then factor E0/RT ap ches to 0. w

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1 16

KVPY Question Paper 201 5 Stream : SB/SX

k = Ae° k=A Thus, the rate constant of a chemical reaction at very high temperature will approach Arrhenius frequency factor.

46. (b) According to electrochemical

series, more is negative emf value, stronger is the reducing agent and thus more will be the reducing strength. Therefore, the reducing strength of the metals follows the order Cr > Pb > Cu > Ag

reaction and the resultant product is an alkene. This type of elimination is known as B-elimination, where hydrogen is eliminated from B-carbon and the halogen is lost from a-carbon atom. 50. (d) When H2S gas is passed through a hot acidic aqueous solution containing AJ3 • , Cu2+ , Pb2+ and Ni2• , precipitate of only group 11 cations is formed. Among the given cations, Pb 2+ and Cu2+ lie in group 2 cations. + Cu 2+ + H2S - CuS + 2H

47. (b) Optical activity can be exhibited

only by those molecules, which have chiral centre. (a) CH3 -CH2 -CH2 -Br 1-bromopropane

*

No chiral centre present thus, it will not exhibit optical activity.

I

Br

( 3 -bromopentane)

l t is optically inactive. (d) Br

6

O Styrene

I

-

+ 2H•

Electronic configuration

1s 2s 2p 2

2

6

2

2

6

1

2

2

6

2

2

2

1

1� 2s 2p

Group number 18 1

2

2

2

13

As electronic configuration, Is 2s 2p 6 corresponds to h'TOUP 18, i.e. noble gases, thus electron removal from fully-filled stable configuration is very difficult. Therefore, it will have the largest difference between 1 and 2 ionisation energies.

L9J Polystyrene

CH:JCH20-Na •

I

(33.3%s-character)

( 25%s-character)

53. (b) The most abundant transition

CH 2

� k

CH:i -C- Cl ____, CH3 -C CH:i

PbS

51 . (a) The group number of given electronic configuration are as follows :

(50%s-character)

1

This polystyrene is clear, hard and rather brittle. lt is a rather poor barrier to oxygen and water vapour and has a relatively low melting point. It is a thermoplastic polymer. 49. (b) The major product of the reaction between CH:i CH20Na and (CH3 )3 CCI in ethanol is 2 - methyl propene. CH3 CH:i

I

( Black ppl.)

directly proportional to the s-character in hybridised state, i.e. more is the s-character more will be the electronegativity. Thus, the order of electronegativity of carbon will be sp > sp 2 > sp3

l t is also optically inactive.

Dibenzoyl pero ide :

( Black ppt.)

52. (a) Eleetronegativity of carbon is

(Bromucycluhexane)

CH2

+ H2S --

1s 2s 2p 3s

2-bromobutane

CH

Pb

1� 2s 2p 3s

As there is the presence of one chiral centre ( *), so it will exhibit optical activity. (c) CH3 -CH2 -CH-CH2CH:i

48. (a)

2+

= CH2

As (CH:i )3 CCI is tertiary halide and CI-1:i CHp-Ka + is a strong base, so elimination is favoured over substitution

metal in human body is iron. iron is present in haemoglobin. It is the iron­ containing oxygen transport metalloprotein in the red blood cells. Haemoglobin in the blood carries oxygen from the lungs or gills to the rest of the body.

54. (a) Molar conductivity always increases with increase in concentration of electrolyte. Among the given compounds, the molar conductivity depends upon the concentration ofH+ ions. Thus, stronger the acid more will be H + ions and more will be the molar conductivity. The order of acidic strength is HCI > CI-1:i COOH > NaCl > CH:i COONa

Strong acid

Weak acid

Neutral salt

Salt of weak acid

Thus, molar conductivity at infinite dilution follows the order. HCl > CH:i COOH > NaCl > CHg COON a

55. (c) Spin magnetic moment of a complex, can be calculated, µ = Jn(n + 2) where, n is the number of unpaired electron. The oxidation state of Z in [ZC14 ] 2- is +2 . The spin magnetic moment of given metals in +2 state are as follows: (a) Mn 2+ Electronic configuration = 3d5 4s0 3d

--- 11 11 11 11 1 1 1

µ = .,)5(5 + 2) = 5.9 BM (b) Ni 2+ Electronic configuration = 3d 8 4s0

11�11�11�11 11

I µ = .,)2(2 + 2) = 2.9 BM (c) Co2• Electronic configuration = 3d7 4s0

11�11�1 1 11 11 1

µ = .,)3(3 + 2) = 3.87 BM (d) Cu 2+ Electronic configuration = 3d 9 4s0

11�11�11� 11� 11 1

µ = .Jl(l + 2) = 1.9 BM Therefore, Z in [ZC1 4 ]2- is Co, where the spin only magnetic moment is 3.87 BM. 56. (d) Ifo-D-glucose is dissolved in water and kept for a few hours, the major constituents present in the solution is a mixture ofa-D glucose and B-D glucose. These two cyclic form exist in equilibrium with open chain structure.

I 1

II

0

H - C1

H - C - OH H HO

2

3

H H

OH H

2

H

_____::,.

� HO

H

6 CH OH 2

6

I

u-D-(+) glucose

H

4

H

OH 5

OH

3

OH

5

OH

CH2 OH

Open chain structure

HO -C - H

� HO ---'-- H

H ---'-- OH

_____::,.

H

--'- -

OH

H --1----�-D-(+) glucose

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117

KVPY Question Paper 2015 Stream : SB/SX +

57. (c) According to Bronsted-Lowry

concept of acids and bases, more is the stability of the conjugate base stronger will be the acid. The conjugate bases of the given solutions are :

er

HCl � H+ +

base

Conjugate

Aniline

NaNO2 + dil. HCI (HNO,,)

-

� O N

�

Benzene diazonium ch]oride

HCOOH � Hcoo - + H+ Conjugate base CH3 COOH � CH3 Coo - + H+ Conjugate base Stability of conjugate base follows the order Maximum stable as -ve charge is dispersed overlarge size of Cl

Cl-

II

II

> H CO - > CH3 CO 0

(Least stable because of -/- effect)

0

Thus, the acidic strength follows the same order, i.e. HCl > HCOOH > CH3 COOH We also know that pH is inversely proportional to the concentration ofH+ ions (acidic strength) 1.e.

1

pH oc -[H+ ]

Thus, the pH of lN aqueous solution follows the order CH3 COOH > HCOOH > HCI

58. (a)

Addition of water to alkene in the presence of an acid gives alcohol, which is in accordance with Markownikoffs rule. According to this rule, negative part (OH-) of water gets attached to that carbon atom, which possesses lesser number of hydrogen atoms.

59. (c) Reaction of aniline with NaN0 2 +

di!. HCl at 0° C gives benzene diazonium salt, which then reacts with CuCN to give cyanobenzene with the evolution of N2 gas.

65. (c) Sympathetic and

NCI

Cyanobenzene

CuCN

60. (a) Schottky defect is basically a vacancy defect in ionic solids, where the number of missing cations and anions arc equal in order to maintain electrical neutrality. This type of defect decreases the density of the substance and are shown by ionic substances, where the cation and anion are of almost similar sizes, e.g. KaCl, AgBr, etc. 61 . (a) Cyclosporine works by suppressing the immune system to prevent the white blood cells from trying to get rid of the transplanted organ. Its mode of action is inhibition of the production of cytokines involved in the regulation of T-cell activation, In particular, cyclosporine inhibits the transcription of interleukin 2,

62. (b) Glucose represses lac operon. This is because the enzymes to metabolise glucose arc made constantly by E. coli. When both glucose and lactose are available, the genes for lactose metabolism are transcribed at low levels. Maximal transcription of the lac opcron occurs only when glucose is absent and lactose is present. 63. (a) The amino acids in an a-helix are arranged in a right-handed helical structure where each amino acid residue corresponds to a 100° turn in the helix (i.e. the helix has 3.6 residues per turn). lts pitch = 5.4A.. :. The number of amino acids in 3.6 polypeptide = ---� x 27 x 1 0 -6 5.4 X 10- lO = 1.8 x 104 amino acids = 18000 amino acids 64. (c) dNTPs have phophoric acid anhydride bonds. dNTPs stand for dcoxyribonuclcotidc triphosphatc, Each dKTP is made up of a phosphate group, a deoxyribose sugar and a nitrogenous base.

parasympathetic nervous system work antagonistic to each other. During exercise, the parasympathetic control decreases and the sympathetic control increases, Vagus nerve is a spinal nerve which controls the heartbeat during rest as well as during exercise. Stimulation of sympathetic system results in the dilation of pupil.

66. (a) Genetic codes arc triplet. So, the total number of codon = 64. Out of these there is one start (coding for methionine) codon and three stop codons (i.e. do not code for any amino acid). :. Functional codon = 64 - 3 = 61 :. Frequency of encountering stop codon 3 1 = - = - approx. 61 20 67. (b) The two alleles that determine the blood group AB of an individual are located on the same autosomc. Homologous chromosomes have the same genes arranged in the same order but they have slightly different DNA sequences. Different versions of the same genes arc called alleles; homologous chromosomes often contain different alleles.

68. (c) A selectable marker is a gene introduced into a cell, especially a bacterium or to cells in culture, that confers a trait suitable for artificial selection, i.e. it allows the selection of positively transformed cells and to eliminate the non-transformed cells, 69. (a) Primary spcrmatocytcs arc

diploid (2n) cells. After meiosis-I, two secondary spcrmatocytcs arc formed. Secondary spermatocytes are haploid (n) cells that contain half the number of chromosomes. 70. (b) Phospholipids are formed by the cstcrification of one glycerol and two fatty acid molecules. A phospholipid contains a phosphate f,'TOUp in its molecule. These are a major component of all cell membranes. They can form lipid bilayer, because of their amphiphilic characteristic.

7 1 . (a) The hi stones are rich in the amino acids arginine and lysine. Histoncs are highly alkaline proteins found in eukaryotic cell nuclei that package and order the DNA into structural units called nuclcosomcs. They arc the chief protein components of chromatin, acting as spools around which DNA winds and playing a role in gene regulation.

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1 18

72. (a) Molecular weight of polypeptide = 15.3kDa Molecular weight of amino acid = 90 Da .-. Number of amino acids in polypeptide 15.3x Hf = = 170 90 We have one amino acid coded by three nitrogen base .-. 1 70 amino acids would be coded by 170 x 3 = 510 nitrogen bases. 73. (b) The DNA with more cytosine and guanine has high melting temperature, because of the presence of triple bonds between them. There in the question the DNA strand with highest melting temperature is DNA with 30% cytosine. 74. (a) The incorrectly matched immunoglobulin is lgD - viral pathogen. l mmunoglobulin D (lgD) is an antibody isotype that makes up about 1% of proteins in the plasma membranes of immature B-lymphocytcs, where it is usually co-expressed with another cell surface antibody called lgM. In B-cclls, the function of lgD is to signal the B-cells to be activated. 75. (b) Kinhydrin test is used to detect amino acids. This test is due to a reaction between an amino group of free amino acid and ninhydrin, which gives a coloured product. 76. (d) Plciotropy occurs when one gene influences two or more seemingly unrelated phcnotypic traits. Such a gene that exhibits multiple phenotypic expressions is called a pleiotropic gene. 77. (c) Antibodies that specifically target a certain antigen, such as one found on cancer cell. They can then make many copies of that antibody in the lab. These are known as monoclonal antibodies (mAbs). Monoclonal antibodies arc used to treat many diseases, including some types of cancer. 78. (c) Apoptosis is programmed cell death. Apoptosis occurring in a cell can be visualised as a ladder pattern of 180-200 bp due to DNA cleavage by the activation of a nuclear cndonuclcasc by standard agarose gel electrophoresis. 79. (b) Vitamin-B12 is involved in cellular metabolism in two active coenzymc forms-methylcobalamine and 5-deoxyadenosyl cobalaminc. Vitamin-B12 cooperates with folic acid (folate) in the synthesis of Deoxyribonucleic Acid (DNA) .

KVPY Question Paper 201 5 Stream : SB/SX 80. (a) Peptidoglycan also known as murein is a polymer consisting of sugars and amino acids that forms a mesh-like layer outside the plasma membrane of most bacteria, forming the cell wall. The glycan component is typically composed of two amino sugars, N-Acetyl-D-Glucosamine (NAG) and N-Acetylmuramic Acid (NAM). The KAM carries a tetrapeptide of alternating D and L amino acids. 81 . (a) Given, 3 3 X = (M + 7)U _ (M _ 7)U On cubing both sides, we get i" = (M + 7) - (M - 7) - s (M + 7)U3 (M _ 7)U3 M + 7)U3 - (M - 7)U3 l [( 3 x = 14 - 3 (50 - 49) 1/3 (x) ⇒ ⇒ i" = 1 4 - 3x

Eq. (i) + Eq. (ii) w2 + Eq. (iii) w, we get 1 + 22014 _ w 2014w 2 + 22014 ((w) 2) 2014 _ w

⇒

⇒ ⇒

3 = Ui. - a4 + Gi + · · · 1 + 22014 (1)201 6 + 22014 (1)4029 B = ---------3 1 + 22014 + 22014 1 + 2201 5 -----B= 3 3 I B l > IA l = I C I

83. (c) Equation of hyperbola xy = 1 y

Y'

x3 + 3x - 14 = 0 ⇒ (x - 2) (x2 + 2x + 7) = 0 ⇒

y=­ x

82. (d) We have, (1 + x + x2 ) 2014

= Uo + Ui X + Ur2 + ... + U4028X4028 Put x = - 1 1 = ao - Ui. + a2 - aJ + a4 - ... + a402s . . . (i) Put x = - w (1 - (l) + (!)2) 2014 = (2.w) 2014 = 0o - (1i_W + a,JJ} - � + - . . 2

(l - w + w)

X= -W = (2w2 ) 2014

20 l4

On adding Eqs. (i), (ii) and (iii), we get 1 + (2.w) 2014 + (2.w2 ) 2014 = 3(a0 - � + a6 + .. .) 1 + 22014 (f) + 22014 (!) ⇒ ao - � + as + ... = - - - - - 3 1- 22014 ⇒ A = --3 22014 - 1 ⇒ IA l = --and Eq. (i) + Eq. (ii) w + Eq. (iii) w2, WC get 1 + 22014 w2014w + 22014 (w2 ) 2014 w2 3

⇒

⇒

(

. . . (ii)

2 = ao - (1i_W2 + U,jJ)- Ug + U4W + ... . . . (iii)

⇒

Slope of tangent at ( 2,

1 + 22014 (1)201 5 + (f)2014(!)4030 - - - - - - - 3 1- 22014 C = --3 22014 - 1 IC I = - = IA I and Similarly, 3 C=-

dy ) dx

-

i)

2, i � 4

_ - 1 _ -1

-1

Slope of normal = -=-1 = 4

2

Put

1

dy =-I = dx x2

x=2

⇒

4 Let the slope of beam of light = m = 1: �-� :+ 4: ⇒

I

4- m

1+ 4m

l l =

I

l5 ± �⇒m= 8 4

84. (d) Given, C(8) =

f cos (ntl)

n=

n!

O

= 1 1 1 1 C(0) = I_ - = 1 + - + - + - + ... = e 1! 2! 3! n ! = 0 n

n •'

C(1t) = i:, (- lf _!_ n=

0

[·: cos ,m = (- lf]

C( rt) = 1 - _!:_ + _!:__ - _!:__ + ... = e-1 l! 2! 3! (A) C(0) . C(1t) = e . e- 1 = l True 1 (B) C(0) + C(1t) = e + - > 2 True e (C) C(8) > 0 V 8 E R True n sin (n8) (D) C' (8) = n! n= O

f_

C' (8) = 0 ⇒ 8 = 0 False Hence, option (d) is false.

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1 19

KVPY Question Paper 2015 Stream : SB/SX 85. (d) Let L = lim x➔ 2

ax + a3 - x - (a2 + a)

Apply L - Hospital rule g a - a3 -x log ax lo=.. ___ ______-"'--aL = lim __ 1 x➔ 2 - a3 -x log a - - ax/2 1og a 2

L=-

-

2 - = - (l - a)

2

a

Given, area of I region = area of II region f (2x - 4x3 ) dx = 2 (b - a) c a [x2 - x4 = 2 (b - a) c 2 (b - b4 ) - (a2 - a 4 ) = 2 (b - a)c ⇒ (b2 - a2 ) - (b4 - a 4 ) = 2 (b - a)c ⇒ 2 2 2 2 ⇒ (b - a ) (1 - (b + a )) = 2 (b - a) c 2 ⇒ (b + a) (1 - b - a2) = 2c 2 ⇒ (b + a) (1 - (b + a) + 2ab) = 2c [·: C = 2x- 4x3 ] ⇒ c = 2x - 4x3 3 4x - 2x + c = 0

t,

86. (d) We have, f (x) = ax2 - 2 + I X

3 - 2x + l, > 0 x f(x) = ax X

f(x) '?. O ifai3 - 2x + l> 0 Let g(x) = ax3 - 2x + 1 g' (x) = 3ax2 - 2 = 0 ⇒

x=± �

Clearly, x =

⇒

⇒

{I is point of minima.

11 ) (2 3a

(2 ) 3a

112

Let a, b and a are roots.

v 3a

:. a + b + a = 0 , ab + (a + b) a 1/2

-2 (2 ) 3a

+ 1'2

⇒

89. (c) When n

f'(x) = ---x f (x) + 2

On integrating, we get log (f (x) + 2) = -

x2 2

+C

⇒ ⇒

f

f (O) = O = A - 2 2 (x) = 2e-x 12 - 2

⇒

A

=

2

lim f (x) = - 2 x➔ = lim f(x) = - 2 x ➔ -oo f(x) is an even function. f(x) intersect X-axis at one point (0, 0) :. Option (b) is correct.

4a

a2

a2 = 4 ⇒a

⇒

2

- C

8a (a 2

l - a2 + 2o:: 2 - 1 = 80:: 2

⇒

f(x) + f t f (t) dt + x = 0 0 On differentiating, we get f' (x) + xf (x) +2x = 0 ⇒ f'(x) = - x (f(x) + 2) X

1

2

2 ( -¾ ) = = 1 0,

=

✓'7

D

=

2014201 4 20 1 7201 7 20202020

-

4

2

-¾)

let A, be number of

2 0 1 520 1 5 201 3201 8 20212021

= =

9 1 . (c) It is given that observer O can view wall CD. Let h = height of water in vessel at which object P is visible to observer. D

2016201 6 2019201 9 20222022

P

B

ways of selecting r numbers. Number of selection of A is n(A0 ) + n (A1 ) + n (A2) + n (As ) + n(A4 ) = 1 + 1 0 + (7 + 6 + 5 + ... + 1) + (4 + 3 + 2 + 1) + (3 + 2 + 1) + (2 + 1) + 1 = 1 1 + 28 + 10 + 6 + 3 + 1 = 60 = N(p) n (Number of ways 1 is selected) = 1 + 7 + 4 + 3 + 2 + 1 + 1 = 19 N (q) = n (Number of ways 2 is selected) = 1 + 6+ 3 + 2+ 1 = 1 3 13 19 , q = = p So, 60 6O 6 =1 p > q, p - q = - 1 9 - 1- 3 = 6 0 10 6O 90. (d) Let

(24040 + 1) + 2201 7 (5 - 1) 2020 + 1 + 2(5 - 1)1 008 1+ 1+ 2 = 4

- ..:':, 2

a + b = - a, ab = a 2 - -, ab = l - a2 +

87. (b) We have,

⇒

0

( � - 2 ) + 1'2 O 3

< � ⇒ 2 < -�4 3a 1 6

=

aba = _ !:, 4

2

32 a '?. 27

(2020 -2) (2020 + 1) 2021

= 1

b

3

-a - !!:

(2015 + 1) 2016 (2020 - 1) 2019 (2 02 0 +2) 2022

2018

Remainder when divided by 5, is 0 1 1 D = 2201 1 2201 8 -1 0 22022 1

X - a3 -X_ __ _ L = lim _a x ➔ 2 3 X ax/2 -a -

a2

(2015) 2015

(2 0 1 5 - 1) 2014 (2015 + 2) 2017 (2020) 2020

88. (a) We have, y = 2x - 4.x3

/ a3 - x - ax 2

',, , C 1 0 cm --->1

� (h- 1 O) cm

Light from object is refracted and reaches observer. From geometry of arrangement, we have h - 10 sin i = tan i =

h

where, i angle of incidence for small angles, sin i = tan i. Also, from geometry of arrangement, angle of refraction r = 45 °. sm i Now, µ (air with respect to water) = Sill I' (h - 1 0) / h � = h - 10 ⇒ � = ⇒ =

1 1'12

4

h ill h = 22 cm

10 =

⇒ ⇒

1-

�

W

⇒

10

h

=

26 57

h

So, if water is filled upto height of 27 cm, object is visible to the observer.

92. (a) Let O is centre of disc and C is its centre of mass.

y

:a .,__�........_�....1. '

,-

-&'1

ICM R-y - - - - - - - (,: ➔ Axis R+y

�c'" o� : --'"'-------"::__ : '

'

-&3

through centre of

mass

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120

KVPY Question Paper 201 5 Stream : SB/SX

l f Jc = moment of inertia about an axis through centre of mass of disc, 11 = moment of inertia about an tangential axis AB, 13 = moment of inertia about an tangential axis CD, M = mass of disc and R = radius of disc. Then, by parallel axes theorem, we have . . . (i) II = IC + M - (R - y) 2 2 . . . (ii) l3 = IC + M - (R + y) Subtracting Eqs. (i) and (ii), we get 2 2 ⇒ Il - 13 = M((R - y) - (R + y) ) = - 4MRy . . . (iii)

(ii) Weight 171ig and "½g, different on each ball vertically downwards through centre of ball. (iii) Tensions T1 and T2, different on each ball along the string. As, angle between strings is given 30° + 60° = 90° and strings are of equal length, strings forms an isosceles triangle as shown below,

= l. R de X R = l.R 2de

2

2

Centre of mass of this ti.OPQ is at a

distance of � R cos0 from 0. 3 So, position of centre of mass of complete segment is y = ydm l dm o 1 2 2R r2 I - cos0 · p- d0 9 2 y= 0 3

f

l4 '

f

o12

''

'

y)

D ';--�-s----__,

I

r-

p - d0 2 where, p = mass density. J

0

A :�--::,,--t--s.;'----1

:c

Similarly, from above figure b y parallel axes theorem, we have 2 12 = Jc + M (R - x) 2 = + M (R + x) 14 Jc and . . . (iv) 12 - 14 = - 4MRx ⇒ So, from Eqs. (iii) and (iv), we have (11 - 13 ) 2 + (12 - 14 ) 2 = 16M2R2 (x2 + y2) 2 1 2 x2 + y2 = -- Ui - I3 ) + U2 - I4 ) 4 MR

⇒✓

Consider a differentiable are of angle de. 1 Arca of l'iOPQ = - x PQ x OP

✓

But ✓x2 + y2 = distance of centre of mass

from centre of the disc.

93. (c) As the steel rail is contraincd

from expansion, the expansion pressure causes stress in the steel rail. Thermal stress depends upon coefficient of expansion a and rise of temperature l'iT. :. Thermal stress, cr = al'iT ⇒ cr =Y - a - l'iT (where, Y = Young's modulus) . (J = 2 X 1D1 1 . . X ll x 1 0-5 x (40 - 25) = 3.3 X 107 Pa 94. (c) ln given case, angle of refraction is negative. This is possible only when refractive index of material is negative. Negative index Metamaterial (NIM) is a material for which refractive index has a negative value for an electromagnetic wave over certain frequency range. These are artificially produced materials, which found their application in optical signal transmission.

95. (d) Each ball is in equilibrium under three forces. (i) Electrostatic repulsion force Fe , equal on both balls along line joining centre to centre of balls.

m19 Free body diagrams of balls are 1 35 �� f• " r, f• o : F l1 05 m1g

mi)

Now, using Lami's theorem for ball 1 and ball 2, we have ""1 g (For ball 1) sin 150° sin 135° m2tt Fe and (For ball 2) sin 120° sin 135° Dividing these equations, we get sin 120° 171i = sin 150° "½ ° ° 171i = sin (180 - 60 ) ⇒ ° ° m2 sin (180 - 30 ) sin 60° .J3 / 2 = .J3 = = sin 30° 1/ 2 So,

""i � 2.0 ni2

96. (a) A diamagnetic material tries to expel field lines from its volume and therefore experiences a repulsion when placed in region of strong fields. In given arrangement, field is zero in region marked A. So, diamagnetic powder accumulates in region A. 91. (a) Refer to figure below,

2r

J

o 12

- cos0d0 0 Y- 3 -

e12

fo

de

2

!!:_ . sin0/2 3 0/2

_i sin (0 / 2) R 3 e

98. (a) Let v = velocity of object on =

reaching surface of planet.

Applying energy conservation between points A and B, we have Total energy at A = Total energy at B ⇒ Potential energy at separation (4R + R) = Potential energy at separation R + Kinetic energy at surface - GMm - GMm = + l. mv2 ⇒ 5R R 2 where, m = mass of object . GMm _ J:. = l. mv2 ⇒ � GM = v2 ⇒ (1 ) R 5 2 5 R

5 R

:. Velocity at surface, v = 2✓�

GM

99. (a) ln steady state, after the key L1

remains ON for a long time.

-=-

I

I'

I

L2

I'

I

As resistances arc identical and parallel (inductors behaves like closed paths), currents in resistors arc equal as shown above.

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121

KVPY Question Paper 2015 Stream : SB/SX When switch is turned OFF, current in inductors remains same. So, we have following current distribution in circuit.

,�

,�

�1CEJ11

So, currents in resistances are 21 upwards, I downwards and I downwards.

1 00. (a) From given cyclic process, p(atm)

1 0 1 . (a) N2 + 3F2 ------t 2NF3

2 1 04. (a) Zn (s) + Cuz + (aq) ------t Zn + (aq)

+ Cu(s)

Or N=N + 3 F - F ----4 2 N

/ l "'F F F

941kJ mol- 1 1 = 155 kJ mol = 272 kJ mol -1 = t.Hr IB E ,eactants - IBE products = BE N aae N + 3B E F - F - 6BE N = 941 + 3(155) - 6(272) = - 22 6 kJ moJ- 1 For the reaction, N2 + 3Cl2 ------t 2NC13 Or Given, BE N = N B E F -F BE N -F

According Ncrnst equation, [Pl 0.0591 Ecell _ - E 0cell - --- 1og [R] n

=

When cell is completely discharge then Ecell = 0 2+ 0 = 1 . 1 0 - 0.0591 10g [Zn ] 2 [Cu2+ ] 2+ -0.0591 [Zn ] l -1. O log [Cu2+ ] 2 [Zn2+ l = _3_-3_()_ = 3 7 .3 :. log [Cu2+ ] 0.0591

F

1 05. (c)

/ l "'c1 Cl Cl

N-N + 3 Cl-Cl ----4 2 N

V(L) --+-------'2 - 4'-------'-6-➔

t.Hr

Br

Process equation is (p - 4) 2 + (V - 4) 2 = 4 . . . (i) Now, from p V = nRT Wc can say that T is maximum when p V 1s maximum. Now, for given cyclic process, p V maximum occur when p 2V 2 is maximum. Now, p 2V 2 = p 2 (4 - (p - 4) 2 ) [from Eq. (i)] Now, p 2V 2 is maximum when

1 02. (a) Let the initial concentration of X and Z be 1 and a be the degree of

_.!!:_ p 2V 2 = o. dp

Equili. cone.

⇒

_.!!:__ (p 2 - (p 2 - (4 dp

(p -

R

1 6 + 2 + 2 x 4 x ,J2

R

29.32

30

R

R

Dry ether

H

� """'' "I

X � 2Y

Initial cone.

1- a 2a 2 2a) 2 [Y] ( Kl = -- = -[X] 1 - a

Equili. cone.

Initial cone.

0

a.

a

1 06. (b) An organic compound having molecular formula C2 Hi,O can be alcohol or ether, but only alcohol undergoes oxidation to produce carboxylic acid. Thus, alcohol with molecular formula C2H6 0 is ethanol, Clfa CH20H which oxidises to acetic acid, C Ha COOH

[P [ ] K2 = ] Q = � 1-a [Z] K1 1 - a = =4 K2 a2 1-a

CH3 CH2 OH -

2

[MnBr4 1 - is +2 . The electronic configuration ofMn2+ is [Ar] 3d5 4s0 . Mn (Ground state)

11 1 1 11 1 1 1 1 I

4s

3d

I1 I 1 I 1 I 1 I 1 I

DI I I 4p

y �l�I 4s

-

--. CH3 ---C---OH (X)

Confermation of molecular formula of X is as follows:

4p

As, Br- is a weak field ligand pairing of electron will not occur. [MnBr4l 2

II

K2Cr2 0, / H2S04

1 03. (c) The oxidation state of Mn in 3d

+

Bromobenzene, when reacts with Mg in the presence of dry ether gives Grignard reagent, which then reacts with CO 2 in the presence of an acid to give benzoic acid.

4a2

2+

� Mgfu

(Benzoic acid)

Z�P+Q 0 0 1- a

II _

(Grignard reagent)

dissociation.

4) 2 ) = 0

p 2 - 8p + l4 = 0 8± � ⇒ p= 2 = 4± J2 and from Eq. (i), we get p = 4 ± J2, V = 4 ± J2 Taking positive values, we have (pV) max ⇒ P = 4 + J"2 and V = 4 + J2 So, by gas equation, we have ( V) [for l mol of gasj Tmax = p max R _ (4 + J2) (4 + J2)

Mg (x)

BE N = N + 3B Ec1 -c1 - 6 ( BE N -c1l = 941 + 3(242) - 6(200) = + 467 kJ mol -1 =

-,.J

sp3 hybridisation As, the hybridisation of Mn in [MnB14J 2is sp3 , thus its geometry will be tetrahedral and it contains five unpaired electrons.

Symbols

% of element

C

40

H

0

6.7 53.3

No. Simplest Simplest of molar whole moles ratio no. 40/12 3.33/3.33 ; 3_33 ;j 6.7/1 ; 6.7

6.7/3.33 ; 203

= 3. 33

;j

53.3/16 3.33/3.33

1

2 1

Hence, empirical formula of X is CH2 O.

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122

KVPY Question Paper 201 5 Stream : SB/SX

Empirical formula mass = 12 X 1 + 2 + 16 X 1 = 12 + 2 + 16 = 30 Molecular mass of Cl-f:i COOH n= Empirical mass 60 =2 = 30 :. Molecular formula will be (CHP) 2, i.e. C2H4 02 . Thus, the above analysis shows that the compound X has molecular formula C2H4 02, i.e. Cl-f:i COOH

1 07. (b) There can be 3 maximum number of cyclic isomers (positional and optical) of a compound having molecular formula C3 H2CL These arc as follows

intense purple solution of potassium permanganate Y and MnO 2. MnO 2 + KOH + O 2 � K2MnO 4 + 2H2O

1

) (X Disproportionation

KMnO 4 + MnO 2 + Hp (Y

)

1 1 0. (b) Al(X) metal dissolves both in dil. HCl and dil. NaOH to liberate H 2. Addition of NH4 Cl and excess NH4 OH to an HCl solution of Al produces Al(OH)3 (Y) as an precipitate. Al(OH)3 is also produced by adding NH4 Cl to the Na OH solution of Al. Al + HCl� AlC13 + H-o (�

Al + NaOH� NaA1O 2 + H2

1

(X)

NH,Cl

Al(OH)3 f--AlC13 + NH4 Cl (Y ) + NH4 OH(excess) (Optical active due to the presence of chiral centre)

Thus, it will have enantiomeric pair.

1 08. (a) According to ideal gas equation, pV = nRT For 1 mole, n = l

⇒

pV = RT RT p=-

v

For gas X,

V = 50 L, T = 200 K

:. Pressure of gas X = -----

0.082 l x 200 50 = 0.328 atm

For gas Y,

V = 50 L, T = 500 K

:. Pressure of gas Y = -

0.082 l x 500 50 = 0.821 atm

For gas Z,

V = 2 0 L, T = 200 K

:. Pressure of gas Z = -

0.082 1 x 200 - - - 20 = 0.82 1 atm 1 09. (a) MnO 2, when fused with KOH and oxidised in air gives a dark green compound, X, i.e. potassium manganatc (K2MnO 4 ). In acidic solution, K�nO 4 undergoes diproportionation to give an

1 1 1 . (c) On SDS-PAGE oflgG, two (2) principal bands arc visible. IgG antibodies is a monomeric structure consisting of heavy chains and two identical light chains. Thus, on SDS-PAGE under reducing conditions shows two bands.

1 1 2. (a) Penicillin works by bursting the cell wall of bacteria. When a bacterium multiplies, small holes open up in its cell wall as the cells divide. Therefore, in a mixed culture with penicillin, it kills the fast growing bacteria more than the slow growing ones. 1 1 3. (c) Haemophilia is a sex-linked recessive disorder. These kinds of defects occur more often in men than in women, because men are heterozygous for X-chromosomes and if it carries the gene they are diseased. Whereas, for a woman to be affected she needs to be homozygous for the trait.

This is because, there is more chance that fast dividing cells would have taken up thymidine from the media during that one hour. Radioactivity is mostly incorporated in a rapidly dividing cell during S-phase.

1 1 6. (c) In a double-stranded DKA, according to Chargaff's rule, A + T= G + C A= T and G = C Thus, when cytosine = 15% Guanine = 15% Both C+ G= 30% Thus, A+ T= 70% 70 = 35% , T = 35% and A =

2

1 1 7. (b) The electron transport chain and

chemiosmosis take place on this membrane as the part of cellular respiration to create ATP. The cristac increase the surface area of the inner membrane, allowing for faster production of ATP, because there arc more places to perform the process. Thus, option (b) is correct.

1 1 8. (a) Phosphorylation is a biochemical process that involves the addition of phosphate to an organic compound. It is known that eukaryotes rely on the phosphorylation of the hydroxyl group on the side chains of serine, threonine and tyrosine for cell signalling. Thus, any change in these amino acids will affect the function of the molecule. Therefore, a mutation from tyrosine to tryptophan is damaging to cell. 1 1 9. (c) The gel electrophoresis will show 3 bands because even though 4 fragments arc formed but 2 fragments are of 3 kb and they will appear as one single band. e�

1 1 4. (d) Radius of spherical alveoli = 0.1mm Total number of alveoli = 400 x 2 = 800 million Surface area of sphere = 41tr 2 Total respiratory surface area = 41tr 2 x 800 million = 1000 mm 2 (approx)

1 1 5. (a) Among slow dividing and fast

dividing cells, the radioactivity will be maximum shown by fast dividing cells.

"'

-

-

5

1

3

1 20. (c) GTP-X complex ➔ Active cell division GDP-X-complex ➔ No cell division and GTP

hydrolysis

X•enzyme

GDP

Mutation causes the decrease of rate of hydrolysis of GTP. Thus, GTP - X complex remains present for long time and cell division becomes uncontrolled.

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KVPV

KISHORE VAIGYANIK PROTSAHAN YOJANA

OUESTION PAPER 201 4 Stream : 58/SX

M M : 160

Instructions

1 . There are 1 20 questions in this paper. 2. The question paper contains two parts; Part I (1 Mark Questions) and Part II (2 Marks Questions). 3. There are four sections i n each part; Mathematics, Physics, Chemistry and Biology.

4.

Out of the four options given with each question, only one is correct.

� PART-I

(1 Mark Questions)

MATHEMATICS 1. Let C0 be a circle of radius 1. For n 2: 1 , let Cn be a

I. Area (C; ) equals

circle whose area equals the area of a square inscribed in en (a) 7t 2

-

l'

Then,

-2 (b) it 2 7t

i=0

1 (c)1t2

1t2 (d)-1t - 2

2. For a real number r we denote by [r] the largest integer less than or equal to r. If x, y are real numbers with x, y 2: 1 then which of the following statements is always true? (a) [x + y] S: (x] + [ y] (c)[2x J S: � x]

l::_J

(h)[xy] S: [ x] [y] [x] (d) s; y [y]

An = max {(;) IO s; r s; n} Then, the number of

3. For each positive integer n, let

elements n is {1, 2, ..., 20} for which 1. 9 S: � S: 2 is An - 1 (a) 9

(b) 1 0

(c) 1 1

(d) 12

4. Let b, d > 0. The locus of all points P(r,0) for which the line OP (where, 0 is the origin) cuts the line r sin 0 = b in Q such that PQ = d is (a) (r- d) sin 0 = b (c) (r- d) cos e = b

(b) (r ± d) sin 0 = b (d)(r ± d) cos e = b

5. Let C be the circle x2 + y 2 = 1 in the XY-plane. For

each t 2: 0, let L 1 be the line passing through (0, 1) and (t, 0). Note that L1 intersects C in two points, one of which is (0, 1). Let Qt be the other point. As t varies 2 between 1 and 1 + ✓ , the collection of points Qt sweeps out an arc on C. The angle subtended by this arc at (0, 0) is (a) _'.'._ (c) _.::. 3

(b) _.::. 4 31t (d) 8

6. In an ellipse, its foci and ends of its major axis are equally spaced. If the length of its semi-minor axis is 2✓2, then the length of its semi-major axis is (b) z./3 (a) 4 (c) Jio (d) 3

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124

KVPY Question Paper 2014 Stream : SB/SX

7. Let ABC be a triangle such that AB = BC. Let F be the mid-point of AB and X be a point on BC such that FX is perpendicular to AB. If BX = 3XC, then the ratio BC I AC equals (b).J2

(a) J3

(d) 1

8. The number of solutions to the equation 1 . 4 1 . . cos4 X + -- = sin X + - n the 1nterva1 '0 [ , 2 7t] l. S - 2 1 2 COS

(a) 6

X

(b)4

S ln X

(c)2

(d) O

9. Consider the function x+

5 1. f x >' 2 f(x) = ! x- 2 ' 1, ifx = 2

Then, f (f(x)) is discontinuous (a) at all real numbers (b) at exactly two values of x (c) at exactly one value of x (d) at exactly three values of x

10. For a real number x let [x] denote the largest number less than or equal to x. For x E R let f (x) = [x] sin 7t x. Then, (a) f is differentiable on R. (b) f is symmetric about the line x = 0. 3 (c)J_/ (x)dx = 0.

(d) For each real a, the equation f (x)- a = 0 has infinitely many roots.

11. Let / : [0, n] ➔ R be defined as

sin x, if x is irrational and x E [O, rrj f(x) = { tan 2 x, if x is rational and x E [O, rrj

The number of points in (0, rr] at which the function f is continuous is �4 �o �2 W6 00 12. Let f : [O, l] ➔ [O, ) be a continuous function such

that f f (x) dx = 10. Which of the following statements 1

0

is NOT necessarily true?

J l

(a) e-x f(x) dx5. 10 0

(b) J _lM_2 dx 5, 1 0 o (1 + x)

(c)- 1 0 5, J sin (1 00 x) f (x) dx 5, 1 0 0 1

J 1

(d) f (x)2 dx5. 100 0

13. A continuous function f : R ➔ R satisfies the equation

f (x) = x + f f (t) dt. Which of the following options X

true?

0

(a) f(x + y) = f(x) + f (y) (b) f(x + y) = f (x) f (y) (c) f (x + y)= f (x)+ f (y) + f (x)(y) (d)f (x + y) = f (xy)

14. For a real number x let [x] denote the largest integer less than or equal to x and {x} = x- [x]. Let n be a n

positive integer. Then, f cos (2n[x]{x}) dxis equal to 0 (a) 0 (b) 1 (d)2n - 1 (c) n

15. Two persons A and B throw a (fair)die (six-faced cube

with faces numbered from 1 to 6) alternately, starting with A. The first person to get an outcome different from the previous one by the opponent wins. The probability that B wins is (a) � 6

(b) � 7

(c) '!... 8

16. Let n � 3. A list of numbers Xi , x, . . . , xn has mean µ

and standard deviation cr. A new list of numbers Y1 , Y2 , . • . , Yn is made as follows X2 _ -Xi +Xi + X2 , _ • , y2 _ y1 - --- and y - x for J- 3, 4, ... , n. 1 1 2 2 The mean and the standard deviation of the new list are µ and &. Then, which of the following is necessarily true ? (a) µ = µ and a 5, a (b) µ = µ and a � a (c) o = 6 (d)µ ae µ

( �l)

17. What is the angle subtended by an edge of a regular tetrahedron at its centre ?

(a) cos-1 (c) cos- 1

( - l)

(b) cos- 1

(d)cos- 1

( �) (�)

18. Let S = {(a, b) : a , b, E Z, 0 5, a, b 5. 18} . The number of elements (x, y) in S such that 3x + 4 y + 5 is divisible by 19, is (a) 38 (c) 18

3

(b) 1 9 (d) 1

19. For a real number r let [r] denote the largest integer less than or equal to r. Let a > 1 be a real number which is not an integer, and let h be the smallest positive integer such that [ah ] > [at. Then, which of the following statements is always true? (a) k 5, 2 ( [ a] + 1)2 (b) k 5, ([a]+ 1)4 1 (c) k 5, 2( a] + 1 (d)k 5, -- + 1 a- [ a]

20. Let X be set of 5 elements. The number d of ordered pairs (A, B) of subsets of X such that A "# , B"# , A n B"# satisfies (b) 1015. d 5, 1 50 (a) 50 5, d 5, 1 0 0 (c) 1 51 5. d 5, 200 (d)20 b d

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125

KVPY Question Paper 2014 Stream : SB/SX 26. A solid expands upon heating because

PHYSICS 21. An uniform thin rod of length 2L and mass m lies on a horizontal table. A horizontal impulse J is given to the rod at one end. There is no friction. The total kinetic energy of the rod just after the impulse will be J2 2J 2 (c) (a) 2m m m m

22. A solid cylinder P rolls without slipping from rest

down an inclined plane attaining a speed vP at the bottom. Another smooth solid cylinder Q of same mass and dimensions slides without friction from rest down the inclined plane attaining a speed vq at the V bottom. The ratio of the speeds _!!__ is ( ) b R

(c)

VP

( ) d H

R

23. A body moves in a circular orbit of radius R under the action of a central force. Potential due to the central force is given by V(r) = kr (k is a positive constant). Period of revolution of the body is proportional to (b)K 112 (a) R112 (c)K31 2 (d)K512

24. A simple pendulum is attached to a block which slides without friction down an inclined plane ABC having an angle of inclination a as shown below. C

(X

A

B

While the block is sliding down the pendulum oscillates in such a way that at its mean position the direction of the string is (a)at angle a to the perpendicular to the inclined plane AC (b) parallel to the inclined plane AC (c) vertically downwards (d) perpendicular to the inclined plane AC

25. Water containing air bubbles flows without

turbulence through a horizontal pipe which has a region of narrow cross-section. In this region, the bubbles (a) move with greater speed and are smaller in size than in the rest of the pipe (b) move with greater speed and are larger in size than in the rest of the pipe (c) move with lesser speed and are smaller in size than in the rest of the pipe (d) move with lesser speed and are of the same size as in the rest of the pipe

(a) the potential energy of interaction between atoms in the solid is asymmetric about the equilibrium positions of atoms (b) the frequency of vibration of the atoms increases (c) the heating generates a thermal gradient between opposite sides (d) a fluid called the caloric flows into the intcratomic spacing of the solid during heating there by expanding it

27. Consider two thermometers T1 and T2 of equal length,

which can be used to measure temperature over the range01 to 02. Y'i contains mercury as the thermometric liquid, while T2 contains bromine. The volumes of the two liquids are the same at the temperature01 • The volumetric coefficients of expansion of mercury and bromine are 18 x 10-5 K-1 and 108 x 10-5 K-1, respectively. The increase in length of each liquid is the same for the same increase in temperature. If the diameters of the capillary tubes of the two thermometers are d1 and d2 , respectively. Then, the ratio of d1 : d2 would be closest to (c) 0.6 (a) 6.0 (b) 2.5 (d)0.4

28. An ideal gas follows a process described by p V 2 = C from (A , V1 , T1 ) to (p2 , V2 , T2 ) and C is a constant. Then, (a) if p1 > p2 then T2 > T1 (b) if V2 > vl then T2 < Tl (c) if V2 > vl then T2 > Tl (d)if p1 > p2 then V1 > V2

29. A whistle emitting a loud sound of frequency 540 Hz

is whirled in a horizontal circle of radius 2 m and at a constant angular speed of 15 rad/s. The speed of sound is 3 3 0 m/s. The ratio of the highest to the lowest frequency heard by a listener standing at rest at a large distance from the centre of the circle is (b) 1.1 (a) 1 .0 (c) 1.2 (d) 1 .4

30. Monochromatic light passes through a prism.

Compared to that in air, inside the prism the light's (a) speed and wavelength arc different, but frequency remains same (b) speed and frequency are different, but wavelength remains same (c) wavelength and frequency arc different, but speed remains same (d)speed, wavelength and frequency are all different

31. The flat face of a plano-convex lens of focal length 10 cm is silvered. A point source placed 30 cm in front of the curved surface will produce a (a) real image 1 5 cm away from the lens (b) real image 6 cm away from the lens (c) virtual image 1 5 cm away from the lens (d)virtual image 6 cm away from the lens

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KVPY Question Paper 2014 Stream : SB/SX

32. Two identical metallic square loops L1 and L2 are

placed next to each other with their sides parallel on a smooth horizontal table. Loop L1 is fixed and a current which increases as a function of time is passed through it. Then, loop L is 2 (a) rotates about its centre of mass (b) moves towards ½ (c)remains stationary (d) moves away from ½

33. An electron enters a parallel plate capacitor with

horizontal speed u and is found to deflect by angle e on leaving the capacitor as shown below. It is found that tan e = 0.4 and gravity is negligible.

µ

..

✓

- - -- - -�-) eIf the initial horizontal speed is doubled, then the value of tan 0 will be (a) 0.1 (b) 0.2 (c) 0.8 (d) 1 .6

34. Consider a spherical shell of radius R with a total

charge + Q uniformly spread on its surface (centre of the shell lies at the origin x = 0). Two point charges + q and - q are brought, one after the other from far away and placed at x =- a / 2 and x = +a/ 2 (a < R), respectively. Magnitude of the work done in this process is (b)zero (a) (Q + q)2 I 4nc0 a (d) Qq / 4nc0 a (c)q2 I 4nc0 a

35. Two identical parallel plate capacitors of capacitance

C each are connected in series with a battery of emf, E as shown below. If one of the capacitors is now filled with a dielectric of dielectric constant k, then the amount of charge which will flow through the battery is (neglect internal resistance of the battery) C

C

�� E

(a) � - CE 2(k - 1) -2 k · CE (c) k+2

(b)� CE 2(k + 1) +2 k (d) CE k-2

36. A certain p-n j unction, having a depletion region of

width 20 µm, was found to have a breakdown voltage of 100 V. If the width of the depletion region is reduced to 1 µm during its production, then it can be used as a zener diode for voltage regulation of (a) 5 V (b)lO V (c)7.5 V (d) 2000 V 37. The half-life of a particle of mass 1.6 x 10-26 kg is 6.9 s and a stream of such particles is travelling with the

kinetic energy of a particle being 0.05 eV. The fraction of particles which will decay, when they travel a distance of 1 m is (a) 0.1 (b) 0.01 (c) 0.001 (d)0.0001

38. A 160 W light source is radiating light of wavelength

6200 A uniformly in all directions. The photon flux at a distance of 1.8 m is of the order of (Planck's constant = 6.6 3 X 10- 34 J-s) (a) 1 02 m -2s- 1 (b) 1012 m -2s- 1 (c) 101 9 m-2s- 1 (d)1 025 m-2s- 1

39. The wavelength of the first Balmer line caused by a

transition from the n = 3 level to the n = 2 level in hydrogen is A. 1 . The wavelength of the line caused by an electronic transition from n = 5 to n = 3 is 1 25 'A, 375 'A,I (a) (b) l 128 64 'A, 128 'A, (c) � I (d) l 1 25 375

40. The binding energy per nucleon of 5 B 10 is 8.0 MeV

and that of 5 B 1 1 is 7.5 MeV. The energy required to remove a neutron from 5 B 11 is (take, mass of electron and proton are 9.11 x 10-3 1 kg and 1.67 x 10-27 kg, respectively) (a) 2.5 MeV (b) 8.0 MeV (c) 0.5 MeV (d)7.5 MeV

CHEMISTRY

41. When 1.88 g of AgBr(s) is added to a 10- 3 M aqueous solution of KBr, the concentration of Ag+ is 5 x 10- 10 M. If the same amount of AgBr(s) is added to a 10- 2 M aqueous solution of AgN0 3 , the concentration of Br- is (b) 5 x l0- 10 M (a) 9.4 X 10-9 M (d)5 x 10- 11 M (c) l x 1 0- 11 M

42. Aniline reacts with excess Br2/H2 O to give the major product.

(a)

Br Br * NH2

�

Br

I

NH2

0 (b)be > 0 (c)ca > O (d) a+ b + c > O

84. Let n � 3 and let C1 , C2 , . . . , C,, , be circles with radii

'i , r2 , . . . , rn , respectively. Assume that C; and C; + 1 touch externally for 1 ::; i ::; n- 1. It is also given that the X-axis and the line y = 2./2 x + 10 are tangential to each of the circles. Then, r1 , r2 , . . . , rn are in (a) an arithmetic progression with common difference 3+ .J2 (b) a geometric progression with common ratio 3 + .J2 (c) an arithmetic progression with common difference 2+ J3 (d) a geometric progression with common ratio 2 + J3 85. The number of integers n for which 3f - 25x + n = 0 has three real roots is (c) 55 (a) 1 (b) 25 (d)infinite

86. An ellipse inscribed in a semi-circle touches the

circular arc at two distinct points and also touches the bounding diameter. Its major axis is parallel to the bounding diameter. When the ellipse has the maximum possible area, its eccentricity is (a) ..l .J2

(b) 2 2

(c) ..l J3

(d) G. �j

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KVPY Question Paper 2014 Stream : SB/SX 11 / 2

87. Let In = f x' cos xdx, where n is a non-negative 0

2)

In � I integer. Then, L, (_.I!.. + ---- equals n ! (n - 2) 1 n = 2 (b)e"12 - 1

(a) e"12- 1- � 2 (c)e"12 - � 2

less than or equal to x. The smallest positive integer n

n for which the integral J [x] [✓x] dx exceeds 60 is 1

(b)9

(c) 1 0

set {l, 2, ... ,100} . Choose one of the first seven days of the year 2014 at random and consider n consecutive days starting from the chosen day. What is the probability that among the chosen n days, the number of Sundays is different from the number of Mondays? 12 (a) ..!:. (b)� (c) (d)� 175 2 7 49

90. Let S = {(a, b) a, b E Z, 0 :,; a, b :,; 1 8 }. The number of lines in R2 passing through (0, 0) and exactly one (c) 28

(d) 32

91. A solid sphere spinning about a horizontal axis with an angular velocity w is placed on a horizontal surface. Subsequently it rolls without slipping with an angular velocity of (a) 2w / 5 (b)7w / 5 (c)2w/7 (d)w

92. Consider the system shown below. X

A horizontal force F is applied to a block X of mass 8 kg, such that the block Y of mass 2 kg adjacent to it does not slip downwards under gravity. There is no friction between the horizontal plane and the base of the block X. The coefficient of friction between the surfaces of blocks X and Y is 0.5. The minimum value of F is (take, acceleration due to gravity to be 10 rns-2) (b) 1 60 N

(c)40 N

(b) � R 2 (d)R

95. A concave mirror of radius of curvature R has a

circular outline of radius r. A circular disc is to be placed normal to the axis at the focus, so that it collects all the light that is reflected from the mirror from a beam parallel to the axis. For r T2 > T1 ), total energy of atoms also increases (� > E2 > El ) . As a result, mean separation between atoms also increases (r3 > r2 > r1 ). Hence, crystalline solids in general expands on heating.

