KVPY Stream SA Examination for Class XI Practice Test Papers Solved R Gupta's 9789388642798

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Cover
Title
Contents
Practice Paper-1
Practice Paper-2
Practice Paper-3
Practice Paper-4
Practice Paper-5
Practice Paper-6
Practice Paper-7
Practice Paper-8
Practice Paper-9
Practice Paper-10
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R

R

KVPY STREAM–SA

Practice Test Papers (Solved)

Purple Patches of the Book: The book comprises numerous questions in several Practice Test Papers which will prove extremely useful for aspirants to be familiar with the current exam pattern, the type of questions asked, and their answers.

l

All the questions incorporated in Practice Test Papers in the book have been solved by respective subject-experts with due diligence.

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Detailed Explanatory Answers have also been provided for selected questions for Better Understanding of readers for study and self-practice.

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The book will serve well both as practice material & a true test of your studies and preparation with actual exam-style questions.

While the specialised practice material in the form of Practice Test Papers is published with the sole aim of Paving the Way to your Success, your own intelligent study and practice in Harmony with this, will definitely ensure you Success in the Prestigious Exam.

Books to Accomplish A Successful Career

PRACTICE TEST PAPERS (SOLVED)

l

KVPY STREAM–SA

The present book of Practice Test Papers has specially been published for the aspirants of Kishore Vaigyanik Protsahan Yojana (KVPY–Stream-SA) (Class XI) Fellowship Exam. The book is highly recommended for the aspirants to sharpen their problem solving skills, speed and accuracy, and help them prepare well by practising through these papers to face the prestigious exam with Confidence, Successfully.

KVPY Kishore Vaigyanik Protsahan Yojana

STREAM–SA Practice Test Papers (Solved) For Class XI

HSN Code : 49011010

All Questions Solved by Experts with Selected Explanatory Answers

ISBN 978-93-88642-79-8

RAMESH PUBLISHING HOUSE 9 789388 642798

R-2015

12-H, New Daryaganj Road, New Delhi-110002 E-mail: [email protected] Website: www.rameshpublishinghouse.com

YOUR SUCCESS MATE Size: 23×36×8, Date: 07/2019

R. Gupta’s®

KVPY STREAM–SA Examination

For Class XI

Practice Test Papers (Solved)

Ramesh Publishing House , New Delhi

Published by O.P. Gupta for Ramesh Publishing House Admin. Office 12-H, New Daryaganj Road, Opp. Officers' Mess, New Delhi-110002  23261567, 23275224, 23275124 E-mail: [email protected] Website: www.rameshpublishinghouse.com



Showroom Balaji Market, Nai Sarak, Delhi-6  23253720, 23282525  4457, Nai Sarak, Delhi-6,  23918938 © Reserved with the Publisher

No Part of this book may be reproduced or transmitted in any form or by any means, electronic or mechanical including photocopying, recording or by any transformation storage and retrieval system without written permission from the Publisher. Indemnification Clause: This book is being sold/distributed subject to the exclusive condition that neither the author nor the publishers, individually or collectively, shall be responsible to indemnify the buyer/user/ possessor of this book beyond the selling price of this book for any reason under any circumstances. If you do not agree to it, please do not buy/accept/use/possess this book.

Book Code: R-2015 ISBN: 978-93-88642-79-8 HSN Code: 49011010

CONTENTS 

Practice Paper-1 ..................................................................... 1-12



Practice Paper-2 ................................................................... 13-24



Practice Paper-3 ................................................................... 25-36



Practice Paper-4 ................................................................... 37-48



Practice Paper-5 ................................................................... 49-60



Practice Paper-6 ................................................................... 61-72



Practice Paper-7 ................................................................... 73-84



Practice Paper-8 ................................................................... 85-96



Practice Paper-9 ................................................................. 97-108



Practice Paper-10 ............................................................. 109-120    

(iii)

FELLOWSHIP ELIGIBILITY 

The KVPY Fellowships are given to Indian Nationals to Study in India (Students intending to pursue/pursuing undergraduate program under Distance Education scheme/correspondence course of any university are not eligible to apply).



Stream SA: Students enrolled in XI Standard (Science Subjects) and having secured a minimum of 75% (65% for SC/ST/PWD) marks in aggregate in MATHEMATICS and SCIENCE subjects in the X Standard Board examination immediately in the preceding academic year are eligible to appear for Aptitude test.



The fellowship of the students selected under this stream will be activated only if they join an undergraduate course in Basic Sciences (B.Sc./B.S./B.Stat./B.Math./Int. M.Sc./Int. M.S.) after having secured a minimum of 60% (50% for SC/ST/PWD) marks in aggregate in Science subjects in the XII standard/(+2) Board Examination. During the interim period of one year they will be invited for the National Science (Vijyoshi) Camp and their travel expenses and local hospitality will be met by KVPY.



In the Stream-SA all questions are compulsory (Science and Mathematics).