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KVPY Question Paper 2014 Stream : SB/SX 27. (d) Ratio of increase in volume per

unit original volume per degree rise of temperature of mercury and bromine is (Ve - Ve 1 ) Hg { V01 . i'.0 } 'YHg - or T (Ve - Ve1 ls, { V · 0 } e1 i'. We - Ve , lttg ⇒ = Yttg (Ve - Ve 1 ls, Ysr

(M . rcrl.i.2 )

⇒

4

TC:: L

Hg _ 'YHg �--� (M .

⇒ ⇒

[ :. Wa 1 ) Hg = We 1 ls, l

d[

�

Ysr

- Tor

28. (b) p V 2 = C ⇒ ( n�T ) - V 2 = C

[:. for an ideal gas, p V = nRT]

TV = __2___ = a constant nR

T = __!__

31 . (b) When a lens is silvered, it acts like a mirror with focal length f. -1 2 1 . given . It 1s by - = - + f ft fm

where, ft = focal length of lens and fm = focal length of mirror

-- o�"::c::�·- -- -- - --

So, if V2 > V1 , then T2 < T1 .

Minimum frequency

- = - +! 10

-1

2

1

⇒ f = - 5 cm Here, note that negative sign appears in formula because mirror obtained is concave in nature. Herc, u = - 30 cm So, by mirror formula, we have 1 1 1 -=-+! V U ⇒

1

1

1

'\ Maximum frequency

� Vs

Speed of approach of source = Vs = rOJ = 30 ms-I Using formula for Doppler's effect, Maximum frequency,

J.r

v !max = (v - v.

Minimum frequency, v fmin = (V + Vs

)- r

y = u t + --1: a t 2 y

y=-

⇒ Now, tane So,

eEx2

1 =y=2

2mu 2

2

y

-

ll

tan e2 _ tan el

uf

ui

Given that, tan el = 0.4, ll 2 = 2ul tane 2 = o.4 ( _!!i___ ) = 2

(2u1 )

0.4

4

= G.l

34. (c) Work done in the process = Potential energy of the system y

Also from shell theorem, charge Q on shell behaves as a point charge at centre. So, magnitude of work done is IUsystem l = I U12 + U23 + Us , 1 k q kq(-q) k Q (-q) = Q + + I a/2 a a/2 I

=I �

(--Vs

�;!,;,:;

plates = t = � u In this time acceleration of e- in y F eE direction is ay = = m m Deflection y is then,

1

32. (d) As current increases in loop

{0

Time taken by e- to cross region between

- = - + --5 V -30

1 1 1 - 6+ 1 -=-+-= - 30 v -5 30 ⇒ v = - 6cm So, image is real and 6 cm in front. ⇒

29. (c) Given situation is as shown

below.

source it docs not changes during refraction process. Inside the prism wavelength and speed arc reduced than that in air.

⇒

Substituting values in above equation, we have 18 X 1 0 -5 IT = d1 = 0.4 = ⇒ d2 108 X 1 0 -5 �6

⇒

30. (a) Frequency is a characteristics of

- YHg

d1 ✓YHg = d2 Ys, d�

So, ratio of / ax fmaJ< V + V8 330 + 30 = ⇒ m = l2 = 330 30 fmin V Vs /min

½ flux linked with L2 (due to field produced by I,.,) also increases. Now, induced emf, current due to in loop L2 is such that, in nearby arms of ½ and L2 currents are in same direction. So, there is repulision between parallel arms of ½ and ½· As a result, L2 moves away from ½ .

33. (a) Electron i s subjected t o electric force due to field of capacitor plates.

e� � X -----41

(ZQq - q2 - 2Qq)

I

35. (b)

Initially, So, charge that is delivered by cell is CE Q1 = Ceq E =�

2

In series charge remains same for both capacitors.

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KVPY Question Paper 2014 Stream : SB/SX

When one of capacitors is filled with a dielectric then,

I

fi - �

= (_!!__J c

---� >I E �

Cq e

=

k C2 (C + kC)

1+ k

As battery remains connected, V = E. So, charge of combination Q2

= = (_!!__) Ceq V

CE

1+ k

So, extra charge given by cell after insertion of dielectric is

(-k-

- --1: ) CE !:t. Q = Q? - Ql = 1+ k 2 k-1 . CE 2 (k + 1)

36. (a) Breakdown occurs for a

semiconductor at a particular value of field which is independent of amount of doping concentration. Hence, breakdown field is same for both cases.

dV Now using, E = , we have dr

100 V2 V1 V2 ⇒ = = 20 1 d1 d2

37. (d) Kinetic energy of particle is

V2 = 5 V

⇒

K = --1: mv 2 2

⇒ 0. 05 X 16 X 10-19 = _!: X 16 X 10-26 X v2 2 0. 05 X 16 X 1 0 - 19 X 2 ⇒ 16 X 10-26 ⇒ v = 10 3 ms-1 Time taken by particle beam to travel l D _ through lm = t = = _ ⇒ t = 10-3 s u 10 3 Number of half lives occured in 1 0 -3 s t 10-3 s = n = - = -- = 0.0001 T112 6.9s 0.0001 1 So, fraction decayed = 1 - ( )

2

39. (b) Using Balmer's formula, we have For transition n = 3 to n = 2, A.:

=RU

2

-

(ii) NbCl5

) = R· � 9 4 3\

and for transition n = 5 to n = 3, 1 1 16 1 = R ( 2 - 2) = R. 9 X 25 5 3 A. 2 5 A. 2 = _ _ X 9 X 25 So, ;\1 9x 4 16 125 or A. 2 = - A.1 64

40. (a) Consider following reaction, 1 10 11 5 B - 5 B + on So, energy required to remove a neutron from 5 B1 1 is !:t.E = Binding energy of 5 B11 - Binding energy of 5 B10 . = 7.5 x 1 1 - 8.0 x 1 0 = 2 .5 MeV

per unit area per unit time I D.. P),.,. = = = he 4nr2hc (

:c )

160 X 6200 X 10- IO 4 X 7t X (1.8) 2 X 6.63 X 10-34 X 3 X 108 = 122 x 101 9 m-2s- 1

X= + 4 Oxidation state of Mn is MnO2 is + 4. Thus, Cr has the highest oxidation state (+ 6) among the given metals.

44. (b) [CrC1 2 (en)(NHs ) 2] is a type of

M(AA)XJ72 complex. It can have 3 geometrical isomers (2 trans and 1 cis form) as shown below.

41 . (d) AgBr � Ag + + Br-

cis

5 X 10-10 M 10-• M +

K,p = [Ag ] [Br- ] = (5 X 1 0 -10 ) ( 1 0 -3 ) = 5 X 10-1 3

Cl

If same amount of AgBr(s) is added to a 10-2 M aqueous solution of AgN03 , then the concentration of Ag + = 10-2 M 5 x 1 0- 13 = [10- 2 ) [Br- ] 11

[Br-J = 5 x 10- M

42. (a) Aniline reacts with excess

Br2/H 2 O to give 2, 4, 6-tribromoanilinc as a major product. The NH 2 group attached to benzene ring makes it more activating and occupy all the ortho and para­ positions.

0

+ 3Br2

7

H20

NPc1 NH3

trans

trans

45. (c) Amphoteric oxide are those oxides, which shows both the properties of acidic and basic oxides. Among the given elements, Al combine with oxygen to give Al203 , which is an amphoteric oxide. N(s) + 0 2(g) - N0 2 (s) (Acidic oxide)

P4 (s) + 5O 2 (g) - P4 O10 (s)

(Acidic oxide)

Al203 (s)

(Amphoteric oxide)

2Ka(s) + O2 (g) - Na202(s)

(Basic oxide)

B,�B, • :mBr Br

2, 4, 6-tribromoaniline

�

2Al(s) + � 0 2 (g) 2

= 0. 0001

38. (c) Photon flux = Number of photons

x + 5(- 1) = 0 x=+ 5 Oxidation state of Kb in NbCl5 is + 5. (iii) MnO2 X + 2 ( - 2) = Q

43. (d) Let the oxidation state of metals in given compounds be x. (i) K2CrO 4 2(+ 1) + X + 4 (- 2) = 0 x=S-2=+ 6 The oxidation state of K is K2CrO 4 is + 6.

46. (a) The temperature dependance of rate of a chemical reaction is expressed by Arrhenius equation, k = Ae-E0 /RT

RT

In k = lnA - � If we compare the above equation with the equation of a straight line, y = mx + c :. For the plot between log k versus 1 /T, slope = - E0 I R and intercept = lnA :. E1 > En and A1 > An

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KVPY Question Paper 2014 Stream : SB/SX 47. (c) The oxidation state of Ni in

CHO

Ni (CO) 4 is 0. Thus, the electronic configuration of Ni(O) is [Ar] 3d 8 4s0 .

I1� I 1� I1�1 1 1 1 I [1[1 I I I 3d

Ni

(Ground state)

H

4s

4p

As CO is a strong ligand pairing of electrons occur.

3

Ni(CO) 4 has tetrahedral geometry (sp hybridisation) and is diamagnetic due to absence of unpaired electrons.

o,

C(

(Step 1)

4

O

l

,,1

OW (Step 2)

:-\, lntramolecular 0 aldol . condensation

c�t HO CH 3

(�-hydroxy ketone)

- H�

l

HO

H OH

H

H

OH

HO

H

H

OH

HO

H

D-(+)-glucose

CH 20H

L-(-)-glucose

50. (a) The cubic lattice possesses

8 atoms on the corner. :. Fractional contribution by of atoms at

the corner = ..1:. 8 Each atom located at the face centered of

ccp structure is shared between two adjacent unit cell and only ..1: of each atom 2

belongs to a unit cell.

Ozonide

Zn+H20

AA

0

H

CH2 0H

sp3-hybridisation

48. (a)

HO

OH

�

0

0

: ¢r CH3 (X)

(a.�-unsaturated carbonyl)

In ste p I Ozonolysis of alkene involves the addition of ozone molecule to alkene to form ozonide and then cleavage of ozomide by Zn - H 2 0 to smaller carboxyl groups. In ste p 2 This step involves intramolecular aldol condensation to form B-hydroxy ketone, which readily loses water to form a, B-unsaturatcd carbonyl compound. 49. (b) D and L before the name of any compound commonly indicates the relative configuration. For assigning the configuration of monosaccharidcs it is the lowest asymmetric carbon atom, which is compared to glyccraldehyde. For glucose, if -OH on the lowest asymmetric carbon is on the right side. Then, it is assigned D-configuration and if OH is on the left side then it is assigned L- configuration. Both D and L configuration are mirror images of each other.

Thus, fractional contribution of each atom at face = ..1:. 2 51 . (a) For the reaction, 2A � B + C [B] [C] . Kc - 0.5) QC = --(Given, [A]2 If QC > Kc then the reaction will proceed in backward direction. The value of Qc in the given options can be calculated as follows: [10-2] [10-2] 10-4 = = 102 (i) QC = [l0-3 ]2 10-6 As Qc > Kc the reaction will proceed in backward direction. [ 1 0 -2] [10-2] 1 0 -4 = = 10-2 (ii) QC = [l0 -1 ]2 l 0 -2 As Qc < Kc , the reaction will proceed in forward direction. 3 2 [10- ] [10- ] 10-5 = = 1 0 -1 (iii) Q = 2 2 10-4 C [10- ]

Herc, QC < Kc , the reaction will proceed in forward direction. [10-3 1 [10-3 J 10-6 . = = 10-2 (1v) QC = 10-4 [IO-2]2 The reaction will proceed in forward direction as Qc < Kc .

-?c�

CHO

52. (c) Major products formed when t-butyl methyl ether reacts with H l are tert-butyl iodide and methanol.

CH3

I

c�

HI

OCH3 - C�

C�

I -?-

139 I

Clfa

+ C� OH The cleavage of C-0 bond in ethers takes place under drastic conditions with excess of hydrogen halide, HX. This reaction follows SNl mechanism.

53. (d) Given,

2 1 ),NaCl = 126 S cm mol2 1 ),KCI = 150 S cm mol2 1 AN . aOH = 250 S cm mol-

),KOH = /\.KCI + x;laOH - X�aCI = (150 + 250) - 126 = 400 - 126 = 274 S cm2mol- 1

Thus, the molar conductivity of KOH is 274 S cm2mol-1 .

54. (b) 4-formyl bcnzoic acid on

treatment with one equivalent of hydrazine followed by heating with ale. KOH gives 4-methyl benzoic acid the major product.

CHO

NH2-NH2

l9J

ale. KOH

�

+

COOH coo-K This reaction is known as Wolf-Kishner reduction, where the aldehyde or ketone when treated with hydrazine and KOH, get reduced to methylene group to form a hydrocarbon. This reagent doesn't affect COOH group.

55. (c) X with atomic number 33 and Y

with atomic number 17 belong to 15th and 17th group respectively. Thus, the valency of X and Y are 3 and 1 respectively.

x

x+

y-

3

I

= XY3

:. The molecular formula of a stable

compound formed between X and Y is X:fa . Herc, X could be As (Z = 33) and Y could be Cl (Z = 17) which gives compound AsC13 . + 7

-1

56. (d) KMn0 4 + KI+ H2S0 4 +2

0

-

Mn80 4 + 12 + K280 4 + H20

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KVPY Question Paper 2014 Stream : SB/SX

Balance factor for Mn= 5 Balance factor for I = 1 Equivalent of KMnO 4 = equivalent of KI Number of equivalents = y x Kumbcr of moles where, y is the balance factor. = YMn X nMn = YKr X nKr = 5 X xMn = l X l nMn

=5

57. (c) The correct way of reporting the

average value is that it must have exactly same number of digits after decimal, which has least digit after decimal. 10.9+ 1 1 . 4042 + 1 1 . 42 Average value 3

= 1 1 .2414 = 1 1 . 2 A s among the given data least digit after decimal is 1 digit value, so the average value must also be have one digit after decimal, i.e. 1 1.2.

58. (a) Given, latent heat of melting of ice, tiH = 6 kJ mol- 1 tiH tiS = -Trreezing

6 kJ 6 x 10 3 J = �� = ---273 K 273 K

= 2 1 9 = 2 2 J/K

59. (a)

� I

Ullmann reaction

This reaction is known as Ullmann reaction. It is a coupling reaction between aryl halides and copper.

60. (c) The five degenerate d-orbitals of

the metal ion split into different sets of orbital having different energies in the presence of electrical field of ligand. In octahedral metal complexes dx 2 2 -y and d 2-orbitals point towards the axes along zthc direction of the ligand will experience more repulsion and will be raised in energy and the dxy , dzx and dyz -orbitals are directed between the axes will be lowered in energy. ln tetrahedral metal complexes dxy , dzx and dyz experience more repulsion and have more energy than dX 2 2 and dz2 • -y 6 1 . (c) Holocrinc is a term used to classify the mode of secretion in exocrine glands. Holocrinc secretions arc produced

in the cytoplasm of the cell and released by the rupture of the plasma membrane, which destroys the cells and results m the secretion of the product into the lumen.

62. (a) Commensalism does not promote cocvolution. Cocvolution occurs when two or more species reciprocally affect each other's evolution. Each party in coevolutionary relationship exerts selective pressure on the other, thereby affecting each other's evolution. Cocvolution includes many forms of mutualism, host parasite and predator-prey relationship between species, as well as interspecific competition, (i.e. competition within or between species). 63. (b) The process of arrangement of plants into defined vertical layers depending on their height is called stratification. The tropical rainforest is a hot and moist biome found near the equator of the earth. It has a dense growth of flora arranged at different strata of the forest. Thus, the correct answer is option (b). 64. (d) The third ventricle is one of the

four connected fluid-filled cavities comprising the ventricular system within the mammalian brain. l t is a median cleft in the dicncephalon between the two thalami and is filled with Cerebrospinal Fluid (CSF).

65. (c) Gene expression is the process by which information from a gene is used in the synthesis of a functional gene product. These products are often protein, but in non-protein coding genes such as transfer RKA (tRNA) or small nuclear RNA (snRNA) genes, the product is a functional RNA. Thus, option (c) is correct. 66. (a) Divergent evolution occurs when two different species share a common ancestor but have different characteristics from one another. An example of divergent evolution is human's arm, the whale's fin, cheetah's legs and bat's wing. All four forelimbs perform different purposes, but all contain the same bones in different sizes and shapes because all four animals (human, cheetah, whale, bat) evolved from a common ancestor whose legs contained those bones. 67. (d) Conjugation is a type of sexual reproduction in Paramecium. lt is a temporary union of two individuals of

same species for mutual exchange of genetic materials. Conjugation results in rejuvenation and transference of hereditary material (i.e. recombination) in different strains.

68. (b) The chemicals which perceive the photopcriodic stimulus arc called phytochromes, which are present on leaves. Thus, if leaves of a plant arc removed, photoperiodic response will not occur in it. 69. (a) Cells in the testes called Leydig

cells, produce testosterone in response to the production of Luteinising Hormone (LH). Testosterone is the primary sex hormone produced in males. lt maintains typical male characteristics of the body, e.g. facial hair.

70. (d) Transversion refers to a point

mutation in deoxyribonucleic acid, where a single purine is changed for a pyrimidine or vice-versa. A transvcrsion can be spontaneous or can be caused by ionising radiation or alkylating agent. It can only be reversed by spontaneous reversion.

7 1 . (c) Acetylcholine is the

neurotransmitter used at the neuromuscular junction by the ends of axon- in other words, it is the chemical that motor neurons of the nervous system release in order to activate muscles. The synaptic bulbs of axon have vesicles which are filled with acetylcholine.

72. (a) lf a single bacterium is placed in

a nutrient medium, it will consume the nutrients and begin to grow and multiply by repetitive division. This generates thousands to millions to billions of cells that begin to pile up, becoming visible to the naked eye. This pile of cells originates from one cell and is called bacterial colony. Thus, option (a) is correct.

73. (d) Rhinoviruscs arc the causative agents of common cold. The common cold is viral infectious disease that infects the upper respiratory system. lt is also known as acute viral rhinopharyngitis and acute coryza. Common cold symptoms include dry or sore throat, blocked or runny nose and sneezing. 74. (c) Ebola virus is a member of the

Filovirus family and causes a severe febrile disease in humans with high case fatality rates. The non-segmented negative-sense single-stranded RNA genome of cbola virus is 19 kb in length and contains seven genes.

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KVPY Question Paper 2014 Stream : SB/SX 75. (d) The main function of electron transport chain is to produce ATP, during which water is produced as a byproduct. At the end of the chain, the electrons are taken up by the oxygen molecules to make water. This is why, oxygen is called the final electron acceptor. 76. (a) The Benedict's test is used to test for simple sugars. The nitric acid reacts with proteins to form yellow nitrated products, thus used to detect proteins in sample. Therefore, the tube 'P', positive for Benedict's test and tube 'Q' positive for nitric acid contains sugar and protein, respectively. 77. (c) For a circular DNA, 'N' represents the number of fragments of DNA obtained if the plasmid has 'N' recognition sequences for a given restriction endonucleus. There, if a restriction enzyme have 3 sites of action, the number of fragments produced will be 3 for a circular DNA. 78. (b) Xylem moves water from roots to the leaves by the process of transpiration. As the relative humidity of the air surrounding the plant rises, the transpiration rate falls. lt is easier for water to evaporate into dryer air than into more saturated air. Thus, option (b) is correct. 79. (a) For writing complementary sequence just use the base pairing rules, C bonds with G and A bonds with T and for each base in the original sequence. Suppose we have 5'-CGTACTA-3' lts complementary sequence will be 3'-GCATGAT-5' But since the two strands run in opposite directions. So, technically more correct answer is 5'-TAGTACG-3'. BO. (b) If non-disjunction occurs during meiosis-I, this means that atleast one pair of homologous chromosomes did not separate. The end result is two cells that have an extra copy of one chromosome and two cells that arc missing that chromosome. Therefore, if the egg cell has 6 chromosomes produced from 2n = 14 chromosome plant, it would be because of non-disjunction in meiosis-I. 8 1 . (a) We have, n fa< x) = anx n - l + an _ 1 x - 2 + ... + a2 x + Ui a = (Ui , a2 , (¾ , . • . , an ) of (1, 2, 3, ... , n)

S0 = Sum of roots of (0 (x) = 0

S = I.S0

s

r - a,. A - an I - + ... + A - Ui =- A + a a lli n n 1 l -

lj

83. (b) We have, y = ax2 + bx+ c Parabola has two roots one is positive and one is negative.

(2,-2) Clearly, c is negative a> 0

=-!!. = 2 a

From AM 2': HM Ui + a2 + a3 + ... + a,. 2':

n

_ .!?, > 0

1

n

1

1

- + - + ... + a2 an Ui

S $ - n (n - 1) S$ - n! 82. (c) We have,

I, " Ck w

i=0

[·: a > OJ

a b< O

be> O Hence, option (b) is correct.

84. (d) We have, cl , C2 , c; , ... , en be circle with radii r1 , r2 , •.• , rn respectively. q and Ci + 1 touch externally X-axis and y = 2✓ 2x + 10 are tangent of each circle.

k

, where ro is

cube root of unity.

f

"Ck(jl = nco + nc!ro + ne,p:,2 + k�O = (1 + wt = (- w2) n = (- l) n ro2n

I " C ro

e k=O

k

k

[·: l + w + w 2 = OJ

= I e( - t)" m 2• I = I e 0

18. The number of integers n with 100 s n s 999 and containing at most two distinct digits is � 2M W 2� (c) 324 (d) 360

19. For an integer n let Sn ={n + l, n + 2, ...... , n + 18}. Which of the following is true for all n � 10 ? (a) Sn has a multiple of 19 (b) Sn has a prime (c) Sn has at least four multiples of 5 (d) Sn has at most six primes

20. Let P be a closed polygon with 10 sides and

10 vertices (assume that the sides do not intersect except at the vertices). Let k be the number of interior angles of P that are greater than 180° . The maximum possible value of k is � 5 W 3 M7 W 9

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PHYSICS 21. Consider an initially neutral hollow conducting

spherical shell with inner radius r and outer radius 2r. A point charge +Q is now placed inside the shell at a distance r /2 from the centre. The shell is then grounded by connecting the outer surface to the earth. P is an external point at a distance 2r from the point charge + Q on the line passing through the centre and the point charge +Q as shown in the figure. '

where a and b are constant characteristics of the gas. Which of the following can represent the equation of a quasistatic adiabat for this gas (assume that, Cv is the molar heat capacity at constant volume is independent of temperature)? (a) T (V - nbl'cv = constant (b) T (V - nbfv!R = constant (c) + --5!!!._ ) (V - nbl'Cv = constant

(r

V2R

n2ab R 2 ) (V - nbfv' = constant V R

(d)( T +

25. A blackbox (BB) which may contain a combination of electrical circuit elements (resistor, capacitor or inductor) is connected with other external circuit elements as shown below in the figure (A). After the switch S is closed at time t = 0, the current I as a function of time t is shown in the figure (B).

2r /

(A) fJ

The magnitude of the force on a test charge + q placed at P will be 1 1 4 1 9qQ (d)0 (c)__ qQ2 (a) __ qQ (b)__ 2 2 41tE0 25r 41tE0 100r 41tE0 4r

22. Consider the circuit shown in the figure below.

� !c �

-J 1 I

j

All the resistors are identical. The charge stored in the capacitor, once it is fully charged is 5 2 5 (d)- CV (a) 0 (b)- CV (c)- CV 8 13 3

23. A nuclear decay is possible if the mass of the parent

nucleus exceeds the total mass of the decay particles. If M(A, Z) denotes the mass of a single neutral atom of an element with mass number A and atomic number Z, then the minimal condition that the � decay X1 ➔ Yf + 1 + �- + Ye will occur is (m. denotes the mass of the � particle and the neutrino mass mv can be neglected) (a) M(A, Z) > M(A, Z + 1)+ me (b) M(A, Z) > M(A, Z + 1) (c)M(A, Z)> M(A, Z + 1)+ Zm. (d)M(A, Z)> M(A, Z + 1) - me

24. The equation of state of n moles of a non-ideal gas ( p + �� J (V - nb) = nRT,

can be approximated by the equation

(

Ls

B' Jc�

O

t

From this we can infer that the blackbox contains (a) a resistor and a capacitor in series (b) a resistor and a capacitor in parallel (c) a resistor and an inductor in series (d)a resistor and an inductor in parallel

26. In a photocell circuit the stopping potential V0 is a

measure of the maximum kinetic energy of the photoelectrons. The following graph shows experimentally measured values of stopping potential versus frequency v of incident light. Vo(V)

5 4 3

/

2 0 -1 -2

!} 4

1/

0.8

/

/'

1 .2

V

.,/

1 .6

v( 2

,

The values of Planck's constant and the work function as determined from the graph are (taking, the magnitude of electronic charge to be e = 1.6 x 1 0-19 C) (a) 6.4 x 1 0-34 Js, 2.0 eV (c) 6.4 x 1 0-34 Js, 3.2 eV

(b) 6.0 x 1 0-34 Js, 2.0 eV (d)6.0 x 1 0-34 Js, 3.2 eV

27. An engine moving away from a vertical cliff blows a

horn at a frequency /. Its speed is 0.5% of the speed of sound in air. The frequency of the reflected sound received at the engine is (b) 0.995/ (a) 0.990/ (d)1010{ (c) 1005/

c==-=--=-=,.......,..-=,----. ..-.-; ,......,,. w-=-,w w � NI .JEEB o o KS. l�

149

KVPY Question Paper 2013 Stream : SB/SX 28. An arrangement with a pair of quarter circular coils of radii r and R with a common centre C and carrying a current I is shown in the figure.

32. The bulk modulus of a gas is defined as B =- Vdp I dV. For an adiabatic process the variation of B is proportional to p" . For an ideal gas n is (a) zero (c) 5/3

(b) 1 (d) 2

33. Photons of energy 7 eV are incident on two metals A

R/

•'

/ .,'(

The permeability of free space is µ 0 • The magnetic field at C is (a) µ 0 1 (1 / r - 1 / R) I 8 into the page (b) µ 0 1 (1 / r - 1 / R) I 8 out of the page (c)µ 0 1 (1 / r + 1 / R) I 8 out of the page (d)µ 01 (1 / r + 1 / R) / 8 into the page

and Bwith work functions 6 eV and 3 eV, respectively. The minimum de-Broglie wavelengths of the emitted photoelectrons with maximum energies are "-A and 'A B respectively, where "-A ! 'A B is nearly (b) 1.4 (a) 0.5 (c) 4.0 (d) 2.0

34. An electron enters a chamber in which an uniform

magnetic field is present as shown in figure. Ignore gravity.

29. The circuit shown has been connected for a long time. The voltage across the capacitor is 1 kn Magnetic fiel d 6V -=-

(b) 2.0 V (d) 4.0 V

(a) 1 .2 V (c) 2.4 V

30. A wheel of radius R with an axle of radius R/2 is as

shown in the figure and is free to rotate about a frictionless axis through its centre and perpendicular to the page. Three forces (F, F, 2F) are exerted tangentially to the respective rims as shown in the figure.

2F

The magnitude of the net torque acting on the system is nearly (a) 3.5FR (b) 3.2FR (c) 2.5FR (d) l.5FR

31. Two species of radioactive atoms are mixed in equal

number. The disintegration constant of the first species is A and of the second is 'A/3. After a long time the mixture will behave as a species with mean life of approximately (a) 0.70 1 A (b)2.1 0 1A (c)l . 00 I A (d) 0.521A

During its motion inside the chamber (a) the force on the electron remains constant (b) the kinetic energy of the electron remains constant (c) the momentum of the electron remains constant (d)the speed of the electron increases at a uniform rate

35. A ray of3light incident on a glass sphere (refractive

index ✓ ) suffers total internal reflection before emerging out exactly parallel to the incident ray. The angle of incidence was (a) 75° (b) 30° (c) 45° (d)60 °

36. Young-Laplace law states that the excess pressure inside a soap bubble of radius R is given by

b.P = 4cr / R, where cr is the coefficient of surface tension of the soap. The EOTVOS number E0 is a dimensionless number that is used to describe the shape of bubbles rising through a surrounding fluid. It is a combination of g, the acceleration due to gravity p the density of the surrounding fluid cr and a characteristic length scale L which could be the radius of the bubble. A possible expression for E0 is pL2 pgL2 gL2 (c) (a) _ff_ (b) (d) �

�

CT

�

37. A plank is resting on a horizontal ground in the

northern hemisphere of the earth at a 45° latitude. Let the angular speed of the earth be w and its radius re . The magnitude of the frictional force on the plank will be mr 2 mr 2 (c) em (b) em (d)zero (a) mr.m2

-J2

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KVPY Question Paper 2013 Stream : SB/SX

38. The average distance between molecules of an ideal gas at STP is approximately of the order of (a) 1 nm (b) 1 00 nm (c) 1 0 0 cm (d) 1 µ m

39. A point particle of mass 0.5 kg is moving along the X-axis under a force described by the potential energy V shown below. It is projected towards the right from the origin with a speed u. V(in J) 3

-3

-2

-7

different sound waves in air with time at a given position. Both the figures are drawn to the same scale. p

G4 G3

Pressure The gas with the highest value of Henry's law constant is (c) G3 (a) G4 (b) G2 (d) Gl

46. For the reaction, A � nB, the concentration of A

decreases from 0.06 to 0.03 mol 1- 1 and that of B rises from O to 0.06 mol 1- 1 at equilibrium. The values of n and the equilibrium constant for the reaction, respectively, are (a) 2 and 0.1 2 (b) 2 and 1 .2 (c) 3 and 0.1 2 (d) 3 and 1 .2

47. The reaction of ethyl methyl ketone with Cl2/excess

v V V V v v Wave 1

Wave 2 Which of the following statements is true? (a) Wave 1 has lower frequency and smaller amplitude compared to wave 2 (b) Wave 1 has higher frequency and greater amplitude compared to wave 2 (c)Wave 1 has shorter wavelength and greater amplitude compared to wave 2 (d) Wave 1 has shorter wavelength and smaller amplitude compared to wave 2

CHEMISTRY 41 . Among the following the set of isoelectronic ions is (a) Na +, Mg2+ , F-, c1(c)Na +, Mg2+ , F-, 02-

(a) C� CONHCONHNH2 (b) CH3 CON(KH2 )CONH2 (c) C� CONHNHCONH2 (d) C�CH2NHNHCONH2

2 3 4 5 x(in m)

40. The figure below shows pressure variation in two

V

44. Ethyl acetate reacts with NH2NHCONH2 to form

G2, G3, G4 etc.) in a given solvent with pressure at a constant temperature is shown in the plot.

What is the minimum value of u for which the particle will escape infinitely far away from the origin? (a) 'J✓'2, ms- 1 (b) 2 ms- 1 (c)4 ms-1 (d)The particle will never escape

p

produce 2, 4, 6-tribromophenol, is (b) 1 ,3-cyclohexanedione (a) 1,3-cyclohexadiene (c) salicyclic acid (d)cyclohexanone

45. The variation of solubility of four different gases (G 1,

2

- 4

43. The compound which reacts with excess bromine to

(b)Na +, Ca2+, F-, o(d) Na+, K+, s2-, cl-

42. For a zero-order reaction with rate constant k, the slope of the plot of reactant concentration against time is (b)Ii (a) Ii I 2. 303 (c)- h 12. 303 (d) - h

OH- gives the following major product (a) CICH 2 CH 2 COCH 3 (b) CH 3 CH 2 COCCl3 (c) ClCH 2 CH 2 COCH 2 Cl (d)CH 3 CC1 2 COCH 2 Cl

48. The compound that readily tautomerises is (a) CH 3 COCH 2 CO2 C2 H 5 (c) CH 3 COCH 2 CH 2 CH 3

(b) CH 3 COCH 2 CHCH 3 (d)(CH 3 )3 CCOC(CH 3 )3

49. Hydrolysis of BC1 3 gives X, which on treatment with sodium carbonate produces Y. X and Y, respectively, are (a) H3 BO3 and NaBO2 (b) H 3 BO3 and Na2 B4 O 7 (c) B2 O3 and NaBO2 (d)B2 O3 and Na2 B4 O 7

50. The numbers of lone pair (s) on Xe in XeF2 and XeF4 are, respectively (a) 2 and 3 (c) 3 and 2

(b) 4 and 1 (d)4 and 2

51 . The entropy change in the isothermal reversible

expansion of 2 moles of an ideal gas from 10 to 100 L at 300 K is (a) 42.3 J K- 1 (b) 35.8 J K- 1 1 (c) 38.3 J K(d)32.3 J K- 1

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151

KVPY Question Paper 2013 Stream : SB/SX 52. D-glucose upon treatment with bromine-water gives CH � H H HO H (b) H i OH H OH CHO

co �� H HO H (a) H i OH H OH CHpH

! CH

(c)

HO H H

CH

� H H HO H (d) H i OH H OH COOH

�

H OH OH CHpH

53. In the structure of borax, the numbers of boron atoms and B - 0 - B units, respectively are (b)4 and 3 (a) 4 and 5 (c) 5 and 4 (d) 5 and 3

54. The number of peptide bonds in the compound, 0

CH 3

H N

0

NH NH 2 is H 3cA N)ly Y H O �r CH 3 H3C (a) 1 (b) 2 (d) 4 (c)3

55. For the isothermal reversible expansion of an ideal gas (a) !).H > 0 and !).U = 0 (c)!).H = O and !). U = 0

(b)!).H > O and !). U < 0 (d) AH = O and !). U > 0

56. If the angle of incidence of X-ray of wavelength 3A

which produces a second order diffracted beam from the (100) planes in a simple cubic lattice with interlayer spacing a = 6 A is 30°, the angle of incidence that produces a first order diffracted beam from the (200) planes is (a) 1 5° (c) 30 ° (d) 60 ° (b)45°

57. The number of ions produced in water by dissolution

of the complex having the empirical formula, CoC1:J · 4NI-Ia , is �4 W2 Wl W3 58. The spin-only magnetic moments of [Fe (NH3 )6] 3 + and [FeF6] 3 - in BM are, respectively (a) 1 .73 and 1 .73 (b) 5.92 and 1 .73 (c) 1 .7 3 and 5.92 (d) 5.92 and 5.92

59. The order of SN 1 reactivity in aqueous acetic acid solution for the compounds

II

1. HaC-C-CH2

-

Cl

2. HaC----CH2 ----C H2----Cl

18

(b) 1 > 3 > 2 (d) 3 > 1 > 2

(a) 1 > 2 > 3 (c) 3 > 2 > 1

60. An ionic compound is formed between a metal M and a non-metal Y. If M occupies half the octahedral voids in the cubic close-packed arrangement formed by Y, the chemical formula of the ionic compound is (c) M,7 (d)MY3 (a) MY (b) MY2

BIOLOGY

61. Human foetal haemoglobin differs from the adult haemoglobin in that it has (a) higher affinity for oxygen (b) lower affinity for oxygen (c) two subunits only (d)glycosylated

62. Nucleolus is an organelle responsible for the production of (a) carbohydrates (c) lipids

(b) messenger RNA (d)ribosomal RNA

63. The sequences of four DNA molecules are given below (i) TATATATATATATA ATATATATATATAT (ii) TTTCCCGGGAAA AAAGGGCCCTTT (iii) TTGCGTTGCGCC AACGCAACGCGG (iv)GCCGGATCCGGC CGGCCTAGGCCG Which one of these DKA molecules will have the highest melting temperature (Tm )? � ill Wi W ii W� 64. If DNA codons are ATG GAA, insertion of thymine after the first codon results in (a) non-sense mutation (b) mis-sense mutation (c) frameshift mutation (d)silent mutation

65. Genetic content of a cell reduces to half during (a) meiotic prophase-1 (c) meiotic prophase-II

(b) mitotic prophase (d)meiotic telophase

66. Which one of the following techniques is used for the detection of proteins? (a) Northern blotting (c) Southern blotting

(b) Western blotting (d)In situ hybridisation

67. Fission yeasts are (a) archaebacteria (c) prokaryotes

(b) eubacteria (d)eukaryotes

68. In green leaves, the light and dark reactions occur in (a) stroma and grana respectively (b) grana and stroma respectively (c) cristae and matrix respectively (d)both occur in cytoplasm

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KVPY Question Paper 2013 Stream : SB/SX 75. In orange and lemon, the edible part of the fruit is

69. According to Mendel, ............... segregate and .............. assort independently. (a) alleles of a gene, alleles of different genes (b) alleles of different genes, alleles of a gene (c) dominant traits, recessive traits (d)recessive traits, recessive traits

(a) placenta (b) thalamus (c) hairs of the ovary wall (d)succulent mesocarp

76. Which one of the following statements about

nitrogenase is correct? (a) lt is sensitive to CO2 and therefore present in isolated nodules (b) lt requires 02 and therefore functional during the day (c) lt is sensitive to 02 and therefore is functional in anaerobic environments (d) lt is sensitive to light and therefore functions only in dark

70. The two enzymatic activities associated with RuBisCO are (a) oxidase and oxygenase (b) oxygenase and carboxylase (c)oxidase and carboxylase (d)oxygenase and carbamylation

77. Part of epidermis that keeps out unwanted particles

71 . Chlorofluorocarbons (CFCs) are believed to be

associated with cancers because (a) CFCs react with DNA and cause mutations (b) CFCs react with proteins involved in DNA repair (c)CFCs destroy the ozone layer and permit harmful UV rays to reach on the earth (d)CFCs react with DNA polymerase and reduce fidelity of DNA replication

72. Morphogenetic movements take place predominantly during the following embryonic stage (a) blastula (b)morula (c)gastrula (d) fertilised eggs

(b) pancreas (d) muscle

= 100 and /(x) s100 for all

real x Which of the following statements is NOT necessarily true? (a) The coefficient of the highest degree term in f (x) is negative. (b) f (x) has at least two real roots. (c) If x ;t 1 / 2 then f (x) < 100. (d)At least one of the coefficients of f(x) is bigger than 50.

82. Let a, b, c, d be real numbers such that

I,

locus are at Hardy-Weinberg equilibrium. The frequencies are p = 0.6 and q = 0.4. The proportion of the heterozygous genotype in the population is (a) 0.24 (b) 1 (c) 0.48 (d)0.1 2 (b) spleen and stomach (d)liver and muscles

(2 Marks Questions)

81 . Let f(x) be a non-constant polynomial with real

t(½)

habitats show (a) low reproductive ability (b) high dispersal ability (c) slow growth and maturation (d) high competitive ability

(a) heart and blood (c) bones and lymph

MATHEMATICS

coefficients such that

78. Species that are most effective at colonising new

80. In vertebrates, glycogen is stored chiefly in

74. Stroke could be prevented/treated with (a) balanced diet (b)clotting factors (c)insulin (d) blood thinners

� PART - I I

(b) squamous epithelium (d)cuboidal epithelium

79. In a large isolated population, alleles p and q at a

73. The only organ which is capable of producing fructose in human is (a) liver (c) seminal vesicles

is called (a) columnar epithelium (c) ciliated epithelium

(ak3 + bk2+ ck+ d) = n 4, for every natural number n. k�1 Then, l a I+ l b I+ l e I + I d I is equal to (a) 1 5 (b) 1 6 (c) 31 (d) 32

83. The vertices of the base of an isosceles triangle lie on a parabola y2 = 4x and the base is a part of the line y = 2x- 4. If the third vertex of the triangle lies on the X-axis, its coordinates are

(a) rn , o)

(b) (i , o)

(c) rn, o)

1

(d) ( � , o)

84. In a MBC, let G denote its centroid and let M, N be

points in the interiors of the segments AB, AC, respectively, such that M, G, N are collinear. If r denotes the ratio of the area of f.AMN to the area of ABC, then (a) r = 1 / 2 (b)r > l/ 2 (c)4 / 9 $ r < 1 / 2 (d) 4 / 9 < r

85. Let XY be the diameter of a semi-circle with centre 0. Let A be a variable point on the semi-circle and B another point on the semi-circle such that AB is parallel to XY.

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KVPY Question Paper 2013 Stream : SB/SX

The value of L.BOY for which the inradius of MOB is maximum, is

( Jg - 1)

(a) cos-1 --

2

(c)� 3 X

x2 :('

(d) � 5 4

X

86. Let f(x) = 1 + - + - + - + -. The number of real 1! 2! roots of f(x) = 0 is (b) 1 (a) 0

3!

4!

(d) 4

(c)2

87. Suppose that the earth is a sphere of radius 6400 km. The height from the earth's surface from where exactly a fourth of the earth's surface is visible, is (b) 320 0 -f2 km (a) 320 0 km (c) 3200 ./3 km (d) 6400 km

88. Let n be a positive integer. For a real number x, let [x] denote the largest integer not exceeding x and n

+1

[x]

({ x}ix]

{x} = x - [x]. Then, f __ dxis equal to (a) log. (n) n (c)__ n+l

1

The bullet hits the sphere at a height h from the table and sticks to its surface. If the sphere starts rolling without slipping immediately on impact, then Jz_ 4m + 3M fz_ m + M (a) = (b) = R 2(m + M) R m + 2M fz_ 1 0m + 7M fz_ 4m + 3M (c) = (d) = R m+ M R 5(m + M)

92. A small boy is throwing a ball towards a wall 6 m in

front of him. He releases the ball at a height of 1.4 m from the ground. The ball bounces from the wall at a height of 3 m, rebounds from the ground and reaches the boy's hand exactly at the point of release. Assuming the two bounces (one from the wall and the other from the ground) to be perfectly elastic, how far ahead of the boy did the ball bounce from the ground? (a) 1 .5 m (b) 2.5 m (c) 3.5 m (d) 4.5 m

93. In the p- V diagram below, the dashed curved line is an adiabat.

p

1 (b)-n+ 1 1 1 (d) l+ - + ..... + 2 n

89. A box contains coupons labelled 1, 2, ... , 100. Five

coupons are picked at random one after another without replacement. Let the numbers on the coupons be Xi_, x2 , • • . , x5. What is the probability that Xi_ > x2 > X:i and X:i < x4 < Xr; ? (a) 1 /120 (b)1 / 60 (c) 1 /20 (d) 1/10

90. In a tournament with five teams, each team plays

against every other team exactly once. Each game is won by one of the playing teams and the winning team scores one point, while the losing team scores zero. Which of the following is NOT necessarily true ') (a) There are at least two teams which have at most two points each. (b) There are at least two teams which have at least two points each. (c) There are at most three teams which have at least three points each. (d) There are at most four teams which have at most two points each.

PHYSICS 91. A bullet of mass m is fired horizontally into a large

sphere of mass M and radius R resting on a smooth horizontal table.

�-------> V

For a process that is described by a straight line j oining two points X and Y on the adiabat (solid line in the diagram) heat is

(Hint consider the variation in temperature from X to Y along the straight line) (a) absorbed throughout from X to Y (b) released throughout from X to Y (c) absorbed from X upto an intermediate point Z (not shown in the figure)and then released from Z to Y (d) released from X upto an intermediate point Z (not shown in the figure)and then absorbed from Z to Y

94. A singly ionised helium atom in an excited state

(n = 4) emits a photon of energy 2.6 eV. Given that the ground state energy of hydrogen atom is- 1 3.6 eV, the energy E r and quantum number n of the resulting state are respectively, (a) Er = - 1 3.6 eV, n = l (b) Er = - 6.0 eV, n = 3 (c) Er = - 6.0 eV, n = 2 (d)E1 = - 1 3.6 eV, n = 2

95. The figure below shows a circuit and its input voltage V; as function of time t.

2R

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KVPY Question Paper 2013 Stream : SB/SX

Assuming the diodes to be ideal, which of the following graphs depicts the output voltage V0 as function of time t ?

(a)

�

t

(c) � � - V ,t 4

(b)

-�3�»

I

99. A block of mass m slides from rest at a height H on a

frictionless inclined plane as shown in the figure. It travels a distance d across a rough horizontal surface with coefficient of kinetic friction µ and compresses a spring of spring constant k by a distance x before coming to rest momentarily. Then the spring extends and the block travels back attaining a final height of h.

(d) � 'vl-4V , t

96. A ball is rolling without slipping in a spherical

shallow bowl (radius R) as shown in the figure and is executing simple harmonic motion. If the radius of the ball is doubled, then the time period of oscillation

H

1

�

�d �

Then, (a) h = H - 2µ (d + x) (b) h = H + 2µ (d- x) (c) h = H - 2µd + kx2 I mg (d) h = H - 2µ (d + x)+ kx2 /2mg

100. A metallic prong consists of 4 rods made of the same material, cross-sections and same lengths as shown below. The three forked ends are kept at 100 ° C and the handle end is at 0 ° C. The temperature of the junction is

(a) increases slightly (b) is reduced by a factor of 1/2 (c) is increased by a factor of 2 (d) decreases slightly

horizontally and then up to a point X at height h on an inclined plane before rolling down, as shown in the figure below.

T = 0°C ------+--- T = 1 00°C

97. A solid sphere rolls without slipping, first

°

(a) 25 C

X

(b) 50° C

°

(c) 60 C

T = 100° C

(d)75° C

CHEMISTRY 101. The major product obtained in the reaction of aniline The initial horizontal speed of the sphere is (a) ..}lOgh /7 (b).j7gh / 5 (c).j5gh /7 (d).j2gh

98. The three processes in a thermodynamic cycle shown

in the figure are : Process 1 ➔ 2 is isothermal; Process 2 ➔ 3 is isochoric (volume remains constant); Process 3 ➔ 1 is adiabatic.

p

�-------4 V

The total work done by the ideal gas in this cycle is 10 J. The internal energy decreases by 20 J in the isochoric process. The work done by the gas in the adiabatic process is -20 J. The heat added to the system in the isothermal process is (a) 0 J (b) 10 J (c) 20 J (d) 30 J

with acetic anhydride is

NH COCH ,

a ( )

(b)

(c)

()

H 3C

NHCOCH3

0

O

N' COCH, /

�

�

(d)

0

COCH3 NH2

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155

KVPY Question Paper 2013 Stream : SB/SX 102. The maximum number of isomers that can result

from monobromination of 2-methyl-2-pentene with N-bromosuccinimide in boiling CC1 4 is �4 �3 W2 Wl

103. The compound X(C 7 H9N) reacts with

benzene sulphonyl chloride to give Y (C 1 3 H13 N0 2S) which is insoluble in alkali. The compound X is

(c)

C(

(b)

NH2

CH 3

(d) H 3C

110. In aqueous solution, [Co(H2 O) 6 ] 2 + (X), reacts with

molecular oxygen in the presence of excess liquor NH3 to give a new complex Y. The number of unpaired electrons in X and Y are, respectively (c) 3, 3 (a) 3, 1 (6) 3, 0 (d)7, 0

O

111. 1 0 9 bacteria were spread on an agar plate containing

NH 2

dissolved. At 100° C, the vapour pressure of the solution is 750 mm Hg. Assuming that the compound does not undergo association or dissociation, the molar mass of the compound in g mo1- 1 is (b) 1 82 (d) 228

105. The standard electrode potential of Zn 2 + /Zn is

- 0.76 V and that ofCu 2 +/Cu is 0.34 V. The emf (V) and the free energy change (kJ mol- 1 ), respectively for a daniell cell will be (a) - 0.42 and 81 (b)1 . 1 and - 21 3 (c)- lland 213 (d) 0.42 and - 81

106. Consider the equilibria (1) and (2) with equilibrium constants K1 and K2 , respectively. S 0 2 (g) + 1/2 0 2 (g) � S0 3 (g) 2S0 3 (g) � 250 2 (g) + 0 2 (g) K1 and K2 are related as (a) 2K.1 = Ki

reduce 100 mL of 0.25 N iodine solution, is (a) 6.20 g (b) 9.30 g (c) 3.10 g (d)7. 75 g

BIOLOGY

104. In 108 g of water, 18 g of a non-volatile compound is

(a) 1 28 (c) 1 52

109. The amount of Na 2S2 O3 -5H2 O required to completely

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

(b)Kt = J:.__ K2 2 (d) K2 = ­ KI2

107. Aqueous solution of a metallic nitrate X reacts with

NH4 OH to form Y, which dissolves in excess NH4 0H. The resulting complex is reduced by acetaldehyde to deposit the metal. X and Y, respectively, are (a) Cs(NO3 ) and CsOH (b)Zn(NO3 ) 2 and ZnO (d) Mg(N03 )2 and Mg(OH)2 (c)AgNO3 and Ag2O

108. The density and equivalent weight of a metal are

10.5 g cm- 3 and 100, respectively. The time required for a current of 3 amp to deposit a 0.005 mm thick layer of the same metal on an area of 80 cm2 is closest to (a) 1 20 s (b) 1 35 s (c)67. 5 s (d) 270 s

penicillin. After incubation overnight at 37°C, 10 bacterial colonies were observed on the plate. That the colonies are likely to be resistant to penicillin can be tested by (a) measuring their growth rate (b) observing the colour of the colonies (c) checking their ability to grow on another plate containing penicillin (d) checking their ability to cause disease

112. Watson and Crick model of DNA is a

(a) B-form DNA with a spiral length of 34 A and a diameter of 20 A (b) A-form DNA with a spiral length of 1 5 A and a diameter of 20 A (c) Z-form DKA with a spiral length of 34 A and a diameter of 20 A (d) B-form DNA with a spiral length of 28 A and a diameter of 1 4 A

113. Eco RI and Rsa I restriction endonucleases require 6 and 4 bp sequences, respectively for cleavage. In a 10 kb DNA fragment how many probable cleavage sites are present for these enzymes? (a) 0 Eco Rl and 1 0 Rsa l (b) 1 Eco Rl and 29 Rsa l (c) 4 Eco RI and 69 Rsa I (d) 2 Eco RI and 39 Rsa I

114. From an early amphibian embryo, the cells that

would give rise to skin in adults were transplanted into the developing brain region of another embryo. The transplanted cells developed into brain tissue in the recipient embryo. What do you infer from this experiment? (a) Cell fate is permanently determined during early embryonic development (b) Development fate of donor cells is influenced by the surrounding cells (c) Developmental fate of donor cells is not influenced by recipient cells (d) Any cell which is transplanted into another embryo always develops into a brain

115. The presence of plastids in Plasmodium suggests

(a) it is a plant species (b) it is a parasite with a cyanobacterium as an endosymbiont (c) it is a parasite with an archacbacterium as an endosymbiont. (d) it is a plant species with an archaebacterium as an endosymbiont.

w

..--:-a �-=-w Nl Ec= B--= w � .Ja-=E� oc-=oa,-;-K==sa---a. l�

156

KVPY Question Paper 2013 Stream : SB/SX Which one of the following sets of DNA fragments generated by digestion with both Eco RI and Bam HI as shown in (III) is from the gene? (a) 1 kb and 4 kb (b) 1 kb and 2.5 kb (c) 1 kb and 3 kb (d) 1 kb and 3.5 kb

116. The figure below demonstrates the growth curves of two organisms A and Bgrowing in the same area. What kind of relation exists between A and B?