   

(iv)

1

Practice Paper (Solved)

1

Kishore Vaigyanik Protsahan Yojana (KVPY) STREAM – SA

Mathematics 1. What is the remainder when (163 + 173 + 183 + 193) is divided by 70? A. 0 B. 1 C. 2 D. 3

A.

1   2 , 2  

B. [–1, 2]

C.

 1   – 2 , 1  

D.

2233

2. What is the unit digit of 711 ? A. 1 B. 3 C. 7 D. 9

8. If f ( x ) 





4. The domain of definition of f ( x ) 

log 2 ( x  3)

x 2  3x  2 B. (–2, ) D. (–3, ) | (–1, –2)

is:

A. C.

6. If x1, x2, ... xn are any real numbers and n is any positive integer, then: A.

n i 1 n

C.

n

2

  n  xi2  n   xi   i 1  i 1

n

B.

 3  f  ...   1999 

n i 1

xi2

 n     xi   i 1 

 1998  f  1999 

B. 1999 D. 999

10. The adjacent sides AB, BC of a square of side ‘a’ units are tangent to a circle. The vertex D of the square lies on the circumference of the circle. The radius of the circle could be:

5. The slope of the tangent to the curve (y – x5)2 = x(1 + x2)2 at the point (1, 3) is: A. 2 B. 4 C. 8 D. 12

 n     xi   i 1 

 2  f   1999 

9. What is number of digits in 250 (given that log 2 = 0.301)? A. 15 B. 16 C. 17 D. 18

C. f (x) = x2, g(x) = sin x D. f and g can not be determined

xi2

, then find the value of

A. 1998 C. 998

A. f (x) = sin2x, g(x) = x B. f (x) = sin x, g(x) = | x |

n

4  2x

 1  f   1999 

2

3. If g{f (x)} = | sin x | and f {g( x )}  sin x , then:

A. R | {–1, –2} C. R | {–1, –2, –3}

4x x

1   –1, 2   

 a

a 2– 2



2  1.5

B.



D.

 a a

 2  1 2 –1

x

 x – 3 11. For x  R, lim   is equal to: x   x  2  A. e B. e–1 –5 C. e D. e5

2

e2

2

12. The value of the integral

D. None of the above



e –1

3 2 C. 3 A.

7. If x2 + y2 + z2 = 1, then the value of xy + yz + zx lies in the interval of: 1

loge x dx is: x

5 2 D. 5 B.

(2015) KVPY—Practice Paper—1

2 13. The area of the equilateral triangle, in which three coins of radius 1 cm are placed as shown in the figure is:

A. C.

6  4 3  sq. cm 7  4 3  sq. cm

4

D.

4 3 sq. cm

27 sq. units 4

B. 9 sq. units

C.

27 sq. units 2

D. 27 sq. units

  3   5   7   15.  1  cos   1  cos   1  cos   1  cos  8 8 8 8



B.

A.

3 – 6 sq. cm

is

equal to:

14. The area of the quadrilateral formed by the tangents at x 2 y2  1 the end points of latus rectum to the ellipse 9 5 is:

A.

1 2

B.

C.

1 8

D.

cos

 8

1 2 2 2

Physics 16. A particle moves in a circle of radius 5 cm with constant speed and time period 0.2 s. The acceleration of the particle is: A. 5 ms–2 B. 10 ms–2 –2 C. 15 ms D. 25 ms–2 17. A particle starts SHM from the mean position as shown in Fig. Its amplitude is A and its time period is T. At one time its speed is half that of the maximum speed. What is this displacement? O

A

R B

A.

C.

v = vmax/2

1 A 2

2 3

3 A 2

B.

A

3

D.

2

A

18. The load versus elongation graph for four wires of the same material is shown in Fig. The thinnest wire is represented by the line: D C

Load

B

O

A. OA C. OC

A

Elongation

B. OB D. OD

19. If g denotes the value of acceleration due to gravity at point distant r from the centre of the radius of the earth is R. If r < R, then:

1 r C. g  r2 A.

g 

B.

g 

1

r2 D. g  r

20. A dancer on ice spins faster when she folds her arms. This is due to: A. decrease in friction at the skates B. constant angular momentum and increase in kinetic energy C. increase in energy and increase in angular momentum D. increase in energy and decrease in angular momentum 21. A stationary particle explodes into two particles masses m1 and m2, which move in opposite directions with velocities v1 and v2. The ratio of their kinetic energies (E1/E2) is: A. 1 B. m1/m2 C. m2/m1 D. m1v1/m2v 2 22. A light string passing over a smooth light pulley connects two blocks of masses m1 and m2 (vertically). If the acceleration of the system is (g/8), then the ratio of masses is: A. 9 : 7 B. 7 : 9 C. 4 : 5 D. 4 : 7 23. A person slides freely down a frictionless inclined plane while his bag falls down vertically from the same height. The final speeds of the man (vm) and the bag (vb) should be such that: A. vm > vb B. vm < vb

(2015) KVPY—Practice Paper—1-II

3 C. vm = vb D. they depend on their masses 24. To a fish under water, viewing obliquely a fisherman standing on the bank of a lake, the man looks: A. taller than what he actually is B. shorter than what he actually is C. the same height as he actually is D. depends on the obliquity 25. A battery of emf 10 V and internal resistance 3 is connected to a resistor. The current in the circuit is 0.5 A. The terminal voltage of the battery, when the circuit is closed is: A. 1.5 V B. 4.5 V C. 8.5 V D. 10.5 V 26. Starting with a sample of pure 66Cu(7/8) of it decays into Zn in 15 minutes. The corresponding half-life is: A. 5 minutes B. 10 minutes C. 7½ minutes D. 14 minutes 27. An ammeter and a voltmeter are joined in series to a cell. Their readings are A and V respectively. If a resistance is now joined in parallel with the voltmeter, then: A. A will increase, V will decrease B. A will decrease, V will increase C. Both A and V will decrease D. Both A and V will increase

28. White light is incident on the interface of glass and air as shown in the figure. If green light is just totally internally reflected, then the emerging in air contains: Air Green Glass White

A. B. C. D.

yellow, orange, red violet, indigo, blue all colours except green all colours

29. Two particles of masses m and 2m with charges 2q and 2q are placed in a uniform electric field E and allowed to move for the same time. The ratio of their kinetic energies will be: A. 2 : 1 B. 1 : 2 C. 4 : 1 D. 1 : 4 30. A particle goes from point A to point B, moving in a semicircle of radius 1.0 m, in 1.0 s, as shown in the figure. The magnitude of the average velocity of the particle is: 1.0 m A

B

A. 1.0 ms–1 C. 3.14 ms–1

B. 2 ms–1 D. 0

Chemistry 31. 2.5 litre of 1 M NaOH solution are mixed with another 3 litre of 0.5 M NaOH solution. Then, the molarity of the resulting solution is: A. 0.73 M B. 0.12 M C. 0.95 M D. 0.60 M 32. The best coagulant for the precipitation of Fe(OH)3 solution is: A. NaNO 3 B. Na3PO 4 C. Na2SO 4 D. Na2HPO 3 33. The metal used to recover copper from a solution of copper sulphate is: A. Ag B. Hg C. Na D. Fe 34. The correct order of the thermal stability of hydrogen halides (HX) is: A. HI > HCl > HF < HBr B. HF > HCl > HBr > HI C. HI > HBr > HCl > HF D. HBr < HF < HFI < HCl 35. KMnO 4 can prepared from K 2MnO4 as per the reaction 3MnO42– + 2H2O  2MnO4– + MnO2 + 4OH–. The reaction can go to completion by removing OH – ions by adding:

A. HCl C. CO 2

B. NO2 D. SO 2

36. What is the mass per cent of different elements in sodium sulphate (Na2SO 4)? A. 21 g B. 32 g C. 46 g D. 52 g 37. Increasing order of e/m value of electron (e), proton (p) and neutron (n) and -particle is: A. n, , p, e B. n, p, e,  C. n, p, , e D. e, p, n,  38. Considering the elements B, C, N, F and Si, the correct order of their non-metallic character is: A. N > F > C > Si > B B. F > N > C > B > Si C. B > C > Si > N > F D. C > Si > B > N > F 39. The increasing order of hydrolysis of the following compounds is: Br

Br

Br

CH3 CH 3 — C — Br CH3

(a)

(b)

(c)

(d)

4 A. B. C. D.

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

< < <
–3 and for denominator, (x + 1)(x + 2)  0 x  –1, –2 From equation (i) and (ii) Domain is (–3, ) | (–1, –2).



log 2 ( x  3) ( x  1)( x  2)

...(i)

...(ii)

 dy  5. Slope of tangent at the point (x1 y1) is    dx  ( x

1 y1 )

Now

fog(x) = f (g(x)) = f

and

gof(x) = g{f (x)} = g(sin2x) =

Again,

4. Given,

 x   sin

2

x

sin 2 x  | sin x |

Given curve

(y –

x5)2

= x(1 +

x2)2

5  dy 4  2( y – x )  – 5 x  = (1 + x2)2 + 2x(1 + x2)2x dx put x = 1, and y = 3

let f (x) = sin x, g(x) = | x | fog(x) = f {g(x)} = f (| x |)

dy = 8. dx 2

  x 

= sin| x |  sin when,

 x1  x2  ....  x n  x12  x22  ...  x n2     n n

f (x) = x2, g(x) = sin x fog(x) = f {g(x)}



 

= f sin x  sin x and

6. Since x1, x2, ... xn are any real numbers

2



gof(x) = g{f (x)} = g(x2) = sin x 2 = sin| x |  | sin x |.