T

-

ro

-�

' , ;,." ,,. -----·-·,,, ,.

ro

Q) rt)

I I

abundant mitochondria and rich blood supply. Brown fat (a) insulates animals that arc acclimatised to cold (b) is the major source of heat production in birds (c) provides energy to muscles (d) produces heat without producing ATP

· . - - · · · · Organism A in presence of B

_ I /

(.) C

118. Brown fat is a specialised adipose tissue with

Organism B Only , · Organism B in presence of A

rt) rt)

E 0 :0

Organism A Only

• • •

119. In some species, individuals forego reproduction and

Time--+ (a) Competition (c) Commcnsalism

help bring up another individual's offspring. Such altruistic behaviour cannot be explained by which of the following? (a) An individual helps relatives only and gets indirect genetic benefits (b) The individual benefits because it can later inherit the breeding position (c) The individual benefits because it gets access to resources, such as food and security from predators, in return (d) The species benefits from a reduction in competition among offspring

(b)Symbiosis (d) Mutualism

gene into the Eco RI site of a vector of 6 kb size. The cloned DNA has no other Eco RI site within. Digestions of the cloned DNA is shown below.

117. A scientist has cloned an 8 kb fragment of a mouse

EcoRI only

BamHI only

EcoRI and Bam HI

120. Lions in India are currently restricted to Gir, Gujarat.

5.0 kb 5.5 kb 4.5 kb 4.0 kb -

5.0 kb -

(II)

(I)

PART-/ 11

21 31

41 51

61 71

(c)

(a) (d) (b) (c)

(c)

(a) (c)

PART-II 81 91

101

111

(c) (c)

(a) (c)

Efforts are being made to move them to other parts of the country. This is because they are most susceptible to extinction due to infectious diseases under the following conditions when present as (a) several small, isolated populations (b) one large population (c) several large, connected populations (d)several large, isolated populations

4.0 kb 3.5 kb 3.0 kb 2.5 kb 1.0 kb (III)

Answers 2

(a)

3

22

(d)

23

12

32 42 52

62

(a) (b)

13

33

(d)

43

(d)

63

(a)

53

73

72

(c)

82

(a)

83

1 02

(d)

1 03

92

112

(a) (a)

* No options are correct.

93

113

(b) (d) (b)

(d) (c)

(a) (d! (c) (c) (c)

(a) (d)

4

(b)

24

(a)

14

34 44

54

64 74

(a) (b) (c)

(a) (a) (d)

5

15

25 35

45 55

65 75

(b)

(d) (c)

(*)

16

36

(c)

66

56

76

84

(c)

85

(a)

86

1 04

(d)

1 05

(b)

1 06

94

114

(b) (b)

95

115

(a) (b)

(b)

27

(c)

47

67

26 46

(c)

(d)

(a)

(d) (c)

7

6

96

116

(a)

(a) (c)

(b)

17

37

57

8

(d)

(c)

(a)

28

(b)

(b)

(c)

48

(d)

68

(b)

(c)

77

(c)

(a)

87

(d)

(b)

1 07

(d)

(a)

97

117

(a) (c) (c)

18

38

58

78

88 98

1 08 118

(a)

9

19

40

49

(b)

(c)

69

(b)

(c)

(d) (b)

(d)

59

79 89 99

1 09 1 19

(c)

(b)

(a) (b)

30

20

(d)

39

(b)

(d)

29

(a)

10

(c)

(c)

(a) (c) (c)

(a) (a) (d)

50

60 70

(a) (d) (c)

(b) (b)

80

(d)

90

(d)

110

(b)

1 00 1 20

(d) (b)

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Solutions 1 . (c) We have, :x!" + 3x2 + 3x + 3 = 0 Let f (x) = :x!" + 3x2 + 3x + 3 f' (x) = 3x2 + 6x + 3 f' (x) = 3 (x + 1)2 f' (x) > 0, V x E R :. f (x) is monotonic increasing function. Hence, :x!" + 3x2 + 3x + 3 = 0 has only one real root. Now, f (-3) = (-3)3 + 3(-3)2 + 3 (-3) + 3 = - 6 < 0 f(-2)= (-2)3 + 3 (-2) + 3 (- 2) + 3 = 1> 0 :. Roots lie between - 3 and - 2 a + P + r=-3 Let a E (- 3, - 2), i.e. - 3 < a < - 2 a=- 3-P-r -3 < - 3 - p - y < - 2 ⇒ 0 < - $ + y) < 1 ⇒ - 1< $ + y) < 0 :. Sum of non-real roots lie between - 1 and 0. 2. (a) We have, log 2log 2log 2log 2log 2 (n) < 0 < logJog 2logJog 2 (n) ⇒ log 2log 2log 2log 2log 2 (n) < 0 ⇒ log 2log 2log 2log 2 (n) < 2° ⇒ logJog 2logJog 2 (n) < 1 ⇒ log 2log 2log 2 (n) < 2 logJog 2 (n) < 22 ⇒ ⇒ log 2 (n) < 24 ⇒ n < ;21 6 Similarly, for logJog 2log 2log 2 (n) > 0 ⇒ n > 24 24 < n < ;21 6 Hence, :. The minimum number of digits in binary expansion of n is 5 and maximum numbers of digits in binary expansion of n is 16. 3. (b) We have, I a + bro + cro21 , a, b, c E {- 1, l} For maximum put a = 1, b = - 1, c = - 1 ll - ro - ro2 1 I 1 - (ro + ro2 ) I = 1 1 + 1 I = 2 [·: (J) + OJ2 = - 1] 4. (b) We have, ax + 9y = 5 and 4x + by = 3 arc parallel. _cz_ = __\:) ⇒ ab = 36 4 b

AM 2 GM

a+ b 2M 2 a + b 2 2/36

8(1 , 1 )

p

a + b 2 l2

Hence, least possible value of a + b = 12. 5. (b) Let the equation of circle passing through ABCD is x2 + y2 + 2gx + 2fy + c = 0 Polar coordinates of centre of circle is - g = r cos 0

A

x=a Area of !l.CDE = I Arca of MBC 2 ⇒ ⇒

- f = r sin 0

I X CD X DE = I X BC X AP 2

4

I (9 - a) x ( 1 - _cz_ ) = I x 8 x 1

9 4 2 ⇒ (9 - a) (9 - a) = 36 ⇒ 9 - a = 6⇒a = 3 7. (d) Given, in MBC, D is mid-point of BC and AB = AD LB = LADE 0 = 1t - LADB = 1t - B BD = DC = 1 : 1

y

D

A

Y'

4

AB = length of intercept of X-axis ⇒

a2 - = g2 - c 4 CD = length of intercept in Y-axis ⇒ ⇒

B

a = 2.jg2 - C

⇒

. . . (i)

b=�

b = !2 - C 4 2

. . . (ii)

a2 - b2 r 2 cos20 = --4 Hence, locus of centre of circle passing through ABCD in polar coordinates is ⇒

a2 - b2 r 2 cos 20 = ---. 4 6. (a) Given, in MBC A(O, 0), B(l, 1) C(9, 1)

C(9, 1 )

C

Apply (m - n) theorem, (m + n ) cot0 = n cotB - m cotC ⇒ (1 + 1) cot ( 7t - B) = cot B - cot C [·: m = n = l] ⇒ - 2 cot B = cot B - cot C [·: cot (7t - B) = - cot BJ 3 cot B = cot C ⇒ tan B = 3 ⇒ tan C

8. (d) We have, a, p, y are angle of triangle. a + p + y = 180° . . . (i) 2 sin a + 3 cosP = 3-J2 Given, . . .(ii) and 3 sin p + 2 cos a = l On squaring and adding E qs. (i) and (ii), we get 4sin 2 a + 9 cos2 p + 12 sin a cos P + 9 sin 2 P + 4 cos2 a + 12cos a sin P = 18 + 1 ⇒ 4(sin 2 a + cos2 a) + 9 (cos2 P + sin 2 P) + 12 (sin a cosp + cosa sinP) = 19 ⇒ 4 + 9 + 12 sin (a + P) = 19 ⇒ 12 sin (a + P) = 1 9 - 9 - 4

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158

KVPY Question Paper 2013 Stream : SB/SX sin (a + �) = l 2 a + � = 150° y = 30°

⇒ ⇒

9. (c) We have, lim f(x)= M > 0 x➔= (a) }�� x sin (-; ) f (x) = }��

f ((g(x) )2 ) is defined if

( 1)

-----r. Sln -

f (x)

. ( 1) = lim __x_ x lim f(x) = l x M = M x➔ =

Sill

1

X

-

r ➔ oo

X

Hence, option (a) is true. (b) lim sin (f (x)) X➔ = sin ( lim f (x)) = sin M x➔ = Option (b) is also true. x (c) lim X sin v ) f(x) x➔ = sin-(e-x ) . xe- x I1m f(x) x ➔ oo e-x

h➔ 0

2

2

A(t) = ((sin t ) x - (2 cos t ) x + sin t) dx 1

l3 sm. t - x cost + x sm t1Jo 2

A(t) = l. sin t - cos t + sin t 3

.

A(t) = i sin t - cos t 3

A' (t) = .i cos t + sin t 3 A(t) has infinitely critical points. A(t) = O for infinitely many value of t. Hence, II and III statements arc true. 1 1 . (a) We have, f(x) = ✓2- x- x

and

2

g(x) = cos x (g(x)) 2 = cos2 x ---------2 - (g (x))2 - (g (x) ) 4 f ((g(x))2 ) = ✓� = ✓2 - cos x- cos x 2

f(x) = J 21 ' 1 dt, X E (- 10, 10] -10 Let ,. be an integer r E [ - 10, l OJ

J

r-h

Jz1'1 dt

-10

4

n2 - n +

⇒

(n -

r-h

2-10 + 2-9 + 2-8 + ... + 2r - 1

7

. . . (i)

z!- t ! dt RHL lim+ 21 1 1 dt ⇒ lim x➔ r+ h x ➔ r 10 -10 r+ h -8 t f -9 l dt 7 dt + z! l dt + ...+ }�� L J

J

J

r+ h

x

n2

⇒

> n_

⇒

= 2- 10 + 2- 9 + 2- B + . . . + 2r - 1

. . . (ii)

f(r) = J zl t ldt = 2-10 + 2-9 + ... + 2r - l - 10

lim f (x) = lim f (x) = f(x) x ➔ ,x ➔ r+ Hence, f (x) is continuous for all I' E (- 10, 10).

:

✓32209 + l_

. . . (iii)

2 - l/ ✓2 (x - 0) 0 + rt / 4

y - 2 = (8 - 2✓2) X (8 - 2✓2) y= X+ 2

⇒

. . . . of 1·inc 101n1ng E quat10n ( rt , cos rt ) and

4

(0, 2) is y- 2 =

!z 'l

/') U

-D 2

⇒

y- 2 = -8

l. ;;;: 4052 + l 4 4 2 32 09

E quation of line joining ( 4rt , cos ( 4rt ) ) and (0, 2) is

lim [2- 10 + 2-9 + Z-8 + . . . + 2r -l (1- h )J

0

r x3

Domain of

1 2 . (a) We have,

1 -9

A(t) = J p1 (x) dx

A( t ) =

)) =

t t t }�� l L z! l dt + r� ldt + ... + ! � l dt J r

p, (x) = (sin t) x - (2 cost) x + sin t

J

Hence, domain of f ((g(x)

f (g(x)).

2

⇒

2 2 1 79.46 + 1 2 180.46 - -2 ⇒ n 2 90.23 :. Least value of x is 91. 1 4. (a) Given, y = cos x

Similarly, for domain of f (g(x)) is x E R

lim x➔ r - h

Option (d) is also true. 1 0. (b) We have, 1

f (g(x)) = ✓2- cos x - cos2 x

XE R

For LHL lim_ z! t l dt x ➔ r 10 -

= Ox lx M = O Hence, option (c) is false. sm x (d) lim f(x) = 0 x M = 0

I

Given, 1 2 2013 n (n - l) 2 2013 4 ⇒ n 2 - n 2 4052

2 - cos2 x- cos4 x ;;;: 0 ⇒ cos4 x + cos2 x - 2 s 0 ⇒ cos4 x + 2 cos2 x - cos2 x - 2 s 0 ⇒ (cos2 x + 2) (cos2 x - 1) s 0 ⇒ cos2 x E [-2, 1]

⇒

4

.J2 (x - 0)

2- ...!_ 0- �

y=(

-8

2✓2

\

}+ 2

Graph of given curve, y = cosx, and line are

A(O,

2)

1 3. (d) Let I = J [x] {x} dx n

2

3

n

I = J {x} dx + J 2{x} dx + ... + J (n - 1) {x} dx 1

2

I = (1 + 2 + 3 + 4 + . . . + (n - 1)) {x} dx

J 1

0

1

l = n (n - l) J x dx 2 0 _ I-

n (n - 1) f x2 l 2J 0 2

l

(- rt/4, 0)

(rt/4, 0)

n-1

1

n (n - 1) 4

Arca of shaded region = 2 area of curve APCA 7 ni4 f (- 8 + 2✓2 ) =2 x + 2 - cos x dx J rt l '4 2 r In 2 2v2x . xT = 2 -Bx + � + 2x - sm

l l�

Jo WWW.JEEBOOKS.IN]

159

KVPY Question Paper 2013 Stream : SB/SX =

= =

/l� (�)\ ./2 (�)2 4

4 \./2

2 lj 2n: - _J2._ 4

r -12J 4

r n: ( -4 + -12 + 2 1 l4 4 4 n:

2

+

7t

) n: - ./2

1

1

L1J

cr' � cr

(1, n ) = 1nl

r

1] -

Number of favourable outcomes, when . n7 x = ], y = l lS 2 L1

r1

Number of favourable outcomes, when

X = 2, y = 2, X ;,c 1, J ;,c 1= 2 [

1

Similarly, for number of favourable outcomes, when x = k, y = ll but x, y ic {I, 2, ... , ll - l}

Lkr

= 2 1nl

I,

1

So, required probability

r_ri_l - (1 + 1 + ... n times) 2 L k=l h J

n

n

Standard deviation cr =

✓;

2

- (µ ) 2

Mean of other observations Y1 + Y2 + Ya + · · · + Yn - 1 + Yn Mean (µ ') = n

=

v1 = cosai + sinaj

a E (0, 90° ) P E (90° , 180 ° )

v3 = cos y i + sin y j

° ° Y E (180 , 270 )

V1

0 E (270° , 360 ° )

= COS Oi + sinoj

(a) v1 + v2 + v3 + v4 = 0 not necessarily true for all v1 , v2 , v3 and v4 (b) V; + vi (ls i < j s 4) v1 + v2 = (cos a + cosP) i + (sin a + sinp)j lf y coordinate is positive i.e. a in 1st quadrant P in 2nd quadrant. But in this case x-coordinate is not necessarily positive. (c) v, = vi (ls i < j s 4) Let a in 1st quadrant, y in 3rd quadrant V; · Vi < 0 (d) V; . vi = (cos a cosP) + sin a sin P = cos (a - P) is positive 0 < I a - P I < � 2 3 n: = - < la - Pl < 2 11: 2 Not positive for all values ofv1 , v2 , v3 , v4 .

9 X 9 X 8 = 648 n containing at most two distinct digit is Total number - n containing three distinct digits = 900 - 648 = 252

I.x;

(b) Sn has more than one prime for n CH 3 CH 2 CH; > CH 3 -C-CH 2 30

+

I°

[Stability is least due to presence of electron withdrawing group] Thus, the order of reactivity of given compounds in SN lrcaction is 3 > 2 > 1.

60. (b) The number of atoms present in

closed cubic packed arrangement is 4. :. Number of Y-atoms = 4 Number of octahedral voids = N Number of M-atoms = I x N 2

Ix4=2 2 :. The formula of ionic compound becomes

MY2. 6 1 . (a) Foetal haemoglobin has greater affinity for oxygen as compared to the adult haemoglobin mainly to facilitate more oxygen supply to the developing foetus. Foetal haemoglobin lifespan is 80 days as compared to that of adult haemoglobin is 120 days.

62. (d) The nucleolus is a distinct

structure in the nucleus of the cell composed of filamentous and granular material. It is the site of synthesis and processing of ribosomal RNA and the assembly of this RNA with ribosomal proteins into ribosomal subunits.

63. (d) The melting temperature of DKA

is directly proportional to the number of hydrogen bonds in the DNA.

We know, that there are two H-bonds between adenine and thymine and three H-bonds between guanine and cytosine. Thus, the DKA molecule, 'iv', i.e. CCCCCATCCCCC CCCCCTAGGCCC has the highest melting temperature.

64. (a) lf thymine is inserted in between ATC CAA. The new sequence will be ATC TGAA. Since the RNA transcript of new sequence will be

dioxide is converted by plants and other photosynthetic organisms to energy rich molecules such as glucose.

Y t

Chlorofluorocarbons (CFCs) can cause skin irritation and dermatitis. CFCs are involved in the destruction of the stratospheric ozone layer resulting in increased exposure to UV-radiation which is known to cause skin cancer.

AUG UGAA Start Stop codon codon It is a non-sense mutation. A non-sense mutation is the one in which a sense codon that corresponds to one of the 20 amino acids specified by the genetic code is changed to a chain terminating codon.

65. (c) Meiosis is a special type of cell

division that reduces the chromosome number by half, creating four haploid cells, each genetically distinct from the parent cell. The genetic content of cell reduces to half during meiotic prophasc-11.

66. (b) Western blotting (protein

blotting) is a rapid and sensitive assay for detection and characterisation of proteins. l t is based on the principle of immunochromatography where proteins are separated into polyacrylamide gel according to their molecular weight.

67. (d) Schizosaccharomyces pombe, also

called 'fission yeast' is a species of yeast used in traditional brewing and as a model organism in molecular and cell biology. It is a unicellular eukaryote, whose cells are rod-shaped.

68. (b) Light reaction takes place in the grana thylakoids of the chloroplast. Dark reaction takes place in the stroma of the chloroplast. In light reaction, H+ ions are utilised by KADP to form NADPH. In dark reaction, the hydrogen of NADPH is used to combine with CO2• 69. (a) According to Mendel, alleles of a

gene segregate and alleles of different genes assort independently. Mendel's law of segregation states that allele pairs separate or segregate during gamete formation. Mendel's law of independent assortment states that the alleles of two (or more) different genes get sorted into gametes independently of one another.

70. (b) RuBisCO has both oxygcnase and carboxylase activities. Ribulose- 1 , 5-Bisphosphate Carboxylasc/Oxygenase, commonly known by the abbreviations RuBisCO, is an enzyme involved in the first major step of carbon fixation, a process by which atmospheric carbon

7 1 . (c) Dermal contact with

72. (c) Gastrulation occurring during gastrula stage is one of the most important morphogenetic movement in the formation of the basic tubular structure of animals. Morphogenetic movement in the early embryo takes place mainly through the movement and deformation of the epithelial cells, which constitute the embryo and include invagination, extension and locomotive movements. 73. (c) The only organ which is capable

of producing fructose in human is seminal vesicles. The function of seminal vesicles is to produce and store fluid that will eventually become semen. Fructose is a sugar that provides sperm with energy in males.

74. (d) Blood thinners or anticoagulants,

decrease the chances of blood clot formation in the heart, thereby reducing the risk of stroke. Stroke is a medical emergency in which damage to the brain occurs due to interruption of its blood supply.

75. (c) In a fruit, the endocarp is the inner layer surrounding the seed. In hesperidium type of fruit, the epicarp forms the cover, the mesocarp is fibrous and the cndocarp bends inwards giving rise to a chamber from which hair-like structures arc formed. These arc juicy and form the edible part in orange and lemon. Thus, the correct answer is option (c). 76. (c) The enzyme, nitrogenase which is

capable of nitrogen reduction is present exclusively in prokaryotes (e.g. Rhizobium). It is highly sensitive to 02 and gets inactivated when exposed to it, thus docs not require oxygen (i.e. anaerobic condition) for its functioning.

77. (c) Ciliated epithelium is a thin

tissue that has hair-like structures on it. These hairs, called cilia, move back and forth to help move particles out of our body.

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KVPY Question Paper 2013 Stream : SB/SX

78. (b) Dispersal is the process by which individuals move from the immediate environment of their parents to establish in an area more or less distant from them. Because dispersal can enable escape from low quality environments and access to higher quality resources, many species that specialise in colonising disturbed areas tend to have greater dispersal abilities than species that live in relatively stable habitats. 79. (c) The equation for Hardy-Weinberg equilibrium is-p 2 + 2pq + q2 = 1 where p 2 represents the frequency of the homozygous genotype A A, q2 represents the frequency of the homozygous genotype aa and 2pq represents the frequency of the heterozygous genotype A a. Since p = 0.6 and q = 0.4 2pq = 2 X 0.6 X 0.4 = 0.48 BO. (d) Glycogen is a large multi-branched polymer of glucose which is accumulated in response to insulin and broken down into glucose in response to glycogen. Glycogen is mainly stored in the liver and muscles and provides the body with a readily available source of energy if blood glucose levels decrease. 8 1 . (c) We have,

t (½)

=

100

/(x) 5' 100, V x E R f(x) = a ( x - ½ ) [a0x" - 1 + Cli x" - 2 + ... + a,, _ i J + 100 If /(x) 5' 100, V X E R :. a < 0 and f (x) must be even degree polynomial. Since, there may be more value of x at which /(x) attains maximum. 1 .-. lf x * -, then /(x) < 100 may be false. 2 82. (a) We have,

f

(ak3 + bh 2 + cli + d) = n 4 , n E N

k=I

On solving these equations, we get a = 4, b = - 6, c = 4, d = - 1

AB = AC

rn,

O)

l_ area of MBC

2 1

6

=

0,

2

!?. + � + d = O

2

[·: AB = 2R cos0J R sin e cose 1 + cos 0 dr = (1 + cos 0) (cos 20) + sin 2 0 cos0 R (1 + cos 0) 2 d0 R2 sin 20 r = ----2R (1 + cos 0)

dr = O de .-. (1 + cos 0) (cos2 0 sin 2 0) + sin 2 0 cose = 0 ⇒ cos2 0 - sin 2 0 + cos3 8 - cos e sin 2 0 + sin 2 0 cose = 0 ⇒ 2cos2 0 - 1 + cos3 0 = 0 ⇒ cos3 e + 2cos2 0 - I = 0 ⇒ (cos0 + l) (cos2 8 + cos8 - I) = 0 ⇒ cose = - lor cos2 e + cos8 -l = O 1 ± -J5 cos0 = - ---

C

r

Arca ofMBC

=B� + � A �B� 'O�'A� + o 2

For maxima or minima

MGN are collinear. If M and B arc coincide. :. MON arc median of MBC.

=

. . . (i)

.

v'5 - 1 2

.

r 1s maximum, when cos8 = --

,!_BO Y

A

86. (a) We have, M 1---

1--------- N G

y

0

2 r=� =- ��__�2R + 2R c os 0

-

N

=

2

9

85. (a) XY be the diameter of a semi-circle with centre 0. A, B point on a semi- circle such that AB is parallel to XY.

l. R2 sin (!t - 20)

A

n 2+ - + - + d ) n = n4

a 2a 26 a 3h c - -- - = 1, - + - = 0, - + - + 4 4 6 4 6 2

� 5, r < l.

r=

84. (c) We have, 0 is the centroid of MBC, M and N arc interior point on sides AB and AC respectively.

:. Arca of MMN

. . . (ii)

9 From Eqs. (i) and (ii), we get

r = inradius of MBC

Case I When MON arc parallel to BC

c 2

r��

OA = 0B = R

(x - 1) 2 + (2) 2 = (x - 4) 2 + (4) 2 ⇒ x2 - 2x + 1 + 4 = x 2 - Sx + 16 + 1 6 ⇒ 6x = 27 27 9 ⇒ x=-=2 6

1;,,ax =

6

2

ar (MMN) = (� ) = � 9 ar (MBC) 3

X

Coordinate =

2:3

=

MMN ~ MBC

. -. I al + I bl + I d + I di = 4 + 6 + 4 + 1 = 15 83. (c) We have, Equation of parabola . . . (i) y2 = 4x Equation of line y = 2x - 4 . . . (ii) On solving Eqs. (i) and (ii), we get (1, - 2) and (4, 4) Now, ABC is an isosceles triangle

area of MMN . G1ven, -----area of MBC

(b

AG : AD

=

cos- 1

(-J6

-1 2 )

x x2 x3 x4 f(x) = 1 + - + - + - + /(x) =

1!

2!

3!

4!

2

6

24

x2 x3 x 4 1+ X + -+ - + -

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KVPY Question Paper 2013 Stream : SB/SX x3

l

x

, f (x) = 1 + X + - + 6 2 2

2

("(x) = 1 + X + � 2 f"(x) > 0, V x E R :. f ' (x) is increasing. Let f' (x) = O at x = x0

⇒ ⇒ ⇒ ⇒

[·: OP = R + HJ

6

88. (c) Let =

0

f ' (-2) = 1 -2 + 2 - _13 < 0 6 f'(-1) = 1 - 1 + ..:' : _ _:i: > 0 2 6 /' (-2)'/(- 1) < 0 X0 E (-2, - 1)

l + Xo + x5 + x� + Xci > 0 1! 2! 4! 31 :.f (x) has no solution.

f(:x:o )

OP

H = 6400km

x5 fa l + x0 + +2

T,

R

2R = OP 2R = H + R R=H 2

⇒ ⇒1 =

xn x2 x3 + ... + + I = f ( _:1'. + ) dx 1 2 3 n 0

'

I X2 + x3 + X4 Xn+ 7 + ··· + l lx 2 2 x 3 3 x 4 n(n + 1) J

I

0

⇒ J = ( 1 - _l_ ) + ( _l_ _ _l_ ) + ( _l_ _ _l_ ) + ... 2 2 3 3 4

+ (� - n � l)

cos0 = _l_ 2 0 = 60°

In t10PB,

cos60°

=

OP

OB p

T2

T3

T4

T5

4 4 3 3 3 2 2 3 2 3 3 2 ,

2 1 2 3 1 2 2 2 2

0 1 1

0 0 0 0

()

3

2 2

2

1

0

1

1 1

2 1 2 2

1 1 1

0 0 2

Clearly, option (d) is correct.

1 1 1 1 ⇒ l = -- + _ _ + __ + ... + 1x 2 2 x 3 3x 4 n(n + 1)

For a cone solid angles. Total solid angle full round = 4 7t for 71,-2 ➔ n [·:one-fourth of area is covered] w = n = solid angle w = 2n (l- cos 9) 71 = 2n (I - cos 9)

4 4 4 4 4 4 4 3 3 3 3 2

9 1 . (c) Let v0

=

87. (d) Radius of earth = 6400 km

16 5

1 n n + 1- 1 = _ _ ⇒ 1 = 1 _ __ = n+l n+l n+l

89. (c) We have, 100 coupons labelled 1, 2, 3, ... 100 Five coupons arc random selected and arranged. :. Total numbers of outcomes = 10°C x 5 1 5

Five coupons Xi_ , x2 , x.i , x4 , Xr, arc arranged such that Xi. > � > x.i and x.i < x4 < Xr,

Favourable outcomes = 10°C5 x __f_ 2!2! 4! lOO C X 5 2!2! = : . Required probability IOO C X 5! 5

=

linear velocity of centre

of mass after collision and m0 = angular velocity of rotation of composite body about centre of rotation. V

m

h

--

M --,; � lh' R----+-

C

-

--

wo

Linear momentum is conserved mv = (m + M) v0 . . . (i) Angular momentum about axis of rotation is conserved, mu (R - h) = Im0 Substituting for mv from Eq. (i), we get (m + M) v0 (R - h) = ( MR 2 + mR2) mo

i

As pure rolling occurs, v0 = Rw0 So, (m + M) · ROJo · (h - R) ⇒ (m + M) R20Jo

(1-

=

(¾

M+ m

1) = ( ¾ M +

f

2coo

m) R m0 2

( M + m) 2M + 5m + 1 ⇒ !!__ = i + l ⇒ !!___ = R 5(m + M) R (m + M) h 7M + 10m ⇒ R 5(m + M)

92. (a) Motion of the ball is as shown below.

1

20 90. (d) We have, Five teams in a tournament and each team plays against every team. :. Total number of one team = 4 Possible number of distribution

1 .4

0, then the equation has all real and distinct roots. (d) lf 4a3 - 2762 > 0, then the equation has real and distinct roots.

5. All the points (x, y) in the plane satisfying the equation x2 + 2x sin (xy) + 1 = 0 lie on

(a) a pair of straight lines (b) a family of hyperbolas (c) a parabola (d)an ellipse

6. Let A = (4, 0), B = (0, 12) be two points in the plane. The locus of a point C such that the area of l\ABC is 18 sq units is (a) (y + 3x+ 12)2 = 81 (c) (y + 3.x - 12)2 = 81

(b) (y + 3x+ 81)2 = 1 2 (d)(y + 3x- 81)2 = 1 2

7. In a rectangle ABCD, the coordinates of A and Bare (1, 2) and (3, 6) respectively and some diameter of the circumscribing circle of ABCD has equation 2x- y + 4 = 0. Then, the area of the rectangle is (a) 1 6 (b) 2✓1 0 (c) 2J5 (d)20

8. In the XY-plane, three distinct lines � , 12 , � concur at a point (A, 0). Further the lines �, l2, � are normals to the parabola y2 = 6x at the points A = (Xi_ , y1 ), B = (x2 , y2 ), C = (x.i , y3 ) respectively. Then, we have (a) A < - 5 (c) - 5 < A < - 3

(b) b 3 (d)0 < 1s. < 3

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9. Let f (x) = cos 5x+ A cos 4x + Bcos 3x

r = t Sn (l - l') = --- nl 1- i i(l- i n ) _ ni'i + 1 ⇒ Sn = (1- i)2 (1- i) i") - ni'> + 1 � sn = - i(l --⇒ 1- i - 2i ⇒

Let

n+

1- i " _ ni i(l + i) Sn = -2 2 1- i n n i " + 1 (l + i) zl = -- and Z2 = -----2 2 1 n 1n I Z1 I = 1n or o and I Z0 I = - " 2 - 2 v2

'!_ J2 = 1 8J2 ⇒ n = 36

2 3. (d) G iven,

⇒ x2 + 2x sin xy + sin2zy + 1- sin2 xy = 0 ⇒ (x + sin xy) 2 + cos2 xy = 0 :. x + sin xy = 0 and cos2 xy = 0 cos2 xy = 0 ⇒

xy = (2n + 1)�

2

⇒

x + l= 0 [·: cos2 xy = 0 ⇒ sin xy = lj

⇒

x =- 1

y =- (2n + 1)�

2 which represent the pair of straight line. 6. (c) Given, A(4, 0)and B(0,1 2). Let C (x, y) X

y

Arca of !! ABC = � 4 0 2 0 12

1 1 1

⇒ 1 8 = I I {x(0- 1 2)- y (4- 0)+ 1(4 8- O)} I 2 ⇒ 18= I I ( 1 2x- 4y + 48) I 2 4 ⇒ 1 8 = - I (3x + y- 1 2)I 2 ⇒ (3x + y- 1 2)2 = 81 ⇒ (y + 3x- 1 2)2 = 81 7. (a) Given,

P2 = P

(I + Pt = (1 + It [·: p2 = p => p -1 p2 = p -1 p = p = _l] ⇒ (1 + P) n = (2ft = 2n ] = (2 n - l+ 1)I = I + (2 n - 1)1 = 1 + (2n - l) P [·: I = PJ 4. (a) We have, :!' + ax2 + bx + c = 0

Let

5. (a) We have, x2 + 2x sin xy + 1 = 0

f(x) = :!' + ax 2 + bx+ c P'(x) = 3x2 + 2ax + b

D = (2a) 2 - 4(3) (b) D = 4a 2 - 1 26 D = 4(a 2 - 3b) For three roots a 2 - 3b > 0 but given a 2 - 2b< 0 D< O Hence, f (x) has one real roots and two imaginary roots.

Now,

AB = ✓(3- 1)2+ (6- 2)2 5

=.J4 + 1 6 = 2✓ Area of rectangle = AB x BC

J5

= 2J"5 X � = 1 6

8. (b) We have, y = 6x 2

Equation of normal of parabola y2 = 4ax is y = mx- 2am- am3 :. Equation of normal of parabola y2 = 6x IS 3 y = mx - 3m - - m3 2 Normal is passes through (A., 0).

⇒ ⇒

0 = A.m- 3m- �m3 2 2 r..'r.. = 3 + -3 m 2 ⇒ m2 = (' 3) 2 3

m = ✓� (A.-3)

:. m is real.

'A- 3 > 0 ⇒ 'A > 3

9. (c) We have, f(x) = cos5x + A cos4x + B cos3x + Ccos2x + D cosx + E

We know that, cosx = cos(2 n- x) f(x) = f (2 n- x) [·: f (x) contains only cosines terms) 9 = 2n - �) = ;)

r(�) r(

Similarly, ABCD is a rectangle A(L 2)and B(3, 6), equation of one of the diameter of circle circumscribing the rectangle is 2x - y + 4 = 0.

Slope of diameter = 2

6- 2 4 and slope of line AB = -- = - = 2 3- 1 2

:. Side AB is parallel to diameter of circle

equation of line AB y-2 = 2 (x- 1) ⇒ 2x- y = 0 Distance between diameter of circle and line AB,

r(

t( Z51t ) = t( 85n } t( 351t ) = t (75re )

r( 5re ) = r( 6sre ) 4

Let T = f(0)-

t ( �) + t( Z51t ) - t( 35re )

r = r ac > aA Hcncc, B is accelerated more than C and A is accelerated less than C. So, with time A move upwards and B move downwards relative to the cabin. So, option (b) is correct.

2

⇒

⇒

C•

2

25. (b) Let n = number of moles of gas in

Ks1ee1 = 50Js-1 m- 1 K-1

v 2 + co 2x 2 = co 2A 2 4

C0 X2 2 = co2A 2 v + -2 (0

Kcapper = 385 Js-1 m-1 K-1

2

!:!:.._ = oi2A 2 v2 + (02

a2 v 2 + --=1 2 2 co A co4A 2

This is an ellipse (

x2

2

+ Y2 = a2 b

1) with

centre at origin.

24. (a) Expansion is in vacuum, so work done by expanding gas is .6. W = 0. Container is insulated, so heat exchanged with surroundings is .6. Q = 0. Now, from first law of thermodynamics, we have .6. Q = .6. U + .6. W .6. U = 0 ⇒ = Ur ⇒ Hence, initial and final internal energies of gas are equal. As, the gas is not ideal expansion results in reduction of pressure, so intermolecular potential energy increases. This results in reduction of kinetic energy, so temperature reduces.

ui

Then, assuming no heat loss and steady state, Heat flow through copper rod per second = Heat flow through steel rod per second KcapperA (lO0 - T) K,,eeiA(T - 0) = ⇒

l

l

⇒ Keopper (100 - T) = Ks,eel · T :. Area and length of both rods are equal. Substituting values, we get 385 (100 - T) = 50 T 385 x 100 = (385 + 50) T ⇒ X 100 = 880 C 385 ⇒ T= 435

27. (b) Compression of a gas in a

compressor is nearly an adiabatic process. So, by using p �: r · T� = P ��,1 · T!u, We get, (0.28)1 -1 (233) 1 = (1) 1 -Y (T) Y Here, for air, y = 14= 7__ 5 Hence, T = (233) (0.28)-217 =

233 (0.28)217 This temperature is much higher than 298 K or 25 ° C. So, an air-conditioner is needed to cool the air coming out of compressor.

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KVPY Question Paper 2011 Stream : SB/SX

(-v l

28. (b) When observer is stationary and source is moving, observed frequency is f1 = fo

v- u)

where, v = speed of sound and u = speed of source. When source is stationary and the observer is moving towards source, then measured frequency is Zl f2 = fo ( V : ) So, difference in frequencies observed is v+ u v J - __ f2 - fi = fo ( V

V- U)

u 2fo v(v - u)

⇒

(2 - t;_ < 0 or f2 < t;_ 29. (c) Let intensity of incident light is 10 . From Malus' law, intensity of light after first polaroid is 11 = 10 . Here note that, light is plane polarised and first polariser's axis is parallel to plane of polarisation of incident light. Intensity of light after second polaroid is 2 12 = 10 cos 30°

� 10 4 Similarly, intensity of light after fifth polaroid will be 15 =

(1f

:. p = � , keeping other parameters 'J.., = cl f

As,

f

constant. So, when frequency f is doubled, fringe separation P is halved. 1 p = - mm = 0.5 mm 2

32. (a) For a domestic AC supply of 220 V at 50 cps, Peak voltage = Vnns X .J2 = 220 .J2 v Angular frequency, co = 2rcf = 2rr x 50 rad = 100 re s So, instantaneous potential difference between terminals of a two pin outlet is V = Vmax coscot = 220 .J2 cos 100 rct volt 33. (a) Given circuit is

, ,, , -

: : 6on 30a', 4on:, ',,

R = 20+20 = 40Q Above circuit can be reduced to following equivalent circuit. 40n

So, intensity of light at exit is nearly 30% of the incident light.

A

30. (c) For an open end pipe, nu Frequency, f = � = 'A. 2L nu L= 2f

3V

+

40Q

value, potential drop across inductor (= L

�!)

tends to zero.

, :L

So, correct variation is

0

t

36. (a) For an electron and nucleus pair, . K(- e) (+Ze) Potential energy = - - - � r -KZe2 r So, potential energy of electron is negative and it tends to zero as separation r increases. Hence, correct variation of potential energy with r is as shown in graph (3). For a neutron, force outside nucleus is zero. Hence, potential energy of neutron is zero as r > r0 . So, correct variation of potential energy with r for a neutron is as shown in graph (1). For a proton, as r > r0 , force is repulsive. Hence, potential energy is positive. So, correct variation of potential energy with r is as shown in graph (2).

37. (d) Given transitions are

As, length of tube remains same in both cases. = nu1 = nv2 L So, 2ft 2f2 f2 = V2 . t;_ U1 Here, u2 = 1.04 v1 and t;_ = 450 Hz l.04vi x 450 So, /2 = V1

= l .04 x 450 = 468 Hz 3 1 . (b) In Young's double slit experiment, fringe width or fringe separation is lD = cD P=

d

, -_--t::.

40Q

=

10 = 0.31 10

60x30 = R= 20Q 60+30

With time, current begins to grow in circuit and so potential drop across resistor increases, while potential drop across inductor reduces. When current reaches a maximum steady

fd

So, power dissipation is maximum in 40 Q resistor marked A as current through it is maximum.

34. (b) Part of kinetic energy of incoming electron is absorbed by molecule and it gets into excited state. So, kinetic energy after collision is reduced (collision is inelastic). Linear momentum is conserved as momentum conservation holds correct for elastic as well as inelastic collision. K' < K and p' = p So,

35. (d) At t = 0, as the switch is closed, current in circuit is zero and potential drop across resistor (= iR)is zero. So, potential drop across inductor is 10 V.

n = 1 --�---"---As given, 13 > 'A. 2 > 'A.1 ⇒ fs < f2 < fi Hence, 'A.1 corresponds to n = 3 to n = 1 transition, 'A.2 corresponds to n = 2 to n = 1 transition and 'A.3 corresponds to n = 3 to n = 2 transition. Clearly, � ➔ 1 = E3 ➔ 2 + E2➔ 1

he

⇒

=

he

+

he

12 11 ¾ 1 1 1 -=-+A1 ¾ 'A. 2

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KVPY Question Paper 2011 Stream : SB/SX

38. (a) Radiative power of a spherical body is given by P = o AT 4 = o4nR 2T 4 ⇒ p = R2T 4 When temperature is halved and radius is doubled, then power becomes P' = P =P

X

(2)2 x ( ¾ r

C:

)=

As,

R=

. . . (i)

2y-3 A-2 ) [RH ] = [ML h = E- t [h] = [ML2r 1 ] ⇒ e = I · t ⇒ [e] = [A· TJ Substituting above values in Eq. (i), we have 2 2 b [ML2J' -3 A- ] = k [ML r 1 [AT] a a+ = k [M L2aT- bAb ] Equating dimensions, we get a = l and b = - 2 RH =

k( ) )

h

. e2

4 1 . (c) The oxidation state of Ki in

Ni(CO) 4 is 0. It's electronic configuration 2 will be [Ar]3d 8 4s . 3d

I1lI 1lI 1l I 11 11

-

[Zn(CN4 ) 2 ] + Ag

[ill 4s

1',.Hmix = 0, I', Vmix = O It means that no heat is absorbed or evolved when the two liquids are mixed. Also, the volume of solution would be equal to the sum of the volumes of two components.

c

40. (a) Precision refers to the limit of resolution. Precision is determined by the least count of measuring instrument. A smaller least count means greater precision. So, more precise readings have least spread of readings as in case I.

Ni Ground state

r

is given as, 7t = CRT At constant temperature, n=C �= 1 0.01M = = 10 ⇒ n 2 C2 0.001M

So, dimensions of hall resistance are same as that of

42. (a) Extraction of silver is achieved by initial complcxation of the ore (argcntitc) with cyanide ion (X) followed by reduction with Zn (Y). This process is known as cyanide process. Zn i Ag 2S + CW ----t [Ag(GN') 2

spontaneous process takes place, it is always accompanied by an increase in total entropy of the universe (system and surrounding) . Thus, the correct option is (d). Surface area 4nr 2 = 49 _ (b) Volume � 1tr3 3 nd2 6 or 3

4p

II

As CO is a strong field ligand, pairing of electrons will occur. 3d 4s 4p Ni(C0)4 1f 1f 1l 1l 1l

IIIIII□ II

sp3-hybridised The oxidation state of Cr in [Cr(HP) 6 ]2+ is +2. It's electronic configuration will [Ar]3d 4 4s0 • s 3d 4p 4 Cr2+

ID II

45. (b) The rate of gas phase chemical reaction generally, increases rapidly with rise in temperature. This is because on increasing the temperature, the fraction of molecules also increases, having the energies greater than activation energy. This is according to Arrhenius equation K = Ae-EJRT , where the factor e-E.IRT corresponds to the fraction of molecules having kinetic energy greater than Ea . 46. (d) ln free radical polymerisation,

initiation is the first step. During initiation, an active centre is created from which a polymer chain is generated. In case of given options, compound (d), � OCffa docs not undergo polymerisation under initiation because of the presence of electronegative O-atom.

47. (a)

tt

COOH

� H OH As the compound is optically active, because it contains chiral centre (*), thus it will have two mirror images, which arc non-superimposable on each other. Thus, the two possible stcrcoisomcrs (for given compounds) are enantiomers (d and l form).

48. (d) According to second law of thermodynamics, whenever a

d

nd

6 When d1 = 30 nm and d2 = 10 nm

(� l ( SA )

= 1� = 3 6 30

V 1

50. (e) Me

Me

OH • --➔ rr

□

OH M e 2� Me

tt

44. (c) The osmotic pressure of a solution

t

Hence,

I ID II

43. (a) Assuming ideal behaviour,

a

RH = kh i' V

I

(X)

�

So, power is reduced to _!th. 4

39. (b) Let

As H20 is also a strong field ligand, pairing of electrons will occur. 3d 4s 4p [Cr(H20) 6l 2 + 1l 11 11 d2sp3.hybridised

4 ring membered

3

1-

LJ. QIH2 4

H20

•. (highly strain) OH Mer (:OH OH 2 1 1e Ring + Me �I+ r:re 2 expansion e +, _ 2 M o Me 3 4 5 e 3 4 3 4

=

)t,Me � l.J'Me This reaction is known as Pinacol-pinacolone reaction, where diol is converted to a carbonyl compound. Also, the given compound gives a four membered ring carbocation, which changes to five-memered ring carbocation, so as to minimise the strain.

5 1 . (b) Br

�N,ci

Br

u,,o, 6

• No,1

The best reagent is ffaPO 2 because on reduction, N;c1- is reduced to H and it is an effective means of removing amino (nitro) group and unlike nucleophillic substitution reactions, this reaction probably proceeds by free radical mechanism.

52. (a) Given,

AONaCl = 126.4 S cm 2 mol- 1 A 0 cl = 425.9 S cm 2 mol-1 A0NaOAc = 91.0 S cm 2 mol- 1 tt

+ AO NaOAC - A,0Nad = 425.9 + 9 1 - 126.4 = 390.5 S cm2 mo1- 1

A,HOAc = A, HCI 0

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205

KVPY Question Paper 2011 Stream : SB/SX 53. (c) According to Bragg's equation, As,

n1., = 2d sin0 )., = d [Given] n=l )., = 2 1., sine

sine = � 2

58. (d) According to Langmuir adsorption isotherm X

°

e = 30

1 mole of Zn gives 2 electrons

hp

:. 2 moles of Zn gives 4 electrons ⇒ n=4 T = 298 K L'1G 0 = - nF'E 0 = - 4 x 96500 x l . l = - 424600 J = - 424.6 kJ

Thus, x increases with p at low pressure and then remains the same at high pressures.

59. (C) H

oxidation of K aBH4 by ½ arc Nal, B2H6 and H2. NaBH4 + ½- Na! + B2H6 + H2

56. (c) The oxidation state for Mn in

Mn[(CN) 6 ]2- is + 4. The electronic configuration for Mn 4 + is [Ar]3d3 .

� 1 1�11�11�1�I □ I I 4s

4p

As CN- is a strong field ligand, pairing of electrons occur [Mn(CN)6] 2-

IH 1 1 I I 1 0 1 1 1 µ =J n (n+2) 3d

4s

4p

In case of [Mn(CN) 6 ]2- , n=l µ = ✓1(1+ 2) = Js = 173 BM The oxidation state of Mn in [Mn Br4 ]2- is +2 . The electronic configuration for Mn 2+ 0 is [Ar]3d5 4s . Mn 2+

1 11 11 11 11 11 □ I I 3d

4s

4p

As Br- is a weak field ligand, therefore pairing of electrons will not occur. In this case,

n=5

µ = ✓5(5 + 2) = J35 = 5.92 BM

57. (a) Time required to reduce the

concentration of the reactant to half of its initial value is known as half- life. For zero order, Ao ly z = -

2k

If

2(:�)

Ao = 2Ao tl/ 2 =

h

If pressure is very low, then bp 1 ap = !!:_ x=

54. (c) Given, E;ell = l .lV

Mn2+

So, if the initial concentration of reactant is double, the time required for half the reactant to be consumed increases by two fold.

0

� + CHCls

Na �'"- � �

� Chloroform Phenol

CHO

� Salicylaldehyde

This reaction is known as Reimer-Tiemann reaction. The above reaction takes place through following mechanism. Mechanism CHCl3 +

0H

mr ,;==' C Cl3 + Hp : cc1� - :cc12 + c1-

0-0& OJ

k 0

o-

H

•

'

o-

0 &c•o�&""'

+·: CCl2 - I ,-0

N CHCl2

1-1 -

low

60. (c) The boiling point of a compound

depends upon the H-bonding present in it. More is the H-bonding more will be the boiling point. In case of phenol (I), there is only H-bonding, so it's boiling point will be least. In compound (Ill) there is intermolecular H-bonding and compound (II), there is intramolecular H-bonding. Thus, compound (III) has maximum boiling point. Thus, the correct order is I < II < Ill.

61. (b) Keurofilaments are intermediate filaments found in the cytoplasm of neurons. They are protein polymers measuring approximately 10 nm in diameter and many micrometres in length. 62. (b) In G2-phase of the cell cycle, the sister chromatids arc available as template for repair. Since, whole DNA is replicated during the S-phase of cell cycle, and it leads to a multiple Double Strand Break (DSB) which needs to be repaired. So, a cell must properly choose the specific pathway to repair broken DNA molecule which occurs in G 2-phase. 63. (d) A person has difficulty in

breathing at higher altitudes because overall intake of 02 by the blood becomes low. Although the percentage of oxygen in inspired air is constant at different altitudes, the fall in atmospheric pressure at higher altitudes decreases the partial pressure of inspired oxygen and hence the driving pressure for gas exchange in the lungs.

64. (a) A zygote is formed by the fusion of male and female gametes thus it is diploid. ln human, the composition of a zygote that will develop a female will be 44A+ XX and for a male is 44A+ XY. 65. (a) On isolation of different cell

organelles by differential centrifugation the set of fraction we will find is nucleus, mitochondria, chloroplast, cytoplasm. Differential centrifugation is a procedure to separate organelles and other sub-cellular particles on the basis of sedimentation rate. The sequence in which a cell fraction is generated is Kuclei ➔ mitochondria, chloroplast, lysosomes and peroxisomes ➔ Plasma membrane, microsomal fraction and large polyribosomes ➔ Ribosomal subunits, small polyribosomcs ➔ Soluble part of cytoplasm.

66. (b) Since, the genetic code is a

triplet, 64 triplets of nucleotides will be involved (43 = 64). If the genetic code is quadruplet it consists of 256 triplets of nucleotides (44 = 256). 64 triplets coded for 100 amino acid residues. So, when 256 triplets arc involved in coding the genes that code for the proteins would have been longer in size by 25%.

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206 Calculation (100 x 256) / 64 = 400 amino acids. Hence, the percentage change will be (100/ 400) X 100 = 25%. Thus, the correct answer is option (b).

67. (b) DNA is made up of two

polynucleotide chains having antiparallel polarity. Therefore, when one chain has 5' ➔ 3' polarity the other chain will have 3' ➔ 5' polarity. Thus, if the sequence of the bases in the coding strand of a double-stranded DNA is 5'-ATGTATCTCAAT-3', its complementary strand will have 3'- TACATAGAGTTA-5'. But in the case of bases in the transcript (i.e. when mRNA is formed). The sequence will be 5'-UACAUAGAGU UA-3', i.e. exactly similar to the coding strand excepting at the place of thymine (T) which will be replaced by uracil (U).

68. (d) The sodium-potassium pump, an

important pump in animal cells, expends energy to move potassium ions into the cell and a different number of sodium ions out of the cell. The action of this pump results in a concentration and charge difference across the membrane. 69. (c) Carbon dioxide in gaseous form diffuse out of the cell into the capillaries, where it is transported in three ways, i.e in dissolved state (7% of all CO2 transported), in the form of bicarbonate (70% of CO� and as carboxyhaemoglobin (23% of CO�.

70. (c) Patients undergone organ

transplants are given anti-rejection medications to prevent T-lymphocytcs proliferation. lmmunosuppressive are drugs or immunosuppressive agents or anti-rejection medications are drugs that inhibit or prevent activity of the immune system.

7 1 . (d) Severe diarrhoea, vomiting,

watery stools arc the chief symptoms of cholera. All these lead to dehydration. Actually, cholera toxin secreted by Vibrio cholerae causes, through a series of metabolic reactions, the continuous activation of adenylate cyclase of intestinal epithelial cells. The resultant high concentration of cAMP triggers continual secretion of Cl- , HCO3 and water in the lumen of the intestine. Administration of saline not only supports the sodium-potassium pump through which water in cell is restored, but glucose is also symported along with sodium.

KVPY Question Paper 2011 Stream : SB/SX 72. (d) A single water molecule can participate in a maximum of four hydrogen bonds because it can accept two bonds using the lone pairs on oxygen and denote two hydrogen atoms. 73. (d) Sleep-wake and other daily patterns arc part of our circadian rhythms which are governed by the body's interval or biological clock, housed deep with in the brain. The average length of a person's intrinsic circadian rhythm has been shown to be around 24 hours and 1 1 minutes. 74. (b) Modern evolutionary theory

consists of the concepts of Darwin modified by knowledge concerning Mendel's laws. Mendel's work provided a way for Darwin's beneficial traits to be preserved. Instead of mixtures that were blended, Mendel proposed particles that could be recombined. In simple terms, Mendel's theory says that individual traits arc 'coded' by pairs of particles.

80. (b) The resting membrane potential

of a neuron is about - 70 mV. This means that the inside of the neuron is 70 mV less than the outside. Therefore, if an electrode is placed in the axioplasm of a mammalian axon and another electrode is placed just outside the axon then the potential difference measured will be - 70 mV. This is because the neuron is at resting stage. 8 1 . (b) We have, AB = BA

⇒

82. (c) We have, 22 + 42 + 62 + ... + (2n) 2 < 1.01 1 + 32 + 52 + ... + (2n - 1) 2 2

:un 2 �(4n

separation of polytcne chromosomes (e.g. chromosomes of Drosophila melanogaster) that represent a site of active RNA synthesis, i.e. transcription which, when very large is known as Balbiani ring.

77. (b) Apoptosis is a form of

programmed cell death that occurs in multicellular organisms. Biochemical events lead to characteristic cell changes and death. These changes include blcbbing, cell shrinkage, nuclear fragmentation, chromatin condensation, DNA cleavage and global mRNA decay.

78. (d) JD Watson and FHC Crick (1953) proposed a double helix model for DNA molecule. According to this model, the DNA molecule consists of two strands which arc connected together by hydrogen bonds and helically twisted. 79. (c) During pachytene stage of

meiosis-I, a tctrad of the chromosomes form known as a bivalent. This is the stage when homologous recombination, including chromosomal crossover (crossing over) occurs.

2

-

4n + 1)

< 1.01

n(n + 1) (2n + 1) 6 n(n + 1) (2n + 1) _ n(n) (n + 1) 4 + n 4 6 2 < 1.01 n(n + 1) (2n + 1) 4 2(n + 1) 101 < -----'6'----- < lOl = n(2n + 1)(2n - 1) 2n - 1 100 3 4

75. (a) The ectoderm differentiates to form the nervous system (spine, peripheral nerves and brain), tooth enamel and the epidermis (the outer part of integument). It also forms the lining of mouth, anus, nostrils, sweat glands, hair and nails. 76. (a) Chromosome puff is a localised

A k = I, B1 = 0 k 1 1A l = J, I B 1 = 0 IB l = 0 det (AB) = IAI I BI = 0

200n + 200 < 202n - 10 1 301 = ⇒ 2n > 301⇒ n > 150.5 2 :. n > 151 =

83. (c) We have, J!l + ax:2 + bx + a = 0 Let o:,�, y are roots of equation and a,�, y are in AP. a+�+ y=-a a � + �Y + ya = b From Eq. (i)

a �y = - a 3� = - a

. . . (i) . . . (ii) . . . (iii) [·:a + y = 2�]

⇒

�=-� 3 (-a) = and ay - a ⇒ ay = 3 3 Put the values of� and ay in Eq. (ii), we get �(a + y) + ya = b ⇒ $ 2 + 3 = b ⇒

2(

2

a 2 J + 3 = b ⇒ 2a + 27 = 96 9

⇒ 2a2 = 9b- 27 :. Locus ofP(a, b) is 2x2 = 9y - 27 which represent the parabola whose vertex on Y-axis.

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207

KVPY Question Paper 2011 Stream : SB/SX 87. (d) We have, 4 9 f(x) = x1 2 - x + x - x + 1 f(O) = f(l) = 1 :. f is not one-one. 11 8 f' (x) = 12x - 9x + 4.x3 - 1 f '(x) is never vanishes and f (x) has not a real root x > l and x :S: 0

84. (b) We have equation of ellipse 9x2 + 16y2 = 144 2

2

x - +L = l 16 9 Equation of tangent of ellipse is

⇒

✓a2m2 + b2 y = mx ± ✓16m2 + 9 y = mx ±

x12 - x9 + x4 - x + l> 0

:. ✓16m2 + 9 = 5 ⇒ 16m2 + 9 = 25 Now, given y-intercept = 5

[·: x12 > x9 , x4 > x]

16m2 = 16 ⇒ m = ± 1 ⇒ :. Positive slope = 1

85. (a) We have, A = {0 E R I cos2 (sin 0) + sin 2 (cos 0) = l} cos2 (sine) + sin 2 (cos0) = 1 :. sine = cos0 ⇒ tan0 = 1

0 = mt + � 4 B = {0 E R l cos(sin S)sin (cos0) = O} cos (sin 0) sin (cos 8) = 0 cos(sin 0) = 0 or sin( cos 0) = 0

⇒

sine = (2n + 1) � or cos0 = 2nrr 2 :. Clearly A n B = qi. :. A n B is an empty set. 86. (b) We have, f(x) = x3 + ax2 + bx + c /(2) = 8 + 4a + 26 + C = 0 f '(x)= 3x + 2ax + b /' (1) = 3 + 2a + b = o

3

1

-1

-1

3

1

=-2 l 3

+

r (1- X) n+ l 71 1

l (n + ) n ! J1/2 -'------'---

1 + 2x

1

J0

14 = - 2 [ --1: + 2] = 3 3

-lXJ = lOb + a 2 xy = (lOb + a) 2 2 ⇒ (x + y) - (x - y) = 4xy (x - y) 2 = (x+ y) 2 - 4xy ⇒ (x - y) 2 = 4(10a + b)2 - 4 (106 + a) 2 ⇒ (x - y) 2 = 4 (lOa + b + lOb + a) ( 0a + b - 06 - a) ⇒ (x - y) 2 = 4(1 1) (a+ b) · 9 (a - b) ⇒ (x - y) 2 = 4 x 1 1 · (a + b) • 9 (a - b)

1

1

4 x 1 1 (a + b) x 9 (a - b) must be a perfect square. a + b = l l, a - b = l On solving these equations, we get a = 6, b = 5 x + y = 2 (10 a + 6) x + y = 2(60 + 5) ⇒ ⇒ x + y = 130

9 1 . (b) Electric field E of a solid sphere varies with distance from centre ,. as 1 · r, r < O E - C2 >O ') ' r -

· - {C

,.-

where, C1 and C2 are constants. So, correct graph of E versus r is E

. . . (iii) ) = 0]

- x2 - x - 2) dx

.x3

1/Z

r XIH- l 7 (n + l)n ! 0

. . . (i)

= - 2 J (x2 + 2) dx 0 3 [·: x and - x arc odd functions] r

l = l-J n

From Eqs. (ii) and (iii), we get a = - 1, b = - l Putting the values of a and 6 in Eq. (i), WC get C = - 2 2 f(x) = x3 - x - x - 2 1

x" (1 - xt . . fn (x) = mm1mum ( - , --- ), X E (0, lJ n! n!

⇒ ⇒

. . . (ii) [·: f' (1) = OJ

[·.- r(�1

2a = --1: + b=o 3 3

J f(x) dx = J (x

88. (a) We have,

[·: f (2) = O]

2

r(=-I)

When x E (0, 1) 1 - x + x4 - x9 + x1 2 > O (·: 1 - x > 0, x4 - x9 > OJ :. f (x) has only positive value.

90. (c) We have, Let two-digits numbers are 10a + 6. Given, 10a + 6 is AM of x and y and 10b + a is GM of x and y. x + y = lOa + b

= 2 ✓e - 3

89. (b) We have, I x + yl + I x - yl = 4 ⇒ x+ y+ x- y = 4 [·: X + y O} in the plane. Then, the area of the region A n Bis

(b)

1 00

(c)

503

(d) 1 0 06

20. Let a 0 = 0 and a n = 3a n l + 1 for n � 1. Then, the -

wo

remainder obtained dividing a2010 by 11 is � 3

� 7

W 4

PHYSICS M placed on a rough table. If the coefficients of

21. A pen of mass

m is lying on a piece of paper of mass

friction between the pen and paper and the paper and the table are µ 1 and µ 2 , respectively. Then, the minimum horizontal force with which the paper has to be pulled for the pen to start slipping is given by (b) (mµ 1 + Mµ 2 ) g (a) (m + M) (µ 1 + µ 2) g (c) (mµ 1 + (m+ M)µ 2) g (d) m( µ 1 - µ 2 ) g

22. Two masses m1 and m2 connected by a spring of

spring constant k rest on a frictionless surface. If the masses are pulled apart and let go, the time period of oscillation is -

-

-

Tl½ m2 (a) T = 2n: k ll½ + m2 __!_ (

)

-

(c) T = 2n: f!f 3 (a)

2

0

(b) 3

(c)4

(d) 5

14. Let In = f (log xf dx, where n is a non-negative

23. A bead of mass m is attached to the mid-point of a tant, weightless string of length l and placed on a frictionless horizontal table.

1

integer. Then, 12011 + 2011 12010 is equal to (b)/889 + 890 /89 1 (a) /1000 + 999 1998 (c)/100 + 1 00/99 (d) /53 + 54152

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213

KVPY Question Paper 2010 Stream : SB/SX Under a small transverse displacement x, as shown in above figure. If the tension in the string is T, then the frequency of oscillation is 21t

(a) _2__

f--;;J

{2T

21t

(b)_2__

f--;;J

{4T

f--;;

{4T

21t

(c)_2__

21t

(d)_2__

{2T v----;;;

24. A carnet (assumed to be in an elliptical orbit around

the sun) is at a distance of 0.4 AU from the sun at the perihelion. If the time period of the carnet is 125 yr. What is the aphelion distance? (AU : Astronomical Unit) (a) 50 AU (b)25 AU (c) 49.6 AU (d)24.6 AU

29. A point source of light is placed a t the bottom o f a

vessel, which is filled with water of refractive index µ to a height h. If a floating opaque disc has to be placed exactly above it, so that the source is invisible from above. The radius of the disc should be h h h (b) -- (c) (d)� (a) -2 2 2 µ -1 )µ-=-i µ -1 µ -1

✓

✓

30. Three transparent media of refractive indices

µ 1 , µ 2 , µ 3 , respectively, are stacked as shown below. A ray of light follows the path shown. No light enters the third medium.

25. The circuit shown consists of a switch S, a battery B of emf E, a resistance R and an inductor L.

7

Then,

The current in the circuit at the instant. The switch is closed to (a) E I R (b)EIR(l - e) (c)= (d) 0

26. Consider a uniform spherical volume charge distribution of radius R. Which of the following

graphs correctly represents the magnitude of the electric field E at a distance r from the centre of the sphere?

(a) �

R

r

E ( ) 0

l �'

E

E

6

R

r

(b )

E

R

r

(d) lc:

R

r

27. A charge +q is placed somewhere inside the cavity of a thick conducting spherical shell of inner radius Bi and outer radius R2 • A charge +Q is placed at a distance r > Rz from the centre of the shell. Then the electric field in the hollow cavity (a) depends on both +q and + Q (b) is zero (c)is only that due to + Q (d) is only that due t o + q

28. The following travelling electromagnetic wave Ex = 0, Ey = E0 sin (kx + cut), Ex = - 2E0 sin (kx- wt) is (a) elliptically polarised (b) circularly polarised (c)linearly polarised (d) unpolarised

(b) µ 2 < µ 1 < µ 3 (d)µ 3 < µ 1 < µ 2

(a) µ 1 < µ 2 < µ 3 (c) µ 1 < µ 3 < µ 2

31. A nucleus has a half-life of 30 min. At 3 PM its decay rate was measured as 120000 cps. What will be the decay rate at 5 PM? (a) 120000 cps (b) 30000 cps (c) 60000 cps (d)7500 cps

32. A book is resting on a shelf that is undergoing

vertical simple harmonic oscillations with an amplitude of 2.5 cm. What is the minimum frequency of oscillation of the shelf for which the book will lose contact with the shelf ? (Assume that, g = 10 ms-2 )

(p + �� ) (V- nb) = nRT. Its internal energy is given (a) 20 Hz

(b) 3.1 8 Hz

(c) 125.6 Hz (d) 1 0 Hz

33. A van der Waal's gas obeys the equation of state by U = CT-

n 2a

V

. The equation of a quasistatic

adiabat for this gas is given by (a) TC!nR • V = constant (b) r< C + nR )lnR . V = constant (c) TC!nR • (V - nb) = constant p< (d) C+ nR)lnR · (V - nb) = constant

34. An ideal gas is made to undergo a cycle depicted by the p-V diagram given below. The curved line from

A to Bis an adiabat. p

�-

-

-

-

-> - V

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214

KVPY Question Paper 2010 Stream : SB/SX

Then, (a)the efficiency of this cycle is given by unity as on heat is released during the cycle (b)heat is absorbed in the upper part of the straight line path and released in the lower part (c) ifT1 and T2 arc the maximum and minimum temperatures reached during the cycle, then the T efficiency is given by 1- 2 Tl

(d)the cycle can only be carried out in the reverse of the direction shown in above figure

35. A bus driving along at 39.6 km/h is approaching a

person who is standing at the bus stop, while honking repeatedly at an interval of 30 s. If the speed of sound is 3 3 0 m�- 1 , at what interval will the person hear the horn? (a) 31 s (b)29 s (c)3O s (d)The interval will depend on the distance of the bus from the passenger

36. Velocity of sound measured at a given temperature in oxygen and hydrogen is in the ratio (a) 1 :4 (b)4: 1 (c) 1 : 1 (d) 32: 1

37. In Young's double slit experiment, the distance

between the two slits is 0.1 mm, the distance between the slits and the screen is 1 m and the wavelength of the light used is 600 nm. The intensity at a point on the screen is 75% of the maximum intensity. What is the smallest distance of this point from the central fringe 9 (a) 1 .0 mm (b) 2.0 mm (c)0.5 mm (d) 1 .5 mm

38. Two masses 711.:t and m2 are connected by a massless spring of spring constant k and unstretched length l.

The masses are placed on a frictionless straight channel, which we consider our X-axis. They are initially at rest at x = 0 and x = l, respectively. At t = 0, a velocity of v0 is suddenly imparted to the first particle. At a later time t, the centre of mass of the two masses is at v t nl (b)x = l' i + n12 o l'ni + m2 m1 + m2 n1i vo t mi (d) x = + + l'ni m2 l'ni + "12

40. A piece of hot copper at 100 ° C is plunged into a pond

at 30 ° C. The copper cools down to 30 ° C, while the pond being huge stays at its initial temperature. Then, (a)copper loses some entropy, the pond stays at the same entropy (b)copper loses some entropy and the pond gains exactly the same amount of entropy (c)copper loses entropy and the pond gains more than this amount of entropy (d) both copper and the pond gain in entropy

CHEMISTRY 41. The number of isomers of Co (diethylene triamine) Cl3 is

w2

�s

(a) p-

(b) NHa

W4

W5

(c) CN-

(d)i-

(c) 1 .5

(d) 1

42. Among the following, the it-acid ligand is 43. The bond order in (a) 2

(b) 3

o:- is

44. The energy of a photon of wavelength (A) = 1 m is

(Planck's constant = 6.626 x 10-34 J s, speed of light = 3 x 10 8ms-2) (a) 1 . 988 X 1 0-23 J (b) l.988x 1 0-23 J 30 (c) 1 . 988 X 1 0 - J (d)1 .988 X 1 0-25 J

45. The concentration of a substance undergoing a

chemical reaction becomes one-half of its original value after time, regardless of the initial concentration. The reaction is an example of a (a) zero order reaction (b) first order reaction (c) second order reaction (d) third order reaction

46. The shape of the molecule CIF; is (a) trigonal planar (c) T • shaped

(b) pyramidal (d)Y • shaped

47. Friedel-Craft's acylation is (a) a. • acylation of a carbonly compound (b) acylation of phenols to genarate esters (c) acylation of aliphatic olefins (d)acylation of aromatic nucleus

48. The order of acidity of compounds I-IV, is

39. A charged particle of charge q and mass m, gets

deflected through an angle0 upon passing through a square region of side a, which contains a uniform magnetic field Bnormal to its plane. Assuming that the particle entered the square at right angles to one side, what is the speed of the particle? B B (b) q a tan e (a) q a cot e m m B (c) q a cot20 m

H,C -OOH III (a) I < Ill < II < IV (c) IV < I < II < Ill

IV (b) Ill < I < II < IV (d)II < IV < Ill < I

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215

KVPY Question Paper 2010 Stream : SB/SX 49. The most stable conformation for n-butane is

(a) H "

$ CH 3

(b) H "

H H

$

CH0

H

H

Cl I ,", F (c) Cl-P< I �F Cl

CH;i Q)CH, (c) H H H

H (d) H H� H

57. The enantiomeric pair among the following four structures is

50. In the nuclear reaction, 2

X is (a) e (c)H

-1

56. The correct structure of PC13 F2 is

g!Th----, 2�f Pa + X, (b) p e (d) f H

51. A concentrated solution of copper sulphate, which is dark blue in colour, is mixed at room temperature with a dilute solution of copper sulphate, which is light blue. For this process (a) entropy change is positive, but enthalpy change is negative (b) entropy and enthalpy changes are both positive (c) entropy change is positive and enthalpy docs not change (d) entropy change is negative and enthalpy change is positive

(I) HO �

(II) �OH

(III) HO �

(IV)

H

(a)I and II (c) II and III

(b) I and IV (d)II and IV

associated with the three species in the reaction,

r

X y

53. In metallic solids, the number of atoms for the face

centered and the body centered cubic unit cells, are respectively. (a) 2, 4 (c)4, 2 (b)2, 2 (d)4, 4

54. From equations 1 and 2, CO2 � CO+½ 02 [Kc1 = 9.1 x 10- at 1000 C] (Eq. (i)) 12

\/ �

58. Consider the reaction, 2NOi(g) ➔ 2NO(g) + 0 2 (g). In the figure below, identify the curves X, Y and Z

52. Increasing the temperature increases the rate of reaction, but does not increase the (a) number of collisions (b) activation energy (c)average energy of collisions (d) average velocity of the reactant molecules

H

/\H

H HO H

1

H2 0 � H2 + ½ 02 [K0 = 7.1 x 10-12at 1000 ° C] (Eq. (ii)) 2 The equilibrium constant for the reaction, CO2 + H2 � CO + H2 0 at the same temperature, is (a) 0.78 (b) 2.0 (c) 1 6.2 (d) 1.28

55. For a first order reaction R� P, the rate constant is k, if the initial concentration of R is [R,, ]. The concentration of R at any time t is given by the expression. (a) [R0 Ji' (b)[R0 ] e-k' (d) [R0 ] (1- e- kt ) (c) [R0 ] (1- e- kt )

Time ---4 (a)X = NO, Y = 02 , Z = NO2 (b)X = 02 , Y- NO, Z = NO2 (c) X = NO2 , Y = NO, Z =02 (d) X = 02 , Y = NO2 , Z =NO

z

59. The aromatic carbocation among the following is (a)

0 O@ o ) @ (b

�

(c)

I _

©

(d)

o I _

(fl

60. Cyclohexene is reacted with bromine in CC14 in the dark. The product of the reaction is, r---------r---Br (a) �Br

Br ) (b [:::::J__Br

6 Br

rYBr (c) �

( ) d

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216 BIOLOGY 61 .

62.

63.

Ribonucleic Acids (RNA) that catalyse enzymatic reactions are called ribozymes. Which one of the following acts as a ribozyme? (a) Ribosome (b) Amylase (c) tRNA (d)Riboflavin

In 1670, Robert Boyle conducted an experiment wherein he placed a viper (a poisonous snake) in a chamber and rapidly reduced the pressure in that chamber. Which of the following would be true? (a)Gas bubbles developed in the tissues of the snake (b)The basal metabolic rate of the snake increased tremendously (c)The venom of the snake was found to decrease in potency (d)The venom of the snake was found to increase in potency Bacteria can survive by absorbing soluble nutrients via their outer body surface but animals cannot, because (a)bacteria cannot ingest particles but animals can (b)bacteria have cell walls and animals do not (c)animals have too small a surface area per unit volume as compared to bacteria (d)animals cannot metabolise soluble nutrients

64. A horse has 64 chromosomes and a donkey has 62. Mules result from crossing a horse and a donkey. State which of the following is incorrect? (a)Mules can have either 64, 63 or 62 chromosomes (b)Mules are infertile (c)Mules have well-defined gender (male/female) (d)Mules have 63 chromosomes

65.

If the total number of photons falling per unit area of a leaf per minute is kept constant then which of the following will result in maximum photosynthesis? (a) Shining green light (b) Shining sunlight (c)Shining blue light (d) Shining ultraviolet light

66.

Path-finding by ants is by means of (a) visually observing landmarks (b) visually observing other ants (c)chemical signals between ants (d)using the earth's magnetic field

67. Sometimes urea is fed to ruminants to improve their health. It works by (a)helping growth of gut microbes that breakdown cellulose (b)killing harmful microorganisms in their gut (c)increasing salt content in the gut (d)directly stimulating blood cell proliferation

KVPY Question Paper 2010 Stream : SB/SX

68.

If you compare adults of two herbivore species of different sizes, but from the same geographical area, the amount of faeces producted per kg body weight would be (a) more in the smaller one than the larger one (b) more in the larger one than the smaller one (c) roughly the same amount in both (d)not possible to predict which would be more

69.

Fruit wrapped in paper ripens faster than when kept in open air because (a) heat of respiration is retained better (b) a chemical in the paper helps fruit ripening (c) a volatile substance produced by the fruit is retained better and helps in ripening (d)the fruit is cut off from the ambient oxygen which is an inhibitor to fruit ripening

70.

When a person is suffering from high fever, it is sometimes observed that the skin has a reddish tinge. Why does this happen9 (a)Red colour of the skin radiates more heat (b) Fever causes the release of a red pigment in the skin (c)There is more blood circulation to the skin to keep the body warm (d)There is more blood circulation to the skin to release heat from the body

71. Bacteriochlorophylls are photosynthetic pigments found in phototrophic bacteria. Their function is distinct from the plant chlorophylls in that they (a) do not produce oxygen (b) do not conduct photosynthesis (c) absorb only blue light (d)function without a light source

72.

Athletes often experience muscle cramps. Which of the following statements is true about muscle cramps? (a)Muscle cramp is caused due to conversion of pyruvic acid into lactic acid in the cytoplasm (b)Muscle cramp is caused due to conversion of pyruvic acid into lactic acid in the mitochondria (c)Muscle cramp is caused due to non-conversion of glucose to pyruvate in the cytoplasm (d)Muscle cramp is caused due to conversion of pyruvic acid into ethanol in the cytoplasm

73.

A couple went to a doctor and reported that both of them are carriers for a particular disorder, their first child is suffering from that disorder and that they are expecting their second child. What is the probability that the new child would be affected by the same disorder? (a) 100%

(b) 50%

(c) 25%

(d) 75%

74. Of the following combinations of cell biological processes, which one is associated with embryogenesis? (a) Mitosis and meiosis (b) Mitosis and differentiation (c) Meiosis and differentiation (d)Differentiation and reprogramming

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KVPY Question Paper 2010 Stream : SB/SX 75. Conversion of the Et protoxin produced by Bacillus thuringiensis to its active form in the gut of the

78. The fluid part of blood flows in and out of capillaries in

insects is mediated by (a)acidic pH of the gut (b)alkaline pH of the gut (c)lipid modification of the protein (d) cleavage by chymotrypsin

76. If you dip a sack full of paddy seeds in water

overnight and then keep it out for a couple of days, it feels warm. What generates this heat? (a)lmbibition (b) Exothermic reaction between water and seed coats (c)Friction among seeds due to swelling (d) Respiration

77. Restriction endonucleases are enzymes that cleave

DNA molecules into smaller fragments. Which type of bond do they act on? (a)N-glycosidic bond (b)Hydrogen bond (c)Phosphodiester bond (d) Disulphide bond

MATHEMATICS

� PART- I I

in decreasing

82. Let r be a real number and n E N be such that the polynomial 2x2 + 2x + 1 divides the polynomial

(x + I)" - r. Then, (n, r) can be

(b)( 4000,

1�00

4

)

1 (d) ( 4000, --)

_l_ ) 41000

4000

83. Suppose a, b are real numbers such that ab ;t 0.

Which of the following four figures represent the curve (y- ax - b) (bx2 + ay 2 - ab) = 0?

Fig 1 (a) Fig 1

Fig 2 (b)Fig 2

pairs is 0.34 nm. If the length of a chromosome is 1 mm, the number of base pairs in the chromosome is approximately (a)3 million (b) 30 million (c) 1.5 million (d) 6 million

80. Estimate the order of the speed of propagation of an action potential or nerve impluse. (a)nm/s (b) micron/s (c) cm/s

(d) m/s

(2 Marks Questions)

2� 4 r powers of x. Suppose the coefficients of the first three terms form an arithmetic progression. Then, the number of terms in the expansion having integer power of x is (a) 1 (b)2 (c)3 (d) more than 3

(c)( 41000 '

79. The distance between two consecutive DNA base

85. The following figure shows the graph of a

81 . Arrange the expansion of (:t11 2 +

(a) (4000, 41000 )

tissues to exchange nutrients and waste materials. Under which of the following conditions will fluid flow out from the capillaries into the surrounding tissue ? (a) When arterial blood pressure exceeds blood osmotic pressure (b) When arterial blood pressure is less than blood osmotic pressure (c) When arterial blood pressure is equal to blood osmotic pressure (d) Arterial blood pressure and blood osmotic pressure have nothing to do with the outflow of fluid from capillaries

Fig 3 (c)Fig 3

Fig 4 (d) Fig 4

84. Among all cyclic quadrilaterals inscribed in a circle of

radius R with one of its angles equal to 120 °. Consider the one with maximum possible area. Its area is (d)2,Js R 2 (c)2 R 2 (b),Js R 2 (a) /2 R 2

differentiable function y = f(x) on the interval [a, b] (not containing 0).

f(x)

A

,,� Ir: ,,� ,,�

Let g (x) =

X

_ Which of the following is a possible graph

of y = g(x)?

Fig 1

Fig 2

Fig 3

Fig 4

(a) Fig 1 (c) Fig 3

(b) Fig 2 (d)Fig 4

86. Let V1 be the volume of a given right circular cone

with O as the centre of the base and A as its apex. Let V2 be the maximum volume of the right circular cone inscribed in the given cone whose apex is O and whose base is parallel to the base of the given cone. Then, the ratio v2 f vl is 4 4 (c) (d)� (a) 3 (b) 27 27 25 9

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218

KVPY Question Paper 2010 Stream : SB/SX

87. Let f : R ➔ R be a continuous function satisfying

f (x) = x + J f (t) dt, for all x E R. Then, the number of

94. Consider the infinite ladder circuit shown below. L

X

elements in the set S = {x E R : f(x) = O} is �4 �2 �3 Wl 0

21t

88. The value of J min{ Ix- rrl , cos- 1 (cos x)} dx is 0

7t 2 (a) 4

2

(b)� 2

7t 2 (c)8

(d) n 2

89. Let ABC be a triangle and P be a point inside ABC ----> ----> ----> such that PA+ 2 PB+ 3 PC = 0. The ratio of the area of l>ABC to that of t,,,APC is W2

�� 2

�3

�� 3

90. Suppose m, n are positive integers such that

6 m + 2 m + n . 3 w + z n = 3 32. The value of the expression m2 + mn + n 2 is (c)1 9 (d)21 (b) 1 3 (a) 7

PHYSICS

91. A ball is dropped vertically from a height of h onto a hard surface. If the ball rebounds from the surface with a fraction r of the speed with which it strikes the latter on each impact, what is the net distance travelled by the ball up to the 10th impact? 2 r10 r .� ) (b)h (a) 2h ) 1- r l- r0 l- r : l- r 2: (c)2h ( (d)2h ( )- h )- h 1- r 1- r

( 1-

( 1-

92. A certain planet completes one rotation about its axis in time T. The weight of an object placed at the equator on the planet's surface is a fraction f (f is close to unity) of its weight recorded at a latitude of 60 ° . The density of the planet (assumed to be a uniform perfect sphere) is given by (a)

(4-/)-� 1 - f 4GT 2

3/ l _ � 1 - f ) 4 GT 2

(c)( 4 -

(4-/)-� f 4 GT

(b)

l r l

J

2 2 711

2 7 2/3

1 (c) __ g_ 6rrE0 k J

For which angular frequency w will the circuit behave like a pure inductance? 1 LC 2L 2 (c) (a) ✓2 (b) __ (d) .Jrc .JLc Jc

95. A narrow parallel beam of light falls on a glass sphere of radius R and refractive index µ at normal

incidence. The distance of the image from the outer edge is given by R(2- µ ) R(2- µ ) R(2 + µ) R(2 + µ ) (c) (b) (a) (d) 2(µ - 1) 2(µ - 1) 2(µ + 1) 2(µ + 1)

96. A particle of mass m undergoes oscillations about x = 0 in a potential given by V (x)- ½ kx2 - V0 cos (�),

where V0 , k, a are constants. If the amplitude of oscillation is much smaller than a, the time period is given by �-(b) 2rr J¥

v�

(d)21t �

97. An ideal gas with heat capacity at constant volume

Cv undergoes a quasistatic process described by p v a in a p- V diagram, where a is a constant. The heat capacity of the gas during this process is given by (a) Cit

(b) Cy + nR (c) Cy + -

nR

1- a

- (d)Cit + -

nR

1- a 2

-

98. An ideal gas with constant heat capacity Cv = � nR is made to carry out a cycle that is depicted by a triangle in the figure given below. p

2

2/ ) . � 1 - f ) 4GT 2

(d) ( 4 -

vertices of an equilateral triangle centred at the origin. They are held in equilibrium by a restoring force of magnitude F(r) = kr directed towards the origin, where k is a constant. What is the distance of the three charges from the origin?

r

L

L

2

l+

93. Three equal charges +q are placed at the three

l_ g_ (a) _ 6nE0 k

L

r

J

113

27 r;c; v u3_ g_ (b) __ 12nE0 k

l r l

J

r;c; 3u g2_7 v_ (d) _ 4rrE0 k

2/3

�-----�...... v V1

V2

The following statement is true about the cycle. (a) The efficiency is given by 1- A V1 P2V2 (b) The efficiency is given by 1- I. A V1 2 P2V2 (c) Net heat absorbed in the cycle is (p2 - A ) (V2 - V1 ) (d)Heat absorbed in part AC is given by 2(pzVz- Pi Vi )+ (A V2- P2V1 )

i

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219

KVPY Question Paper 2010 Stream : SB/SX 99. Two identical particles of mass m and charge q are shot at each other from a very great distance with an initial speed v. The distance of closest approach of these charges is 2 2 2 q q q (a) (d) 0 • (c) • (b) 2ne0 mv2 4ne0 mv~ 8ne0 mv~

1 00. At time t = 0, a container has N9 radioactive atoms

with a decay constant 'A. In addition, c numbers of atoms of the same type are being added to the container per unit time. How many atoms of this type are there at t = T? (a) � exp ( 'A.T)- N0 exp (-'A.T) A, (b) � exp (-'A.T)+ N0 exp ( 'A.T) A, (c)� (1- exp ( 'A.T)) + N0 exp ( 'A.T) A, (d)� (1 + exp ( 'A-.T))+ N0 exp ( AT) A,

CHEMISTRY

101 . 2.52 g of oxalic acid dihydrate was dissolved in 100 mL

of water. 10 mL of this solution was diluted to 500 mL. The normality of the final solution and the amount of oxalic acid ( mg/mL) in the solution are respectively (a) 0.1 6 N, 5.04 (b)0.0 8 N, 3.60 (c) 0.04 N, 3.60 (d) 0.02 N, 1 0.08 1 02. Two isomeric compounds I and II are heated with HBr.

OH

6 CH ,0H

OH

1 03. The number of possible enantiomeric pair(s) produced from the bromination of I and II, respectively, are CH 3 H 3C

>=
_!_ ✓ r2 + h 2 µ

r

r =-

--

µ - 1 \I�

;1!

So, frequency of oscillations is

⇒

1=

=

i=!C = 2 x 10 2� H

__-1:_ . f= 27! V 2.5 x 1 0-2

27!

⇒

f � 3.1 8 Hz 33. (c) Process is quasistatic adiabatic. So, dQ = O

⇒ ⇒

dW = - dU p!:J.V = - nCv dT

. . . (i)

U = CT - ri__ µ3 Also µ 1 ,;, µ 3 and µ 1 > µ 3 , because there is no emergent light. Hence, µ 3 < µ 1 < µ 2 is correct order.

r< R

So, correct graph is option (a).

Now, if a = acceleration of bead. Then, Restoring force = Mass x Acceleration 21'cos 8 = - ma ⇒ 2 T (2x) = - ma ⇒ l ⇒

(0.4 +x)3 = (56 ) ('i ) 0.4 + X = (52 ) (2) x = 50 - 0.4 = 4R6AU

1 l lr.·. µ = sin ic J

30. (d) In first incidence, light ray is

bends towards normal so, it must be going into a denser medium.

� �

µ 2 > µ1 1.e. As, in second incidence, no light goes into third medium so, total internal reflection takes place at second incidence.

and

P=

V

2

nRT n a V - nb V 2

From Eq. (ii), d U = CdT +

. . . (iii)

n2 � dV

V

So, from Eq. (i), we have dW = - dU ⇒

2

pdV = - ( CdT + : � dV J

Substituting for p from Eq. (iii) in above equation, we have ⇒(

nRT _ n 2� ) dV V - nb V 2 2

= - CdT - ri__ O h a = l, b = 0 1 2. (b) Let a point B on parabola is 2

B( � K) and A(0, 3). ,

A (0,3)

( d 2Z J

⇒

dK 2 K = 2

·: Z is minimum at K = 2 AB =

= 1000(210

-

1)

= 1000(1024 - 1) = 1023000 We have given total weight is 10228 7 0.

:. 1022870< 1023000

Extra amount of weight = 1023000 - 1022870

n

n

0

1 6. (b) We have, y = x2 and y = 1- x2 y

f

f

f cos rr xdx - f cos rr xdx 0

1/2

l

1/2

l_rr_Jo _ l_7l_L 0

1/2

. 7112 r Sill. rrx71 _ r Sill rrx ⇒ I_

1f . 1 . 1t . rr l . 07 - ⇒ J = - fL s1n-s1n S111 1t - S1112 2J 1t J 1t L 2 1 1 ⇒ I = -[1- 0J - - [0- 1] = -

7l

7l

Y'

Intersection point of y = x2 and y = 1- x2 1 1 -1 1 . A (1s 2 , - ) and C(- , - )

✓ 2

J2 2

Area of shaded region 1/✓2 = 2 J [ (l- x2) - (x2 )]dx 0 1/✓2 = 2 (1- 2x2 )dx

f 0

=

21 = 128 + 2

1/2 l cos(rrx) cos0dx + cos rrxcos rrdx

rr

+

1

10 + -J x9 cosnxdx no o 1 9 10 r x si nx 7 � nl n- J0

lim l = 0

rx - 2x 711✓2

J1 = L

3

3Jo

1 7 1 J2 - 3./2 J

2, 8. 1 4. (c) Let 1 1 = f cos( rrx) cos[2xjrrdx

⇒ I=

n

n

:. The trucks have the lighter bags arc

⇒ I=

cosn

1

8 _ 90 j x si nxdx � n O n 1 cosn lOsin n 90 s . ⇒ 1- - X Sin nxdX - - -- + --2 f

= 130

0

y2 =4x

⇒ I=

n

J

16 + (1)2 = ./2 16

1000(1 + 2 + 22 + . . . + 29 )

+

l

°

>O

1 3. (d) If all trucks had packets of 1000 g, then total weight is

130 = 27

0

r-.x1 cosnx7 ----

3K 2 4

dK

Y' Clearly, from graph total number of solution = 10 1 1 . (b) We have, / : R ➔ R by

⇒ I=

AB = Z

f

1 5. (a) Let l = x1° sin nxdx

2 3- 11 = 2✓ = 22 rs sq umts J

✓L

3

1 7. (b) We have, a = 3i - 4k and b = 5j + 12k We know that, angle bisector of vector a and b a b7 =/ + l� lblj 31 - 4k 51 + 12k 7 -+ 13 L -5

,J

-l

j

_ / 391 - 52k + 251 = A.[391 + 25J

65

+

SkJ

+

60k 7 J

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246

KVPYQuestion Paper 2009 Stream : SB/SX

1 8. (a) An envelope has space for at most 3 stamps. Three stamps of denomination is 1 and three stamps of denomination is a. The value of stamps of envelope must be 1 + 1 + 1 = 3, a + a + a = 3a, a + a + 1 = 2a + ], a + 1 + 1 = a + 2 a > l, lf a = 2, then maximum value of stamps is 3x 2= 6 :. Least positive integer has not stamp value is 7.

As battery remains connected during this activity, potential difference between plates remains same. So, the charge on plates (Q = CV) decreases. Also, energy of capacitor ( U = 1 CV 2 ) decreases as capacitance decreases. So, correct option is (d).

24. (d) From triangle law, in given vector pentagon,

n;

->m

I_ d;

m

⇒

n

>

d lm d

20. (a) Set A have m-elements, Set B have n-clcmcnts (a, I\ ) E R, (a, 62) E R ⇒ (I\ = 62) By condition relation is a function. :. Total number of function (Relation) = nm

2 1 . (b) Mayer's relation,

CP - Cv = R is true only for ideal gases whether they arc monoatomic, diatomic or polyatomic as for real gases. 2k cp - Cir = where, k = isothermal compressibility and 13 = isobaric thermal expansion coefficient. So, correct option is (b).

m

22. (b) Gas molecules docs not settle in

room because they have too much kinetic energy (k8 T > mgh). lf heat energy is dissipated or extracted from that room, gas molecules docs settle down. For example, earth's atmosphere is more denser near the surface. So, correct option is (b).

A4 B 2 + A3 = B3 B;i + A4 = B 4 B 4 + A,; = B5 B5 + Al = Bl

. . . (i) . . . (ii) . . . (iii) . . . (iv)

Adding these equations, we have ⇒ (B 2 + � + B 4 + B5 ) + (A;i + A4 + A,; + Al ) = (B;i + B 4 + B5 + B 1 ) ⇒ B 2 + B;i + B 4 + B5 + (- A2) = (- B 2 ) . . . (v) As from polygon law of vector addition, A:i + A4 + A,; + Al = - A2 and R3 + B 4 + B5 + B1 = - B 2 So, from Eq. (v), we have B 2 + B3 + B 4 + B5 = A2 - B 2 = - B 1

- B,

25. (c)

pa

decreases.

eA � )

V

=

f dQ = 1i i kQ = Q -R 4rre0R

28. (c) Heat supplied at constant

pressure is AQ = nCP AT. Work done by the gas, A W = pAV = nRA'l' Ratio of work done to heat supplied is AW nRAT --- - AQ nCPAT = !!:__ = cp cp

- Cir

cp

=1-� y

(7 / 5)

7 7 29. (c) Lyman-alpha line sometimes =

1 1 - -- = 1 - � = �

denoted as Ly-o. line is obtained when an electron falls from n = 2 to n = 1 orbital in an one electron ion or atom. Similarly, Balmer-alpha radiation is emitted when an electron makes a transition from n = 3 to n = 2 quantum state. Now, using __!_ = R ( -; - -; )

A

n1

n2

Above system is equivalent to two capacitors in parallel,

23. (d) As plates arc moved far away further, capacity of capacitor ( C =

phase of wave changes by rr radians and its direction of travel is reversed. So, resultant wave is y = a cos (kx + wt + rr) ⇒ y = - a cos (kx + wt)

or

I_ d;

d ln d

26. (c) Due to hard boundary reflection,

=

[·: m < n]

n

d

Potential due to complete charge distribution at centre is k V = f d V = f �Q

We have,

I_ i > I_i

2EoA

R

d is positive division of m.

I_ D; 1 1 1 1 I- = - + - + . . . + - = - DI D2 D, n 1 1

=

d

charge dQ at centre of sphere is kd Q dV =

Let d1 , d2 , f¾ , . . . , dk are divisor of m and D1 , D2, D3 , • • • , D, are divisor of n I_ d; = I_ D; (given) I_ d; 1 1 1 1 Now, I, - = - + - + . . . + - = -m cl; d1 d2 dk =

d

27. (d) Potential due to an elemental

1 9. (c) We have,

(! )

So, capacitance of system is E E A Ceq = C1 + C2 = {0 + o

( ) + P v q

As, electric force on positively charged particle is also in same direction. So, a positively charged particle will be deflected upwards. Hence, no particle can get through the hole in slit. So, option (a) is correct.

40. (a) From conservation of energy, energy of particle at height H over inclined plane must be equal to energy of particle at point 0.

____,r2--➔x ---+--� ----:r1 ro

3 1 . (d) As the loop is moving in the

e

Clearly, force on block by inclined plane is m (g + a) in upward direction.

2 4 _ 3 kq

-8?

'y'

\-1/

m (g+a )

36. (c) Given, potential energy curve closely represents potential energy of a simple harmonic oscillator 2 (U = ¾ kx , shown in dotted line }

Now, the final condition is

2

:

Equatorial line

of A and C will be 2..

� �

, s'c 2Nl-f:iKc = 4.1 ... (i) ]:_ N2 + � H2 s===>c NI-f:i ... (ii) 2 2 We divide equation (i) by 2 to get equation (ii) . :. The new equilibrium constant, K � for Eq. (ii) will be the square root of K � Eq. (i), i.e. K = fKc � K' =✓41 = 6.4 54. (d) For secohd order reaction, 2A- Product Rate ofreaction, r = k [A]2. If the concentration is reduced to half, . A Le. -. 2

1 or r ' = - r

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249

KVPY Question Paper 2009 Stream : SB/SX 55. (b) The conjugate base differ by a proton from its respective acid. The conjugate bases for HCO3 and NHs respectively arc co;- and NH2 . + HCO:i � co;- + H Conjugate base

Acid

Nl-ls � NH2 + H

+

Acid

Conjugate base

56. (d) The aromatic compounds are

those compounds, which follow three given conditions (i) the compound has to be planar and cyclic. (ii) delocalisation of electrons should occur. (iii) follows Huckcl's rule [( 4n + 2)1t].

CH2

6

10 1t electrons (Aromatic)

(I)

6 1t electrons (Non-aromatic due to localisation of electrons)

(III)

(II)

s

51

6 1t electrons (lone pair takes part in conjugation (Aromatic)

Thus, correct option is (d).

(IV)

57. (a) ln compound l and compound l lI the central atom N is sp3 -hybridised, so their basicity will be more than compound II and IV. Compound III will be less basic as compared to I because of the presence of more electronegative atom, 0. Between 11 and IV, II will be more basic because the lone pair on N - atom will not take part in resonance and will be available for donation while the lone pair on N-atom of compound l V will take part in resonance and will b e least available.Thus, the correct order of basicity will be I > III > II > IV. 58. (d)

59. (b) Given, t112 of first order reaction = 30 min 0693 ·� For first order, Ii = t11 2

k=

0.693 = 0.0231 30

Also, if the reaction is 75% completed. .303 log ___Cl-_ t=2

k

a-x

100 2.303 t= l og 0.0231 25 = 99.69 x log 4 = 99.69x 0.602 = 60 min

60. (a) Number of moles of H2SO 4

6 1t e)ectrons Non-aromatic as the bond is exocydic which makes it non-planar

0

conformers.This can be obtained by rotating one of C2 or Ca carbon atoms roughly at an angle of 60° .

Me Me Hr +
c2 + b2 - 2 be ⇒ b2 + c2 - a 2 < 2bc Similarly, a2 + b2 - c2 < 2ab

2 2 2 a + b +c ⇒ ----- < 2

ab+ be+ ca

t n 4 > n 2 > n1 (a) n 2 > n3 > n 4 > n1 (c)n3 = n1 > n 4 > n 2 (d) n 2 > n1 = ng > n 4

✓

90. If If(x) I+ 1 + cos2nx = tan 2 ( �x)t(x), then f(3) is equal to 1 (a) .f2

1 (b)-2.fi

-1 (c)­ .fi

1 (d)--_ 2.fi

+

6V (b) 24 W (d)72 W

(a) 1 2 W (c) 36 W

95. Moment of inertia of a rod of mass M and length L that rotated about its centre along an axis, which makes angle of 30 ° with the length of rod will be

PHYSICS 9 1 . Which of these transitions are not frequently possible? (a) 3p----. 1 s (c) 3p ----. 2p

(b) 2p----. l s (d) 4p----. 3p

92. A gas is contained in a small container which is

connected to a vaccum chamber via a stop value.

(a)

ML2 12

(c)

ML2 48

(d)

ML2 16

96. Two nucleus are projected in a region of magnetic Vaccum

Volume (150 L)

field B, such that their velocity vectors v are parallel as shown below. B

Volume (2 L)

The 2L container contains 5 moles of gas at 300 K. Given, ln 76 = 4.3. At time t = 0, stop value is opened and gas expands in 150 L vaccum chamber. Now, Choose the correct option.

+

+

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265

KVPY Practice Set 1 Stream : SB/SX Now, choose the correct option. (a) Trajectories of isobars are similar circles (b) Trajectories are always similar helix (c)Trajectories of isobars arc similar helix (d)Heavier isotope traces a helix of larger radius

97. A uniform rod of length L is placed over a horizontal frictionless surface. It is pulled over the surface by applying a force F on one of its end. Area of rod is A L ---------

� L/3 � Stress in the rod at a distance L 1 3 from the end, where the force is applied is F (a) "f... (c)'!:. "f... (d)� (b)_!_ "f... A 4A 3A 3A

98. A rigid massless rod o f length L connects two

particles each of mass m. The rod lies over a horizontal frictionless table. Another particle of mass m and velocity v0 struck one of two particles attached with rod.

(b) Length B is equal to length D as average distance between two liquid molecules is much less than distance between two gas molecules (c) Length B is less than length D as average distance between two liquid molecules is much less than distance between two gas molecules (d) Length B can be less or more than D depending on temperature

CHEMISTRY 101. Which of the following molecules is least resonance stabilised?

(a)

O

(b) Q

(c) O I � N

102. The pair having the same magnetic moment is [at. no. Cr = 24, Mn = 25, Fe = 26 and Co = 27] 2 (a) [ Cr(H20)6 ] + and [Fe(H20)d+ 2+ (b) [Mn(H2O)6 ]2+ and [Cr(H2O)6 ] 22+ (c) [CoC1 4 ] and [Fe(H20) 6 ] (d)[Cr(H20) 6 ]2+ and [CoC1 4 ]2-

103. The hybridisation and geometry of XeF4 is (a) dsp 2, square planar (c) sp3 , tetrahedral

----+- ---------------------- ·

Angular velocity of rod about its centre of mass just after collision will be (take, collision elastic) v v v (a) J2 v0 (b) 212 0 (c) 3-./2 0 (d) 4J2 0 7L 7L 7L 7L

99. A square loop of area 100 cm with sides parallel to X 2

and Y-axes is moved with velocity 8 cms- 1 in positive direction of X-axis. A magnetic field exists in region which points in z-direction. Magnetic field decreases along negative x-direction at rate of 0.1 Tm- 1 and also decreases in time at a rate of 1 0 -3 Ts- 1 • Emf induced in the loop is (c)90 µ V (b)70 µ V (a) lO µ V (d)lOO µ V 100. Heating curve for water is shown below. T(°C)

.--D---+

(b) sp3 d 2, octahedral (d) sp3 d 2, square planar

104. Which of the following cannot be made by using Williamson's synthesis? (a) Di - tert butyl ether (b) Methoxy benzene (c) Benzyl p-nitro phenyl ether (d)Methyl tertiary butyl ether

105. Which of the following complex ion will exhibit

t

geometrical isomerism? (b) [Pt(NI-fa )3 Cl] (a) [ Cr(H20)4 Cl2 (d)[Co(CN)5 Clf (c) [ Co(N� )6 f +

106. An organic compound (A) with molecular formula C9H10 O forms an orange-red precipitate with 2, 4-DNP reagent and gives yellow precipitate on heating with iodine and NaOH. It does not reduces Tollen's reagent or Fehling's solution nor it decolourises bromine water as Baeyer's reagent. On drastic oxidation with chromic acid, it gives a carboxylic acid having molecular formula C7 H602 . Identify the compound (A) .

I S:.4. ISo, more entropy gain occurs in B than entropy loss in A. Total entropy of universe increases due to this process.

26. (d) As air passes around plane wing,

= S x 10-4 kgm- is- 1

Flow rate using Poiseuille's law is 4 . _ 1t r (P; - po )

J-

8TtL

7t X l x 10-4 X 5000 4 8 x 8 x l 0 - x 0.2 51t 5 = x 1 0- m3 s- 1 = l0mL s- 1 128

23. (a) From gas laws, we have m=

M- ( ;� )

Mass of hydrogen gas present is

Due to shape of aerofoil, velocity of air layers over above surface of wing is more than lower surface of wing.

As, � p v2 + p = constant

2

When v velocity increases, pressure will fall. So, a pressure difference acts on lower surface of wing causing lift of plane.

T

C�-)

MH PH x 100 mH = -----�-�----(MH pH + Mco Pco + Mctt ctt 2 2 4 4 +

Afc

p

Le.

mvi = � ml2 - w

⇒

2

⇒

W=

3 3mu -

4lm

3J2mf{

ii =_ _

w·

29. (a) With !lx = 5 x 1 0- 15 m, =

4lm

✓K

mz2

By uncertainty principle, we have h !l p · !lx ?. 41t h (- ) 34 � 1054 X 1 0 !lp ?. ?. ⇒ 15 2Ax (2x 5 x l 0 - ) !lp ?. llx 10-20 kg ms-1 ⇒

1 1 . energy, K = . me2 =- pc Kmet1c

2

2

= � x ll x 10-20 x 3 x 108 2 = 2.6 X 1 0-12 J

30. (d)

Percentage of mass of hydrogen is

3

As there is no external torque, angular momentum about P is conserved.

27. (d)

mH = MH ·PH ( .: )

2

6

= - A Q = - mCy A T Heat gained by cold gas in B

T2 - T1 - F = 0

⇒

ml2 ml2 2 = -- + -- = - ml2

X

X

X

P

2

tt 6 c 2H6 > ( ;T ) P

2 x 200x 100

2 x 200 + 44x 150 + 1 6 x 320 + 3 0 x 105

= 0.026 x 100 = 2.6%

24. (b) Kinetic energy of a molecule at

temperature 1' is 1 3 KE = - m v2 = -KB T 2 rms 2 The molecule can rises until whole of its kinetic energy is changed to potential energy. Let this height is h. Then, 3 mgh = -KB T 2 3KB T 3 x 138x 10-23 x 300 = ⇒ h= 2mg 2 X 4.5 X 10-26 X 10 = 27.6km = 28 km

If applied force is just sufficient to hold the block over incline, friction acts in upwards direction (so treated positive). F = F; = mg(sin0 - µ s cos 0)

⇒ F; = mg sin 0 -/ or / = mg sin 0 - F When F = mg sin 0, friction f is zero.

If applied force exceeds mg sin0, then block moves upwards and friction now acts downwards.

:. Fr = mg(sin0 + µK cos 0)

Hence, friction in now negative.

As µ s > µK , graph must be (d).

28. (c) Moment of inertia of cube about an edge is (by parallel axis theorem) l = lcM +

m(�r

Direction of force on an electron using Flemming's right hand rule is towards centre of wire. Force on electron is

F =Bqvd

where, ud = F=

- . nqA µ ol . 21ta

q ·-I nq1ta2

F =� 3 2

2 1t na Substituting values in above equation, we have 7 2 4 x 1t x 10- x 10 F= 3 3 2 28 2 x 1t x 5 x 1 0 x (0.5 x 10- ) = 81t X 10-26 N'

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271

KVPY Practice Set 1 Stream : SB/SX 3 1 . (c) ln rotating disc, let a potential difference dV occurs across radial distance dr.

ln equilibrium, electric field balances centrifugal force on an electron.

ed

V

dr

= =

m u} r ⇒ d V

=

m 2 oi rdr e

:. integrating between limits, we have

V

JV(R)

V (R) =

o

dV

-mw2R2 e- 2

=

JR m

- w2rdr o e

⇒

32. (c) For adiabatic process,

=

Tr y y - 1 ⇒ f

V =10 Here, _!.