2

[Using mth Power theorem] 2

n



n i 1

xi2

 n     xi  .  i 1 

8 7. Due to symmetry, we can say that the maximum value of xy + yz + zx will be at x=y= z Now, x2 + y2 + z2 = 1 

1

x=y= z=

3  xy + yz + zx  1 which is present only in one option.



r = a 2– 2



B

a

a

O

A

r



C

D

11. For x  R,

 x – 3 lim   x   x  2 

x

= lim

x 

 3  1 –  x 2   1   x



12.

e

–1

loge x dx = x

1

 e

–1

e



e

e2

loge x dx – x

e



1

x 2 y2  =1 9 5 To find tangents at the end points of latus rectum, we find ae

14. Given,

i.e.,

ae =

and

b2 (1 – e2 ) =

a2 – b2  4  2  4 5 5 1 –    9 3

Y L (ae, b2 (1 – e2)

X

O

1

log e x dx x

log e x for +ve and –ve x

1 1 2 1 2 e = –  loge x   –1   log e x   e 1 2 2 5 1 1 2 2 = – [0 – (–1) ]  [2 – 0] = . 2 2 2





2

log e x dx x



3 3 (BC)2  4(1  3)2 4 4

e2

values]

log e x dx  = – x –1

3 cm BC = BD + DE + EC = 2  2 3

= 6  4 3 sq. cm.

X'

[since, 1 is turning point for

1

EC =

–3

= e–5. e2

and

By symmetry the quadrilateral is a rhombus

x

x

DE = O1O2 = 2 cm

 Area of  ABC =

OD = r, OB = r 2 r  r 2 = 2a

Hence,

Also, Now,

9. Taking log gives us 50 log2 = 50 × 0.301 = 15.05 Number of digits will be immediate next integer if the result obtained is not an integer. In this case, we obtained the product = 15.05. Hence, the number of digits = 16. 10.

O1 D BD BD = 3 cm

In O1BD, tan 30° =

2

13. Since tangents drawn from external points to the circle subtends equal angle at the centre  O1BD = 30°

L'

Y'

So, area is four times the area of the right angled triangle formed by the tangent and axes in the Ist quadrant.  5  Equation of tangent at  2,  is  3 2 5 y x . = 1 9 3 5 x y  =1  9/2 3  Area of quadrilateral ABCD = 4 [area of  AOB] 1 9  = 4  . .3  27 sq. units. 2 2 

A

  3   5   7   15.  1  cos   1  cos   1  cos   1  cos  8 8 8 8 O3 O1

O2

1 cm

B

30° 3 cm

D

  3   3     =  1  cos   1  cos   1 – cos   1 – cos         8 8 8 8

1 cm

2 cm

E

30° 3 cm

C

  3   =  1 – cos2   1 – cos2     8 8

9

=

1   3   2 – 1 – cos   2 – 1 – cos  4 4 4



m1 v = 2 m2 v1

=

1   3   1 – cos   1 – cos  4 4 4

Now,

E1 1 / 2 m1 v12 m v  m m  = = 1  1  1  2 2 E2 m v m   1 / 2 m2 v2 2 2 2  m1 

=

1 1  1  1  1 1 1– 1      1 –   .    4 2 2  4  2 8



E1 m = 2. E2 m1

2

r = 5 × 10–2 m and T = 0.2 s In the uniform circular motion, the accn of the particle,

16. We know that,

22. In the given system

 m – m2  g g a=  1  8  m1  m2 

2

 2  ac = 2r =   r  T  2   5  10 –2 ac =   0.2   = 5 ms–2.

17. Let the particle be at R, when its velocity v = vmax/2 = A/2 and its displacement from the mean position O be y As,

v =  A2 – y2



y=

Given,

m1 – m2 1 = m1  m2 8



2



 

8m1 – 8m2 = m1 + m2 7m1 = 9m2

m1 9 = . m2 7 23. Since both person and bag fall down through the same vertical height and experiencing same accleration due to gravity, hence they aquire same speed, while reaching the ground. 

24.

A2 – v2 / 2 v = A/2, then

O i

Air

y=

A2 –

A2 2

3  A. 2

4 2 18. For the thinnest wire, the elongation in wire will be maximum for a given load, which is so corresponding to line OA.

19. We know that,

d  g = g  1 –  for (r < R) R r  R – d = g    g R R i.e.,

g  r.

20. On folding arms, distance k decreases, I = Mk2 decreases As, I ×  = constant   increases

1 2 I 2  Due to decrease in I and increase in , there is overall increase in KE. As KE of rotation =

21. According to the principle of conservation of linear momentum m 1v1 – m 2v2 = 0

2

Water

O

r

As incident ray is extended further, O is seen as coming from O. Therefore, it appears to the fish as if the man is taller than what he is, he looks taller. However, it depends on r i.e., the obliqueness also. Therefore, that person appears taller for the fish but how tall he looks, also depends on the obliqueness. 25. The terminal voltage, V = E – ir = 10 – 0.5 × 3 = 10 – 1.5 = 8.5 V. 26. Sample becomes (1/8)th in 15 minutes n 3 N  1  1 =      or n = 3 N0  2  2 1  n=3 = T 15 or T= = 5 minutes. 3 27. Current of ammeter = A and voltage of voltmeter = V. We know that, when a resistance is joined in parallel



(2015) KVPY—Practice Paper—2

10 with the voltmeter, their equivalent resistance will be less than both the resistances. As a result of this, current in the ammeter (A) will increase and the voltmeter reads the potential difference of the resistance. We know that potential difference of cell is more than the potential difference of a resistance. Therefore, voltage in voltmeter (V) will decrease. 28. High  implies less , Now critical angle, –1  1  C = sin     Hence, in the sequence VIBGYOR, C increases from violet to red. In this case, the angle of incidence is such that it is equal to critical angle for green colour i.e., angle of incidence is less than critical angle for yellow and red colours. Therefore, rays of yellow, orange and red colours are refracted into air. Hence, (A) is correct answer.

2 qE 2 qE and a2 = m 2m a1 = 2a2

29. We have,

a1 =



1 1 m1 v12  m1 (a1t )2 2 2 1 1 K2 = m2 v22  m2 (2 a2 t )2 2 2 K1 =

K1 m a2 m 4 2   . = 1 12  K2 m2 a2 2 m 1 1



30. Given, R = 1 m, t = 1s 1m A

R

R

B

Displacement Time interval 2R 2  1  vav = = 2 ms–1. t 1

Average velocity = 

31. We have, M 1V 1 + M 2V 2 = M 3V 3 1 × 2.5 + 0.5 × 3 = M 3 × 5.5 [ Total solution = 5.5 L]  2.5 + 1.5 = M 3 × 5.5

4 = 0.73 M. 5.5 32. Since Fe(OH)3 is a positive solution and hence the best coagulant is PO43– ion. or

M3 =

33. Iron (Fe) can displace copper from CuSO 4 solution and hence Cu can be recovered from the solution by adding iron.

34. Order of the thermal stability is inversely proportional to the bond length. Hence, HF > HCl > HBr > HI. 35. CO 2 is neither oxidising nor reducing agent. It will provide acidic medium and can shift the reaction in forward direction and the reaction can go to completion. 36. Atomic weights: Na = 23, S = 32, O = 16. Molecular weight of Na2SO 4 = 2 × 23 + 32 + 4 × 16 = 142 Mass of one mole of Na2SO 4 = 142 g One mole of Na2SO 4 contains 2 moles of Na, 1 mole of S and 4 moles of O atoms. Mass of 2 moles of Na = 2 × 23 = 46 g. 37. Magnitudes of charge on, e, p, n and  are 1, 1, 0 and 2 units respectively. Masses of e, p, n and  are approximately 0, 1, 1 and 4 amu respectively. Hence, e/m ratio can be easily found and compared. If the order of e/m values is found by including the sign of the charge, none of the alternate matches with this order. Thus, the sign of the charge is not considered and magnitude of charge has to be taken for finding e/m values. Hence, the increasing order of e/m values is n, , p and e. 38. Non-metallic character increases from left to right across the period and decreases on going down a group. It is impossible to answer of this question on the basis as non-metallic characters of B and Si cannot be compared. Since, electronegativity  Electronegative character  non-metallic character.  Electronegatives of F, N, C, B and Si are 4.0, 3.0, 2.5, 2.0 and 1.8 respectively. Hence, (B) is correct answer. 39. This is in accordance with the stability of the carbocation as: + +

(CH 3)3 C > (d )

+ >

( c)

+ >

(b )

(a )

Hence, (C) is correct answer. 40. We have, HNO3 + H2SO 4  NO2+ + H2O + HSO4– Here, HNO 3 losses OH– to give NO2+ Hence, it acts as a base. 41.

CH 2 OH(CHOH)n CHO

Br2 H 2 O

CH 2OH(CHOH) nCOOH

Glucose

Gluconic acid

42. Neoprene is a polymer of chloroprene n CH2 = CH – C = CH2 Cl

Polymerisation

CH2 – CH = C – CH2 Cl

n

11

43. As,

m=

or

T f kf



P

Tb kb Q

Tb = 100.18 – 100 = 0.18°

T

Tb  kf kb 0.18  1.86  0.654 = 0.512  Freezing point of solution = 0 – 0.654°C = –0.654°C. 

Tf =

44. We have, Rate = k[A] 1.8 × 10–3 mol L–1 min–1 = k(0.3 mol L–1) 

k= =

1.8  10 –3 = 6 × 10–3 min–1 0.3

6  10 –3 = 1 × 10–4 s–1. 60

45. The central atom in all these compounds in sp 3 hybridised, because of the largest electronegativity and smallest size of N atom. The bond pairs in NH3 are closed to N and to each other and hence face maximum repulsion. Bond angle in NH3 is therefore maximum.