'!.l T;_

=(

y v; ) -

1

Vt

Vt

So, T1 = T;_ (lO)Y - 1

For monoatomic gas, y = .fi and for 3 7 . . 1c gas, y = - . drntom 5

� -1

4

T Hence A = � = 1 015 '

1,

B

2- - 1

10 5

33. (b) Given is a refrigeration cycle. In process BC, pressure is decreasing at constant volume. i.e. heat is being extracted from the working medium (gas).

34. (a) We can throw a ball up with any velocity but it cannot have velocity greater than its terminal velocity during free fall. So, option (a) is correct. 35. (b)

⇒ . . . (i)

cos(r1 ) max � _!

n

sin a sin(a) max :::; =n ⇒ Also, n .sm 'i sin(r1 ) max sin 90 < 1 n ⇒ ____ :,; n sin(r1 ) max sin(r1 ) max . 1 ⇒ Sill( rl ) max > --

n

2 =1

k = 8.62 x 10-5

nmin =J2 36. (c) First reaction is 14 N + 2 H- 15 N + 1 H+ 8.53 McV

⇒

We can rearrange above equation as, 14 N - 15 N - p - d + 8.53 ⇒

where, p = 1 H , d = 2H

K - 13 C = a - d + 7.58

. . . (ii)

C - 1 1 B = a - d + 5.16

. . . (iii)

13

Adding above equations, we have 14 N - 11 B = p + 2a - 3d + 2 127 11 ⇒ B - 1 4 N+a- n

= 3d - a - p - n - 2127

Now, 3d - a - p - n = 3 x 2.0140- 4.0026-1.0078- 1.0087 = 0.0229u

= 0.05 McV

37. (d) Using Ampere's law, magnetic

=1:1t�2

field of a solid conducting rod is B

· r , r:::::R , r>R µ 01 21tr

So, correct graph is (d).

= 143 x 104 K = 14300 K

39. (c) To keep situations similar, we

must keep reynolds number of flow same in both cases.

⇒

So, rewriting given reactions, we have 14 N _ l5 N = p - d + 8.53 . . . (i) 15

eV

� X 1 3.6 4 8.62 x 10-5 x log 4000

n2

So, Q value for reaction is 11 B(a , n)1 4 N = 0.229x 93 1 - 2 127

n

E1 = - 1 3. 6 eV

Taking log in Eq. (i), we have 3 (- E1 ) 4 T =�k log(4000)

1 - _2_2 = __.!:._2

⇒

. . . (i)

where, N2 and N1 are number of possible electron states in n = 2 and n = l quantum states.

. . . (ii)

From Eqs. (i) and (ii), we have 2 1 - cos2 (r1 ) max = sin (r1 ) max n

N 1 ) _ _ = 2 . (e- (E2 - E1 ! kT ) 1000 Nl

Also, E2 and E1 are energies in n = 2 and n = l quantum states. E Here, N2 = 8 , N1 = 2 , E2 = 1 and 4

But a max = 90°

In practice the potential difference is very-very small.

' '

state 2 to number of atoms in state 1 is 0.001. n(E2) 1 =-So, n(E1 ) 1000

n

Note That outer rim will be negatively charged.

rvy- 1

38. (d) Given ratio o f number of atoms in

From figure, we have r2 = 90 - r1 (r2 ) mi n = 90 - (rl ) max Also, for TlR, (r2 ) min � ec (critical angle) sin(12) min � sin ec ⇒ sin(90 - (Ii ) max ) � sin 8c

- puD ) NR - ( �

⇒ ⇒

model

- ( puD )

-

� car

=_____:�I �

40. (a) Due to induced charge produced

by polarisation of dielectric, charge q1 will be acted upon by an additional charge.

� �I

As negative induced charge is now more near to q1 than induced positive charge, attraction is more prominent. Hence, net force on q1 will be more. 4 1 . (c) Total number of stcrcoisomcrs = number of geometrical + number of optical isomers C� CH = CHCH2 CHBrC� is an alkcnc, so it can have two geometrical forms cis and trans. Since, the given compound has a chiral centre, therefore each geometrical isomer has a pair of cnantiomcr. Thus, in all there arc 2 x 2 = 4 stereoisomers possible for given

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272

KVPV Practice Set 1 Stream : SB/SX

n -1 cm 2

42. (b) Given,

A';:iaCI = 126.45 A';;CI = 426.16 Q- l cm 2 2 1 AC: , 2 H 5 COONa = 91 Q- cm A c2 H 5 COOH = "-c:2 H 5 COONa + "-7ic, - "-';:iac1 = 9 1 + 426.16 - 126.45 = 390.71Q- l cm-2

43. (c) ti.O ° = - 262 kJ = - 26200GJ ti.G = - 2.303RT log k ti.G

logk

-2.30RT

26200 2.303x 8.314x 298 logk = 45.918 45 k = 8.279 X 10

44. (b) Ions of elements which have

paired electrons or no electrons in d-orbitals are colourless. The electronic configuration of the elements are as follows: Zn 2+ = [ArJ 3 d 1 0 4s0 ⇒ Paired d- electrons. Cr + = [Ar] 3d10 4s0 ⇒ Paired d-electrons. Ti 4+ = [Ar] 3d 0 ⇒ Ko d-electrons. y2+ = [Ar] 3d 0 ⇒ No d-clectrons. Thus, the set of above ions arc colourless.

45. (a) The depression in freezing point

is a colligative property. Thus, it depends upon number of particles or ions. More is the number of ions, minimum will be the freezing point. For 0.01 M KaCl, Number of ions = 0.01 x 2= 0.02 For 0.005 M C2 H5 OH, Number of ions 0.005 x 1 = 0.005 For 0.005 M Mgl 2 , Number of ions = 0.005 x 3 = 0.0 1 5 For MgSO4 , Number of ions = 0.005 x 2 = 0.01 As, 0.01 M NaCl has highest number of ions, so it will have minimum freezing point. 46. (a) 90 Th 232 ---, s2 Pb2os + Xi a + Y1�

Let the number of a-particles be x. Let the number of�-particles be y. In radioactive disintegration atomic number as well as mass number must be equal to two sides of equation. :. 208 + 4x + Oy = 232 4x = 24

x= 6

and 82 + 2x - y = 90 2x - y = 8 ⇒ y = 4 :. Number of a-particles = 6 Number of�-particlcs = 4 .

47. (b) Given, conditions should be

followed for an electron in an orbital. (i) The value of I should ranges between 0 to n - 1 (ii) The value of m should ranges between -I to + I. (iii) The value of s = + � or -�. 2 2 The electronic configuration of Ti is [Ar] 3d 2 ,4s2. 1 Therefore, n = 4, I = 0, m = 0, s = + - . 2

48. (c) Natural rubber is made from

monomer isoprene (2 -methyl, 1, 3-butadiene). lt resembles synthetic rubber, neoprene, which is produced by the monomer chloroprene. CH 2-CH = C(Cl)-CH2 +,;_

+

Neoprene

Natural rubber

Polymerisation

49. (a) pn - dn overlapping is formed when p-orbital of one atom and d-orbital of another atom overlap laterally. ln case of Pd!-, P is sp3 -hybridised, it forms four P-0 cr bonds by overlap of sp3 -hybridised orbitals of P with p - orbitals of 0. The bond is however formed by overlap of p-orbitals of O and d-orbitals of P. N and C on the other hand cannot form pre - dn bond, since they do not have d-orbitals. 0 �+---prr-d11 bond

-o/ 1 '-00-

50. (c) Total number of voids = Kumbcr

of octahedral voids + Number of tetrahedral voids. Number of octahedral voids = Total number of atoms (N) is a closed structure (0.5 mo!) = 0.5 X 6.022 X 1023 = 3.011 X 1023 Number of tetrahedral voids = 2N = 2 X 3.011 X 1023 = 6.022 X 1023 Total number of voids = 3.0 l l x 1023 + 6.022 X 1023 = 9.033 X 1023 51 . (d) For first order reaction 2.303 a log-_ t= a-x k For 90% reaction, 2.303 100 100 = log 100 - 90 k

k=

AlSO,

2.303 log 10 100

k = 0.023 0.693 0.693 -=- = 30 , . 1 ID!. n t112 = k 0.023 = 30 min

52. (d) Cu produces NO (X) gas when

treated with di!. HNO3 . This gas is reduced to NH2OH- HCl(Y) by SnCl 2 /HC! and then oxidised by HNO2 to give N2 O (Z). The reaction is as follows: SnCl 2 /HCI - =-- -' Cu + Di! . HNO3 ---, NO HN0

(X)

2 N2O NH2 OH · HCI --�➔ (Y)

53. (b)

HO

CH20H

(Z)

+ HCI

4-hydroxymethyl phenol

Heat CH2CI + H 20

HO This reaction undergoes SN l reaction. Where, the OH of benzyl group gets substituted by the chloride group through carbocation formation. Herc, the phenolic OH group does not get substituted by Cl group.

54. (b) The structure of given acids arc as follows (a) Aspartic acid 0 O �OH OH NH2 It has one chiral centre . (b) Arginine NH

0

H2�N�oH H NH 2 It has two chiral centres . (c) Lysine 0 H2 �0H NH 2 It has one chiral centre . (d) Histidine (� OH NH +NH 3 It has one chiral centre . Thus, arginine has more than chiral centre.

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273

KVPY Practice Set 1 Stream : SB/SX 55. (c) Pyridinium hydrochloride is a

salt of pyridine (weak base) and hydrochloric acid (strong acid). pH of a solution of strong acids and weak base can be calculated as 1 pH = 7 - - (pK6 + log c) 2 pK6 _ log 0.02 3.44 = 7 _ 2 2 pKb = 8.82 pKb = 8.82 = - log Kb Kb = 15 x 10-9

56. (c) Among the given compounds, III

is most basic as one of the nitrogen atom is sp3 -hybridised. Between I and II, I is more basic as the lone pair on N atom is relatively more available for donation as compared to 11. Also, in compound 11, the other N-atom shows -I -effect which makes it a weaker base. Thus, the correct increasing order of basicity is II < I < III.

57. (a) A mixture of bcnzaldchydc and

formaldehyde on heating with aqueous NaOH solution gives bcnzyl alcohol and sodium formate. This reaction is known as Cannizzaro reaction, where the formaldehyde gets oxidised to sodium formate and bcnzaldchydc gets reduced to benzyl alcohol.

�

I

/n\_CHO + H-C-H NaOH Benzaldehyde

H20

0

Formaldehyde

@-cH20H+ HCOONa

Benzyl alcohol

Sodium formate

58. (b) For the reaction, 11 C3H7 NO,(s) • + -Oi (g) 2

- 3CO2 (g) + J:. N,(g) + 2 H,O(g) 2 2 7 t:i.Hcomb = 3t:i.Hr CO,_ + 0 + - t:i.Hr (H2O) 2 - t:i.Hr (C3H7 N02 ) + 0 = 3 (-94.05) + '!._ (-68.32) - (-133.57) 2 = - 387.70 kcal/mo!.

59. (b) Given, T1 = 25° C = 298K T2 = T

E0 = 104.4 kJ mol- 1 = 104.4 X 1 d3 J mol- 1 hi_ = 3 x 10-4 s -1 , k2 = ? According to Arrhenius equation.

[-1:_ - -1:_]

k log 2 = � hi_ 2 303R 1;

T2

log

k2

3 x 1 0 -4

----- X --

1 04.4 x HY 2303 X 8.314

k2 log = 18.297 3 x 10-4 k2

3 X 1 0-4

1 298

1l as T � oo '

64. (d) Nuclear division would have been

.!.. � ol T

J

= 198 x 1018

k2 = 198xla18 X 3 xl 0-4 = 5.94 X 104 z 6 X 1014 S-l 60. (c) ln BF3 boron is sp 2 hybridised. lt has a vacant 2p -orbital, each fluorine atom in BF3 has completely filled unutilised 2p-orbitals. So, prr - prr back bonding occurs, where a lone pair of electrons of F-atoms gets transferred into vacant 2p-orbital. Therefore, B-F, bond has some double bond character. That is why all the 3B - F bonds are shorter than the usual single B-F bond. Whereas, in [BF4J, boron is sp3 -hybridised and thus docs not have empty 2p-orbital, so no such p11 - prr back bonding can occur. BF_;- has pure B-F bonds. Hence, B - F bond length in BF3 are shorter than BF_;-.

Back bonding in BF3

61 . (a) Acctylcholinc released from an excited terminal axon, depolarises the membrane and is then very rapidly inactivated by the enzyme cholinesterase, which is present in great concentrations in the neuromuscular junction. The inhibition of cholinesterase will result in the muscle being unable to restore the state of polarisation. 62. (b) Before pollination, the pollen grain cytoplasm divides between the generative cell and vegetative cell. The generative cell divides to form two male gametes. 63. (b) In a case, when SA-node or the pacemaker becomes non-functional, then there will be no origin of heartbeat and will be no transmission of impulses of atria. Also, the ventricles will fail to receive atrial impulse by obstruction in AV conduction. Thus, overall conducting system of heart is disrupted. Hence, the complete cardiac muscles fail to contract rhythmically in a well-coordinated manner.

completed by telophase, resulting in the amount of DNA in the nucleus being halved, compared to the amount present at interphase, prophasc and metaphasc of cell divison cycle.

65. (c) Haemoglobin is a globular protein

that binds to a non-proteinaceous substance, haem, which actually carries the oxygen. This material is known as a prosthetic group and influences the shape of the polypeptide.

66. (c) Only transformed bacteria

containing plasmids coding for resistances to antibiotics will be able to survive in a medium containing antibiotics. Bacteria lacking the plasmids will be destroyed by the antibiotic.

67. (b) Any agent or factor that induces

cancer is called carcinogen or carcinogenic factor, e.g. cadmium oxide, which induces prostate cancer. Chromium and nickel compounds induce lung cancer, whereas vinyl chloride and aflatoxins induce liver cancer.

68. (c) The experiments demonstrated

that genes carry information for making proteins. Beadle and Tatum proposed one gene one enzyme hypothesis on the basis of biochemical genetics in Neurospora crassa . Beadle and Tatum showed that each kind of mutant bread mould lacks a specific enzyme and were unable to form the protein.

69. (c) In a nucleotide, the 5'-OH of a nucleoside (rather than a nitrogen base) is linked to a phosphate group. DNA acts as a genetic material in most organisms. lt is a long polymer of deoxyribonucleotides. The number of nucleotides present in the DNA determines its length. A nucleotide is composed of a nitrogenous base, pcntosc sugar and a phosphate group. Rest all other statements are correct for DKA. 70. (c) Option (c) cannot be explained on

the basis of Mendel's law of dominance. When F1 hybrids exhibit a mixture or blending of characters of two parents, the case is considered as that of incomplete blending inheritance. It simply means that the two genes of an allelomorphic pair arc not related as dominant or recessive, but each of them can express itself partially.

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274 7 1 . (c) Gene mutations are spontaneous events which result in the alternation of existing genes and the occurrence of new sequences. These mutations are thus responsible for variations occurring in the gene pool of a population. 72. (b) Pure culture without any contamination is called axcnic culture. It is not associated with any living organisms and is raised under sterile conditions. 73. (a) Photosynthesis occurs most efficiently at wavelengths of red and blue light, because the chlorophyll pigments absorb light maximally at these wavelengths. Thus, for every 100 units of sunlight falling on a chloroplast of a green plant, so units are not used for photosynthesis as the wavelength are inappropriate. 74. (c) Pyruvate is a three carbon compound (CH:i C[Oj C [OjOH), which on complete oxidation releases three moles of CO2. 75. (b) The correct matches arc as follows Cup - shaped - Chlamydomonas Girdle-shaped - Ulothrix Stcllatc - Zygnema

KVPV Practice Set 1 Stream : SB/SX

80. (c) Embryonic stem cells arc not totipotent but are multipotent cells. They have the ability to differentiate into almost any cell type to form any organ or type of cell. 81 . (c) Suppose the green ball goes to bin i, for some i 2'. L The probability of this occurring is

77. (b) DKA polymerase binds to the single-stranded RNA primer and catalyses the formation of a phosphodicstcr bond between the free nucleotide and the preceding nucleotide, thus extending the new strand of DNA. In the event that the primase is defective, RNA primer will be absent and DNA polymerase will be unable to add nucleotides at the origin of replication. 78. (c) AIDS symptoms appear in the most advanced stage of HIV disease. Some people show HIV symptom shortly after incubation period but it usually takes more than 10 years. HIV attacks T-helper cells which regulate both hormonal and cellular immunity. This causes their depletion due to which patients immune system becomes very weak and susceptible to even minor infections.

-l 2'

Given that this occurs, the probability that the red ball goes in a higher numbered bin is

1

1

1

.- + . - + ... = � 2' 2 •+ 1 2' + 2 Probability that the green ball goes to bin i and the red ball goes in a bin greater than i is (

½

r

2�

-

QC = � AQ 2

⇒

16 = 6 x BR ⇒ BR =

16 � = 6 3

83. (d) We have, 1 12 + 9x2 a = f (e9x + 3 tan- x )( ) dx 1 + x2 o tan-1 x ) 9 + � dx 9x a = f (e + 3 ) ( 1+ X O Put 9x + 3tan- 1 x = t ⇒

(a + 1) = 31t log.(a + 1) - - = 9 4

84. (a) We have, n1 , n 2 , n3 , n 4 , n5 are positive integer and n5 > n 4 > ns > n 2 > n1 .". nl n4

90. (a) We have,

I/ (x) I + .Jl + cos2 7tx = tan 2( �

)r (x)

M x2

x · J mM L 4

m

L 4

2

-

· dx

dx = � ML2 48

96. (d) Velocity of nucleus can be resolved into parallel and perpendicular components. Due to parallel component nucleus moves in direction of B and due to perpendicular component, it rotates in circular path. mu Radius of circular path, r = 1_ ⇒ r = m

Bq 2nm and pitch of helix, p = · v 1 1 ⇒ p oc m Bq

As, both isotopes have same charge q, heavier traces a path with larger radius and larger pitch.

n4 = l

n; = sin 2 7 + cos2 7 + 2 sin 7 cos 7

n1 = .J1 + sin 1 4 = n3

M dx L

1

I(--- X ----li

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276

KVPV Practice Set 1 Stream : SB/SX H= l [V + MO ± anion / cation J

For L - x length, T is only force, so

2

T = Mass x Acceleration = - X (L - x) X -

M

F

L

M

where, H = Hybridised orbitals

V= Valence

2

= Mass per unit length x Length

of remaining part x Acceleration

c (L� x) F {) } c iF

ffi =

99. (c) Induced emf, E = � q>B dt =a

2F :. Stress, cr = - 3A

98. (d) Linear momentum is conserved, mv0 = 2mq - mv2 2 vl - V2 = Vo

v2 = 2v1 - v0

. . . (i)

MO = Number of monoatomic atoms

4 ✓ vo

=a

7L

rJoa( v_l!!!_ + aB d ax dT )

2

( v dB + dB )

x

ax ar

= (0.1) 2 (0.08 X 0.1 X 10-3 ) = 9 X 1 0 -5 V = 90µV

mv0 ( i sin 45 ° ) = 2m( iJ m

⇒

- mv2 ( ½ sin 45° )

v0 = ✓2 Lm - v2

. . . (ii)

As, collision is elastic.

Relative velocity of separation

2(0

= Relative velocity of approach L 1 V2 + Vi + ... (iii) X 2 = Vo ✓

Putting the value of v2 from Eq. (i) in Eq. (iii), we get Lm 2v1 - v0 + q + = v0 r;; 2-v2 Lm 3v1 - 2v0 + 2✓ 2

=

0

H = _l_ [8 + 4] = 6 2 spa d 2

It has 4 bond pairs and 2 lone pair, where 2 lone pair occupies axial positions and four F occupies equitorial positions. So, its geometry is square planar and not octahedral. F c$J F

1 00. (c) Latent heat is used in

increasing molecular separation. Length D shows heat absorbed during liquid to vapour change. So, length D is large.

1 0 1 . (d) Aromatic compounds are stable

Angular momentum is also conserved, so

For XeF4

than non-aromatics are due to resonance. Aromaticity can be determined by Huckel's rule according to which (or 4n + 2 rule), "a planar, cyclic compound is aromatic, when its 7t cloud must contain (4n + 2)7t electrons, where n is any whole number.

o.

0

and

Q (6rre- system) (4rre- + 2e- system)

0

N (6rre- system)

is non-aromatic, hence,

least stabilised by resonance.

1 02. (a) The complexes that has same

number of unpaired electrons will have same magnetic moment. The number of unpaired electrons in complexes are as follows: Complex ion

[Cr(H2 0)6] 2+

Electronic Number of configuration unpaired of metal ion electrons (n) Cr2+ ; [Ar] 3d4

[Fe(H2 0)6] 2+ Fe2+ ; [Ar] 3d6

[Mn(H20) 6] 2+ Mn2+ ; [Ar] 3d5 [CoC1] 2-

Co2+ ; [Ar] 3d7

1111 I 1 I 1 I l : 4 1,,111 11111 1 ;4 11111111111 ;5 11,11,1111111 ;3

Thus, [Cr(H20) 6 ] 2+ , [Fe(H20) 6 ]2+ have same magnetic moment.

1 03. (d) The hybridisation of a compound can be calculated using formula.

Xe

F

F

1 04. (a) Williamson's synthesis is an organic reaction used to convert an alcohol and an alkyl halide to an ether using a base such as NaOH. This reaction proceed via SN l mechanism.

Among the given compounds, Di-tert butyl ether cannot be made by Williamson's synthesis, since tert-alkyl (3°) halides prefer to undergo elimination rather than substitution. i.e.

I I - - Br + CH - - 0-Na

CH3

CH3

3

,

CH3

fl � - NaBr

CH3 I

CH 3

+

,

CH 3

CH 3 I

+ CH3 - C - O H

cH3 - c =cH2

CH 3 I

1 05. (a) Octahedral complexes of the type [ MA4 B2 ] exhibit geometrical isomers. The type of complexes in the given options arc as follows: (a)

(b) (c)

(d)

Complexes

t

[Cr(H20) 4 Cl2 [Pt(NH3) 3 CI]

[Co(NH3)6] 3+ [Co(CN) C1] 35

Types of complexes [.MA4B2] [.MA3B] [.MA6] [.MA5B]

Hence, the given options only [Cr(H20) 4 Cl2 j+ exhibit geometrical isomerism, cis and trans form.

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277

KVPY Practice Set 1 Stream : SB/SX 1 06. (a) The organic compound (A) with molecular formula C9 H 100 forms an orange-red ppt. with 2, 4-DNP reagent and gives yellow ppt. on heating with 1 2 and NaOH. This confirms the presence of carboxyl group in compound (A). Also, it does not reduces Tollen's reagent or Fchling's solution, this shows it is a kctonic group and not an aldehyde. As it does not decolourises bromine water it indicates the absence of unsaturated bonds. The reactions arc as follows:

II

0 CH2-C-CH 3 �

(A) 1-phenyl propanone

0 II

CH2- CNa

� + CHI�

(Yellow ppt.)

IifNaOH

j H2NHN�N02 (2, 4-DNP)

O2N

@-cH 2-h =NH-N�N0 2 (Red-orange ppt.)

1 07. (a) Pbl PbCl 2 I I AgCIII Ag

The cell reaction is

P b + 2AgCI - PbCl 2 + 2Ag t,G = - nFE =- 2 X 9650 0 x 0.5 J =- 96500 ,J or -96.5 k J dE t,S = nF ( )

ar

p

= 2 X 96500 X- 2 X 1 0-4 Also,

[Ag(NH:,)2]'0HTollen's reagent (x)

CHO

¢0

COOH

cH _ ,_ oH _ ._ H· _ ➔ Esterification

(y)

reagent

l

COOH 3

Excess (z)

Grignard CH,.Mg Br

H

CH 3 X +OH

02N

CH 3

¢

1 08. (a)

=- 38.6 JK- 1 t,G = t,H - TD.S =- 96500 = t,H- 300 (-38.6) t,H = - 96500- 300x 38.6 =- 108080 J =- 10 8.08 k J

Reagent x [Ag(NH3 )2t Olf" is a Tollen's reagent . It will only oxidises aldehyde group and not ketone group. Reagent y, CH3 0H is an alcohol, which then reacts with formed carboxylic acid to give ester. ln presence of reagent z, CH3MgBr (excess) it will reduce both ketone and ester group to alcoholic group. 1 09. (a) At A Temp. = 300 K, volume = 1 0 L Let pressure at this point = p 1 At C Temp. = 600 K, volume = 20 L Let the pressure at this point = p2 According to ideal gas, pV = RT (For 1 mole) R = pV

x10 :. For system A, R = Pi 300 x 20 For system C, R = P2 600 p1 X 10 p2 X 20 = 300 600 = p i 2 P Thus, it is an isochoric process, when system moves from A ➔ C. 1 1 0. (b) Molar mass of CsBr = 212.8g mol- 1 molar mass Molar volume = density 212.8 = = 47.39cm3 4.49

6.023 x 1 023 molecules are there in 47.39 cm3 47.39 . 1 molecule 1. s there 1n 6.023 X 1 0-03 = 7.868 x 1 0-2 3 cm3

:. Volume of the unit cell

= a3 = 7.868 x 1 0-23 cm3 Side of the unit cell = a = (7.868 X 1 0-23 )113 = 4285 x 10-s cm = 4285A

1 1 1 . (c) A normal red-green colour vision man with Rh- blood group is XY Rh- Rh -. A normal woman with Rh + blood group with heterozygocity at both red-green colour vision and blood group is XXCRh + Rh- . The inheritance of colour vision in the progeny 1s

i.e. 50% boys arc colourblind also 50% of offspring are boys. Thus, 0.25 is the probability the first boy is colourblind. The inheritance of Rhesus blood group in the progeny is

Rhesus positive

Rhesus negative

Rhesus positive

Rhesus negative

i.e. 50% arc Rhesus positive and 50% Rhesus negative or 0.5 is the probability of the progeny being Rhesus negative. Probability of their first child being a rhesus negative = 0.5. Probability of their first child being a red green colourblind boy = 0.25. Therefore, probability of their first child being both = 0.5 x 0.25= 0.125. 1 1 2. (d) The net increase in population will be Birth rate (B) + Immigration (I)- Death rate (D)+ Emigration (E), i.e. 250- 20- (240 + 30)= 0

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278 1 1 3. (a) A suitable restriction cndonuclcasc should cut the plasmid at only one position in order to allow the DNA containing the gene of interest to be inserted into the circular plasmid without loss of any part of the plasmid. Thus, in the map of the plasmid given, the most suitable enzyme will be Bam HI. 1 1 4. (c) As there arc 2 types of base, 2n

where n = number of bases per codon. Thus, 26 would provide 64 combination, which is more than sufficient. Thus, 6 are the minimum number of bases per codon that could code for proteins.

1 1 5. (d) Option (d) is correct. Other statements can be corrected as primary productivity varies for all ecosystems. Net primary productivity is the amount of biomass available for consumption by heterotrophs. Secondary productivity is defined as the rate of formation of new organic matter by consumers.

KVPV Practice Set 1 Stream : SB/SX 1 1 6. (a) Since, there are different amino acids, the number of different polypeptides that can be formed, each with r amino acid is n r .

1 1 7. (d) Out of the statements given, I and V are the correct options, remaining three arc wrong. These can be corrected as in bacteria, transcription and translation do not take place in same compartments they arc instead coupled in the cytosol itself as there is no separation of nucleus and cytosol in bacteria therefore, translation starts much before the mRNA is fully transcribed. RNA polymerase-I is responsible for the transcription of hnRNA (not tRNA). When hnRNA undergoes capping process, methyl guanosine triphosphate are added at 5' end in a template independent manner. 1 1 8. (d) With increasing substrate concentration in the presence of a

competitive inhibitor, the rate of reaction only remains constant at a later stage. This is because the additional substrates can compete with the inhibitor for the enzyme's active site. Thus, graph (d) is correct.

1 1 9. (a) The correct matching is Potassium - Involved in stomata! movement Sulphur - Constituent of ferredoxin Molybdenum - Component of nitrogenase Zinc - Needed in the synthesis of auxin 1 20. (c) Parent 3 is certain to have the genotype rr. Since, the male 10 is also having the genotype rr, this means that parent 4 must have the genotype Rr. Therefore, female 9 must have the genotype Rr. As female 13 has the genotype rr, both her parents (7 and 8) must have the genotypes Rr. Female 15 must have the genotype Rr because her father (male 10) has the genotype rr and her mother (female 11) is normal.

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KVPV

KISHORE VAIGYANIK PROTSAHAN YOJANA

PRACTICE SET 2 Stream : 58/SX

MM : 160

Instructions 1 . There are 1 20 questions in this paper. 2. The question paper contains two parts ; Part I (1 Mark Questions) and Part II (2 Marks Questions) . 3. There are four sections in each part; Mathematics, Physics, Chemistry and Biology. 4. Out of the four options given with each question, only one is correct.

� PART-I

MATHEMATICS

(1 Mark Questions)

..n - 3 + . . . 1 . If equat1. 0n xll - nxn - 1 + a 2x..n - 2 + a 3 x

+ a,, _ 1x + ( -1 )" = 0 has n positive roots, then the least

value of n for which a 2 + a3 is negative is

W 2

� 6

� 4

� l

(a) 3ab

(b)3ab2

(c)3a2b

(d) 3

2. If in an isosceles triangle with base 'a' vertical angle 20 ° and lateral side each of length 'b' is given, then the value of a3 + b3 equals

3. The number of triangles with each side having

integral length and the longest side is of 11 units is (b) 34 (a) 30 (d) 36 (c) 35 l - ab . 4. If log245 175 = a and log1715 8 75 = b, then -- 1s equal to (a) 2 .

5. Given

s-"

2-

a-b

(b)3 -

(c) 5

Sill X -

- dx = K,

o 1 + sin x + cos x

definite integral

J%

0 1

dx

(d)7 then the value of the

. X + COS X is equal to + Sill

K (a) -

(b) � - K 2 (d)�- 2K 2

(c) � + K 2

6. If the line y = ./Sx cuts the curve

x4 + ax2y + bxy + ex + dy + 6 =0 at A,B,C and D, then

the value of _!__ OA • OB • OC • OD (where O is origin) is 12 equal to (c) 1 2 (a) 6 (b) 8 (d) 1 6 x2 dx 7. Let f (x) = f and /(0) = 0, (1 + x2 )(1 + v� l + x2 ) then f(1) is (a) log(.J2 +1)

(b) log(.J2 + 1)- � 4 (d) None of these

(c) log(.J2 +1)+ � 4

8. The number of ways in which the letters of the word 'ABBCABBC ' can be arranged such that the word ABBC does not appear in any word is (a) 420

(b) 392

(c) 361

(d) 360

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280

KVPVPractice Set 2 Stream : SB/SX

9. If f(x) is a function satisfying the relation

20. A ten-digit number is formed without repeating any

x1'f(x)- 2f(-�) = g(x), where g(x) is odd function, then the value of f(2) is (a) -/ (2)

(b)-/ ( 2)

(c)

(d)

r(½)

-t(½)

10. If A,Band C are n x n matrices and det(A) =2, det (B) =3 and det (C) = 5 , then the value of det (A2BC- 1 ) is 12 18 24 (a) � (b) (c) (d) 5 5 5 5

3 (a) 2

(-�))

= 0 in (0, 2rc) is

9 (c)2

(b)4

(d)

13 3

12. If the chords of contact of tangents from two points x2 Y2 ( - 4, 2) and (2, 1) to the hyperbola 2- 2 = 1 are at b

J7 (a) 2

a

right angle, then the eccentricity of the hyperbola is

JI

�

13. If Z is a complex number satisfying Z3 + z-3 :s: 2, (b) 3

(c) 2

(d)-/2

(a) 2

z-1 is 1

I

then the maximum possible value of I Z + (b)3-/2

1

(c)2-/2 (d) 1 14. If /(x) = x1' + a x1' + �x + y, where a. , �, y, are rational numbers and two roots of f (x) = 0 are eccentricities of a parabola and a rectangular hyperbola, then a + � + y is equal to (a)-1 (b)0 (c) 1 (d)2

15.

rn ( cosa.x-sin �x) dx is equal to (a) 0

(b)� 2

2

(c) 1t

(d)2n

16. If a , b, c and d are four positive real numbers such that abed = 1, then the minimum value of (1 + a)(l + b)(l + c)(l + d) is (a) 4 (b) 1 (c) 1 6 (d) 18

17. There are three coplaner parallel lines. If any point p taken on each the lines, the maximum number of triangles with vertices at these points is (a) 3p2(p- 1) + 1 (b) 3p2(p-1) 2 (c)p (4p- 3) (d) p 2 (2p- 3) 11 n

18. The value of lim n 2 f lS

(a) 0

n➔=

(b) 1

-1/ n

(20 1 8 sin x + 2019 cosx) l xl dx (c) 201 8

(d)2019

19. The value of tan 6 40 °- 3 3 tan4 40 ° + 2 7 tan 2 40 ° is equal to (a) 2 (c)4

(b) 3 (d) None of these

PHYSICS 21. A number of particles start simultaneously from

same point in all possible directions with same speed in a vertical plane. Now, choose the correct option. (a) After time t they all lie on a parabola (b) After time t they all lie on a circle (c) After time t parabola described by particles has focal distance ut (d) After time t circle described by particles has radius 2ut

11. The sum of all the roots of the equation sin( re log3

digit the probability that the difference of digits at equal distances from the beginning and the end is always 1, is 1 17 4 (c)(d)J!±_ (b) ( a) 945 27 1 944 243

22. Two particles each of mass m are attached to ends of a

light string of length a and placed on a horizontal turntable at distances a and 2a from centre of rotation. Let turntable rotates with a maximum angular speed w, such that particle remains over it without any slip. Necessary friction coefficient must be 2 aw2 2aw2 3aw2 (d) _':(1)__ (c) , (a) (b) 2g 2g 3g g

23. A chord of length 90 m is used to tie a 200 kg

astronaut (with his space suit and instruments) with his spaceship. Tension in the chord is nearly (a) 0 .0084 N (b) 0.084 N (c) 0.84 N (d)8.40 N

24. A non-uniform rod of length l has a linear mass

density A = 'A, 0 + kx, is placed over X-axis with its one end at origin. The x-coordinate of centre of mass of rod is + 2kl2 3 + 2kl2 (a) A 0 1 (b) 11, 0 ! (2A 0 + kl) 3(2A 0 + Id) 31c 0 l + 2kl2 411, 0 ! + kl2 (c) (d) 3(2A 0 + kl) 3(211, 0 + kl)

25. Two masses m1and m2 (m1 > m2 ) are connected by a

light spring. This system is placed on a horizontal frictionless table. At t = 0, blocks are pushed closer by compressing the spring and then released. The system oscillates with a time period T. Time period of oscillation of centre of mass of system is (a) 111:i'lli m2 . T -T (b) 111:i 'lli + m2 + m2 (c) 111:i "½ · T (d)zero + m2

26. A particle moves around a circular path of radius r

with a constant speed v. If particle describes angle 8 in 4 s, then average acceleration of the particle is 2 22 (a) � sin (� ) (b) � sin (I) (c) �; cos( ) i

(d)f sin ( � )

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281

KVPY Practice Set 2 Stream : SB/SX

-

27. A block of mass m1 is placed over a plank of mass m2 . Block and plank are given initial velocities v1 and v2 relative to ground (v1 > v2 ).

Tµ =0

2

1 V

±= v2

µ� There is friction between block and plank, but there is no friction between plank and ground.

Correct variation of velocity of block vb and velocity of plank vP is

Z

(a} � l (b) � I

(d)

(c)

---------V

0

vp

b

t

28. If you are pulling one end of an ideal spring of mass m with velocity v, while keeping other end fixed to a

wall, then kinetic energy of the spring at this instant lS

1 (a) -mv2 2

1 (b)- mu2 4

1 (c)- mv2 6

1 (d)- mu 2 8

29. Consider given pulley and mass system consisting of

ideal pulley, ideal springs and ideal identical strings.

(c) thin water covered surface reflects light in one direction only (d) water layer acts like a converging lens

31. XOY-plane is a boundary between two transparent

point O in medium 1. If refracted ray in medium 2 is r = a i + bj + c k. Then, correct relation is

4b (a) a = 3 5a (c) c = - 3

(b) c =

=

( i mz )g ln : 4

(d)a1

=(

mJ -

- m2, } g ::

30. Colours of moist objects seems deeper and richer than those of dry ones. This happens because (a) water oxidises the dye of colour (b) interference occurs from reflections from top and bottom layers of water film

3

- 4b (d)a = 3

jet as shown alongside. Vertical elevation -Z depends on d = sphere diameter, g = gravitational constant, p 1 = fermi density, p 2 = mass density of ball and F = force of fluid on ball. Then, which of these graphs is correct experimental observation of

-z-

F

.

- versus x, where x = 3 1s d d g (pz- P1 ) (a)

....-.,

-�-, ..�........

r:;:-

-· •......

'

(b)

X

•.-:f

i:• ..

(c)a1

5a

32. When a spherical ball is suspended in an air or water

(c )

Let accelerations of m1 , m2 , � and m4 are a1 , a 2 , a3 and a 4 , respectively immediately after cutting thread A which keeps the system stable. Now, choose the correct option. + m4 lni - mz (a) a4 = 0 (b) a4 = ( 111:J ) It �

2

media. Medium 1 with ,e � 0, has a refractive index ✓ and medium 2 with ,e :S: 0, has a refractive index Js. A light ray AO = 6$i + S--/3j - 10 k is incident over

: ,

X

(d)

� '•.•·•.� �.\II ,....\II'•. / �.... , J ...... . . ,... ..

..

..... . .. ...

X

•· ·· ....-...•..� ...•:•·-�,--, , , -

_

,

X

33. An unstable radioactive substance decays into a

daughter product, but this in turn will probably itself decay until, a stable non-decaying isotope is formed. The list of all members is called a radioactive series. An example of a series is

p p '' 28 ? 2 a 22 8Ra � 89 Ac � go3 Th � ss s � 22 1Ra � 220 Ra � 21 6p0 22 00 Th � W M Now, choose the correct option. (a) 232Th and 228 Th are isobars (b) 22 8Ra and 228Ac arc isotopes (c) 232Th and 220 Ra are isotones (d) After some time all the members of series are in equilibrium

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282

KVPVPractice Set 2 Stream : SB/SX

34. Which region of binding energy per nucleon versus

mass number curve indicates that the nucleus also have some kind of its own structure. >, [2' C (l) 0 C T3 > T4

(d) Data insufficient

pulled into region of magnetic field B at an angle of 45° with the horizontal as shown below. 4

! X

5i

X

X

X

X

___ { X ___ X __ X __ X ___ X __

°

X

X

X

X

X

X

X

X

X

X

X

B

X

X

X

X

X

X

IV. A bound electron can absorbs two photons (one after other). Which of the above statement is correct? (a) Only I (b) Only III (c) Only IV (d)Only II

CHEMISTRY 41. In the compound CH2 = CH - CH2 - CH2 - C == CH, the C2 - C3 bond is of (a) sp- sp2 (b) sp3- sp3 (c) sp- sp3 (d)sp2- sp3

42. When hydrochloric acid gas is treated with propene in the presence of benzoyl peroxide, it gives (a) 2-chloropropane (b) ally! chloride (c) n-propyl chloride (d) isopropyl chloride

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283

KVPY Practice Set 2 Stream : SB/SX 43. Plots showing the variation of the rate constant (k) with temperature (T) are given below. The plot that follows Arrhenius equation is

(•l t L

(b

(

(

T-

T-

d

T-

) t=�

44. For a linear plot of log (xlm ) versus log p in a

-

Freundlich adsorption isotherm, which of the following statements is correct? (I� and n are constants) (a) 1 /n appears as the intercept (b) Only 1/n appears as the slope (c) log (� ) appears as the intercept (d)Both k and 1/n appear in the slope term 45. A gaseous substance dissolves in water giving a pale blue solution which decolorises KMn0,1 and oxidises KI to 12 • Gaseous substance is (d) HN03 (c) N 203 (b) NH3 (a) N 205

46. The geometry of Ni(C0)4 and [Ni(PPh3 )2 ]Cl2 are (a) both square planar (b) tetrahedral and square planar respectively (c) both tetrahedral (d) square planar and tetrahedral respectively

47. Given, E�r>+-icr =- 0. 74 V;

array of oxide ions with two out of every three octahedral holes occupied by iron ions. Derive the formula of the iron oxide. (b) Fe 203 (a) FeO (d) All are possible (c) Fe3 0 4

51. If at 298 K the bond energies of C-H, C-C, C=C

) /L

)L

50. Iron oxide crystallises in a hexagonal close-packed

E�n0 /Mn2+ 4

= 1.51 V

E�,,0 2_ 1c,>+- = 1 . 33 V; E�, ic, - = 1 . 3 6 V - 7

and H - H bonds are respectively 4 1 4, 347, 615 and 4 3 5 kJ mol-1, the value of enthalpy change for the reaction, H2C = CH2 (g) + H2 (g)� fis C - CH3 (g)

at 298 K will be (b) -250k J (d)- 1 25 k J

(a) + 260 k J (c) + 1 25 k J

52. The oxidation state of chromium in the final product formed by the reaction between KI and acidified potassium dichromate solution is (a) + 3 (b) + 2 (c) + 6 (d)+ 4

53. Phenol is treated with excess bromine water and

shaken well. The white precipitate formed during the process is

& OH

(a)

B (b) � ,

LSJl_ Br

Br

¥

OH Br� Br Q (c) Br

54. The correct representation of pre - drt bonding is

Based on the data given above strongest oxidising agent will be (d) MnO� (c) Mn 2+ (a) Cl (b) Cr3 +

48. Consider the following compounds:

I

CfisCH2NH2 ,

0

II

CH3 �NH2 ,

0

III

PhlNH2

Basic nature of above compounds is in the order as: (b) I > Ill > II (a) I > II > III (c) II > Ill > I (d) Ill > II > I

49. Element A burns in nitrogen to give an ionic compound B. Compound B reacts with water to give

(d) �

55. 2-hexyne gives trans-2-hexene on treatment with (a) Pt / H2 (c) Pd / BaS04

(b) Li / NI-ls (d) LiAIH4

56. In which of the following polymers ethylene glycol is one of the monomer units?

C and D. A solution of C becomes 'milky' on bubbling carbon dioxide. The element A is (a) Na (b) Mg (c) Ca (a') Be

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284

KVPVPractice Set 2 Stream : SB/SX

64.

(d) -f--O-CH-CH2-C-O- CH-CH2 -C+,;

Which one of the following would exocrine cells be expected to contain as a result of their function? (a) Increased amounts of DNA (b) Increased amounts of rough endoplasmic reticulum (c) Increased numbers of lysosomes (d)Large mitochondria

65.

2 3 Na is the more stable isotope of Na. Find out the process by which �{Na can undergo radioactive decay

66.

How many fatty acid molecule(s) is/are normally present in a fat or oil molecule? �2 W3 W4 Wl The specificity of enzymes is due to (a) their high molecular weight (b) their hydrogen bonding (c) their pH sensitivity (d)their surface configuration

Q

(c)---f-CH2 - CH = CH---CH�H-CHrt,

I

II

O

CI-f:i

57.

(a)�--emission (c)�+ -emission

0

58. The reduction,

CH2 CH3

II

0

(b)a-emission (d) K-electron capture

0

1 -01COCH 1

HC

I

3

- HOH2C

0

-011COCH

67.

What is the role of centrioles during meiosis in animal cells? (a) Breaking down the nuclear membrane during prophase (b) Helping homologous chromosomes to pair and form bivalents (c) Holding the two chromatids of a chromosome together (d)Organising microtubules to form spindle fibres

68.

How does DNA synthesis along the lagging strand differ from that on the leading strand? (a) Okazaki fragments, synthesised 5' ➔ 3' are linked by Dl\"A ligase (b) An RNA primer is needed on the lagging strand, but not on the leading strand (c) Deoxyribonucleotides are added to the 5' end instead of the 3' end (d) Helicase synthesises Okazaki fragments, which are then linked together

69.

Insulin is a protein containing 51 amino acids. These include 1 7 of the 20 different amino acids commonly occurring in proteins.

3

can be achieved by using (a) NaBH4 (b) LiAIH4 (d) None of these (c) CuO CuCN20 4

59. Which of the following compounds is formed when glucose react with Br2 water at pH 5-6? (a) Glucaric acid (b)Saccharic acid (c)Gluconic acid (d) Acetic acid 60. Which of the following are sets of diamagnetic species? (a) TiC1 4, [Ni(H20)4 ]2+ (b)TiC1 4, 02 2 (c)TiC1 4, [Ni(CN)4 )2(d) [Ni(CN)4 )2- [ Ni (HP)4J +

BIOLOGY 61 .

Sliding filament theory can be best explained as (a)when myofilaments slide pass each other actin filaments shorten while myosin filaments do not shorten (b)actin and myosin filaments shorten and slide pass each other (c)actin and myosin filaments do not shorten but rather slide pass each other (d)when myofilaments slide pass each other myosin filament shorten while actin filaments do not shorten

62.

Which of the following cells during gametogenesis is normally diploid? (a) Primary polar body (b)Spermatid (c)Secondary polar body (d) Spermatogonia

63.

An artificial pacemaker is implanted subcutaneously and connected to the heart in patients (a) having 90% blockage of the three main coronary arteries (b) having a very high blood pressure (c)with irregularity in the heart rhythm (d) suffering from arteriosclerosis

What is the minimum number of different kinds of tRNA molecules involved in the synthesis of insulin? (c) 17 (b) 51 (d) 20 (a) 3

70.

During transcription of the DNA fragment shown, a single base is paired incorrectly. - C- T- A- A- C- T Sense strand - G- A- T- T- G- A Anti-sense strand Which mRNA strand results? (a) C- A- T- T- G- A (b) G- A- U- U- C- A (c) G- A- U- U- G- A (d)G- T- U- U- G- U

71.

Which one of the following statements about gene mutation is incorrect? (a) It can occur in both somatic and sex cells (b) It can cause Down's syndrome in humans (c) It can change a dominant allele into a recessive one (d) It can be brought about by exposure to ionising radiation

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285

KVPY Practice Set 2 Stream : SB/SX 72. Promoters and control elements work together to

regulate transcription. What shows the possible locations of these in relation to a transcription start site on the DNA molecule? Promoters

(a) Downstream (b) Upstream (c) Distal or proximal ( ) Distal or proximal d

Control elements

this supply of protein was only half of the minimum required to supply its energy needs, the mammal would show an increase in the (a) concentration of amino acids in the blood (b) concentration of insulin in the blood (c) synthesis of glycogen (d) hydrolysis of glycogen

73. Which of the following would cause phenotypic

variation among organisms of the same genotype ? (a) Continuous variation within the species (b) Different varieties of the same species (c)Different sexes (d) Exposure to different environments

78. The diagram shows an action potential R

+

74. During the light phase of photosynthesis, the photoactivated pigments remove an electron from the hydroxylation derived from the water molecule. The fate of the free hydroxyl radical is that it (a) is broken down into oxygen and a free radical of hydrogen (b) is used to raise the activation level of chlorophyll by donating a positive charge (c) is used to produce adenosine triphosphate from adenosine diphosphate (d)reduces carbon dioxide to sugar

75. When mitochondria are extracted from cells for

mV O t---+-+----,,Tacie m-- Q

p

s

Resting potential

At which point on the graph is the membrane most permeable to sodium ions?

�R WP �Q WS 79. The frequency of a mutant gene in a population would be likely to increase if (a) the genes were selectively advantageous (b) the genes were dominant (c) the genes were sex-linked (d)the population increased

biochemical study, they are usually kept in a 0.25 mol dm -3 sucrose solution. Why is the sucrose solution used? (a) To act as a solvent (b) To provide a source of food (c) To assist in the extraction of enzymes (d) To prevent the mitochondria from changing in structure

80. Which genetic modification could increase the yield of crop measured as mass of crop produced per unit area per year? (a) Herbicide resistant plants (b) Delayed ripening in the fruits (c) More essential amino acids in the seeds (d)More vitamin-A in the grain

76. Four identical samples of plant tissue, each with a

water potential (\/1) of sign - 700 kPa, are placed in four different solutions. Which solution induces full plasmolysis within the tissue?

(2 Marks Questions)

MATHEMATICS 8 1 . A triangle has base 1 0 cm long and the base angles of 50 ° and 70 ° . If the perimeter of the triangle is x + y cos 2°, where 2 E (0, 90 ° ), then the value of x + y + z equals to (c) 50 (a) 60 (d) 40 (b)55

82. If g(x) is a differential real valued function satisfying g" (x)- 3 g (x) > 3, x � 0 and Ff (0) =- 1, then g(x) + x for x > 0 is (a) an increasing function (b) a decreasing function

(\jl)

-700kPa -lOOO kPa -400kPa -200 kPa

77. Protein was the only food available to a mammal. If

Distal or proximal Distal or proximal Downstream Upstream

� PART-II

Solution (a) A (b) B (c) C (d)D

(c) a constant function (d) None of these

83. The number of integers between 1 and 10000 with at least one 8 and at least one 9 as digits is (b) 974 (a) 972

� ew

w ew

84. If the expression 2 - 32 can be factorized into linear 5

and quadratic factors over real coefficient as (25 - 32) = (2 -2)(22 - p2 + 4)(22 - qz + 4), then (p2 + 2q) is equal to (b) 3 (c)4 (d)S (a) 2

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KVPVPractice Set 2 Stream : SB/SX

85. If the line x + y = 1 is a tangent to a parabola with focus (1, 2) at A and intersects the directrix at Band tangent at vertex at C respectively, then AC • BC is equal to 1 (a) 2 (b) 1 (c)2

86. The line 2x- y + 1 = 0 is a tangent to the circle at the

x -2 y = 4 . The radius of the circle is

point (2, 5) and the centre of the circles lies on (a) 3-/5

(b)5J3

(c)2-15

(d) 5J2

87. X ={x E R : cos(sin x) =sin(cos x)}. The number of elements in X is (a) 0 (c)4

(b)2

88. There are 6 boxes labelled B1 , B2 , Rs , ... , B6 . In each trial two fair dice D1 , D2 are thrown. If D1 shows j and D2 shows k, then j balls are put into the box BK , After (d) Not finite

93. A ladder PQ of length 5 m is

inclined to a vertical wall is slipping over horizontal floor with velocity of 2 ms- 1 . Velocity of centre of mass of ladder, when Q is 3 m from the wall is (b) 1 .25 ms- 1 (a) 2 ms- 1 1 (d)2.75 ms- 1 (c) 175 ms-

94. A voltmeter with resistance 500 Q is used to measure the emf of a cell of internal resistance 4 Q. Percentage error in reading of voltmeter is around below (a) 1 % (b) 0.8% (c) 0.1% (d)8%

95. A gas undergoes two processes as shown below. p

-t----------->V

n trials, what is the probability that Bi contains at most one balJ?