1 SQ.SR 2 SQ.SR

 2

(iii)

2

x  x 1  1, From (ii) and (iii), 0  0 2  x + x+ 1=  x2 + x =  x(x + 1) =  x=

since domain of

sin–1

function.

x2 + x + 1  1  x2 + x  0 x2 + x + 1  1  x2 + x + 1  1 1 0 0 0, x = –1.

62. Let a straight line through the vertex P of a given  PQR intersects the side QR at the point S and the circumcircle of  PQR at the point T. Points P, Q, R, T are concyclic, then PS.ST = QS.SR PS  ST Now, > PS.ST [ AM > GM] 2 and Also,

1 1  > PS ST

SQ  QR > 2 QR > 2

2 PS.ST

SQ.SR SQ.SR



2 QS.SR

2 QR

>

4 QR 2



4 QR

QS.SR 63. Here to find the least value of   R for which 1 4x 2   1 , for all x > 0 x i.e., to find the minimum value of  when 1 y = 4 x 2  ; x x > 0 attains minimum value of  dy 1  = 8x – 2 ...(i) dx x

d2 y

2 = 8  dx 2 x3 dy = 0, then 8x3 = 1 dx

Now, when,

Function is defined, if (i) x (x + 1)  0, since domain of square root function. (ii) x2 + x + 1  0, since domain of square root function.

>

1 1  > PS ST

61. Given function is

tan –1 x ( x  1)  sin –1 x 2  x  1 

R

S

 1 x=    8 

At,

...(ii)

1/ 3

d2 y

= 8 + 16 = 24 dx 2 Thus, y attains minimum when

 1 x=    8 

1/ 3

;> 0

 1  y attains minimum when x =    8  i.e.,

 1 4    8 

1/ 3

2/3

 (8)1/ 3  1 1/3 + 21/3  1 31/3  1



1 27 1 Hence, the least value of  is . 27    2  64. Now, a  b  c = d 2  2 a.b  0      | a |  | b |  | c |  1  2 a.b  –3     Now,  | a – b |2 = 2  d 2 – 2  a .b  2(3) + 3 = 9. 





12

k1  – 25 k2 = 100 –  2  – 25 = 3 100 –  200 – 2 = 3 – 75 5 = 275 or  = 55°C.

1

 f (1  x )  x y=    f (1) 

65. Let

 or

1  log f (1  x) – log f (1)  x  1  f (1  x )  lim log y = lim   x  0 f (1  x ) x 0   [Using L’ Hospital’s Rule] 

log y =

=

or  70.

f (1) 6  2 f (1) 3

Since the refractive index is changing, the light cannot travel in a straight line in the liquid as shown in options (B) & (C). Initially, it will bend towards normal and after reflecting from the bottom, it will bend away from the normal as shown in the figure.

 

log lim y = 2



x 0

lim y = e2.

x 0

66. Time taken by the projectile to reaches the maximum height, 1 10  u sin  2  1 s  0.5 s  t= g 10 2 Initial height of the second ball = Maximum height of projectile + distance covered by the second ball in 0.5s

=

u  sin 2  1 2  gt  2.5 m. 2g 2

71. We For For i.e.,

73. We have,

M2 =

1000  K f  w2

w1  T f Tf = 0 – (–0.465) = 0.465

1000  1.86  1.8  180 40  0.465  Empirical formula mass = 12 + 2 × 1 + 16 = 30

200 = 8 ms–1 25 1 2 KE = (m1  m2 )v 2 1 2 = (20  5)8 2 1 =  25  64  800 J. 2

=

Molecular mass Empirical formula mass 180 6 = 30  Molecular formula (CH 2O)6 = C6H12O6. 

68. Orbital velocity of spacecraft while orbiting close to the earth =



v=

Now,

pKa = –log Ka pKa = 4  Ka = 10–4 pKa = 5  Ka = 10–5 is 10 times more than that of B.

72. Higher the value of K H, the lower is the solubility of gas in the liquid. Hence, the order of increasing solubility is Ar < CO 2 < CH 4 < HCHO KH value: 40.39 1.67 0.413 1.83 × 10–5.

67. Applying principle of conservation of linear momentum (20 + 5)v = 20 × 10 + 5 × 0 

have, A, B, Ka for A

6

gR  9.8  6.4  10 = 7.92 km/s

 Escape velocity =

2 gR = 11.19 km/s  Velocity needed to escape = 11.19 – 7.92 = 3.25 km/s. 69. Let °C be the temperature of junction C. If two rods are connected in series, the rate of flow of heat through each rod is same, k1 A(100 – ) k2 A( – 25) So, = x x   

n=

74. The presence of electron withdrawing group like–NO 2 at p-position will decrease the basic character. So (B) will be least basic. Presence of electron donating group like –OCH 3 at p-position will increase the basic character and therefore (D) is the most basic. The order is II < V < I < III < IV. H

75. +H

O – CH 2 – R

+

O

+

O

+ O

CH 2

H O

O

Acetal

CH 2

R

R

–H

+

13

Practice Paper (Solved)

2

Kishore Vaigyanik Protsahan Yojana (KVPY) STREAM – SA

Mathematics D

1. Solve the inequality for x: 2.log10(x – 4) < log10 (x + 8): A. x > 4 B. –8 < x < 4 C. 4 < x < 8 D. x < 8

I

2. Find the maximum value of y = 5 – | x + 1 | – | x – 3 |. A. 1 B. 5 C. 4 D. 

A.