(::J(�) (c) (:: )+n(:::: )(�) (d) (:: ) +n ( :::: ) (:2 ) (a) ( :::: ) +

(b)( :: )+ ( ::::: )(�)

89. The minimum distance between a point on the curve y = ex and a point on the curve y = loge x is (b).J2

(a) }z

(c).J3

Ratio of molar heat capacities for process AB and BA lS

96. A regular hexagonal mesh using equal resistances each of 2 Q is made as shown below.

(d)2)2

90. Let [x] and {x} be the integral part and fractional part of a real number x respectively. Then the value of the 5

integral f [x]{x}dxis o

(a) � 2

(b) 5

(c)

69 2

(d)

71 2

PHYSICS water at an angle of 60 °. Total deviation of light ray after a single reflection from internal surface of drop (a) 1 80 ° (c) 1 38°

i,3 value of 8 will be

(b) 1 45° (d) 1 76 °

92. A cart of mass m0 starts moving right due to a

constant horizontal force F. At same time, sand starts falling into cart from a stationary hopper. Rate of loading is constant and equal to µ kg s- 1 . Velocity of cart at time t is Ft (b) (a) ___I!!__ 2( mo + µt mo + µt) Ft (d) (c)___!!!__ 2 (m0 - µt) m0 - µt

B

A

Then, equivalent resistance between terminals A and Bis 8 (c) __'l n (d)2Q (a) - Q 8 5

97. A

91. A light beam is incident over a spherical drop of is let 8. Then, for µwater =

(b) greater than 1 (d) Data insufficient

(a) less than 1 (c) equal to 1

238

U nucleus emits a 4.2 MeV a-particle. If we take account of recoil energy of 234 Th daughter nucleus formed in the process, then Q value (disintegration energy) of the reaction will be 234 (a) 4.2 ( 1 + __±._ ) MeV (b) 4.2 ( ) MeV 238 238 238 (c) 4.2 ( (d)4.2 (1- __±._ ) MeV ) MeV 234 234

98. Two towers AB and CD are situated d distance apart.

A

C

AB is 20 m high and CD is 30 m high. From top of AB a particle of mass m is thrown towards CD with speed u = 10 m/s.

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287

KVPY Practice Set 2 Stream : SB/SX At same instant, another particle of mass 2 m is thrown from CD towards AB at an angle of 60 ° with the horizontal in same vertical plane with same speed of 10 ms- 1 . Distance between towers is (a) &/3 m (b)6-J3 m (c)8,/3 m (d)lo/3 m

99. An ideal gas undergoes an adiabatic process in which internal energy U is U = a + b(p V), where a and b are constants. Value of y for this process is b b+ l (a) __ (b) b (c) (d)1 + _!: b l+ b b+ 2 1 00. A measurement establishes the position of a proton with an accuracy of ± 1.00 x 10- 11 m. Uncertainly in the proton's position 1 s later is (take, u ------ ---E- --- • I

µmg

,) A

In above figure,

��/

B

1' = tension in string when masses about

to slip. Accelerations of A and B are aA = aw2 aB = 2aco2 Frictions of A and B arc fA = µ mg !iJ = µmg

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KVPVPractice Set 2 Stream : SB/SX

Net force on A is FA = µmfi? - T = Net force on B is

mao:i

2

fs =µmg + T = m(2a)ro2 For no slip; FA + fs

⇒

It, v I = .Jv; + v� - 21-'J. v2 cos 8

Herc, ½. = v2 = v -___--______ ⇒ It, v I = ✓�v + v _ 2u 2 co s 0 2

=

3aro2 2 fA + f,B = 3amro ⇒ µ = -2g

2

2u ( 1 - 1 + 2 sin

Astronaut

⇒ mg ( 1 - �

g + T = � (r + L)

Taking r � R (radius of earth) 2 mgL mgL ⇒ mg _ + T = mg + ⇒T

r

R

R

=

3mgL = 3 x 200 x l0 x 90 = 0. 084 N R 6400 x Hf

x

24. �_)fi-=�==��-)���:�f_m�E-:��A:��: We have, dm = A dx = (A 0 + kx)dx l (A. J x 0 + kx) dx J x dm XcM = ____ = -______ �� . °s . J dm Q o + kx) dx

2v2 ( 2 sin 2

A ol +

hl2 2

_- 3A0 l + 2hz2 3(21- 0 + kl)

25. (d) As system is released at t = 0,

momentum of system at any instant t is p(t) = Msystem X !.JcM= Psystem (t = 0) = 0 :. VCM = 0 So, centre of mass of system is always at rest and is not oscillating. 26. (d) As particle moves by an angle 0.

%)

2

=

It, v I = 2 v sin _()_ 2 Time duration of motion is given 4 s So, L',t

=

4=

Re

4v

⇒ 0=

--;:;-

R

So, average acceleration of the particle is . e 2 v - sm Jt, v l = 2 = _!: sin 2v a l avg 1 ( ) L'lt 4 2 R 27. (d) For ½. > v2, mi slows down and m2 speed up till their velocities are equal at some time t0 • l v1 - v2 I where, to = ( l + �) µ k g

28. (c) Mass dm of spring at a distance x moves with velocity is

= -· V

V

l

X

dx

�v

0

12 kl3 Ao 2 + 3

%)] %)

2u2 [1 - ( 1- 2 sin 2

and astronaut is ro.

m Forces on astronaut provides necessary centripetal pull. GM m e ⇒ + T = m (r + L)ro2 (r + L)2

2

.J2v2 (1 - cos 0)

23. (b) Let angular speed of spaceship

T

Kinetic energy of mass d m is 1 1 v2x2 dK = - (dm)v"2 ⇒ dK = - · dm 2 2 z2 Kinetic energy of complete spring is l 2 I K = f dK = - �2 f x2 dm Here,

dm

=

m

2

dx

l 2 m 2 x K = ..!: � x dx = ..!: m v J 2 t2 I O 6 l

2

29. (b) Here, mi + � > n-13 + m4 otherwise equilibrium is not possible. Force balance for mass m3 is V2 Then, change in velocity is t,v = v 2 - v 1 . Magnitude of change in velocity is It, v I

=

.J(v2 - vt )2

Also, force balance for mass mi is

���

Here, T1 = m2g So, T = (mi + �)g ma g + 12 - T = O ⇒ T2 = T - ma g or T2 = (7ni + � - ma )g When lower thread A is cut equations of motion for all masses can be written as 7ni� = 7ni g + T1 - T mp2 = 'T½g - T1 m3Cl:J = T2 + ma g - T m4 a4 = m4 g - T2 Solving these equations, we have and :

=r�:·:.�."' -

" g ½l

30. (c) A thin water film covering a moist

object reflects the incident light in one definite direction.The surface of object no longer diffuses white light in all directions and its own colour becomes prominent. The diffused light is not superimposed on the light reflected from the object and for this reason the colours seems more richer. 3 1 . (c) Let refracted ray is r = ai + b) + cf< Then, normal to plane of incidence is i j k

N=

6J3

sf§ -10 1 0

0

= sfs i

- 6,/3)

This must be normal to refracted ray r·N= 0 &J 3a - BJ3b = 0 ⇒ b = � a Also, cos ( 7t - "1 )

=

6J3i + , 6J3 l i+

1 2 ⇒ cos(1t - i) = cos120° As, .J3 sin r = .J2 sin i

sf§) - 101'B = dt R dt

⇒ Total charge passing through coil is independent of time. d Also, (j> can be determined only when dt velocity is given. 38. (a) Effective emf of circuit is

E

Eeff = 2

Effective resistance of circuit is Reff = '!!_ R 2 So, time constant of circuit is , = Leff = 2L Reff 3R Current through inductor at time t is

;�

;[, - /('.:)]

iR = §_ (1- e-tl a ) [·: iR = V ] 3 V = §_ (1- e-110 ) 3 Potential drop across inductor is

di §_/(!�) VL = L = dt

2

In process B,

kJ -200 0 = _ 2.5 ii Ssource = BOO K 200 0 = kJ S = 2.7 ii s'nk 750 K kJ li 8i,0,a1 =- 2.5 + 2.7 = 0.2 -

K

As, iiSr0,a1 is less for process B. So, it has less irreversibility or B can be more easily reversed. 36. (b) At low pressure and at high temperature the compressibility

coefficient .Z = p V , is always equal to one

RT

for even real gases. But at high pressures and low temperatures gases begun to deviate from ideal gas behaviour. So, T4 is greatest and T1 is least temperature.

Total energy = Kinetic energy + Potential energy ____![_ _ __i:__ = e2 - 2e2 = d 81tE0 r 81tE0 r 81tE0 r 7tE0 r

40. (b) A free electron cannot absorbs a photon. In an electron-photon collision, when electron is bound. i.e. with in a metal surface, electron absorbs the photon and it cannot further absorbs photon.

4 1 . (d) 1

2

4

5

6

C= C H

Hybridisation a t C2 = sp 2 Hybridisation at ½ = sp3

Thus, the C2- ½ bond is of sp2 • sp3 type.

42. (a) Peroxide effect is observed only

in case of HBr. Therefore, addition of HCl to propene even in the presence of benzoyl peroxide occurs according to Markownikoffs rule. =C=1CH:i CH = CH2 - �H ➔ Propene

(C 6 H 5C00) 2

Cl

Cfl:i -CH - CH3 2 •chloropropane

43. (a) According to Arrhenius equation, rate constant increases exponentially with temperature: k = Ae-E0 1RT

Thus, correct plot is given in option (a).

44. (b) According to Freundlich adsorption isotherm, _::, = kp lfn m On taking logarithm of both sides, we get log _::, = log k + log plfn m 1 X or log = log h + - log p n

m

39 . (d) Centripetal force on electron is mv2 e2 -- r 41tE0 r 2

3

C H2 = CH-CH2 -CH2 -

t

.,,/(lope =l S n d_J

log k

log p --

So,

e2 mu2 = -41tE0 r

e2 . . energy = -1 mu 2 = -Kinetic 2 81tE0 r 1 e e2 Potential energy = -- - ( e)=- -41tE0 r 41tE0 r

On compairing the above equation, with equation of straight line, i.e. y = c + mx, we get y = log -=:., m c = intercept = log k 1 m = slope = -and x = log p n Thus, option (b) is correct.

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KVPVPractice Set 2 Stream : SB/SX

45. (c) N203 dissolves in water giving a pale blue solution of Hl\0 2 which decolorises KMnO4 and oxidises Kl to 1 2• N203 + H20 -- HNO ? Pale bl;;:e HN02 is a reducing agent HNO Mn04 � Mn2+ HN02 is also an oxidising agent HN0 2 1- 12

Ni(C0) 4 ; Oxidation state of Ni = 0 Ni = 1s2 ,2s2 2p 6 , 3s2 3p 6 , 4s2 , 3d 8 As CO is a strong ligand and hence pairing of electrons occurs. 3d

Ni= I

1l I 1l I 1l I 1l I 1l I

□

4p

4s

I I

sp -hybridisation 3

Oxidation state of Ni in [Ni(PPh3 ) 2 ] Cl 2 is X + 0 X 2 + (- 2) = 0 x=+2

1_ 1 �½ M�=1111111111 1 r�□__ 3d

4s

A

B

M3 N2 + 6H20 -- 3M(OH) 2 + 2Nl-la B

M(OH) 2 +

CO2

C

4p

sp3-hybridis.ation

As both the complexes has sp3 -hybridisation, thus their geometry is tetrahedral. 47. (d) Higher the standard reduction potential, better is the oxidising agent. Among the given options, EMn0_ ; Mn 2+ is 4 highest, hence Mn04 is the strongest oxidising agent.

48. (a) Basic nature of a compound depends upon the ease with which it can donate its unshared pair of electrons.

cj . . CH3CH -+NH2, CH -C+-NH 3

if7Q_C+I ..NH2

II

2

()

� III

-- MC03 + H20 D

magnesium because Mg(OH) 2 has very low solubility in water. Thus, the correct option is (c).

II

CH3-C is an

0

I. electron donating group. (Maximum basic) II. electron withdrawing. The lone pair of N is involved in resonance with

c1-1a co.

III. Lone-pair of K-atom is also used in dclocalisation of n-electrons, resonance of carbonyl benzene nucleus (least basic)

n hond

The type of molecular orbital shown in given representations arc as follows : (,)ff

(b)

�

�

d-d cr boatli,g

� p-d s boodlng

50. (b) There is one octahedral hole for

each atom in hexagonal close packed arrangement. If the number of oxide ions (02-) per unit cell is 1, then the number of Fe3 + ions = 2 / 3 x octahedral holes = 2 1 3 x 1 = 2 / 3. Thus, the formula of the compound = Fe213 01 or Fe 203 .

(c) 4 - d-d cr ,,tlbo,dl,g (d)

�

5 1 . (d)

CH2 = CH2 + H2 - Cl-Ia - C l-Ia

t,.H = (BE) reactants - (BE) products

= 4(BE)c - H + (BE)c =C + (BE) H-H -[6(BE)c- H + (BE)c-c l

= [4 x 414+ 6 1 5 + 435] - [6x414+ 347] = -125 kJ 52. (a)

0i +

14W + 6X-

Cr2

3+

- 2Cr

0

2

D

C

o-bond

�

49. (c) 3M + N2 -- M3 N2

M may be either Ca or Ba. It is not

46. (c)

C>O

Thus, the correct order of basic nature of given compounds arc as follows: I > II > Ill.

+ 7H20 + 312 3+

Since, Cr20;- is reduced to Cr , thus final state of Cr is +3 , which is formed by the reaction between Kl and acidified potassium dichromate solution.

55. (b) 2 hexyne gives trans-2-hexene on treatment with Li/NH3 • (b) Ll/ NH 3 CH CH CH - C = C - CH 3

2

2

OH ©

OH

I3 Excess

Br2 ,

Phenol

r * Br

3

2 -hexyne CH H ""' / 3 C = C �H CHaCH2CHz/ trans-2-hexene

The products obtained by the other given reagents on treatment with 2-hexyne arc as follows: (a) Pt/H2 n • hexane 2-hexyne -

(c) Pd/BaS0 4 - - - ---> cis - 2 -hcxenc (d) LiAIH4

53. (c) Phenol on treatment with

bromine water produces a white precipitate of 2, 4, 6-tribromophcnol.

- p-d n sntlbondlng

no reaction

56. (a) Given polymer can be obtained by

condensation polymerisation of ethylene glycol and phthalic acid with the elimination of water molecule.

I3r 2,4,6-tribromophenol (white precipitate)

54. (b) This problem includes basic concept of bonding. It can be solved by using the concept of molecular orbital theory. + ve phase

Q

-+-

O._

-ve phase When two same phase overlap with each other, it forms bonding molecular orbital otherwise antibonding.

420-460 K

Phthalic acid

Zn (OCOCH,),+Sb,O, - nl-!,O

0

CH,

II

C: H,OOCO

Terylene or dacron n

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295

KVPY Practice Set 2 Stream : SB/SX 57. (a) In stable isotope of Na, there are

1 1 protons and 12 neutrons. In the given radioactive isotope of sodium (Na 24 ), there arc 13 neutrons, one neutron is more than that required for stability. A neutron rich isotope always decay by � - -emission as 1 0 1 o n � -1� + i H

58. (a) 0

HLOJ-OCH1

NaBH4 is a selective reducing agent, it only reduces aldehyde and ketone group. Groups like ester, cyanide remains unaffected.

59. (c) When glucose reacts with Br2

water at pH 5-6, then gluconic acid is formed.

I

( HOH) 4 I

Brifwater pH 5-6

CH2 0H

I

COOR ( HOH) 4 I CH2 0H Gluconic acid

Glucose

60. (c) TiC1 4 Oxidation state of Ti = + 4 , the electronic configuration is [Ar] 3d 0 .

Thus, all electrons are paired and hence, diamagnetic. 0 2 No. of electron in 02 = 16. Electronic configuration : 2 crls2cr * ls2cr2s2cr * 2s2cr2p2 112p/ = 112p/11 * 2p} = 11 * 2p/

Two electrons are unpaired (by MO therory). Thus, paramagnetic. [Ni(CN)4 ] 2 - Electronic configuration of Ni 2+ = 3d 8 4s0

As CN- is a strong ligand, pairing of electrons occurs. Ni2+ I

3d

1l I 1l I 1l I 1l I

□ □ 4s

2--

[Ni(H20)4] 2+ I

3d

1l I 1 l I 1l 1 1 1 1

□ [I] 4s

I

4p

sp -hybridisation as two electrons arc 3

unpaired. Thus, paramagnetic in nature. Therefore, option (c) is correct. 61 . (c) Sliding filament theory is explained as actin and myosin filaments do not shorten but rather slide pass each other.

0

CHO

Thus, [Ni(H 2 O)4 ] Ko. pairing occurs as H20 is a weak ligand.

4p

Hybridisation of [Ni(CN) 4 ] is d 2 and sp shape is square plannar. As all electrons paired thus, diamagnetic in nature.

62. (d) Spcrmatogonia arc produced from the undifferentiated primordial germ cells which divide continuously by mitosis to form diploid cells, i.e. primary spermatocyte. These are the precursors of sperms (the haploid cells). Primary polar body, secondary polar body (in oogenesis) are produced after first meiotic division as haploid cells. Spermatids are haploid cells produced by second meiotic division of secondary spermatocytes (n). 63. (c) SA node (Sino-Atrial node) sets the basic pace of the heartbeat, hence it is called pacemaker which is a bundle of modified cardiac muscles. It is generally implanted subcutaneously and connected to heart in patients who have irregularity in their heart rhythm. 64. (b) Secretory cells always have

abundant RER and Golgi apparatus. Newly made secretory proteins are localised to the lumen of the rough ER. The proteins then migrate to the Golgi complex in the membrane bound vesicles.

65. (a) The typical storage forms of fatty acids in cells are triacylglycerols, which is formed from one molecule of glycerol by esterification of fatty acids to each of the 3 hydroxyl groups. HOCH 2 - CH (OH) - CH 2 0H 66. (d) The specificity of the enzymes is

due to their surface configuration. It means that the active site of enzyme consists of certain amino acid side chains whose linear arrangement and the appropriate folding of the peptide chain give the enzyme its specificity. This site recognises and binds the substrate(s) and catalyses the reaction once the substratc(s) have been bound.

67. (d) Ccntriolcs migrate to the poles at mid prophase-1 and start to form the spindle fibres at late prophase-I.

68. (a) Option (a) is correct as the

lagging strand is synthesised discontinuously, the newly synthesised DNAs exist as small fragments (called Okazaki fragments), which are linked together by DNA ligase. Other statements can be corrected as: RNA primer is needed on both the lagging and leading strands. Dcoxyribonuclcotides are added to the 3'-end. DKA polymerase synthesises Okazaki fragments.

69. (b) Since, there are 17 different

amino acids that constitute insulin and each amino acid is coded by a triplet codon attached to a tRNA molecule, the minimum number of different tRNA molecules is 1 7 x 3 = 5 1 . 70. (b) The correct mRNA strand should be G - A - U - U - G - A, in order to be complementary to the sense strand of the DKA fragments. However, with a single cytosine base incorrectly paired the mRNA could end up as G - A- U - U - C - A.

7 1 . (c) A gene mutation cannot cause a

dominant allele to become recessive, but can disallow it from being expressed phcnotypically.

72. (b) The promoter is made up of a basal promoter which is usually located within 40 base pairs from the transcription start site and an upstream promoter (proximal control clement) which may be located as far as 200 base pairs from the transcription start site. Control clements can be divided into two kinds, depending on the distance from the basal promoters, proximal control elements and distal control elements. Enhancers and silencers arc distal control elements. 73. (d) Some phenotypes are influenced

by the different environment even if they are individuals of the same species, e.g. height and skin colour in humans. 74. (a) The water molecule is split as shown in the equation H 2 0 � � 02 + 2H + + 2e-

2

forming oxygen and hydrogen radicals. 75. (d) The osmotic pressure of 0.25 mol dm -3 sucrose solution approximates that of the internal environment of the mitochondria thus this solution is applied to prevent the mitochondria from undergoing structural changes.

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76. (b) The tissue has a water potential of - 700 kPa. 'A' is isotonic to the tissue. Addition of solute tends to lower the water potential. Therefore, the most negative (hypertonic) will have more solutes in it, causing water to move out of the tissue into the solute. This would be the answer (b) ( - 1000 kPa). (c) and (d) are both hypotonic to the tissue, causing water to move in, making it more turgid instead. 77. (d) Glycogen is stored in the liver and muscles. When the external nutrition source is inadequate, glycogen is hydrolysed to provide the energy required in its metabolic processes. 78. (b) The opening of sodium ion gates increases the permeability of the axon membrane to sodium ions. As more sodium ions enter the axon, the membrane potential increases until it reaches a positive value. 79. (a) Each gene probably has its own characteristic mutational behaviour. Some genes undergo mutations more frequently than others in the same organisation and are called unstable or mutable. This gene is likely to mutate if its mutant form will allow the population to adapt successfully in the environment. Thus, frequency of a mutant gene in a population would be likely to increase if the genes were selectively advantageous. BO. (a) Genetic modification of plants into herbicide resistant plants can improve crop yield. This is because herbicides arc used to get rid of weeds which compete with crops for soil nutrients. Transgenic plants are not inhibited by herbicide. Thus, farmers may apply a specific herbicide to control weeds population, without damaging their herbicide tolerant crops. This leads to an increase in crop yield because of less competition from weeds. 8 1 . (d) In MBC, 60c

B

./3

b _g_Q_ = __ sin 50°

20

5,

6rt . . 6rt z.i = 2 ( cos + 1 sm ) 5

sin 50°

c =_g_Q_ sin 70 °

J3

b + c = � (sin 50° + sin 70° )

⇒

2 7t 2 + 4) ( 2 - 4cos

= � sin 60° cos 10° 40 J3 = - x -coslO0 ./3 2

b + C =X + ycos2 °

Perimeter of MBC

= 10 + 20cosl0° = x + ycos2° x = 10, y = 20, 2 =l0

:.X + y

b

85. (a) Using power of C

+ 2 = 10 + 20 + 10 = 40

BC - AC = CS2

82. (a) We have,

g" (x) - 3g' (x) > 3 (

1 ' (x) > 3 g' (x) + 1

⇒

On integrating, we get log(g' (x) + 1) > 3x g' (x) + l > e3 x

⇒

g' (x) + 1> 0 \;/ x> 0 :. g(x) + x is an increasing function. 83. (b) 9' s

S's

other

1

1

2

1

2

1

_±! X 8 = 96 2!

1

3

£ =4 3!

2

1

()

2

2

3

1

Number of ways 4 C x 2 x 8 x 8 = 768 2

1

4! - X 8 = 96 2!

()

__£_ = 6 212 1

0

£= 4 3!

Total number of ways = 97 4 84. (c) We have

⇒

c sin 70

_= =_ °

c =_ _ sin 70°

Z·1

2/m . . 2hn ) -= 2( COS-- + ! Sln 5 5

where Z; = 0, 1, 2,3, 4

C K

=

K

[·: (2 -a)(2 - a) = (z -(a + a)2 + aa)] 2 5 p 2 + 2q = 16sin218° + 8sin 18° = 8(1 -cos36° + sin 18° ) 1 -1 _ = { 1 + 15 Js/ ) = 4 4

p = q = 4cos 7t = 4cos 72 ° = 4sin 18°

b + c = 20cosl0° =a+

5

2

2

10

b_ ___!Q_ = __ sin 60° sin 50°

J3

2 25 -32 = (2 - 2)(2 - p2 + 4)(2 - q2 + 4) 25 -32 = (2 - 20 )(2 - 21 )(2 - 22)

A c

b=

⇒

(z - 23 )(2 - 24 )

BC - AC = (

1

+1- l)

2

=2

86. (a) Given, 2x - y + 1 = 0 is tangent of circle at (2, 5). Centre of circle lies on x - 2y = 4

Solving 2x - y + 1 = 0 and x - 2y = 4, we get intersecting point P ( -2, - 3)

PA = ✓(2 + 2) 2 + (5 + 3) 2

= .J16 + 64 =.Js6 = Angle between lines arc 1 2 -2 -3 tane 4 1 + 2G)

4-J5

ln t..PAO,

5'

2rt . . 2rt + ! sin Zo = 2' Z1 = 2 ( cos ) 5

tan0 = ...!_ ⇒ ⇒

PA r

4 - 4-J5 r = 3J5 3

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297

KVPY Practice Set 2 Stream : SB/SX 87. (a) We have,

91 . (c) Refraction and reflection occurs as shown below

cos(sin x) = sin(cosx)

⇒ ⇒

sin ( % ± sin x ) = sin(cosx)

cosx ± sinx = mt + (-1)" �, n E I 2 J2 As LHS E [- , "'"21 and it docs not satisfies RHS. ⇒

Hence, no solution exists.

�r

Case I D1 never show 1

(�r - l (i )} (� )

89. (b) y

x

e and y

=

=

4 sin 60° We have - = --' 3 sin r sin r = _:3 x sin 60° = 0.66 4 ,. = 40° 30' So, deviation, 8 = 180° + 2 x 60° - 40° 30' X 4 = 138° When sand particle falls onto cart, its' relative velocity with respect to cart is - u (backwards), so force of sand on cart is backwards.

6n - 1

52

log, x are inverse of

0

0, 1

J

⇒

at (0, 1) and (1, 0).

1n (

90. (b) Let

93 .

Ft

T� L

I = J:[x] {x} dx

⇒

⇒

I = fo {0}dx + Ji{x} dx + 2J {x} dx+ i=

1=

3 2

2

• 4 3J {x} dx+ J; 4{ x} dx 3 I (l+ 2 + 3 + 4)J xdx [·: {x} is periodic with 1]

1

1ol2Jo r x2 1

I = lO = 5 2

0

-x _ dx = = dt ✓1 L2 - x2 dt

✓i) - t -3

"2

32

x2

=

1- ). 2' 2

is at ( � (ucM )x

=

ux.

=

z Now, from loops 1 and 2, we have

2x - y - z = 0

2x - 3z = 0

Here, r = 2Q, so R = !! Q _

97. (c) Uranium nucleus is initially at rest. Conservation of momentum gives 0 = Pa + PTh ⇒ PTh = - Pa Also, Q = Ka. + KTh ⇒

.!i_ � or u ( ) x dt 2

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

Let R = Equivalent resistance between A and B, then VAB = R (2x + y). 2 ry = � ⇒ 2 ry = R(2x + y) or R = r 2x + y 5

3 - - ms-1 dt 2 At any instant, position of centre of mass dy or -

2>1 Cl

y = 4k

At any instant, y = .JL2 - x 2 dy

C2 > cl ⇒

So, � = .=. = k (let) ⇒ x = 3k, z = 2k 3 2 Substituting these in Eq. (i), we get

+ µt

m0

� 0 X

⇒

c

L',.U2 and L',. W2 > L',. W1

t

U = ---

(b)

=

follows

m µt � ) = 1n ( o� ) F µu

=}

-Jl + 1 = J2

I

J

E) E)

96. (a) We take current distribution as

l l _ _ . du = ___ . dt F- µu m0 + µt V 1 1 -- . du = --- . dt o F - µu O m0 + µt

⇒

Minimum distance of the curve

So,

du m- = F- µu dt du (m0 + µ t) = F - µu dt

⇒

500E

instead of E. 504 Hence, per cent error in reading is 500 (£ 5o4 Per cent error = x 100 = 0.8% 500 ( 504 So, voltmeter reads

L',.U1

Net force on cart is Fner = F - F' = F - µu

⇒ 1 ,0

0

�

⇒

Potential drop across terminals of voltmeter is _ E 500E V = iR = 500 x _ _ = 500 + 4 504

t-.Q = t-. U+t-.W 95. (b) W e have, C = t-.T t-.T For paths A and B,

�F

5n

each other

⇒

E E i. = -= --R + r 500 + 4

60° \

( 5)n + n (---l )(-1 )

Total probability = 6

1

94. (b) Let E = emf of cell. Current in the circuit is

92. (a) Let velocity of cart in u at time t.

Case II D2 shows 1 (one time) then D1 shows 1 Probability = { nc1

✓

u = � = 1 + � = � = 125 ms-1

cosx = mt + (-1)" ( % + sin x } n E I

Probability = (

. . 1ar Iy, v = 1 3 = 3 ms-1 S1m1 - (- ) y 2 2 4 So, velocity of centre of mass of ladder is

=

dx

__! 2 dt

=

l ms-1

Q = Ka + piJ, 1 2�

2 2ma. Ka = K + ____l}_Q_ = K + a 2�h a 2�h

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298

KVPV Practice Set 2 Stream : SB/SX

4

= K ( l+ � ) = .2 ( l + a 2 238 = .2 ( ) MeV 23

4

4

!4

)

1 0 1 . (b) According to Raoult's law, relative lowering of vapour pressure, . . . (i) WB IMB WB W + A MB MA

98. (d) Let particles collide in mid air at sometime instant t after t = 0.

lfJ D

B

Then, 20 - y = _! gt 2 2

and 30 - y = lOsin 60° t + _! gt2 2 [taking, B as origin] 2 Solving, we get t = - s

J3

Now, Xi = distance covered horizontally 2 m '\/ 3 x2 = horizontal distance of second particle by first particle = lOt =

i

= 10cos60° x t = _!_Q_ m

J3

d = Xi_ + x2 = lM m

:. Distance between towers,

99. (d) As process is adiabatic, A Q = 0.

⇒ ⇒ ⇒

+ pd V = 0 d (a + bp V) + pd V = 0 d d (b + l) V = - b p dU

V

P

Integrating and rearranging, we get ⇒ (b + 1) In V + bin p = constant +1 (b ⇒ p V b = a constant b+ l y = - - = l+ -1 ⇒ b b 1 00. (d) Let �o = uncertainly in position

l

.

at time t = 0 s, so uncertainly in momentum 1s Ap � -

h

4itAx0

As v�0'f � :>B�F -- :>B- F-- YB-F

E 0 Mg2• / Mg 1 06. (b) E 2 / Mg + Mg -

0.059 1 - -- log -2 Mg 2+ 0.0591 1 2 Or E 2 +/ = E o 2 + 0g Mg + Mg Mg Mg •tMg 2

The given equation is of y = mx + c type, which is equation of straight line. So, graph of E 2• (y) vs log [Mg2+ ] (x) Mg 1 Mg _ . a strmght . . with slope - 0 . 059 - and lmc 1s 2 intercept E 0 2+ Mg I Mg

1 07. (a) Formation ofa, �-unsaturated

carboxylic acid by the action of acetic anhydride and sodium acetate on aromatic aldehyde is known as Perkin reaction. The other Perkin like condensation involves condensation of aromatic aldehyde and a-hydrogen containing compound. The final product X of the given reaction is C 2H 5O ­ CH2(COOEt)2 - --

1 04. (c) Buffer solutions arc aqueous solutions consisting of a mixture of a weak acid and its conjugate base or

vice-versa.

When the concentration of NH4 OH (weak base) is higher than the strong acid (HCI), a mixture of weak base and its conjugate acid is obtained, which acts as basic buffer.

--=.-H 2 0 H

©( 0

oCH-CH(COOEth OH

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299

KVPY Practice Set 2 Stream : SB/SX

r

�"

cPo

!m5i�

@;f

�

H+

COOEt

Reimer-Tj.emann reactrnn

� oA-o (X)

1 08. (d) Since, the compound A gives white precipitate with AgNO3 , which is soluble in dilute ammonia solution, it contains at least one Cl atom outside the coordination sphere. Thus, its formula (A) is [Cr(NI-la ) 4 BrCl] CL [Cr(NI-la ) 4 BrCl] Cl - [Cr(NH3 ) 4 BrClt + c1AgNO3 + c1- - AgCl + N03 White ppt.

Further, the compound B on reaction with AgNO3 produces a pale yellow precipitate soluble in cone. NI-Ia solution. Thus, 'B' is [Cr(NI-ls ) 4 Cl 2]Br l Cr(NI-ls ) 4 Cl 2JBr [ Cr(NI-ls ) 4 Cl2

AgNO3 + Br- - Ag Br + N03 Pale yellow ppt.

t + Br-

1 09. (b) On reaction with Cl 2 , phosphorus

forms two halides PC15 (A) which is yellowish powder and PC13 (B) is colourless oily liquid. 'A' is PC15 and 'B'is PC1a ­ P4 + 10Cl 2 - 4PC15 P4 + 6Cl2 - 4PC13

PC15 + 4H 2 0 - H 3P0 4 +5HC1 [A] Phosphoric

When 'A' and 'B' are hydrolysed. (a)

(b)

Phosphorus pentachloride

acid

PCI5 + 3H2 0 - H 3P0 3 +3HC1 [B] Phosphoric

Phosphorus trichloride

acid

& OH

1_ _ _ _ ..

1 1 0. (a) In step 1, when phenol reacts with chloroform in presence of NaOH then salicyaldehydc is formed. This reaction is known as Reimer-Tiemann reaction In step 2, the formed salicylaldchydc then undergoes Cannizzro reaction with formaldehyde in presence of strong base to give an oxidised product (minor) and reduced product (major).

l

CHO

Cann.izzaro react10n

+ HCHO

OH

�+CH2 OH+HCOONa Minor

lS::JJ Major

�

nn 1 1 1 . (c) N = i 2

(N = population size of rats in the field) Here, n1 = 80, m2 = 20, n 2 = 100 80 x 100 - 8000 = 400 Hence, N = 20 20

1 1 2. (c) First prey were eliminated from the clear fluid medium by the predators and then afterwards the predators starved to death in the absence of prey. The sand sediment served as a refuge for the prey from where they emerged later, increased its number via reproduction and finally reached the stable equilibrium. 1 1 3. (c) It is already mentioned that the couple has a child with blood group 0 and the child is diseased. Thus, the parents are heterozygotes for the diseased allele (as parents arc normal and the disease is recessively inherited) and for the blood group. The probability of O blood group in progeny is IA i X I8i J, IA I8 rA iI8i ii i.e. _!: individual is blood group 0. 4 Similarly, for the disease to be inherited, the individuals should have both the recessive alleles, thus Dd x Dd = dd (probability to get the disease is _!:) _ Therefore, probability that

the second child has the disease = P (the child is blood group 0) x P (the child has both the recessive alleles). Thus, the P (diseased child) = _!: x _!: = J_ . 4 4 16 1 1 4. (c) Upon digestion with Hind III: For allele Q, there is the presence of 3 restriction sites, 2 restriction fragments will be formed. For allele q, there is the presence of 2 restriction sites, 1 large restriction fragment will be formed, resulting in a band that is nearest to the well. From

the analysis, there is only 1 band (nearest to the well) observed. This indicates that individual Z carries allele q. The band is more intense as there is twice as much DKA in it proving that individual Z is a homozygotc carrying two copies of the allele (qq). 1 1 5. (d) The presence of U indicates a RNA virus. A and G are purines while U and C arc pyrimidines. When ratio of purines: pyrimidines = 1, this sugges�s . that there is complementary base pa1rmg between the bases. Thus, this must be double-stranded RNA virus. 1 1 6. (c) Since, the y-axis is time, the graph obtained would be a vertically inverted reaction rate us temperature graph. As the temperature increases, the time needed decreases at first until a minimum because an increased number of molecules have sufficient energy to pass over the energy barrier and react to form products. A further increase in temperature results in a more rapid increase of time required because of temperature induced denaturation of the enzyme. 1 1 7. (c) Palaeozoic era is divided into the periods, Cambrian, Ordovician, Silurian Devonian, Carboniferous and Permia� from past to recent, respectively. The' mesozoic era is divided into three periods, Triassic, Jurassic and Cretaceous from past to recent, respectively. 1 1 8. (d) Integrin Transmembrane cell adhesion proteins that act as extracellular matrix receptors. Cadherin Mediates Ca2+ dependent strong homophilic cell-cell adhesion. lg super family Contains extr�cellular lg like domains and are mamly mvolved in the fine tuning of cell-cell adhesive interaction during development and regeneration. Selectin Lectins that mediate a variety of transient cell-cell adhesion interactions in the bloodstream. 1 1 9. (b) Daughter 4 : xx\ XX where x h is the gene for haemophilia. Probability of her being a carrier = 0. 5 Kormal male XX

I

XX

XhX TXY

I

XhX

Carrier

I

XY

I

XhY

Haemophilia

Probability of a hacmophilic = 0.25 Therefore, probability of daughter 4 having a haemophilic child = 0.5 X 0.25 = 0.125 1 20. (d) The total amount of energy used in forming 38 ATP = 38 x 30.6 = 1 162. 8 kJ Thus, the efficiency of aerobic respiration is 1 1 62 · 8 X 100% = 40% 2880

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KVPV

KISHORE VAIGYANIK PROTSAHAN YOJANA

PRACTICE SET 3 Stream : 58/SX

MM : 160

Instructions 1 . There are 1 20 questions in this paper. 2. The question paper contains two parts; Part I (1 Mark Questions) and Part II (2 Marks Questions) . 3. There are four sections in each part; Mathematics, Physics, Chemistry and Biology. Out of the four options given with each question, only one is correct.

4.

� PART-I MATHEMATICS 1. Let a , b, p , q E Q and suppose that

f(x) =:< + ax + b = 0 and g(x) = :l' + px + q = 0 have a common irrational roots. Then, (a) f(x) divides g(x) (c) g(x) = (x - b - q) f(x)

(b) g(x) = xf(x) (d) None of these = =

2. lf 0 < 6, < n: / 2 and x = I, sin 2n 6, y = I, cos2 " and =

n

n

n=O

z = L, cos (e + qi) cos (e - qi), then n=O

(a) xyz + l = yz - zx (c) xyz - xy = yz - zx

n=O

(b) xyz - l = yz + zx (d) xyz + l = yz + zx

3. Solution set of the inequality

log3 (x + 2) (x + 4) + log113 (x + 2) < ½ log./3 7 is (a) (- 2, - 1) (c) (- 1, 3)

(b) (- 2, 3) (d) (3, oo)

(l Mark Questions) 4. The number of ways in which a mixed double game can be arranged from 9 married couples if no husband and wife play in the same game is (a) 756 (c) 3024

(b) 1512 (d) Kone of these

*

5. Let f (x) :

tan[e2].:f - tan[- e2].:f , x 0 ( [·] represents . 3x sm greatest integer function). The value of f(0) for which f(x) is continuous, is

(a) 1 5

6. If J =

I

equals

TC 1 2

0

(a) 2- n + l

(b) 1 2 n

.

cos xsm

flt xdx = A

(b) 2n + l

O

(a)

X

=Y

/2 •

n

sm xdx, then A

(c) 2- n - l

7. If x = tan 278 - tan 8 and y = then

(d) 1 4

(c) - 12

n

(d) 2- n

sine sin 3 e sin 98 -+-+- , cos 3 e cos 9e cos 278

(b) x + y = 2 (c) x = 2y

(d) y = 2x

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301

KVPY Practice Set 3 Stream : SB/SX ' y2 = 1 and A and Bare 8. C is the centre of ellipse :< +

16 9 two points on the ellipse such that LACE = 90", then 1 1 . - 1s equal to -+9 CA- CB2 1 44 (a) � (b) 1 44 25 12 (d) (c)J_ 7 12

9. The shortest distance from the line 3x + 4 y = 25 to the circle : satisfies

23. A homogeneous circular plate of radius r has a

16. The number of integral values of k for which the equation 7 cos x + 5 sin x = 2k + 1 has a solution is (a) 4

(b) 1 (d)infinite

(c) __!l_cos a g

is a

15. In a MBC, bisector of LC meets the side ABat D and circumcentre at E. The maximum value of CD · DE is equal to 2

3

(a) 0 (c) 2

(a) __!l_ sina g

14. The radius of an arc given by locus of z if (c)7

0

with horizontal. Its velocity is perpendicular to initial velocity after time

(d)- _! 2 k2

1 such that f (x) < x2 for all x and J f (x) dx = I_ is

22. A particle starts from origin with speed u and angle a

0

(a) 0

18. The number of continuous function f : [0, 1] ➔ [0, 1] ,

(b) 3/5

(c)4/5

Coordinates of centre of mass of plate with respect to given origin is (a) (0, 0)

(b)

(-%, 0 )

(c)

(- i,

0)

(d)(- !__ , !__ ) 6 6

24. A system with two 3 kg mass particles having

velocities 21 + 3] ms- 1 and 41 - 6] ms- 1 is subjected to a constant force of 24l N for 5 s. Final velocity of centre of mass is (a) 3i - 1.5) ms- 1 (b) 60i - 30}ms- 1 1 (c) 20i - 3)ms(d)23i -1 .5}ms- 1

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302

KVPVPractice Set 3 Stream : SB/SX

25. A block of mass 0.5 kg is placed over top of a wedge of mass 50 kg.

(a) [ML-21-KJ (c) [ML-2T-3 K]

(b) [M- 1 L-21-KJ (d)[M- 1 L -2T3 K]

30. A rectangular glass wedge is partially dipped in 1m 0=37°

M

Angle of inclination 8 is 37 ° with horizontal. Block is released at t =0 and it slide down the incline and reaches bottom with a speed of 1.2625 ms- 1 . There is no friction. Speed of the wedge with respect to ground, when block reaches at bottom is (b)0.02 ms-1 (a) 0.01 ms- 1 1 (c)0.03 ms(d) 0.05 ms- 1

26. A small block of mass 1 kg is pushed with a velocity

2 i ms- 1 over a horizontal plane which itself is moving with a velocity of - 2j ms- 1 . Coefficient of friction between block and plane is 0.25. Force of friction acting on the block (in newton) nearly is (b)17(- i- ]) (a) l7(i + )) (d) l7(- i + J) (c)l7(i- J)

27. A pendulum bob is released from angular position 8,

water ( nglass =!} A beam of light entring face AB normally reaches entirely to AC.

(a) 0 = cos- 1 (�)

(c)e = tan-1 ( �)

(b)0 = sin- 1

(d) e = cos-1

rn)

( ¾)

28. A metal sphere having radius r, charged to a

potential V is enveloped by a thin walled conducting spherical shell of radius 4r asymetrically.

B

Now, value of angle a must be (b) less than 30° (a) less than 60° ° (d)between 30° to 60° (c) more than 60

31. A thick transparent slab is made such that its

refractive index changes from n1 to n 2 nearly linearly. This slab is used by a student to measure lateral displacement in lab. She draw following diagram using four pins. Angle � must be a

such that the magnitude of its initial acceleration and acceleration at lowest position are equal.

The angular position 8 is

a

A

(a) a

32. The word "excellent" is written on a sheet of white

paper with a red pencil and the word "good" is written with a green pencil, on a school notice board. A student look at this notice board through a green and red piece of glass. (a) He can sec "excellent" through red glass (b) He can see "excellent" through green glass (c) He can see "excellent" through both glasses (d)He cannot sec "excellent" through any of the glasses

33. In an experiment of suspending spherical plastic

balls in vertical turbulent jets of air and water it is found that if sphere is displaced slightly to left or right (from its suspended position), the air or water stream again pushes it back to its original position.

Sphere is connected with shell by a conducting wire for a short time, then connection is removed. Potential of sphere after removing connecting wire is (a) V (b) V (c)4V (d) V 2 4

29. In steady state, heat conduction equations governing heat flow are identical to the electric current flowing through a conductor. What are the dimensions of quantity analogous to electrical resistance?

- -(,

.

-'r-

;i�

--,r,

\,

, / ,r -

This can be explained by using (a) Archcmcdcs' principle (b) Bernoulli's principle (c) Pascal's law (d)Stoke's theorem

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303

KVPY Practice Set 3 Stream : SB/SX 34. If 8 x 1 010 atoms of Radon are separated from

Radium, how many disintegrations will occur in 11.46 days? (a) 8 x 1 01° (b)8 x 1 06 (c)7 x 1 01° (d)7 x 1 0 9

35. From examining binding energy per nucleon versus

�a

mass number curve shown below. C 1 (s)

s (m )

s(m) (a)

3

0

( b) 4

s (m )

(d ) 0

4

t(s)

7

3

� 0

- � - ➔ t(s) 4

95. Force of repulsion between two charges of + 2 µC and + 3 µ C, separated by a distance of 10 cms- 1 is F1 . If a dielectric slab of dielectric constant k = 4 is placed, such that it fills 5 cm distance (as shown), then force between charges will be

4 (a) - F; 9

9 (b)- F; 4

-

-

- - - ►- m

(d )

---...!===::±======::,_______, m

0

-

-

-

32 32

16

32

0

II. He places a large similarly charged sphere near S and then measures force per unit charge. Let his recordings are E1 and E2 • (a) E2 > E1 , when charge on S is negative (b) El = E2 (c) E1 > E2 (d)E2 > Ei , when charge on S is positive

100. An automatic rotating water sprinkler is mounted over top of a 5 m high pillar in a field.

7 (c)- F; 5

',

\ \ \ \

(d)� F; 7

surface of a lake. Fraction of light energy that escapes from the lake depends on (a) h (b)h 2 (c)h- 2 (d)h 0

97. A diatomic ideal gas undergoes following two steps

process. Step 1 Constant volume heating to triple pressure of gas. Step 2 Constant pressure heating to double volume. Molar heat capacity of the gas for whole process is 27 3 (a) �R (b) �R (c) R (d) o R 25 32 4 30

98. Instantaneous position of a particle is given by t 3 - 12 t 2 + 36t

v �',=............ ---:-............ ....' , Spray

(±) 2µC

96. A point source of light is placed at depth h below the

X=

-

----+-----+-----+-------> m

-

0

\

(±) 3µ C

-+ - -

(c)

-

I. He places a very small similarly charged sphere near S and then measures force per unit charge.

s(m )

(c)

-

-

99. A student measures magnitude of electric field due to a charged sphere S by two different ways.

t(s)

7

-+- -

(b)

2 Its distance-time graph is

----+--------+------------> m 32 0

Motion of the particle for first 8 s is best illustrated by

Water in --➔

I 1

\

'

....

\

'\

', ', ',

I 1

\ \

I 1

\ \

\

\ \ \

,

I 1

=====:::::It::::::::::=========

Water is flowing at a rate of 5 x 10-4 m3 in 1 s through a nozzle of area 1 cm 2 • Maximum possible area of field that can be irrigated by the nozzle is (take, g = 10 ms- 2) (a) 1 51t m2 (c) l001t m2

(b) 251t m2 (d)751t m2

CHEMISTRY 1 0 1 . Which of the following statement(s) is/are incorrect? (a) The coordination number of each type of ion in Cs Cl crystal is 8. (b) A metal that crystallises in bee structure has a coordination number of 12. (c) A unit cell of an ionic crystal shares some of its ions with other unit cells. (d) The length of the unit cell in NaCl is 276 pm. (rN + = 95 pm; r , - =1 81 pm). c a

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308

KVPVPractice Set 3 Stream : SB/SX

102. Which one of the following complexes shows optical

isomerism? (b)trans-[Co(en)i Cl 2 ]Cl (a) cis-[Co(en)i Cl 2 ]CI (c)cis-[Co(NH3 )4 Cl 2 ]CI (d) trans-[Co(NH3 )4 Cl 3 ] 103. The major product of the following reaction is NII 2

6

2. Br2 3. KOH/C2HoOH 1.CH3COCI

H,

H

� c&°: :, ) )

" c&°: ( ) ,

¢'

( ) d

, &B

Br

1 04. On treatment of 100 mL of 0.1 M solution of CoC13 .6H2 O with excess of AgNO3 ; 1 .2 x 1 022 ions are precipitated. The complex is (a) [Co(Hi0)4 Cl 2 ] Cl-2H20 (b) [Co(H20hCl 3 ] -3H20 (c)[Co(H20)6 ]Cl 3 (d) [Co(Hp)sCI] Cl 2 · Hp

1 05. Identify the set of reagents/reaction conditions X and Y in the following set of transformations CH 3-

CH2 -

CH2 Br � Product � CH3 -

(a) X = dilute aqueous KaOH Y = HBr/acetic acid (b) X = concentrated alcoholic NaOH Y = HBr/acetic acid (c) X = dilute aqueous NaOH, Y = Br2 /CHCl 3 (d) X = concentrated aqueous NaOH Y = Br/CHC1 3

H - CH3 1 Br

106. For one mole of a van der Waals' gas when b = 0 and T = 300 K, the p V vs 1/V plot is shown below. The value of the van der Waals' constant a (atm L mol- 2)

-----=�

24.6

I.0 23.1 E 2 1 .6

ro E

::::'..

I

I

I I

I I

I

I

I I I I

0

(a) 1 .0

(b) 4.5

1 09. One mole of an organic compound A with the formula C 3 H8 O reacts completely with two moles of HI to from X and Y. When Y is boiled with aqueous alkali it forms Z. Z answers the iodoform test. The compound A is (b) propan-1 -ol (a) propan-2-ol (c) ethoxyethane (d)methoxyethane 1 1 0. Consider the following conversion:

II

0 o-CC:H3

?

CH 1

I

O-cH HCH2C:H3 1 OH Which of the following sets of reagents, would successfully accomplish the conversion shown? (a) CH3CH2CH2 MgBr; Hp· ; PCC,CH2Cl2 (b) CH3CH2CH2 MgBr; H30+ ; H2S04 , heat; PCC, CH2Cl2 ----'-------+

(c) (C6 H5 hP - C HCH2CH3 ; B 2 H6 ; H202 , HO+

.-.

(d) (C6 H5 ) 3 P - C HCH2CH3 ; H2S04 , Hp

BIOLOGY

I I I I

2.0 3.0 1/V (mol L-1 )

(c) 1.5

2 1 A (mJ (H Coo- J = 54.6 8 cm mor )

Column I Sap vacuole Contractile vacuole Food vacuole Air vacuole Spherosome

A. B. C. D. E. (d) 3.0

1 07. The molar conductivity of0.25 mol G 1 formic acid is 40.1 S cm2 mor 1 . Calculate its dissociation constant. 0 [Assuming, A m(H + ) = 349.6 S cm2 mol - 1 ; 0

1 08. A hexapeptide has the composition Ala, Gly, Phe, Val. Both the N-terminal and C-terminal units are Val. Cleavage of the hexapeptide by chemotrypsin gives two different tripeptides, both having Val as the N-terminal group. Among the products of random hydrolysis is a Ala-Val dipeptide fragment. What is the primary structure of the hexapeptide? (a) Val-Galy-Phe-Val-Ala-Val (b) Val-Ala-Phe-Val-Gly-Val (c) Val-Gly-Ala-Val-Phe-Val (d)Val-Phe-Val-Ala-Gly-Val

1 1 1 . Match the following columns and choose the correct option from the codes given below.

________ l�

20.1

(b) 3.74 x 1 0-3 mo! L-1 (d)5.27x 1 0 -3 mo! L- 1

(a) 2.73 x 1 0-3 molL-1 (c) 4. 75 X 1 0 -3 mo! L -l

Codes

(a) (b) (c) (d)

A 5 2

5 5

B 3 3 2 3

C 1 4 3 2

D 2 5 1 4

1.

2. 3. 4.

5.