F

(1, 0)

B.

G

B. (x + y)2 D. None of these

(1 – cos 2 x )(3  cos x ) is equal to: x tan 4 x A. 4 B. 3

6. lim

(0, 1)

(–1, 0)

C

B

A. (x + y) (x – y) C. (x – y)2

3. The set {(x, y); | x | + | y |  1} is represented by the shaded region in which one of the four following figures? (0, 1)

E

A

H

x 0

(–1, 0) (1, 0)

(0, –1)

C. 2

1 2

7. What is the area of the inner equilateral triangle if the side of the outermost square is ‘a’? (ABCD is a square)

(0, –1)

(0, 1)

D.

(0, 1)

A

C.

(–1, 0) (1, 0)

D.

(0, –1)

(–1, 0) (1, 0)

B

(0, –1)

D E

4. If A and B are two events such that P(A  B) 

3 and 4

C a

1 3  P(A  B)  , then: 8 8 A.

11 P(A)  P(B)  8

C.

P(A)  P(B) 

7 8

B.

F

3 P(A).P(B)  8

A.

3 3a 2 32

B.

3 3a 2 64

C.

5 3a 2 32

D.

3a2 16

x

8. If f ( x ) 

D. None of these

(1 

5. All the three quadrilaterals ADEC, ABIH and BCGF are squares and ABC = 90°. If the area of ADEC = x2 and the area of AHIB = y2, (x2 > y2) then the area of BCGF is:

1 n n x )

for n  2 and

g( x )   fo fo... of  ( x ) . Then  f occurs n times

13

x

n –2

g( x ) dx equals:

14

A.

1 1  nx n n(n – 1)

B.

1 1  nx n n –1

1–

1 n

C.

1 1  nx n n(n  1)



D.

1 1  nx n n 1







1–





1



12. How many zeroes will be there at the end of 36!36!? A. 7 × 6! B. 8 × 6! C. 7 × 36! D. 8 × 36!

C

13. When x is divided by 6, remainder obtained is 3. Find the remainder when (x4 + x3 + x2 + x + 1) is divided by 6? A. 3 B. 4 C. 1 D. 5

C 1



1 n

1 n

1 n

C

C

9. If a, b, c, d are positive real numbers such that

a a a  b a  b  c a  b  c  d , then    b  2c  3d 3 4 5 6 is: 1 2 C. 2 A.

14. In the following figure, the diameter of circle is 3 cm. AB and MN are two diameters such that MN is perpendicular to AB. In addition, CG is perpendicular to AB such that AE : EB = 1 : 2 and DF is perpendicular to MN such that NL : LM = 1 : 2, the length of DH in cm is: C M

B. 1 A D

D. not determinable

10. How many positive real numbers x are there such that

x

x x



 x x



x

E

O

H

L

B F

G N

?

A. 1 C. 4

A.

B. 2 D. infinite

11. Find the domain of the function y = f (x) which is 1

defined as f ( x ) 

| x | x

A. (–, +) C. (0, –)

.

C.

B.

2 2 –1

3



2 –1 2

D.

2



2 –1 2

2



2 –1 3

15. The number of solutions of the equation sin (ex) = 5x + 5–x is: A. 0 B. 1 C. 2 D. infinitely many

B. (0, +) D. (1, +)

Physics 16. A player caught a cricket ball of mass 150 g moving at a rate of 20 ms–1. If the catching process is completed in 0.1 s, the force of blow exerted by the ball on the hands of the player is equal to: A. 30 N B. 40 N C. 50 N D. 60 N 17. A force F acting on an object varies with distance x as shown in figure. The force is in N and x in m. The work done by the force in moving the object from x = 0 to x = 6 m is:

18. A perfect gas at 27°C is heated at constant pressure so to double its volume. The temperature of the gas will be: A. 127°C B. 227°C C. 327°C D. 217°C 19. A particle is subjected to two mutually perpendicular SHMs such that x = 2 sin t and y = 2 sin (t + /4). The path of the particle will be: A. a circle B. an ellipse C. a straight line D. a parabola 20. A car moves from X to Y with a uniform speed vu and returns to Y with a uniform speed vd. The average speed for this round trip is:

F(N) 3 2 1 1

2

3

4

5

6

A.

vu vd vu  vd

B.