Column II

Contains digestive enzyme Stores metabolic gases Osmoregulation Stores Iipids Stores and concentrates mineral, salts and nutrients E 4 1 4 1

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309

KVPY Practice Set 3 Stream : SB/SX 112. The restriction fragment shown below contains a

gene whose recessive allele is lethal. The normal allele has restriction sites for restriction enzyme pst I at sites I and II. The recessive allele lacks restriction sites I. An individual who had a sister with the lethal trait is being tested to determine if he is a carrier of that lethal trait. II

Which of the band patterns would be produced on a gel if he is a carrier?

e� I I I Well

(a)

ep I (c) ep I (d) ep I (b)

(a) Birth weight is undergoing stabilising selection (b) Birth weight is an example of discontinuous variation (c) Birth weight is inversely proportional to mortality (d)Birth weight is genetically linked to mortality

115. ATP can be formed by oxidative phosphorylation in

the electron transport system and by glycolysis. In the complete oxidation of one molecule of glucose, approximately what percentage of ATP is formed by oxidative phosphorylation? (a) 1 0 % (b) 25% (c)75% (d)90%

116. The diagram shows the inheritance of a form of

Iii!)

breast cancer associated with the presence of just one allele of the autosomal gene BRCA 1.

l(r) l(r) l(r)

Male

■

113. The Galapagos islands are a group o f volcanic islands

in the Eastern Pacific Ocean about 1000 km from South America. Thirteen species of finch are found on the islands, they resemble each other closely but differ in their feeding habit and in the shape of their beaks.

Assuming that an ancestral stock of finches came from the mainland, what is the most likely explanation for the existence of similar but distinct species of Galapagos finches? (a)Finches developed different kinds of beak in order to feed on different kinds of food (b)Finches evolved separately according to the habitat in which they settled (c)Finches from the mainland bred with a resident population of a related species and produced new genotypes (d)Finches underwent convergent evolution to produce very similar species

114. The diagram below represents the proportions of a

population of newborn deer calves falling into various birth weight classes. The graph superimposed on the diagram represents mortality in relation to birth weight. JOO

70 ..µ 50 � 30 � 20 ;;l 10 ,::,.

'cl 10

5 3

5 1

2

3 4 5 6 7 8 9 Birth weight/Kilograms

10

From the information given which one of the following interpretations is correct?

11

...,>, t:0

�

Key Female . No cancer Q Breast cancer developed

What is the probability that the woman X inherits the BRCA 1 allele associated with breast cancer? (a) 0.00 (b) 0.25 (c) 0.50 (d) 1 .0 0

117. The diagram represents a sequence of reactions

taking place in a bacterium in which amino acids are produced from one another by the action of specific enzymes. Number 1 to 6 represent different amino acids, letters V to Z represent different enzymes. All the amino acids are essential for survival. 1 � 2 ....!!:'.H ....1'....+ 6

lx lz 5 3

The original strain of the bacterium required only amino acid 1 . A mutant strain of this bacterium could only survive when produced with amino acids 1, 2 and 5 in its culture medium. Which enzymes were missing in the mutant strain? (a) V and Z (b) W and Z (c) X and Y (d)V , W and Z

118. The table shows the result of an analysis of

percentage concentration of three bases in nucleic acids from four sources. Three of the sources are DNA and one is RNA. Which source is RNA? (a)

(b)

(c)

(d)

A

Source

B

D C

Adenine

19.7 25.5 26.7 31.1

Cytosine

30.4 24.6 28.8 18.3

Guanine

30.2 23.8 22.3 18.7

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310

KVPVPractice Set 3 Stream : SB/SX

119. The graph shows the amount of product formed b y a

20

standard concentration of enzyme and a standard concentration of substrate at a temperature of 20 ° C.

Amount 15 of product formed/ 10 (c) Arbitrary units 5 0

20

Amount 15 of product formed/ 10 Arbitrary units 5 0

Time

20 Amount 15 of product formed/ 10 (d) Arbitrary units 5 0

Time

Which graph shows the effect on the activity of the enzyme of decreasing the temperature to 15° C? 20

Amount 15 of product formed/ 10 (a) Arbitrary units 5

Time

120. Match the five (A- E) group of organisms with their correct taxonomic rank (1-5) and choose the correct option from the codes given below.

()

A. B. C. D.

Time

20 Amount 15 of product formed/ 10 (b) Arbitrary units 5

E.

Codes

A

()

(a) 3 (c) 4

Time

Group

B 1

3

C

5 2

Order Domain Class Phylum Family

2.

3. 4.

D E 4 1

Taxonomic rank

1.

Crustacea Hominidae Dermaptera Ctenophora Archaea

2 5

5.

A B

(b) 1 (d) 3

2 5

C 3 1

D

4 4

E

2

Answers PART-I 11 21 31 41 51 61 71

(a)

(b) (c)

(a)

(a) (c)

(c)

(c)

PART-II 81 91 1 01 111

(b) (c)

(b,d) (a)

2 12 22 32 42 52 62 72 82 92 1 02 112

(c)

(a)

(d)

(b)

(d)

(b)

(a) (c)

(c)

(a)

(a) (d)

3 13 23 33 43 53 63 73 83 93 1 03 113

lb)

le)

le!

lb) Id)

le)

Id)

lb)

lb)

le)

le)

la)

4 14 24 34 44 54 64 74 84 94 1 04 1 14

(b)

(c)

(d)

(c) (c)

(b)

(b)

(c)

(c)

(d)

Id)

la)

5 15 25 35 45 55 65 75 85 95 1 05 115

(a)

(c)

(a)

(c)

(a)

(d)

(d)

(c)

(a)

(a)

(b)

(d)

6 16 26 36 46 56 66 76

(d)

86 96 1 06 116

Id)

(d)

(b) (c)

(a) (b)

(b)

(a)

Id)

le)

la)

7 17 27 37 47 57 67 77 87 97 1 07 117

(c)

(a)

(a)

(a)

(b) (a)

(b)

(c)

(b)

(b) (a)

(a)

8 18 28 38 48 58 68 78

(d)

88 98 1 08 1 18

la)

(a) (a) (d)

(a) (a) le)

(c)

le)

la)

le)

9 19 29 39 49 59 69 79 89 99 1 09 119

la) le)

Id)

Id)

la) lb)

Id)

la) le)

le)

Id)

le)

10 20 30 40 50 60 70 80 90 1 00 110 1 20

(d) (c) (c)

(d) (b)

(b)

(d)

(b)

(a)

(b) (c)

(d)

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Solutions 1 . (a) Let a E R - Q be a common root of f (x) = 0 and g(x) = 0, then, a 2 + aa + b = 0 ⇒ a 2 = - aa - b On putting this in ri + pa. + q = 0, WC get (a 2 - b + p) a. + ab + q = 0 As a is irrational and a, b, p, q E Q, p = b - a 2, q = - ab This give, g(x) = (x - a) f(x) :. f (x) divides g(x).

We can select two men out of 9 in 9 C2 ways. Since no husband and wife are to play in the same game, two women out of the remaining 7 can be chosen in 7 C2 ways. If M1 , M2, W1 W2 arc chosen, then a team consist of M1 W1 and M2W2. Thus, the number of ways of arrangement is 9C x 1 x 2 = 15 12. 2 2

c

5. (a) We have,

2. (c) Given,

f(x) =

x = I, sin 2n 8 n=O

⇒ ⇒

= 1 + sin 2 8 + sin 4 8 + . . . 1 1 x = ---1 - sin 2 8 cos2 8

= 1 + cos cjJ + cos4 2 nl where, n 2 = refractive index of water = ± 3 and n1 = refractive index of glass = _:3_ _ 2 4 :. sina > => sin a > � = 0.89 9 2 a > 60° sin 60° � 0.86

j

3 1 . (a) We divide slab into a series of

no

parallel slabs with different refractive indices.

We have

a

3.82 days 8 X 1 01 0 --..--- 4 X 1 0 1 0 Nucleus

2 X 1 01 0 Nucleus

¾

sin 3 _ n 4 sin 4 _ n 0 sin 4 ;;-• sin P So, we have sin a sin 1 sin 2 x sin 3 x sin 4 x x =1 sin 1 sin 2 sin 3 sin 4 sin P sina = sin p or a = p

32. (b) When student look through green

glass "excellent" appears as black against the green background of the paper. Red colour of "excellent" docs not pass through green glass.

33. (b)

m rlr '

2

zp Work done, dW = qV = - 4nz 2dz - p 3E 41tp2 4 or dW = z · dz 3E Work done in assembling complete sphere R a

JdW

41tp 4 =faR � z · dz 3 2

E

< 3. 82 days I

3.82 days

41t X

1 x 1 01 0 Nucleus

So, only lx 1()1° active nuclii remains after 3-half livcs. So, 7 x 101° disintegrations would occur in 11.46 days. 35. (c) In region B, we observe that binding energy is not increasing with mass number. This indicates a nucleon possibly interacts with other nucleons which lies near to it but not with those which are far off from it. This indicates about saturation property of nuclear forces. and compress the spring is

sin 1 _ n sin 2 _ 113 sin 2 n1 sin 3 n 2

4 - nz3 · p 2 q 3___ = � li Potential of sphere = =_ 4lt E Z r 3E To increase charge layer by thickness dz,

Nucleus

36. (d) Work needed to raise the piston

sina _'Ii_ = ' sin 1 n 0

2 -- - -, -- - -

=>

The sphere shows stable equilibrium for small horizontal displacement because of velocity distribution of approaching fluid. Near edge, velocity is low, so pressure is high. Hence, when ball is slightly displaced, it is pushed back to centre where velocity is more. As, by Bernoulli's principle where v is high, p is low. 34. (c) As, 1 1 . 46 days is three times the half-life of Radon. So, decay scheme is as follows.

2 W = mgh + ..!: kx + Patm A · h 2

= 60 X 9.8 X 0.05 + _!: X 50000 X (0.05) 2 2 100000 X 1t X (0.2) 2 + - - - - - - - X O. o ;:)4 = 250 J By first law of thermodynamics, we have LlU =LlQ - ti W = 200 - 250= - 50J

31 . (a) System undergoes an internally reversible isothermal process. So, entropy change of system is kJ LlS =-Q - = 750 = 25 300 K T,ystem 38. (d) Consider a sphere of radius z charge p· V, where p = charge density (cm-3 ) and V = volume (m3 ) . dz

2 2Rs s - � (z ) /� = � 3 X 5X E 3 X 5E

(_§_g_ ) XR 4nR3 2

5

3 x 5x E 3q2 3 q2 20 n E R 20 n E a hR

39. (d) When an excited atom at rest in

lab frame emits a photon of energy hf, the hhoton also carries a momentum p = _j__ Conservation of momentum C requires that atom also recoils with same momentum. So, energy carried by atom is 2

.!!___ Hence, in emission we can write, 2m LlE = hf + recoil energy of atom. So, emission energy of photon is smaller in case of recoil of atom. As E oc ..!:, emitted wavelength is larger. 'A So, 'A 2 > 'A1 and

!i < 1 . "-2

40. (d) Using, Q = 9315 (mx - my - ma ),

we get

Q = 5.40MeV This energy is distributed between a-particle and daughter nucleus. Energy is shared in inverse proportion of masses. So, ratio of kinetic energies is Ek (Th) _±_ = 228 Ek (a) So, Ek (Th) = 0.lO MeV and Ek (a) = 5.30MeV But if 22 8 Th is in an excited state, then Ek (a) is less.

4 1 . (a) A species is amphotcric, if it is

soluble in acid (behaves as a base) as well as in base (behaves as an acid). SnO2 + 4HC1 ---t SnC14 + 2H2O asic Acid B SnO 2+ 2NaOH ---t Na 2SnO3 + H2O Acid

Base

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315

KVPY Practice Set 3 Stream : SB/SX 42. (d) When salicylic acid is treated,

with zinc dust, bcnzoic acid is obtained as a product. OH Zn dusl

Q

o: acid

Salicylic

acid

�zoic

COOH

COOH

43 (d) For a zero order reaction, unit of

rate constant is mole L- i s- i . Hence, we can easily calculate concentration of B after 20 minutes by the following formula X = kt x = kt = 0.6 x 10-3 x 2 0 x 60= 0.72 M

44. (c) Given, weight of CuC1 2 = 13.44 g

Molecular weight of CuC1 2 = 134.4 g number of moles of solute Molalitv • = mass of solvent (kg) 1 3.44 ·: No. of moles of solute, CuC1 2 = 134.1 13.44 Molality = 134.1 = D.1 1 Now, CuC1 2 � Cu 2+ + 2c13 particles

"-----,------'

i=3 t.. Th = iKh · m = 3 X 0.52 x 0.1 = 0.1 56 � 0.16

45. (a) The reaction in which oxidation

and reduction occurs simultaneously arc termed as redox reaction or where there is change in oxidation number on both reactants and product sides. +4

+6

+1

0

(a) (Xe) F4 + 0 2 (F2) -- X eF6 + 0 2

Since, Xe undergoes oxidation while 0 undergoes reduction. So, it is an example of redox reaction. �

+5

(b) Xe F2 + PF5

--

+5

�

[XeFt + PF6

As the oxidation state of P is same on both sides. Thus, it is not a redox reaction. +6

+6

(c) X e F6 + HP - X eF4 Kot a redox reaction. +6

+6

(d) X eF6 + 2H2O - X eO 2F2 + 4HF Kot a redox reaction.

. = Z . = M. Z Density 46 • (a) D ensity 3 a NA x a3 In case of fcc, Z = 4, a = 3.5A 4 di = (3.5)3 ln case of bcc Z = 2, a = 3A 2 di = 3 (3) ⇒

3

i(2)3 = 1.26

di N1 a2 = ) = 2 3 .5 d2 N2 ( C½

47. (b) NO2 is a brown coloured gas and imparts this colour to concentrated HNO3 during long standing. 4HNO3 -- 2H2O + 2NO2 + 302 48. (a) According to spcctrochcmical series, lower the value of E 0 more easily, it can loose electron. Thus, the order of decreasing value of reduction potential of corresponding metals is Y (£0 = - 3.03 V) > Z(E 0 = - 118 V) > X (E 0 = 0.52 V). Thus, the order of reducing power of the corresponding metal is Y>Z>X 49. (a) For first order reaction i. _ 2.303 a __ log -,, _ ( a - x) t

O

D* � H

& H

f H D

-

kt2

-

2.303

= log a - log (a - x2 ) at time t2 ak_ _ (t2 - ti ) = log ( 2.303

Xi )

a - X2

Rate, 11 = k (a - Xi_ ) r2 = k (a - Xz ) (a - Xi_ ) .5.. 0.04 = ⇒ (a - x2 ) r2 0.03 3 k (20 - 10) = log 3 2.303 ;,, _ 2.303 4 _ 2.303 , ----in log ----in log 133 ,,

i

3-

-

=i

2.303x 0.125 = 0 _ 02878 min-I 10 0· 693 Also, for 1st order t112 = ---

k

0 · 693 T.50 2 4 1 min - 0.02878 50. (b) ln acidic medium, phenol exists in the following resonaning structures.

�v -�·(;6 - 6) l?--- 6 OH (

()-II

co-H +

0-H

H

�

Since, o/p-positions arc electron rich sites, so electrophile will attack on these sites, i.e. hydrogen of these sites get exchanged by D (deuterium) . Hence, the final product of the reaction is

I

O

H

.#

OD

6< �cY" Repeat at

the o _and

OD

D* II

p-pos11lon I I

1

.#

�

D

H

1 1 5 1 . (c) -12 (s) + - Cl2 (g) -- IC! (g) 2 2 1 fl t,.H = -t..Hs--; g +- t..Hdiss (Cl2 ) l2 2

1

+ 2 t..Hdiss (I2)

7

H r t.. 1c1 1 1 1 = ( -X 62.76 + - X 242. 3 + - X 151.0) 2 2 2 - 211.3 = 228. 03- 2 1 1 . 3 t..H=1 6.73 :::; 16.8kJ mol -i

52. (b) The compound 2-methyl butane on monochlorination gives following ISomcrs: CH3 - CH2 - H - CHa � 7 C Ha 2-methylbutane .r - CH -CH - CH c1 ci"3 2 2 I CHa (I) Cl + C Ha - CH2

·O ••

:OH

0 6C = C
JBc > 1Ac

33. Water is filled to a height H in a dam-with a stopping wall of width m.

28. Angular position0 of a body changes with time as shown below.

8

T 2 ---- 2 - ---� - ----= - -- -= - -- 4 - - �1 H

Now, choose the correct option. (a) Body is initially rotating anti-clockwise, then it starts rotating clockwise (b) Body is rotating anti-clockwise (c) Body is rotating clockwise (d) Cannot be said

29. A particle of mass length L.

m is confined to a narrow tube of

If L = 0.5 nm, then possible energies of electrons are (a) 1 .5 eV, 6 eV, 1 3.5 eV . . . . . . , etc (b) 2 eV, 4 eV, 6 eV . . . . . . , etc (c) 1 .5 eV, 3 eV, 6 eV . . . . . ., etc (d) 3 eV, 5 eV, 7 eV . . . . . ., etc

30. Angular momentum of earth with time varies as (•)

ll ,

(c) �

( ) b �,

(d ) I

t=-_,,

Force on face of dam is (take, g = 1 0 ms-2)

(a) ropgH2

(b) '!:_ ropgH2 (c) ]:_ pgroH2 (d)]:_ropgH2 3 2 3

34. Consider two rectangular wave pulses each of which

is moving towards the other with a speed of 1 mms- 1 . V

8 mm ------)1\'4 mm� V

Now, choose the correct option. (a) After 8s energy is 100% kinetic (b) After 1 6s energy is 1 00 % kinetic (c) After 6s energy is 1 00% kinetic (d)After 12s energy is 100% kinetic

35. A parallel plate capacitor has a plate area A = 200 cm 2

� I �:

and a plate separation of 2 mm.

1 mm

K- 2 mm ---11

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325

KVPY Practice Set 4 Stream : SB/SX A dielectric slab of dielectric constant k = 7 and width 1 mm fills the capacitor. Percentage of energy stored in dielectric is (a) 1 2.5% (b) 25% (c)50% (d) 100%

Wire loop in vertical plane

36. A positively charged ion is accelerated by using a

voltage V and then it is allowed to enter a region of perpendicular magnetic field. Frequency of rotation of ion is x. If accelerating voltage is increased four times, then frequency becomes (b)�

(a) 2x

(c)4x

(d)x

37. Stationary wave over a stretched string is

y (x, t) = 0.01 sin 62.Sx cos628t. If at mid point of string there is an anti-node, then (a) fundamental frequency = 1 0 Hz (b) fundamental frequency = 1 5 Hz (c)fundamental frequency = 20 Hz (d)fundamental frequency = 25 Hz

38. A mass m is connected to four ideal springs of same force constant k as shown below.

k

(b) 21t (m

V3k

41 . Which of the following molecules has the maximum dipole moment ? (b) CH4 (a) CO2

(c) NH 3

(d)NF3

42. Which one of the following does not undergo iodoform (b) CH3 - CH- CHa I OH

species is (a) I\'1\ > PC13 > BC13 (c) BC13 > PC13 > NHa

(b) BC13 > NH3 > PC13 (d)PC13 > BC13 > NHa

44. The complexes [Co(NH3 ) 6 ] [Cr(CN) 6 ] and

[Cr(NHa ) 6 ] [Co(CN) 6 ] are the examples of which type of isomerism? (a) Ionisation isomerism (b) Coordination isomerism (c) Geometrical isomerism (d) Linkage isomerism

(d) 2n {2m V5k

k

CHEMISTRY

43. Correct order of bond angles for the following

This system is placed on a horizontal frictionless surface. Time period of oscillation when mass is slightly displaced is

(c) 2n

Now, choose the correct option. (a) Time to slide down along .flP4 is maximum (b) Time to slide down along .flfli is minimum (c) Time to slide down along Pi P2 is more than time to slide along J=;_P3 (d) All arc incorrect

reaction? (a) C2H5 - CH- CHa I OH

k

(a) 2n R

Ground

39. Two particles are thrown from top of a tower from same point in opposite directions as shown below. 4 ms -1 � � 3 ms-1

45. Which of the following compound is aromatic? H H (,) Q �)

0

(c) o (d) 6

46. Which of the following represents physical adsorption?

Their velocities are perpendicular after a time interval of (a) ../3 s

(b) ../3 s 2

../3 s (c)-

(d) ../3 s 7

40. A bead is free to slide down a smooth wire tightly stretched between two points over a vertical wire loop (as shown below).

(a)

�1

l b_

Temp �

(,JL Temp �

1U Temp �

(d) l �

Temp �

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326

KVPVPractice Set 4 Stream : SB/SX

47. In the reversible reaction, 2NO2

55. Which of the following complexes has the least

N 20 4 the rate of disappearance of NO2 is

ki equal to 2 (a) ¾ [NO2]2 k2

(b)2hi_ [NO2]2 - 2hz[NzO 4 ]

(c)2 hi_ [NO z]2 - k2[K2O 4 ] (d) (21;; - k2)[NO 2] 48. XeF4 reacts with water at - 8 0° C to give (a) XeOF2

(b)Xe0F4

(c)Xe03

(d)Xe02 F2

49. Calculate the emf of the cell in which the following reaction takes place. Ni(s)+ 2 Ag + (0.002 M) - Ni2+ (0.160 M)+ 2Ag(s) Given that E;e ll = 1 .0 5 V (a) 0.61 V

(b) 0.91 V

(c) 0.82 V

(d) 0.0 23 V

50. The structure of diborane (B2 H6 ) contains

(a) four 2C - 2e- bonds and four 3C - 2e- bonds (b) two 2C - 2e- bonds and two 3C - 3e- bonds (c) two 2C - 2e- bonds and four 3C - 2e- bonds (d) four 2C - 2e- bonds and two 3C - 2e- bonds

51. Acetophenone when reacted with a base, C2 H5 ONa, yields a stable compound, which has the structure.

= CH - C --© (a) (§r ? CH3

�

CH3

�

wavelength of light in visible region? (a) [Co(CN)6 ] 3(b) [Co(Hp)6 ] 3+ (c) [Co(NH3 )6 ] 3 (d)[CoF6 ] 3 -

56. ClfsCH2OH can be converted into ClfsCHO by (a) catalytic hydrogenation (b) treatment with LiAIH4 (c) treatment with pyridinium chlorochromate (d) treatment with KMn04

57. Which of the following represents the correct order of decreasing number of S = 0 bonds? (a) H2 S 04 > H2S03 > H25i08 (b) H2Si08 > H2S 03 > H2S04 (c) H2 5i08 > H2 S 04 > H2 S 03 (d)H2 S 03 > H2 5i08 > H2 S 04

58. A unit cell with edge length a a. = � = y = 90° is called (a) cubic (c) orthorhombic

I

CaO + 2HCI - CaC12 + H2 O 1.23 g of CaO is reacted with excess of hydrochloric acid and 1.85 g of CaC12 is formed. What is the per cent yield? (a) 76.l (b) 86.3 (c) 95.1 (d)None of these

0

60. The major product in the given reaction is

OH

OH

CH3 CH3

(c)

©r,-fl§J

r H - fH --© (d) (§ ? OH

(b) tetragonal (d) hexagonal

59. For the reaction,

- CH2 - C --© (b) ©r ?

I

* b * c and axial angles

(i) Sn/HCI (ii) CH3l

r:0 CH r , (h) (a)

OH

52. If one strand of DNA has the sequence ATGCTTGA,

the sequence in the complementary strand would be (a) TCCGAACT (b) TACGTAGT (c)TACGAACT (d) TACGTAGT

53. Two Faraday of electricity is passed through a

solution of CuSO 4 . The mass of copper deposited at the cathode is (at. mass of Cu = 63.5 u) (b)63.5 g (a) 0 g (d) 127 g (c)2 g

54. Which of the following is a redox reaction?

(a) NaCl + KNO3 - NaNO2 + KC! (b) CaC2O4 + 2HC1 - CaCl2 + H2C2O 4 (c)Ca(OH)2 + 2NH4 Cl - CaC12 + 2NI--1:i + 2H 2 0 (d)2K[Ag(CN)2J + Zn - 2Ag + K2[ Zn(CN)4 J

BIOLOGY

61. Single step large mutation leading to speciation is also called (a) founder's effect (b) saltation (c) branching descent (d)natural selection

62. Which of the following is correct for

immunomodulators? (a) They always suppress immune system (b) They never suppress immune system (c) They always stimulate immune system (d) Specific immunomodulators stimulate the immunoresponse of immune system, whereas some other immunomodulators inhibit it

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327

KVPY Practice Set 4 Stream : SB/SX 63. A transplant between individuals of the same species but with different MHC/HLA alleles is (a) autograft (b)isogr aft (c)xenograft ( d) allograft

64. Which of the following is not used as a biopesticide? (a) Bacillus thuringiensis (b) Trichoderma harzianum (c)Nuclear Polyhedrosis Virus (NPV) (d)Xanthomonas compestris

65. Which is correct regarding genetically engineered insulin using E.coli? (a) Difficult to purify (b) Obtained in large unlimited quantities (c)Possibility of transmission of animal disease (d)Insulin obtained varies in chemical structure

66. If birth rate is 100, death rate is 10 and number of

individuals in population is 1000, then what will be the percentage of natural growth rate? (a) 0.09% (b) 9.0% (c) 0.9% (d) 90%

67. Which of the following is expected to have the highest value (gm/m 2/yr) in a grassland ecosystem? (a) Secondary Production (SP) (b) Tertiary Production (TP) (c) Gross Production (GP) (d)Net Production (NP)

68. Recent reports of acid rains in big industrial cities

are due to the effect of atmospheric pollution by (a)more release of N02 and S 02 by burning of fossil fuels (b) more release of CO2 by burning of coal/wood cutting of forests and increasing populations (c) excessive release of NHa by coal gas/ industries (d) excessive release of CO by incomplete combustion of carbonaceous fuels

69. Identify the bacterium that appears violet after Gram staining. (a) Salmonella enterica (b) Escherichia coli (c)Mycobacterium tuberculosis (d)Rhizobium meliloti

70. The nitrogenous metabolic waste in Hydra is mostly (a) ammonia and is removed from whole surface of body (b) urea and is removed mainly by tentacles (c)urea and is removed from whole surface of body (d)uric acid and is removed from whole surface of body

71 . Halophitic archaebacterium, e.g. Halobacterium salinarum found in great salt lake and dead sea cannot live in (a) less than 3 M NaCl concentration (b) less than 6 M NaCl concentration (c)more than 4 M NaCl concentration (d)more than 3 M NaCl concentration

72. In the sieve elements which one of the following is the most likely function of P-proteins? (a) Deposition of callose on sieve plates (b) Providing energy for active translocation (c) Autolytic enzymes (d) Scaling mechanism on wounding

73. Autonomic genome system is present in (a) ribosomes and Golgi bodies (b) Golgi bodies and mitochondria (c) mitochondria and chloroplast (d)chloroplasts and ribosomes

74. Which statement regarding coenzyme is incorrect? (a) Every coenzyme is a cofactor and every cofactor is a coenzyme (b) Every coenzyme is a cofactor but every cofactor is not a coenzyme (c) Most of the coenzymes are nucleotides and are composed of vitamins (d) Coenzymes are the active constituents of enyzmes

75. Most cytogenic activities occur during (a) interphase (c) prophase

(b) telophase (d)anaphase

76. Which of the following statements regarding cyclic flow of electrons during light reactions is false? (a) This process takes place in the stromal lamellae (b) ATP synthesis takes place (c) NADPH + Wis synthesised (d) Takes place only when light of wavelength beyond 680 nm is available for excitation

77. Enzyme enolase catalyses the conversion of 2 PGA to phosphoenol pyruvic acid in the presence of which of the following cofactors? (c)Mg2+ (d)Zn 2+ (a) Mn2+ (b) Fe2+

78. In human beings, the three pairs of salivary glands and numerous buccal glands produce about (a) 1 .0 dm3 of saliva per day (b) 1.5 dm3 of saliva per day (c) 2.0 dm3 of saliva per day (d) 2.5 dm3 of saliva per day

79. Coronary heart disease is due to

(a) Streptococci bacteria (b) inflammation of pericardium (c) weakening of the heart valves (d)insufficient blood supply to the heart muscles

80. At their resting stage, the body cells exhibit a potential of -5 to -100 mV known as (a) polarisation (b) resting potential (c) repolarisation (d)depolarisation

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� PART- I I

(2 Marks Questions)

MATHEMATICS

81 . The tangent to the curve y = e" drawn at the point

(c, e ) intersects the line j oining the points (c- 1, e - l ) and (c + 1 , ec + 1 ) is c

c

(a) on the left of x = c (c)at not point

(b)on the right of c (d) at all points

(x- 1)" · 0 < x < 2 m and n are log cosm (x- 1) ' integers, m -ct 0, n > O, and let P be the left hand derivatives of Ix- 11 at x = 1. If lim g(x) = P, then

82. Let g(x)

(a) n = L m = 1 (c)n = 2, m = 2

has

(d)2

85. Let w be a complex cube root of unity with w -ct 1. A

fair die is thrown three times. If r1 , ,2 and r3 are the numbers obtained on the die, then the probability w'l. + w'2 + w"3 = 0 is 2 (c)(a)_!_ (b)_!_ (d)__!_ 18 9 9 36

86. Let P, Q, R and S be the points on the plane with position vectors- 2i- j, 4i, 3i + 3j and- 3i + 2j respectively. The quadrilateral PQRS must be (a) parallelogram, which is neither a rhombus nor a rectangle (b) square (c)rectangle but not a square (d)rhombus but not a square

87. Let (x0 , y0 ) be the solution of the following equations (b)_!_ 3

(c)_!_ 2

88. Let f : (0, 1) ➔ R be defined by / (x) = constant such that 0 < b < 1. Then, (a) f is not invertible in (0, 1 ) l (b) f ctc 1 on (0, 1 ) and f ' (b) = __ f (O)

r

(b) 64

1

(c) 90

(d)91

(b) 5x4 (d)1 + (g(x))°

PHYSICS

84. The number of 3 x 3 matrices A whose entries are

(a) I_ 6

n

value of x for which J [x]{x} dx exceeds 2020 is

l+x

lines (1 + p) x - py + p(l + p) =0, (1 + q) x - qy + q(l + q) = 0 and y = 0 where p -ct q 1s (a)a hyperbola (b)a parabola (c)an ellipse (d) a straight line

(2x)'n 2 = (3 y)ln 3 , 3 lnx = 2 'n y, then Xo is

89. For a real number x let [x] denote the largest integer less or equal to x and {x} = x- [x]. The smallest integer

then g' (x) is equal to (a) 1 + x5 1 (c) 1 + (g(x))5

83. The locus of orthocentre of the triangle formed by

exactly two distinct solutions, is (c) 1 68 (a) 0 (b)29 - 1

r1 is differentiable in (0, 1 )

(d)

90. If g is the inverse of a function f and /' (x) = �,

(b)n = L m =- 1 (d) n > 2, m = n

H� ]

l_ r1 on (0, 1) andf'(b) = _ f'(0)

(a) 63

X � 1+

ciLhcr O m 1 =d foe which Lhc sysLem A [ �

(c) f =

(d) 6

b- x , where b is 1- bx

91. There is a hole in centre of a thin circular biconvex lens of focal length 4 cm. Diameter of hole is half of diameter of the lens. A point source of light is placed 9 cm from the wall and lens is placed in between, such that a single circular illuminated spot with sharp bright edges is formed on the wall. Distance of lens and source must be (a) 1 cm (b) 3 cm (c) 5 cm (d)4 cm

92. A tumor on a person's leg has a mass of 3g. What is

the minimum activity a radiation source can have, if it is to deliver a dose of 10 Gy (Gray) to the tumor in 14 min. Assume each disintegration provides an energy of 0. 7 MeV to the tumor. (a) 3. 7 x Hf Bq (b) 3.7 x 1 0 8 Bq (c) 3.7 Bq (d)3.7 x 1 01° Bq 93. An electron typically spends about 10-8 s in an excited state before it drops to a lower state by emitting a photon. Number of revolutions made by an electron in n = 2, Bohr's orbit in 1.00 x 10-8 s is (a) about 1 0 6 (b) about 1 01 4 (d)about 1 02

(c) about 1 0 15

94. A system undergoes a cycle consiting of three processes.

p

-+------� V

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329

KVPY Practice Set 4 Stream : SB/SX Process involved are listed as, Process l➔ 2 2➔ 3 3➔ 4

6.Q (kJ)

a b

100

6.U (kJ) 100

100 -50 d

6.W (kJ)

Values of a, b, c and d, respectively are (a) 200 kJ, 50 kJ, 100 kJ, 3 00 kJ (b) 200 kJ, 50 kJ, 300 kJ, 100 kJ (c) 50 kJ, 100 kJ, 200 kJ, 3 00 kJ (d) 300 kJ, 200 kJ, 200 kJ, 50 kJ

95. Position of a particle is s = t3

- 6t

2

-200 C

+ 9t m. Distance

travelled by the particle in first 5 s is (a) 20 m (b) 24 m (c) 28 m (d) 23 m

96. Consider a partially filled capacitor.

Half of the volume is filled with a dielectric of succeptibility XE = 3. Percentage of total energy stored in the dielectric is (a) 100% (b) 60% (c) 20% (d) 50%

97. Let R = radius of a spherical refracting surface and l = least distance between conjugate focii. Then, refractive index of medium of refracting surface is (a)

R+ l R-l

2(R - l) (c) R+ l

(b)

(R + / )

2

l )2

CHEMISTRY (i) CF3COOH (ii) NaOH

Find the organic acid produced from the above reaction. H3CyYOH

(a)

it is easier to lift block at 0° C it is easier to lift block at room temperature it is easier to lift block at 4° C lifting block is equally difficult at every temperature

99. For an ideal gas, let CP = a + cT and Cv = b + cT. where, a, b and c are constants. T - V relation that holds in adiabatic expansion is cT b (a) v a - b . T . e = constant a - b . vb . ecT = constant (b) T cT (c) Ta - b . v a . e = constant cT a . e T va = constant b (d)

1 00. A driver is caught crossing a red light. The driver

claims to the judge the colour she actually saw was green (/0 = 5.6 x 1 014 Hz) and not red

LJ

(d) None of the above

1 02. In a decay series, ��

8

i�6 Pb is obtained at the end from

U. How many alpha and beta particles must have

been emitted? (a) 7a , 1013 (b) Sa, 613

(c) 5a , 613

(d) 6a, 813

1 03. In the given reaction identify X and Y.

R(d) 2 ( R+ l R-l

98. A heavy block is immersed in water, then (a) (b) (c) (d)

per hour she exceeded the speed limit of 100 km/h. Find charged will be (a) < 164, 999,999 (b) < 165, 000, 000 (c) � 1 74, 999, 900 (d) < 164, 999, 900

�

(a)

(c)

W N,OWW

X

CHC>o • •o N,OH

OH �

CHO OH

C � OOH

y

OH

(b)

(d)

C OOH ©r OH ©(CHO

1 04. The variation of concentration of the prodcut P with time in the reaction, A ➔ P is shown in following graph.

( fR = 4.80 x 1014 Hz) because of the doppler effect.

The judge accepts this explaination and instead fines her for speeding at the rate of 1 < for each kilometer

Temp

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330

KVPVPractice Set 4 Stream : SB/SX

The graph between type

- d [A] and time will be of the dt

- d[A] � (a) dt l__

Time (C) d( - d[A] L

- d[A] I \ (d) di

Species 1 ·c1 Cf) (l) (l)

H2 0(l) and glucose(s) at 25° C are - 400 kJ/mol, - 300 kJ/mol and - 1300 kJ/mol, respectively. The standard enthalpy of combustion per gram of glucose at 25° C is (a) + 2900 k J (b) - 2900 k J (c) - 1 6. l l kJ (d) + l6 . l l kJ

107. Chromium metal crystallises with a body centred

cubic lattice. The length of the unit edge is found to be 287 pm. Calculate the atomic radius . What would be the density of chromium in g/cm3 ? (a) 7.3 (b)8.1 (c)5.1 (d) 1 1 .3

108. The hybridisation and magnetic moment of complex (b)d 2sp3 , 4.89 BM (d) sp3 d 2, 3.39 BM

109. Calculate the standard free energy change for the

formation of methane at 298 K. The value of 1:i,H 0 for CH4 (g) is -7 4.81 kJ mol- 1 and S values for

C(graphite) H2 (g) and CH4 (g) are 5.70, 130.7 and 186.3 JK- 1 mol- 1 respectively. (a) - 80.8 (b) - 98.91 (c) - 50.74 (d) - 40.4

110. Compound(s) that on hydrogenation produce(s) optically inactive compound(s) is/are

(c)

Br

H3C � ' --->

CA C B = 0

⇒

AB - AC = I AB I I AC l cosA

⇒

B C · BA = I BC I I BAI cosB

⇒ AB · AC+ B C · BA+ CA · C B ⇒ I AB I I ACI cosA + I B C I I BA! cosB + O ⇒ I A B I ( I AC I cosA + I BC l cosB) ⇒ 7

f(x) · f(l) = f(x) + f(l) + f(x) - 2

f(0) = f(l) = 0

By Rolle's theorem

2

=

Put x = y = 1, we get

+

C

=

f(x) · f(y) = f(x) + f(y) + f(xy) - 2

n

L.C = 90°

=

2. (b) We have,

+

n+l

⇒ x = � onlv " one solution 2

5. (d) We have,

=e

9. (b) ABC is right angle triangle AB is hypotenuse and AB = p

0 � [x] - _1_ < l ⇒ [x] = 1, {x} = _1_ 2 2

Also, AG : GD = 2 : 1

e - nln - i

2n

[x] - _1_ = {x} 2

⇒

=

In + nln - i

8. (c) Consider the function n_ xn a�rn - l xn + l _ ao__ -1 _ _ _ 'L:' · _ _

j

[xj - {x} = _1_ 2

3

In

110 + 1019 = e

+-+ 1 2

4. (a) We have, 5{x} = x + [x]

G

G = [-

-=" % - 1-)j

n2 11 n 2 r-=== � (

Total number of function from A to B = 5 Total number of onto function from A to B lS 7! 1 , 7! n (E) - 5.I + - X - X 5 I. 3!2! 2!2! 3!4! 7! X 20 6 7! X 2 0 -7' x 2 . :. Rcqmrc d prob ab·1· 1 1ty = --6-- = -3 X 56 5'

IAB I I AB I = IABl 2 = p 2

[·: p = bcosB + ccosA] 1 0. (c) Tangents at Pi_ and P2 arc

mutually perpendiculars [1 [2 = - 1

=

7. (c) We have,

SP1 = .j(a - at12 ) 2 + (2at1 ) 2

SP1 = a + atf Similarly, SP2 a + at; =

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333

KVPY Practice Set 4 Stream : SB/SX 1=

2 1r (1 + 4x)312 7 x+ 2l e

l = ½ [( 2

1 1 . (c) We have,

⇒

⇒ ⇒

⇒

sec (a + 2)x + a2 - 1 = 0 tan 2 (a + 2)x + a 2 = 0 tan(a + 2)x = 0 and a = 0 tan(2x) = 0 and a = 0 2x = nn and a = 0 2

x=

nn

and a =

X E (- 71, 7!) 7l 7l x = 0, , 2 2

0

Number of ordered pair of (a, x) is 3

i.e.(0, 0) ( 0,

%)

and ( 0,

- %}

1 2. (b) ln MBC, L'.A = 3 0

BC= 2 + -J5 A °

a �- - �-�c a=2+-fs

2 + -J5 a_ = _ 2R ⇒ sin A sin 3 0 R

=

2+

-J5 2 x ..!:

°

=

=

2 + -J5

2R

Now, AH = 2RcosA AH = 2(2 + --/5) cos30° ⇒

AH = 2(2 + --/5)

⇒

2

AH = -J§ (2 + --/5)

1 3. (a) We have,

Let

./3 = (2 + --/5) (-./3)

I = f ✓x +

y = .Jx +

✓x + � oo dx

✓ x + -Jx + . . . y = -Jx + y ⇒ l - y - x = 0 l ± j4x+ i 2 l + -J4x + l y = -�-- y

x < 0 and x + 1 >

X E (- 1, 0) 0

AC = 1 - y

BC = l AB 2 = AC2 + BC2

+ 267 ) - (o+ ½)]

.!: 1 2 + 2 7 - 1 J=. ( )

2

AB = y + 1

Jo

19

(y + 1) 2 = (1- y) 2 + (1) 2 2 (1 + y) - (1 - y) 2 = 1

6

0

- 2x = (x + 1) 2

4y = l ⇒ y

x> - 1

Hence, radius of circle T x2

x + 4x + 1 = 0 -4± 2 x= -./3 = - 2 ± -./3

So, x = - 2 + -./3 only one solution lies in (- 1, 0).

1 5 . (a) All AAAAABBBDEEF can be

. - 12! - ways. arranged m 5! 3!2! Between the gaps C can be arranged 13 G,i ways 12 1· Total ways = 13 G,i x --5! 3! 2! -./3 secx + cosecx + 2(tan x - cotx) =

⇒

⇒

⇒ ⇒ ⇒

⇒

sin x _ cosx = 1_ -./3 + _ + 2( ) 0 cosx s 1n x cosx s1nx

.J3

0

sin x + cosx + 2 (sin 2 x - cos2 x) = -./3 sinx + cosx = 2cos2x cos ( x

- %)

cos2x

=

=

0

cos2x

cos(x - 1 )

2x = 2 n11

±(

x - %)

x = (6n - 1) _1t_ or (6n + 1) _1t_ 9 3

11 11 71t 5 11 . (- 11 rr) x = - - , - , - and - - m 3 9 9 9 7 rr 571 = :. Sum = -=--_7t_ + _1t_ + O 9 9 9 3 ⇒

1 7 . (d) In MBC

A

(1 , 1 ) C ( 1 ,y)

=

..!:

1 8. (c) Equation of ellipse,

2

1 6. (c) We have,

1 4

=

+ al = 6

Equation of tangent of ellipse with slope m is

✓6m2 + 2

y = mx ±

h m=-k

and tangent passes through (h, k) ⇒

(-!!_ )

k=h

✓6h2

±

+ 2

2

✓6h2 + 2k2 k

2 2 k + h =±

k

2 2 ⇒ (h 2 + k2)2 = 6h + 2k 2 2 2 So, locus is (x + y ) = 6x2 + 2y2.

1 9. (c) We have,

7 2 (10) 9 k = 1 09 + 2(11)(1 0) 8 + 3(1 1) (1 0)

⇒ k = l + 2( �� ) + 3 ⇒

(��r

+ ... + 10(11) 9

+ ... + 10 G� r

_!2 k 11 ,l l l 9 11 = + ) + ... ( ) 10 1 0 l. 10 10 = l+

2

+(

9

ll io k ) 10 10 l°

11 11 11 + ( ) + . . . . ( ) - (..!:2 ) 1 0 10 10 10 10

:. k = 1 00

k =

10

20. (b) We have,

log( a x _ 1/x - 2) = log

10

9

2

(

11

10

)

10

-1

11 _ l 10

2 (9x - 6x + I)

-(

1 11 0 ) 10 10

(2x2 - lOx- 2)

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334

KVPVPractice Set 4 Stream : SB/SX

⇒ log(3 x

- l)

(x - 2) 2

log (2x - lOx - 2) ( 3 x -1 )2 ⇒ 2log(3 x l) (x - 2) = log (2x2 - lOx - 2) (a r -l) ⇒ log(3 x - 1 ) (x - 2) 2 = log (2x2 - lOx - 2) (3 x - l ) ⇒ (x - 2) 2 = 2x2 - lOx - 2 ⇒ ⇒ ⇒

=

x2 - 4x + 4 = 2x2 - l0x - 2

Here two tails and one head can occur in three different ways. So, entropy, S = kB · In 3

24. (c) Position of particle is

4

2 1 . (c) Cone with angle a and source S is

2

A

�A

=

10 log(

�B

=

lO log(t )

�A - �B

=

1 0 ( 1og

s

E

B

D

=

1 0 log(

C ⇒

!

3 1 . (a) E =

' , l's, From above construction, image of S is S'. Now, for rays to reflect only once point S' must be below line OB and OCC'. 3a So, LS'OB + ,t,. £L + _!)_J._ ⇒ P 2 > Pi + P� 2m 2m 2m 8 < 90° �7. (d) Ray diagram for given situation 16 8

I

a(xi + yj + zk) x2 + y 2 + z 2

K

=

=

_ci_ _ r

r

K

top

slope ofB - t line

which is positive in given case so, body is rotating anti-clockwise.

29. (a) de-Broglie waves will resonate

with a node at each end of tube.

+

botton

2P.

1 X Area = 2 pgcoH2

completely.

: _ _ _ _ _ _ _ _: So, there is no potential energy as Ynet = 0. Hence, energy is 100% kinetic after 6 s. Uo 35. (a) Uo = p uga + UD UTotal UD X (Vol)o

E 2 .!_ E 0 -'--'ga Ti' pA(d - I) + � E 0 !P A 2 2 K

l/K (d - I) + I / K

Note That this is real and erect. dt

ar ,. 2

P. ) 34. (c) After 6 s, both pulses overlapps

First image A' B' is formed by lens. image of A" B" is formed in plane mirror as A" B". image of A" B" is formed by lens as A 1 1 1 B1 1 1 • dB

=

33. (c) Force = Pavg X Area

� A � I--+- �---++--+ �] B' B"

28. (b) As w =

1, 2, 3, . . . , etc

side is maximum because effective distance of mass distribution about this side is maximum.

=(

P; = P1 + P2 For a general collision, some energy is lost 2

=

32. (d) Moment of inertia about shortest

�A - �B = 10 log 50 = 1 0 x 1.70 = 17 dB

2

n 2h 2 8L m

--, n 2

⇒ qi = f dqi = 41taR :. Qenclosed = Eo 'P = 4 Eo aR 7t

- log

t) Io

=

dqi = E ds = ( i · r ) (ds - r)

J

26. (d) By conservation of momentum,

C' :

Ka. = (

IA

2

.!!__ and (KE),, = P 2m Pn

dt

25. (d) Sound levels of A and B are

Now, we draw reflected rays as shown below.

"'- n

remains constant with time. So, correct option is (d). dL 1.,, = 0 = - ⇒ L = constant

x2 y2 -+-=1 9 16 This is an ellipse.

t

As,

30. (d) Angular momentum of earth

2

So, we have � + L = cos2 wt + sin 2 wt 16 9 or

s

n

=

With m = 9.1 x 10-31 and L = 5 x 1 0-10 m (KE) n = 2.4 X 1 0- 19 n2J = 15 n 2 eV :. KE = 1.5 eV, 6 eV, 13.5 eV, . . . . . . , etc

r = 3 cos wti + 4 sin w t j

3

= 3 + ✓15, X > 2

n L = "'-1 , ;,,,2, � ¾ · .. A. " , etc 2 2 2 2L So, A. 11 = - , n = l, 2, 3- ..

We have, (KE) n

⇒ :: = cos wt and 1'. = sin wt

x = 3 ± ✓15

as shown below.

ways the state can occur.

So, x = 3cos wt, y = 4sincot

x2 - 6x - 6 = 0 X

23. (d) We use Boltzmann relation S = kB · In n, where Q is the number of

-

l (d - l) K + l

1 -1 - lx 7 + 1 8 -

:. Per cent.age of energy stored =

� X 100 = 12.5 %

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335

KVPY Practice Set 4 Stream : SB/SX 2rrm and it is qB independent of accelerating voltage or velocity.

36. (d) Frequency, f =

37. (c) From given equation, W = 2rr/ = 628 628 ⇒ f = - = lOO Hz 2rr This may be third, fifth or seventh harmonic as there is an anti-node at mid-point. So, fundamental frequency may be 100 100 100 -, -, . . . , etc. -, 7 5 3

38. (b) For displacement x,

41 . (c) CO2 and CH4 have zero dipole

moment as these are symmetrical in nature. Between NH 3 and NF3 , NH 3 has greater dipole moment, though in NH3 and NF3 both, K possesses one lone pair of electrons. This is because in case ofNH3 , the net N-H bond dipole is in the same direction as the direction of dipole of lone pair, but in case ofNF3 , the direction of net bond dipole moment of three -N-F bonds is opposite to that of the dipole moment of the lone pair. o � c ++ o ( a) µnet = 0

Thus, the correct order of bond angle is BC13 > NJ-I:i > PC13 .

44. (b) The complexes

[Co(NI--1:i ) 6 ] [Cr(CN) 6 J and [Cr(NJ-I:i ) 6 ] [Co(CN) 6 ] arc the examples of coordination isomerisms. This isomerism occurs only in those complexes in which both cation and anion are complex. l t occurs due to exchange o f ligands between cation and anion.

45. (d) The compounds which have all sp 2-hybridised carbon atoms and follows Huckel' s rule l (4n + 2) 7t electrons] are aromatic compounds.

Force on mass m is

3k

Fnet = ma = - (kx + kx + 2k ( � )

⇒

(a)

v; = V2 + g t2

v; l. v; = v; · v; = 0 As, v1 and v2 are in opposite directions V1 · V2 - V 1 V2 (v; + gt; ) - (v; + g t2 ) = 0 =

- VI V2 + g 2t 2 = 0 .ju1 U2 ✓ 4 X 3 .js

(d )

J't

(c)

µ� N l�F F

F µ2

µ net

= 0.2 D

42. (c) Iodoform reaction is given by only

I

those compounds that have either

CI--1:i - C = 0 or CI--1:i C H - OH group. I (a) C2H,; - ? H- CH3

⇒ t = -- = --- = 10 5 g

OH

40. (d) Acceleration of the bead down the wire is g cos e.

(s-butyl alcohol gives iodoform reaction)

(b) c1--'1-'3 - CH - CH3 I OH

(lsopropyl alchohol

(c)

gives iodoform reaction)

C21-I;; - T - C21-I;; 0

Diethyl ketone (does not give iodoform rear.tion)

Also, length P1 P2 is 2R cos 0. So, v2 = u 2 + 2as Gives, u2 = 2g cos 8 x 2R cos 0 2-fiR cos e g co s 0

= 2-.[iiR cos8

Time,

t=�= a

(d)

fgg

{E

g

2fg

0

Cycloheptatriene (6ne- but all

0 §

Cyclooctatetraene

(8ne-) (non-aromatic)

6 H

(d)

carbons are not

Cycloheptatrienyl cation (6rre- and all the carbons are

sp2 hybridised)

sp2 ·hybridised (aromatic)

(non-aromatic)

46. (d) The physical adsorption isobar at constant pressure shows a decrease in x Im with rise in temperature. Thus, the correct plot is given in option (d). k, 47. (b) 2N0 2 �Np 4 hi. d[N0 2 ] Rate of reaction = - _1_ 2 dt = ki (N0 2 J2 - ki (Nz0 4 ] :. Rate of disappearance of NO 2 . -d [ N0 2 ] = 2 k1 [N0 2 ]2 - 2k2 [N p 4 ] i.e. dt

48. (a) Xenon tetrafluoride reacts with

OH

Which is same regardless of where P2 is located. Time of free fall, 1 4R , ⇒ t = (R 2R = gt 2 ⇒ t 2 =

2

H I CI--1:i � H

Ethyl alcohol

2

(b)

(4ne-) (non-aromatic)

9 t µ4

T = 2n &

39. (c) Velocities of the particle after time t seconds is v; = V1 + g t1

⇒

Cyclopentadiene cation

a = --·X m

So,

0 H

(gives iodoform reaction}

43. (b) BCl 3 is trigonal in shape where °

bond angle is 120 . In NH3 and PCl 3 (both are having pyramidal geometry), though the central atom has equal number of lone pairs and bond pairs, the valence shell of P is relatively bigger, so electrons suffers lesser repulsion and thus bond angle is lower.

water at - 80 ° C to give xenon oxyfluoride and hydrogen fluoride.

° -80 C XeF4 + Hp XeOF2 + 2HF Xenon oxyfluoride

49. (b) From the given cell reaction and Kernst equation, o 0.0591 [Ni2+ ] log Ecell = Ecell _ n [Ag + J2

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KVPVPractice Set 4 Stream : SB/SX =

=

= =

0591 [0.160] 105V _ Q log 2 [0.002]2 105 -

o.o 59l log ( 4 x 104 )

1< 4.6021)

2 0 · 059 1 .05 2

1.05 - 0.14= 0.91V

Ecell = 0.91V

50. (d) The structure of diborane (B )1 6 )

has four 2C - 2e- bonds and two 3C- 2e­ bonds.

Bridging bonds C le-l � H � � . 1 o / 0,,. Terminal Term1na . _ uonds •B• • j bond f B (2C- 2e ) • o • lC-le0 H o H � Bridging bonds • Boron electron (JC- le-) 0 Hydrogen electron

5 1 . (a) Aldchydcs or kctoncs with o:-H-atoms, in presence of dilute base, undergoes aldol condensation to give 13 -hydroxy aldehyde or 13 -hydroxy ketone. On heating, aldols eliminate water molecule to form a, 13 -unsaturated compound.

Acetophenone

Cu 2+ + 2 e- ---t lmol 2mol

W

52. (c) In DNA, adenine (A) and thymine (7) arc joined by two hydrogen bonds while guanine (G) and cytosine (C) are joined by three hydrogen bonds. No other combination of four bases is possible. Hence, the sequence of complementary strand is TACGAACT.

53.(b) According to Faraday's second law

Given, Q = 2F Atomic mass of Cu = 63.5u Valency of the metal Z = 2 We have, CuSO4 - Cu2+ + SO�-

=

ZQ

=

Cu

lmol - 63.5g

E'.. . 2F = 2E = 2 x 63· 5 = 63 .5 .g

Alternatively.

F

2

+l -1

+l - 1

+1

-1

54. (d) (a) Na Cl + K NO:i ---t Na N03 +2 +2 -1 + l -1 -2 (b) Ca Cp 4 + 2 H C1 � Ca Cl 4 -1

+2

+3 + l

+1 -1

+ K C!

+1

+ H 2 C2 -O 4

-1

+2 -1

-3 +I

The oxidation process can be stopped at the aldehyde stage, if Cr(Vl) reagents such as Collin's reagent (CrO3 · 2C5 H,; N), Corey's reagent or pyridinium chlorochromate are used PCC and pyridinium dichromatc l(Ci,H,; NH)�+ crpt] in anhydrous medium arc used as the oxidising agent. So, the correct option is (c).

PCC CH3 CH2OH- CHa CHO

2

(c) Ca(O H) 2 + 2 N H 4 Cl � CaC1 2 +1

Ethanol

H2S2O 8 , H2SO 4 , H2SO 2 0 OH ,f- 0 � S ,f' S =O

-2

O-f' I " o- o/ I OH OH 4S =O bonds H,S,O 8

ln all these cases during reaction, there is no change in oxidation state of ion or molecule or constituent atom, thus these arc simply ionic reactions. (d) 2K[Ag(CN) 2 ] + Zn � 2Ag Ag

+

+ K2 [Zn(CN) 4 ] � Ag gain of e-, reduction

Zn � Zn2+ loss of e-, oxidation. As both oxidation and reduction takes place simultaneouslv. Thus, it is a redox reaction.

55. (a) As we know that, strong field

ligand split the five degenerate energy levels with more energy separation than weak field ligand, i.e. as strength of ligand increases crystal field splitting energy increases. Also, t,, = hc E 'A,

⇒

Ethanal

57. (c) The structures of are as follows:

+ 2 N lfa + 2 H 2 O

⇒

This is an example of self aldol condensation.

2F

2S =O bonds IS =O bonds Thus, the correct order of decreasing number of S = 0 bonds is H2 S2 08 > H2 SO 4 > H2S03

58. (c) A unit cell with edge length a ctc b ctc c and axial angles a: = 13 = y = 90° is called orthorhombic, e.g. rhombic sulphur, KNO3 and BaSO 4 .

t,,E oe -!. "-

A = _!_

t:i.E As energy separation increases, the wavelength decreases. Thus, the correct order of increasing CFSE and decreasing wavelength is 3+ 3+ [CoF6 ] 3 - > [Co(H2O) 6 ] > [Co(Nlfa ) 6 ] > [Co(CN) 6 ] 3 Here, strength of ligand increases, t:i.E increases, CFSE increases and 'A, absorcd decreases. Hence, correct option is (a).

56. (c) Ethanal (CHa CHO) is an oxidised product of cthanol.Pyridinium + - r0 ), PCC chlorochromatc (Ci, H,; N HCl C 3

oxidises primary alcohols to aldehydes. Strong oxidising agents such as KMnO 4 are used for getting carboxylic acid from alcohols.

Orthorhombic

Hp

59. (a) The balanced equation is CaO + 2HCl ---t CaC1 2 + 1 ruol

56 g

1 mol

111 g

56 g of CaO produces CaC12 = 1 1 1 g 1.23 g of CaO will produce

111

CaC1 2 = - X 1 . 2 3 56 = 2.43 g Thus, theoretical yield = 2.43 g Actual yield = 1.85 g 1.85 Per cent yield = x 100 = 76. 1 2.43

60. (b) When nitrobenzene reacts with Sn/HCl, it undergoes reduction to form aniline ,which on reaction with methyl iodide gives N, !\'.dimethyl aniline.

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KVPY Practice Set 4 Stream : SB/SX

0 8 Benzene

(i) Sn/HCl

Aniline

(CH31) 1(2 mol)

c5"'

N. N-dimethyl aniline

6 1 . (b) Single step large mutation

leading to speciation is called saltation . It is the occurrence of a major mutation in single generation, bringing about significant change.

62. (d) lmmunomodulators are natural

or synthetic substances that help to regulate or normalise the immune system. There arc two specific types of immunomodulators viz immuno suppressants, which cause immuno suppression or inhibit the immune system and immunostimulants (immunostimulators) which stimulate immune system by increasing activity of its components.

63. (d) Allograft is transplant between

the individuals of same species, but with different genetical background. lt is generally used as a temporary measure until the damaged tissue is able to repair itself.

64. (d) The bacterium Xanthomonas

compestris is not used as biopesticide. It is the causative agent of a plant disease called black rot of cabbage. Rest all are biopcsticidcs. 65. (b) The only correct option about genetically engineered insulin using E. coli is that it can be obtained in huge amounts and unlimited quantities. The other options given arc not appropriate for recombinant insulin as it is not difficult to purify. It docs not allow transmission of animal diseases and this insulin obtained docs not vary in chemical structure.

66. (b) Birth rate = 100 Death rate = 10 Number of individuals in a population = 1000 Natural growth rate = 100 - 10 = 90 So, percentage of growth rate = � X 100 1000 = 9.0%

67. (c) The rate of total capture of energy or the rate of total production of organic material is Gross Primary Productivity (GPP), while the balance or biomass remaining after meeting the cost of respiration of producers is Net Primary Productivity (NPP). Hence, gross productivity has highest value in grassland ecosystem. 68. (a) More release ofNO2 and SO2 cause acid rain. The acid rain is a mixture of H2SO4 and HNO3 • The SO2 and NO2 produced during the combination of coal and petroleum reacts with water vapour and formed H2SO4 and HNO3 , respectively. These acids combine with air droplets/water vapour causing acid rain. 69. (c) Mycobacterium tuberculosis is a

Gram positive bacterium that appears violet after Gram staining. This is because it retains the colour crystal violet dye due to the presence of a thicker pcptidoglycan layer. 70. (a) In Hydra, the exchange of oxygen and carbon dioxide and the excretion of waste nitrogenous matter (chiefly ammonia) occur directly by diffusion through cell membrane to outside.

71 . (a) Archaebactcrium cannot live in

less than 3 M NaCl concentration as their survival requires very high salt concentrations.

72. (d) Sieve clements arc a component of phloem tissue and arc responsible for conduction of food in plants. A sieve tube element has peripheral layer of cytoplasm without any nucleus. The central part is occupied by a network of canals which contain fibrils of P-protein. The protein is vibratitc and is believed to actively participate in the transport of nutrients. One general property of it is its ability to form a gel and it functions as a puncture repair substance forming a plug at any site of damage in sieve element, thus preventing loss of food materials being translocated by the phloem. Thus, it is believed to have sealing function on wounding. 73. (c) Autonomic genome system is present in mitochondria and chloroplast. These are the autonomous bodies. ln these, small circular DNA particles arc present which can duplicate and express themselves.

75. (a) lnterphase is the period between the end of one cell division to the bcgining of the next cell division. It is the phase where most of the cytogenetic activities take place. The cell becomes metabolically very active and prepares itself for both cell growth and DNA replication in an orderly manner. 76. (c) Cyclic flow of light reaction is the

process of photosynthesis in which light of wavelength more than 680 nm is required. It takes place in the stromal lamella of the chloroplast. During cyclic photophosphorylation only the synthesis of ATP takes place (i.e. not NADPH + Ir) because the stroma lamcllac do not possess enzyme NADP reductase due to which the excited electrons in the cyclic flow do not pass on to !\ADP instead get back to the PS-I complex only.

77. (c) Conversion of 2 PGA to PEP is the 8th step of glycolysis. This conversion is catalysed by the enzyme cnolasc and takes place in the presence of cofactor Mg2+ and R+ with the loss of a water molecule. 78. (b) The saliva is a colourless,

opalasccnt and sticky liquid produced by the salivary glands. It contains salivary amylase for digestion of carbohydrate. In quantity it is about 1.5 dm3 (cubic decimeter), i.e 1.5 L/day.

79. (d) Coronary heart disease occurs due to the insufficient blood supply of blood to the heart muscle. lt occurs due to the hardening of arteries and arteriole because of thickening of fibre tissue and the consequent loss of elasticity. It is also referred to as atherosclerosis. 80. (b) Resting potential is the difference

in electrical potential that exists across the membrane of nerve cells. The resting potential is maintained with the help of sodium - potassium pump.

8 1 . (a) Slope of line joining the points

(c - 1, ec - l ) and (c + 1, ec + 1 ) is equal to e c + 1 - ec - 1 > e" 2

74. (a) Option (a) is incorrect as there

arc three types of cofactors, i.e. prosthetic group, cocnzymcs and mctalions. Thus, every cocnzymc is a cofactor but every cofactor is not a coenzyme.

⇒ tangent to the curve y = ex will intersect the given line to the left of the line x = c.

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338

KVPVPractice Set 4 Stream : SB/SX

82. (c) Given, g(x) =

(x - l)" log cosm (x - 1)

and - !!___ = - 1 ⇒ n m

=

n

S R = 6i + j PS = - i + 3j

⇒

PQ I I BR and QR I I PS

X➔ 1 +

lim g(l + h) = - 1 h--, o . hn ⇒ hm - -=-1 h--, o log cos" h nh n - 1 ⇒ lim -=-1 h--, o m (- tan h) -n hn - 1 = ⇒ lim - -1 m h --, o tan h Which holds of n - l = l ⇒ n

J [x] {x} dx > 2020

QR = - i + 3j

P is left hand derivative of l x - ] P=-1 lim g(x) = P ⇒

89. (d) We have,

86. (a) PQ = 6i + j

S(-3 i + 2j)

P(- 2i - j)

⇒

Q(4i)

m= 2

83. (d) We have, (l + p)x - PY + p (l + p) = 0 . . . (i) (1 + q) x - qy + q(l + q) = 0 . . . (ii) . . . (iii) y= 0 From Eqs. (i) and (ii), we get X = pq, y = (p + 1) (q + 1) From Eqs. (i) and (iii), we get x = - p, y = 0 From Eqs. (ii) and (iii), we get x = - q, y = 0 C(,oq,(1 +p)(1 + q))

=

- 6+ 3

=

- 3 ;,e 0

I PQ I = .j36 + 1 = ff7

.j1 + 9 I PQ l = I QRI =

(1 + 2 + 3 ... + n - l) J x dx > 2020 0 (n - 1) (n)

2

=

✓10

:. It is not a rhombus.

g (x) is inverse off (x).

⇒ ln 2 (ln 2 + ln x) = ln 3(ln 3 + lny)

On differentiating, we get

87. (c) We have, (2x) ln 2 = (3y) ln3

⇒

3lnx = 21ny

⇒

ln x In 3 = ln y ln 2

B(-p ,0) 3

f ' (g(x) · g' (x)) = l

. . . (i)

⇒

1 g' (x) = -f'(gx)

. . . (ii)

⇒

g' (x) =

84. (a) System of equation represent the equation of three planes. But three plane cannot intersect at exactly two points. :. No solution exists. 85. (c) r1 , r2, 7a E {l, 2, 3, 4, 5, 6}

r1 , r2, r3 are forms of (3m, 3m + 1, 3m + 2) :. Required probability _ 31 X 2c1 X 2c1 X 2c1 _ 2

&

-9

⇒ (ln x) [(ln 3) 2 ⇒

� l + (g(x)5 )

ln x · ln 3 ) ln 2

2

⇒ (ln 2) + (ln 2) ln x = (ln 3) ln 2 + (ln 3) 2 ln x

y = � (x + p) l+ q Put x = pq, we get y = - pq h = pq , k = - pq h + k=O Locus of orthocentre x + y = 0 which represent the equation of straight line.

-

(ln 3) 2 + In 3 (

2

> 2020 0

f(g(x)) = X

⇒ (ln 2) 2 + ln 2 - ln x = (ln 3) 2 + ln 3- ln y

=

1

1 f' (x) = -1 + x5

(ln 2) + ln 2 - ln x

x = pq Altitude from B to AC is

27

90. (d) We have,

2

Equation of altitude from C to AB is

l2J

⇒ n (n - l) > 8080 Minimum of x is 91.

From Eqs. (i) and (ii), we get

A(-q,0) y = 0

r!!:__

n (n - l) > 2020 2x2

⇒

:. PQRS is not a rectangle. I QRI

n

n-1

⇒

PQ · QR = (6i + j) (- i + 3j)

2

3

2

R(3 i + 3j)

·: PQRS is a parallelogram =

I

J {x} dx + 2f {x}dx + ... + (n - 1) J {x}dx > 2020 2

(ln 2) 2 ] = ln 2 [(ln 2) 2 - (In 3) 2 ) lnx = - ln 2 1 x=-

g' (x) = l + (g(x))5

⇒

9 1 . (b) To form a sharp image, the light cone that passes the circular hole and light cone that is formed by the refracted light have same base area on screen. This is shown below. Lens with ale Screen E Point

b-x

88. (a) We have,

/(x) = -- , X E (0, 1) and b E (0,1) 1 - bx

b-x y= 1 - bx

Let f(x) = y

⇒ y - ybx = b - x b-y ⇒ x= 1 - by Range f(x) ;,e R

:. f(x) is not invertible in (0, 1).

.____ d --�

Now, from lens equation, we have 1 1 1 V

U

f

_-1: = -1: + _-1:__ = x - f V

f

-X

{X

Image distance = u = --1!__ = __jlJ,_ x-f u -f

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339

KVPY Practice Set 4 Stream : SB/SX Let, h = radius of image on screen, r = radius of hole in lens and R = radius of lens. Now, from similar triangles !:i.OAB and !:i.OCD, we have

h

d . . . (i) ,. u Note That no sign convention is req uired here. And from similar triangles !:i.DCJ and

_

(. _ (_El_ ) - d

_ fu h u + v-d - - ---- . . R - 2r an d u - -- ) u-f R u

MEI, we have

⇒

h 2r

u+ =

...

u- f

( /�

f)

(ii)

From Eqs. (i) and (ii), we have d _ u(u - f) + fu - d(u - f) ⇒ 2u fu ⇒

2

2d ±

✓4d

- 8df

⇒

u=

⇒

u = % ± ( � ..fd (d - 2/) )

Here, f

=

=

=

Number of revolutions made by electron in 1 x 10-8 s is N = f2 X !:i.t = 8.23 X 1014 X 1 X 10-8

8.23 x 106 revolutions 94. (a) From first law of thermodynamics, we have, !:i.Q - !:i. W = !:i.E Applied to process 1➔ 2, we get a - 100 = 100 ⇒ a = 200 kJ For process 3 ➔ 1 , 100 - d = - 200 d = 300 kJ For entire cycle, I.Q = I.W ⇒ 200 + b + 100 = 350 ⇒ b = 50 kJ For a cyclic process, I.!:i. U = 0 ⇒ 100 + c - 200 = 0 ⇒ c = lOO kJ =

=

N

-

268 x 101 1 = 3.7 x 108 decays s -1 14 x 60 :. Source activity required is R = 3.7 X 108 Bq

93. (a) Frequency of revolution in a

where, E1

=

c�)

energy in n

=

1 state.

du = l 1s m1. n1. mum, wh en 0. dx

µ = (�: ff 98. (c) Density of water is maximum at 4° C. So, upthrust is maximum at 4° C.

99. (a) For adiabatic process, !:i.Q = !:i. U + !:i. W = 0 CvdT + pdV = O

⇒

dV =O V dV dT b - + cdT + (a - b) - = 0 T V

(b + cT)dT + (a - b)T . ⇒

bln 1' + cT + (a - b) ln V = a constant So, ln Tb + cT + ln v a - b = a constant

2

V 1 2 V 1 E0 E (2 X 2 E o E + 2 x 2 � )

⇒ V 0 - b . Tb · ecT

f = fo ( where, f =

constant

1 + u/c ) 1 - u/ C

l/2

frequency observed, emitted by source, = v speed of observer moving towards source = c speed of light. =

/0 = frequency

1 + 3 = 4J

=t==i

=

1 00. (d) Longitudinal Doppler effect in light gives,

97. (b)

and

f - f� ) :. We have, u = c ( :

f + fo 2 2 8 (5.6) - (4.8) V = 3 X 10 ( ) 2 (5.6) + (4.8) 2

For a spherical refracting surface, 1 =µ-1 µ _ _ R v (-u) ⇒

RT

CvdT + - - d V = 0 V d Ci,dT + (CP - Ci, )1' · : = 0

Integrating above equation, we have

E5 V K2

1 _ -_ 1 _ 1 K 1 1+ - l+ K 5 K ⇒ U0 = 20 %U [:. K = l + X

((µ - l)u - R) 2

R R As l = u + v l = -- + .Jµ ' .Jµ - 1 .Jµ - 1

U0�Ugap + UD U 0 Vol 0 Vol + U0 Vol 0 Ugap gap 1 2 Eo

µR2

du dx

From Eq (i), we have .Jµ R u= .Jµ - 1

UTotal

corresponds to 10 J of radiation energy. As tumor mass is 0.003 kg energy for 10 Gy dose is El = 0.003 x 1 0 = 0.03 J Each disintegration provides 0.7 MeV. E2 = 0.7 MeV = 0.7 x 106 x 16 x 10-19 J = 112 X 1 0-13 J To provide 0.03 J number of disintegrations required is o.o3 = 2.68 x 101 1 decays N= 112 X 10-13 These must occurs in 14 min,

µRu

(µ - l)u - R

R ... u = �� Jci0 - 1

UD_ = 96. (c) _

92. (b) A 10 Gy dose of radiation

fn = :1

X= U+

⇒

.

t= 5 t=0 t= 1 X= 4 x=O s = 20 :. Total distance covered s = 4 + 4 + 20 = 28 m

So, u = � ± 2 lg (9 - 8) = � ± � 2 2" 2 2 :. u = 6 cm or u = 3 cm So, both arc possible positions.

Bohr's orbit is

Let u + v = x -1:_ = µ - 1 µ ⇒ __ + x- u u R

⇒

9 cm.

:. Number of decays per second = n

- Ei x _! = 0.823 x la15 rps h 23 8.23 x 1a14 rps

So, path of particle is

4

4 cm, d

=

95. (c) Position of a particle is s = 3t2 - l2t + 9= 3(t - 1) (t - 3)

2u - 2du + df = 0 2

/2

So,

So,

= 459 x 107 mis = 165 x 108 km/h fine = (165 x 108 - 100) = i 164, 999, 900

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340

KVPVPractice Set 4 Stream : SB/SX

1 0 1 . (a) The complete reaction can be represented as:

H 3C

"(y

0

II

c', cH

0 frequency bands? (a) 380 to 1 900 MHz (c) Less than 30 MHz

(b) 30 to 40 MHz (d) 30 to 300 GHz

40. Let a radioactive sample undergoes a, � and y decays

(not necessarily in any particular order), then correct order of half-lives of these decays is (b) y > � > a. (a) a. > � > y (d) not following any order (c) � > a. > y

41. Conductivity of a saturated solution of a sparingly

If duration of collision is t,,,t, then torque provided by the wall about origin is (c)

0

CHEMISTRY

b �x , {0J;===:= (b) mua /J.t

0

39. Your mobile phone is operating in which of these

36. A particle of mass m collides elastically with a rigid

(a) 0

(b)

(a)

In above figure, E0 is total energy of the particle. Velocity-position of the particle is V

a

a

energy by friction (b) When h = R, sphere translates without rotation and looses energy by friction

soluble salt AB at 298 K is 1 . 8 5 x 10-6 Sm- 1 . Solubility product of the salt AB at 298 K is [given, A 0m (AB) = 140 x 10-4 Sm-2 mol- 1 )

(a) 5.7 x 1 0- 12 (c)7.5 x 1 0 -12

(b) 1.32 x 1 0-12 (d) 1 .7 4

42. The number of prr.-drr. 'pi' bonds present in XeO3 and XeO4 molecules, respectively (6) 4, 2 (c) 2, 3 (a) 3, 4

(d)3, 2

43. The Langmuir adsorption isotherm is deduced using the assumption (a) the adsorption takes place in multilayers (b) the adsorption sites arc equivalent in their ability to adsorb the particles (c) the heat of adsorption varies with coverage (d)the adsorbed molecules interact with each other

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346

KVPVPractice Set 5 Stream : SB/SX

44. By which of the following processes, pure nitrogen gas is prepared?

(a) (NH4 ) 2 Crp7

�

(c)NI-ls + NaNO2 ti

(b)NH4 Cl+ NaNO 2 � ti

(d) Np + Cu -

II X y 45. CfisCH = CH2 � � CfisCH2 0 H X and Y are respectively 0

(a) I-ls o+ , MnO�/H+ (b) H3 o+ , CrO3 /pyridine (c)BI-ls · THF/H2O 2 , CrO3 /pyridine (d)Bf\ - THF/H2O2, OW, CrO3 /f\ O+

46. In acidic H2 02 reacts with dichromate to give

coloured complex of Cr. Oxidation state of Cr in this complex is (b)+3 (a)+ 5 (d)- 1 0 (c)+6

47. Ge (II) compounds are powerful reducing agents, whereas Pb (IV) compounds are strong oxidants. It can be because (a) Pb is more electropositive than Ge (b) ionisation potential of lead is less than that of Ge (c)ionic radii of Pb2+ and Pb4+ are larger than those of Ge2+ and Ge4+ (d) more pronounced inert pair effect in lead than Ge

48. Vapour pressure of pure benzene is 119 torr and that of toluene is 37.0 torr at the same temperature. Mole fraction of toluene in vapour phase which is in equilibrium with a solution of benzene and toluene having a mole fraction of toluene 0.50, will be (a) 0.1 37 (b) 0.237 (c) 0.435 (d) 0.205

52. Out of the following electronic arrangements for outer electronic configuration. I. II I .

DD 1 1 I 1 I 1 I 1 I 1 I [I] 1 1 1 1 1 1 1 1 1 1 1 DD I 1 I 1 I 1 I 1 I I DD I t I 1 1 1 I I I 4s

3d

II.

4s

3d

IV.

4s

3d

4s

3d

The most stable arrangement is (a) Only I (b) Only II (c) Only III

(d)Only IV

53. Which of the following is the most reactive towards electrophilic reagent? (a)

CH

rY 3 � CH20H CH

(c ) rY 3 � OH

54. A 0. 150 mole sample of an ideal gas is allowed

t

expand at 294 K from 10.00 atm to 1.00 atm. p (atm) 10

1 .0 �-----

If external pressure is kept constant at 1.00 atm work done is (a) -330 J (b) -3.26 dm3 atm (c) Both (a) and (b) (d) None of these

55. Consider the following gaseous equilibria with equilibrium constant K1 and K2 respectively. S02(g)+ _!: 02(g)�803 (g) 2 28O3 (g)�28O2(g)+ 02(g)

49. In the conversion of

The equilibrium constant is related as: (a) 2K1 = K; (b) K12 = __!:_ (c) K; = __!:_ (d) K9 = 2 - K12 - Kl K2

r-r'

56. A magnetic moment of 1. 73 BM will be shown by one

X is (a) H2 / Pt (c)Li / NH3

among the following. (a) [Cu(NH3 ) 4 ]2+ (b)[Ni(CN)4 ]2- (c)TiCl 4

(b)Zn-Hg / HCI (d) NaBH4

50. During change of 0 2 to 02 ion, the electron adds on which one of the following orbitals?

(a) ft-orbital (b) 1t-orbital (c) 5--orbital (d)cr-orbital

51. Total number of voids in 0.5 mole of a compound forming hexagonal closed packed structure are (a) 6.022 X 1 0 23 (b)3.01 1 x 1 023 23 (c)9.033 x 1 0 (d) 4 . 5 1 6 x 1 023

57. Monomer of

,�

CH,

-

(d)[CoCl6 ]4 -

rn

CH3

(a) 2-mcthylpropcnc (c) propylene

(b) styrene (d) cthcnc

58. The reaction, CH3-CH-CH2-O-CH2-CH3 + HI

Heated > • • • •

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347

KVPY Practice Set 5 Stream : SB/SX Which of the following compounds will be formed? Clfs I (a) CH3-CH-C�-I + CH3CHpH (b) CH3- H-CH3 + CH 3CH20H 1 CH3 (c) Cf\ - H - CH2OH + Clfs CH3 f Clfs CH3 (d) CH3 - 6H - CH 20H + CH3 - CH2 - I

59. The following sequence of reactions of A gives CH2CONH2 (( COOCH3

(a)

o:;to 0

0

(c) o;NH 0

(i) Br2/NaOH

(ii) Heat

(b )

(a) Decondensation from chromosome and reassembly of the nuclear lamina (b) Transcription from chromosomes and reassembly of the nuclear lamina (c) Formation of the contractile ring and formation of the phragmoplast (d) Formation of the contractile ring and transcription from chromosomes

65. Lipids, which can be found in oil based salad

dressings and ice-cream, during digestion are splitted into (a) fatty acids and glycerol (b) glycerol and amino acids (c) glucose and fatty acids (d) glucose and amino acids

66. A twig kept in water having some salt remains fresh

ay:o 0

H

(d) 07

0

0

60. C3 H6Cl2 on reaction with NaOH forms C3 H60 which

gives yellow precipitate on heating with NaOH and 12 • Thus, C3 H6Cl2 is (a) 1 ,1 -dichloropropane (b) 1 ,2-dichloropropane (c) 2,2-dichloropropane (d) 1,3-dichloropropane

BIOLOGY

61 . If sexual reproduction takes place between the filaments of Rhizopus of different strains, one with

80 nuclei and another with 24 nuclei, what would be the total number of spores of different strains put together? (a) 24 (b) 48 (c)96 (d)1 1 4

62. All mammals without any exception are

characterised by (a) viviparity and biconcave red blood cell (b) extra abdominal testis and four-chambered heart (c) heterodont teeth and 12 pairs of cranial nerves (d) a muscular diaphragm and milk producing glands

63. Flocculation or coagulation of protoplasm is the (a) interchangeability between sol and gel states (b) ability to scatter that beam of light (c) erratic zig-zag movement of protoplasmic particles (d) ability of protoplasm to increase in size when they lose charges

64. Which one of the following precedes reformation of the nuclear envelope during M-phase of the cell cycle?

for longer period due to (a) decrease in bacterial degradation (b) exosmosis (c) decrease in transpiration rate (d) absorption of more water

67. Farmers in a particular region were concerned that

premature yellowing of leaves of a pulse crop might cause decrease in the yield. Which treatment could be most beneficial to obtain maximum seed yield? (a) Frequent irrigation of the crop (b) Treatment of the plants with cytokinins along with a small dose of nitrogenous fertiliser (c) Removal of all yellow leaves and spraying the remaining green leaves with 2, 4, 5-trichlorophenoxy acetic acid (d) Application of iron and magnesium to promote synthesis of chlorophyll

68. Which of the following is maximum in chloroplast? (a) RuBP carboxylase (c) Phosphatase

(b) Hexokinase (d)Kuclease

69. When breastfeeding is replaced by less nutritive food low in proteins and calories, the infants below the age of one year are likely to suffer from (a) marasmus (b) rickets (c) kwashiorkor (d)pellagra

70. A large proportion of oxygen is left unused in the

human blood even after its uptake by the body tissues. This 0 2 (a) raises the pC02 of blood to 75 mm of Hg (b) is enough to keep oxyhaemoglobin (c) helps in releasing more 02 to the epithelial tissues (d)acts as a reserve during muscular exercises

71. In humans, blood passes from the postcaval to the diastolic right atrium of heart due to (a) pushing open of the venous valves (b) suction pull (c) stimulation of the sinoauricular node (d)pressure difference between the caval and atrium

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KVPVPractice Set 5 Stream : SB/SX

72. Which is a bridge between nervous system and endocrine system? (a) Thalamus (c)Limbic system

(a) histamine and dopamine (b) histamine and kinins (c) interferons and opsonin (d)interferons and histones

(b)Hypothalamus (d) Parietal lobe

78. Palaeontologists unearthed a human skull during

73. Steroid hormones work as (a) they enter into target cells and bind with specific receptor and activate specific genes to form protein (b) they bind to cell membrane (c) they catalyse formation of cAMP (d) None of the above

74. In Mendelian dihybrid cross when heterozygous

round yellow are self-crossed, round green offsprings are represented by the genotype (a) RrYy, RrYy and RRYy (b)Rryy, RRyy and rryy (c)rrYy and rryY (d) Rryy and RRyy

75. If there are 120 adenine molecules in a B-DNA

double helical structure showing 20 coils, what is the number of pyrimidine nucleotides forming three hydrogen bonds in it? (a) 80 (b)1 00 (c) 120 (d) 1 40

76. Which antibiotic inhibits interaction between tRNA

77. An insect bite may result in inflammation of that

spot. This is triggered by the alarm chemicals such as

81 . If log2x 1944 = logx 486�, and x6 = 2° · 3 b , then a + b

✓

(c) 28

(d)30

82. The number of solution of the equation 2

-J3x + 6x + 1 2 + 5x2 + lOx + 9 =4- 2x- x2 is equal to

Wl

�2

w_ n_ 2�

L (-

W4

- - -4r1+ 3-) is equal to

n 1 = r = 0 4r + 1

83. The value of lim n➔

�3

� �2

� �4

w �8

84. MBC is such that a circle touches AB at Band

passes through centroid of MBC and C, if AB = 6, BC = 4, then AC is equal to A&

F

B

(a) m

(b)3--/2

(c)2✓14

(d)2✓1 3

character of a population are of the same value, the following is most likely to occur (a) a normal distribution (b) a bimodal distribution (c) a T-shaped curve (d)a skewed curve during protein synthesis? (a) UAA codon codes for lysine (b) UGG codon codes for tryptophan (c) Cysteine is coded by UGU and UAC codons (d)Tyrosine is coded by UAU and UAC codons

(2 Marks Questions)

MATHEMATICS is equal to (a) 24 (b) 26

79. If the mean and the median pertaining to a certain

80. Which one of the following statements is not correct

and mRNA during bacterial protein synthesis? (a) Erythromycin (b) Neomycin (c)Streptomycin (d) Tetracycline

� PART- I I

excavation. A small fragment of the scalp tissue was still attached to it. Only little DNA could be extracted from it. If the genes of the ancient man need to be analysed, the best way of getting sufficient amount of DNA from this extract is by (a) hybridising the DNA with a Dl\A probe (b) subjecting the DNA to polymerase chain reaction (c) subjecting the DNA to gel-electrophoresis (d) treating the DNA with restriction endonucleases

85. Area enclosed by y = g(x), x = 1 and x = 15, where g(x) is inverse of f (x) = 1' + 3x + 1 is (a) __\) 4

(b) 1 8

86. The remainder when (

I

(d)__\) 2

(c) � 4

k=l

20

6

c2k _ 1) is divided by 1 1, is

�3 W7 Wl �5 87. Let A, B, C be the angles of MBC with vertex A(4,- 1) and x- 1 = 0 and x- y = 1 are internal angle

bisectors through Band C respectively. Let D, E , F be points of contact of sides BC. CA and AB with incircle of MBC. If I.J , E' , F' are images of D, E and F in internal angle bisector of A, B, C, then equation of circumcircle of t:,.I.J E' F is (a) (x - 1)2+ y2 = 5 (b) x2+ (y - 1)2 = 25 2 2 (c) (x - 1) + (y - 1) = 5 (d)x2+ y2 = 25

88. Initially there is 50 gm of salt in tank with 100 L of

water present A liquid at rate of 5 L/min with 2 gm/L of salt is coming into tank. After proper mixing in tank it is running out with 4 L/min. The amount of salt present in tank after time t = 100 min is 1 525 31 25 1 555 (a) 300 gm (b) - gm - gm (d)- gm (c) -

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349

KVPY Practice Set 5 Stream : SB/SX a

89. Let a = i + j + k, b = i- j + k, a x b = b + 'A a and a • c = 1, then which of the following is true 8

1

(a) [ a b c] = - - and A =- 3 3

(b)

(a) � ,

�-----------> X

1

8

(b) [ a b c] = - and A =- 3 3 2 8 (c)[a b c] =- - and A =- 3 3 8 2 (d)[ a b c] =- - and A = 3 3

90. Let f(x) be a twice differentiable function all real

values of x and satisfies f(1) = 1, f(2) = 4, f(3) = 9, then which of the following is definitely true? (a) f"(x) = 2, x E (1, 3) (b) f"(x) = f'(x) = 5, for some X E (2, 3) (c)f"(x) = 3, x E (2, 3)

0

refractive index ✓ rests over a flat horizontal plane mirror.

A horizontal light ray is made incident over right side glass rod I, such that it leaves the rod at a height of 10 cm from the plane mirror. Final emergent ray is found leaving glass rod parallel to the mirror. Distance d between glass rods is nearly (a) 31.5 cm (b) 32.5 cm (c) 37.5 cm (d) 38.5 cm

93. u-x graph of an object is V

a-x graph of body is

--+------..,,.__--, X

_.0

--------> x

94. For a charged circular ring, maximum value of electrical potential is at + + JR + +

B

+

91. A metal bar (length L, mass m) can slides over two

92. Two identical cylindrical rods of radii 10 cm each and 3

X

0

PHYSICS

(c)- Rmg- · tane B 2L2

(d)

(c)

(d)f"(x) = 2 for some x E (1, 3)

connected rails fitted over an inclined plane without friction. Rails are of very less resistance and metal bar is of resistance R. When a magnetic field B is switch ON perpendicular to ground, then terminal speed of the metal bar obtained is Rm� Rmg (b)-· sec0 · tan0 (a) B 2L2 B 2L2

a

a

+ (c) C

(b) B

(a) A

D

(d)D

95. Five identical rods are connected between two large

reservoirs of temperatures 100 ° C and 0 ° C. Each rod has thermal conductivity k, length l, and area A. There is no heat transfer through sides of rods. 4

1 00'C A

CD

®

0°C B

Heat flow rate from reservoir A to Bis 1 0 0 ll A 1 0 0 hA 1 00 kA 1 00 kA (a) (b) (c) (d) l 7l 3l 13 l

96. A resistance R is measured using a voltmeter and an ammeter as shown below.

R

�---\ Vr--�

Ammeter shows 2A and voltmeter shows 120V. Internal resistance of voltmeter is 3000 .Q. Error in measurement of R compared to the reading taken with an ideal voltmeter will be (a) 3.2 Q (b) 4.2 Q (c) 1 .2 Q (d)0.2 Q 97. A free neutron decays into a proton and an electron as, � n � � p + � 1 e +v. If neutron-hydrogen atom mass difference is 840 µu, then the maximum possible kinetic energy of electron will be ( 1 u = 932 MeV) (a) 0.783 MeV (b) 0.840 MeV (c) 0.589 MeV (d)0 .687 MeV

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KVPVPractice Set 5 Stream : SB/SX

98. A chord ACE, 5m long is attached at points A and B to the vertical walls 3 m apart.

------------ BI 1 m

1 03. The given graph represents the variation of

nRT A, B and C. Identify the only incorrect statement.

compressibility factor (Z) = p V , for three real gases

----------

/

, --------------------------,, / 200 N

A pulley of negligible mass and negligible radius carries 200 N load is free to roll over chord without friction. Dimension x in figure, when pulley is in equilibrium is 4 3m (a) � m (d)(c)- m (b)2 m 8 3 5

99. A uniform solid sphere rolls up an inclined plane.

h

m

- - --���e--�-

(c)h

oc

__!_ m

u

2

l (b)h oc __ tane (d) h

__!_2

1 00. Water is filled in a tank upto height h. Bulk modulus oc

oc

r

of water is Band its density at surface is p 0 • Density at depth h is Po Po (a) (b) (1- p0 gh I B) (1- B lp0 gh) (c)Po ( 1 - Po;h )

(d) p0

-

P (atm )

(1- __I!_ ) gh

- -­

------+

(a) For the gas A, a = 0 and its dependence on p is linear at all pressure (b) For the gas B, b = 0 and its dependence on p is linear at all pressure (c) For the gas C, which is typical real gas for which neither a nor b = 0. By knowing the minima and point of the intersection, with Z = ], a and b can be calculated (d) At high pressure, the slope is positive for all real gases

1 04. E'Fe3 • /Fe = - 0.336 V,

0

EFe-

'•

/Fe

= - 0.4 3 9 V. The value of

standard electrode potential for the charge, Fe3+ (aq) + e- � Fe2• (aq) will be (a)- 0.072 V (b) 0.385 V

Now, choose the correct option. (a) h

a�- -

(c) 0.770 V

(d)- 0 .270 V

1 05. Among Ga (Z =64), Lu (Z = 7 1), La (Z = 10 3), Ta

(Z = 73) the elements having half-filled /-orbital is (c) La (a) Ga (b) Lu (d)Ta

1 06. Identify the major products A and Bin the following reaction.

KMn04

A

(i) SOCl2 B ii) H,IPd, BaSO, (

Po

CHEMISTRY 101 . The heat of atomisation of PH3 (g) and P2 H4 (g) are

953 and 1485 kJ mol- 1 respectively. The P - P bond energy in kJ mo1- 1 is (a) 21 4 (b) 426 (c) 31 8 (d) 1272

1 02 . An organic compound undergoes first order

decomposition. The time taken for its decomposition to 1/8 and 1/10 of its initial concentration are t11 8 and t1110 respectively. [ t11 s l x 10? What is the value of l t1110 l (a) 4 (b)8 (c)9 (d) 1 1

1 07. The correct statement with respect to the complexes

[Ni(C0) 4] and [Ni(CN\] 2- is

(a) nickel is in the same oxidation state in both (b) both are paramagnetic in nature (c) have square planar and tetrahedral geometry respectively (d) have tatrahedral and square planar geometry respectively

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351

KVPY Practice Set 5 Stream : SB/SX 108. Compound A, C5 H10 O5 gives a tetraacetate with Ac2 O and oxidation of 'A' with Br2 I H2 O gives an acid, C5 H10 O6 . Reduction of 'A' with HI gives iso-pentane. What is the possible structure of A?

I (a) HO-C- CH2 OH I CHOH I

I I CHOH I

COOH

CHO

(b) HO- C-CH2OH

CH2OH

CH2 OH

I (c) HO-C-H I (CHOHh I

I I CHOH I

CHO

HO-C=O

109. Identify the correct statement about the following pairs of compounds.

//,;.,

V

CO2Me ,�''

113. Strips of plant tissue were immersed in a range of

sucrose solutions of different concentrations. Their lengths were measured before immersion and after 30 minutes in the different solutions. The graph shows the ratio of initial length to final length.

D

(a) A and B diastereomers ; C and D diastereomers (b) A and B enantiomers ; C and D diastereomers (c)A and B diastereomers ; C and D enantiomers (d)A and B enantiomers ; C and D enantiomers

1 1 0. Given H3Cy c113

H Jl

CH, H,C

III. Separation Which combination correctly matches each event with the temperature in the reaction mixture? X X Y Y Z Z (a) I II III (b) II III I (c) Ill I 11 (d) III II I I. Cystic fibrosis II. Severe Combined Immunodeficiency (SCID) The difference between them is (a) I is a transient treatment while 11 can possibly be a permanent treatment (b) I usually uses the ex vivo approach while II usually uses the in vivo approach (c) I usually uses adenoviruses as vectors while (II) usually uses liposomes as vectors (d) l treats a recessive disorder while (II) treats a dominant disorder

CH2 OH

Ph

I. Binding of RNA primers II. DNA synthesis

112. Gene therapy is used for the treatment of

(d) HO- C-CH20H

CH2OH

The main events during one cycle of the reaction are listed below.

1.4 1.2

rn,

y

y CH2 C:H3 CH3 (III) (II) (I) The enthalpy of hydrogenation of these compounds will be the order as: (a) I > II > III (b) III > II > I (c)11 > lll > I (d) II > I > Ill

0 .4 -+----11---+-+---+--+--I-----, 0.1 0.2 0.3 0.4 0.5 0.6 0. 7 0.8 Sucrose concentration/mo! dm--3 Which concentration of sucrose solution has the same water potential as the cell sap? (b) 0.25 mo! dm-3 (a) 0.1 mol dm- 3 3 (c) 0.45 mo! dm(d)0.6 mo! dm-3

114. What is the approximate ratio of ATP synthesised

BIOLOGY

111. The diagram shows the changes in temperature of

the reaction mixture during the Polymerase Chain Reaction (PCR). X (�t to 95"C

z

Heat to 72° C

1

y

Heat to 37 ° C

�

during anaerobic respiration compared with aerobic respiration? (a) 1 :2 (b) 1 :1 0 (c) 1 :20 (d) 1 :30

115. A plant is heterozygous for a pair of alleles that are

codominant. This plant is self-pollinated and the resulting seeds are germinated and allowed to grow. Which ratios are expected in the offspring? (a) (b) (c) ( ) d

Ratio of phenotypes

1:2:1 1:2:1 3:1 3:1

Ratio of genotypes

1:2:1 3:1 1:2:1 3:1

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KVPVPractice Set 5 Stream : SB/SX

116. In fruitflies, a sex-linked gene controls the

development of eye colour. The eyes are either red or white. The male is the heterogametic sex.

118. Match major events in the history of life with earth's geological period.

What will be the expected percentage of eye colours in the progeny when a heterozygous red-eyed female is crossed with a white-eyed male?

(a) (b) (c) (d)

Red eyes

Male

White eyes

Female

25.0 37.5 12.5 0

A. B. C. D.

Male

25.0 37.5 12.5 50.0

Female

25.0 12.5 37.5 50.0

Codes

25.0 12.5 37.5 0

A

(a) 1 (c) 4

(a) p

11 21 31 41 51 61 71

(b) (d) (b)

(d!

(c) (c)

(d!

(b)

(b)

(a)

(c)

(b) 2 (d) 3

B 3 2

D 1 4

C 4 1

120. Albinism in humans is controlled by a recessive

allele. How many copies of this allele will be found at one of the poles of a cell at telophase-1 of meiosis in an albino person') (a) 23 (c) 2 (b) 4 (d) 1

(d) S

Answers 2 12 22 32 42 52 62 72

PART-II 81 91 101 111

iS

(c)R

(b)Q

PART-I (c)

i Ri

1

A

D 4 3

AUCGAAGUUCGU It is transcribed from one strand of DNA What is the sequence of bases on the complementary, non-transcribed strand? (a) ACGAACTTCGAT (b) ATCGAAGTTCGT (c) TAGCTTCAAGCA (d) UGCUUGAAGCUA

T T C A C G A C A A G T Q

C 3 2

2

Geological Period

Quaternary Triassic Carboniferous Devonian

is shown.

terminate the synthesis of a polypeptide. The diagram shows a strand of DNA coding for 4 amino acids. Where would a mutation, involving the insertion of a thymine nucleotide, result in the termination of translation?

i

B

1. 2. 3. 4.

119. The sequence of bases on a messenger RNA molecule

117. The mRNA triplet UGA acts as a stop codon to

P

Event

First reptiles First mammals First humans First amphibians

82 92 1 02 112

(d) (c) (c)

(d) (a) (a) (c)

(b)

(a) (a) (c}

(a)

3 13 23 33 43 53 63 73 83 93 1 03 113

(c)

(a) (a)

(c) (a)

(c}

(a) (a) (c)

(a) (b)

(c)

4 14 24 34 44 54 64 74 84 94 1 04 114

(a) (b) (a) (d) (b) (c)

(a) (d)

(c)

(b) (c) (c)

5 15 25 35 45 55 65 75 85 95 1 05 115

(b) (d! (d) (a)

(c) (b) (a) (a)

(b)

(a) (a) (a)

6 16 26 36 46 56 66 76 86 96 1 06 116

(a) (a) (d) (c) (c)

(a)

(c)

(d) (b)

(c)

(b) (a)

7 17 27 37 47 57 67 77 87 97 1 07 117

(c) (c) (d)

(c)

(d) (a)

(d) (b) (a) (a)

(d) (c)

(c)

8 18 28 38 48 58 68 78

(b) (d) (d) (b) (d) (a) (b)

9 19 29 39 49 59 69 79

88 98 1 08 1 18

(d) (a) (b) (d)

89 99 1 09 1 19

(d) (c}

(a) (a)

(d! (c}

(a)

(a)

(b) (c)

(d) (b)

10 20 30 40 50 60 70 80 90 1 00 110 1 20

(d) (c) (c)

(a) (a)

(c)

(d) (a) (d) (a) (b) (c)

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1 . (c) We have,

=>

By

⇒

loga b + logb a = c 1 log a b + -- = c loga b

q

2 2 ap + bpq + cq = 0 is never possible

a, b, c arc odd.

3. (c) l! + 2! + 3! + 4! + 5! + 6! + 7! + 8!

. . . + 20201 1 + 2 + 6 + 24 + 120 + 720 + 35k [·: 7! onward are divisible by 35] ⇒ 873 + 35k ⇒ 840 + 33 + 35h => 35 X 24 + 35/i + 33 ⇒ 35(24 + h) + 33 :.Remainder = 33

5 2 4. (a) AB = - r ⇒ CD = - r 3

OB - ✓r 2 - -5 r 2

OB = Ju r

- ls; co{ n sin (i cos( rr sin x) ) ) s; 1

⇒

Let .!!_ is roots of ax2 + bx + c = 0

⇒

- It s; 7t sin ( icos(Jt sin x) ) s; 7t

⇒

l log a b + _ ::,, 2 loga b

2. (d) We have,

⇒

- 1s; sin ( %cos(1t sin x) ) s; 1

⇒

AM ::,, GM

6

Solutions

2 36

- ls; y s; l

:. Range of y E [- 1, 1]

0

6. (a) We have,

x tana + y seca = 1 x_ _ _ + Y_ = l cota cosa intercept are cota and cos a . cota · cos a = sin a Given, tan 2 a = 1 ⇒ tan a = ± 1 4

4

Circle passes through (a, 0) and (13, O)

OA = a, OB = B ·: OT is tangent of circle OT 2 = OA · OB ⇒ ⇒

g(x) =

f

1 3'4

(tl3 sin (

fl

dt

1 1 g(x) Jim g'(x) = l x l2 sin-3 4 x O X ➔ l xl / g'(x) . 1 2 • -. 1!ill =0 - = 1lill I X 1 11 Sln x➔ O 1 x➔ a l x l3 /4

a. (c) r = .J3 sin 30" = .J3 I

PA 1 + I PB 1 + I PC 1 + I PD 2

2

2

1

2

5 4 2 . ·1 arIy, OD = r - r = ✓r S1m1

fT,2 9

3

Required probability

2 .2 n ( � , - � x L 20 - 1 1 = --1: 4 36 nr 2

5. (b) We have,

y = co{ 7t sin ( %cos(1t sinx) ) )

⇒

- 1s; sin x s; 1 - It s; It sin x s; 1t - 1s; cos(!t sin x) s; 1

1 0. (d) We have, p2 = P

(0,-1)

(x - .f3) 2 + y2 + x2 + (y - l) 2 2 + (x + ,/3) 2 + y2 + x2 + (y + l) = 4(x2 + y2 + 2)

3 j ')

= 4(r 2 + 2) =

\_4

+ 2 = 11

p- lp 2 = p- lp

=>

⇒

P=I (I + P) n = (P + Pt

⇒

= (2P) n = 2n p n = 2n p

= P + 2" P- P = I + (2" - 1) P

1 1 . (b) We have,

⇒

x2 + l000O[x] = lO000x

x + lOOOO(x - {x}) = lOOOOx 2

=> x + lO000x - lOOOO{x} = lO000x

=>

⇒

OT = ✓cla

⇒

Ct = ± � => Ct = �

7. (c) We have,

OT 2 = aB OT = M

2

x2 {x} = -- => 0 s; {x} < 1 10000 O s; -

x2 -< 1 10000

- 100< x < 100

Total number of x is 199.

1 2. (c) Let

f=

1

dx I= o (l + x 2020 ) (l + x2)

[·: x2 + y2 = ,. 2 ]

9. (d) a, B are the roots of and

ax2 + bx + c = 0

a + B = - bla aB = c la

WWW.JEEBOOKS.IN]

354

⇒ ⇒ ⇒

KVPVPractice Set 5 Stream : SB/SX = x2020 + 1 21 = f ---- dx J o (l + x202 0 ) (l + x2 ) = l 21 = _ dx ⇒ 21 = [tan- 1 x]0 o J (1 + x2 ) 21 = (tan - l oo - tan - l 0)

r_ _

Take dot product with a is a - (a x b) + a · ( b x c) = p(a) 2 + q a - b + r(a • c) 0 + ----1:_ = p + _cJ_ + l'_ . . . (i) 2 2 --/2 Similarly take dot product with b and c 0 = E. + q + ,-_

21 = !!:_ ⇒ 1 = !!:_ 2 4

1 3. (a) We have,

=

2 sinX sinY

⇒ 1 + 1 - 2sin 2 X - 2(1 - 2sin 2 Y) =

2sinX sinY

⇒ 4 sin 2 Y - 2 sin 2 X - 2 sinX sinY

=

0

4b - 2a2 - 2ab = 0 c = ,_l b_ = _ a_ = _ r._. _ _ J sin X sinY sin Z l 2

⇒

l ✓

2 q

M

2ab

. . . (i)

a+ b

and a, 5, q, b arc in AP

b+ 5 q = 10 - a, q = 2 a = l0 - q, b = 2q - 5 ⇒ On putting, the values of a and b in Eq. (i), we get 2(10 - q) (2q - 5) 4= 10 - q + 2q - 5

⇒ ⇒ ⇒ ⇒

2q2 - 23q + 60 = 0 2q2 - 15q - 8q + 60 = 0 (2q - 1 5) (q - 4) = 0 15 q = 4, 2

1 5. (d) We know, [a b c]2

=

a-a b· a

a-b a-c b· b b· c c· b c· c

c· a 1 [a b

1 2 1 2

c]2 =

1 2 1

1 2 1

1 1 2 5 3 1 [a b c]2 = - - - = -

4

1 [a b cJ = 2 ✓

4

2

as given a x b + b x c = pa + qb + re

p

'¼1