2 vu vd vu  vd

C.

vu vd 2

D.

vu  vd 2

7

x in (m)

A. 5.5 J C. 13.5 J

B. 10.5 N D. 15.5 N

15 21. Identify the circuit in which the electrical components have been properly connected. +

+

– – V

R

R

+ +

A

– – A +

+ V –

–

R

A. C.

(i) +

attraction. The speed of each particle with respect to their centre of mass is:

(ii) +

– – V +

+ A –

R

– + A –

+ V –

(iii)

(iv)

A. (i) C. (iii)

B. (ii) D. (iv)

22. Beams of light are incident through the holes A and B and emerge out of the box through the holes C and D respectively as shown in figure. Which of the following could inside the box? A. a rectangular glass slab B. a convex lens C. a concave lens D. a glass prism

A C B D

23. A thin circular disc of radius R is uniformly charged with density  > 0 per unit area. The disc rotates about its axis with a uniform angular speed . The magnetic moment of the disc is:

R 4  B. R2 4 R 2  C. D. 2R2 2 24. Find the equivalent resistance across A and B: A.

4 I1

2 A

I B

I 6 I2

A. 1.2  C. 4.4 

B. 3.4  D. 6.4 

25. Two particles of equal mass M go around a circle of radius R under the action of their mutual gravitational

GM R GM 2R

B. D.

GM 4R 2GM R

26. During lunar eclipse, the visible red colour is because of: A. dust in space B. dust in moon’s atmosphere C. dust in earth’s atmosphere D. None of these 27. Two bodies of mass M and m (M > m) respectively are dropped from the same height. If the repulsive force of air is same for the two bodies, they reach the ground: A. simultaneously B. bigger body reaching first C. smaller body reaching first D. None of these 28. A body of mass 3 kg is under a constant force which causes a displacement s in metres in it, given by the 1 relation s  t 2 , where t is in seconds. Work done by 3 the force in 2 seconds is: 3 4 J J A. B. 4 7 8 13 J J C. D. 3 7 29. Choose the incorrect statement from the following regarding magnetic lines of field. A. If the magnetic field lines are parallel and equidistant they represent zero field strength B. Magnetic field lines are closed curves C. Relative strength of magnetic field is shown by the degree of closeness of the field lines D. The direction of magnetic field at a point is taken to be the direction in which the north pole of a magnetic compass needle points 30. A concave mirror of focal length 20 cm forms a real image at a distance of 40 cm. The position of the object is: A. 10 cm B. 20 cm C. 30 cm D. 40 cm

Chemistry 31. A crystalline solid has a cubic structure in which tungsten (W) atoms are located at the cubic corners of the unit cell, oxygen atoms at the edges of the cube and sodium atom at the cube centre. The molecular formula of the compound is: A. NaWO 2 B. NaWO 3 C. NaWO 4 D. Na2WO 2

32. Vapour pressure of pure A is 70 mm of Hg at 25°C. It forms an ideal solution with ‘B’ in which mole fraction of A is 0.8. If the vapour pressure of the solution is 84 mm of Hg at 25°C, the vapour pressure of pure ‘B’ at 25°C is: A. 140 mm B. 150 mm C. 160 mm D. 170 mm

16 33. Each molecule of a compound contains 9 carbon atoms, 13 hydrogen atoms and 2.33 × 10–23 g of another element X. The molecular weight of compound is: A. 119 B. 127 C. 135 D. 203

40. When 10 ml of 10 M solution of H2SO 4 and 100 ml of 1 M solution of NaOH are mixed. The resulting solution will be: A. acidic B. neutral C. weakly alkaline D. strongly alkaline

34. The Kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [a0 is Bohr radius]

41. The reaction of toluene with Cl2 in the presence of FeCl3 gives X and the reaction in presence of light gives Y. Hence, X and Y are: A. X = m–chlorotoluene, Y = p–chlorotoluene B. X = o–and p-chlorotoluene, Y = trichloro methyl benzene C. X = benzyl chloride, Y = m–chlorotoluene D. X = benzal chloride, Y = o–chlorotoluene

h2 A.

16  2 ma0 2

h2 B.

h2 C.

25 2 ma0 2

32  2 ma0 2 h2

D.

64  2 ma0 2

35. A radioactive sample had an initial activity of 56 disintegration per minute. After 69.3 minutes, it was found to have an activity of 28 disintegration per minute. The number of atoms in a sample having an activity of 20 disintegration per minute (dpm) is: A. 500 B. 1000 C. 2000 D. 6930

42. Arrange the following compounds in order of decreasing acidity:

36. Arrange B, Al, Ga and In in order to their first ionisation energies: A. B < Al < Ga < In B. B > Ga > In > Al C. B > Al > Ga > In D. B > Ga > Al > In 37. Based on lattice energy and other considerations which one of the following alkali metal chlorides is expected to have the highest melting point? A. NaCl B. LiCl C. KCl D. RbCl 38. At 27°C a gas was compressed to half its volume. To what temperature must it now be heated so that it occupies just original volume? A. 127°C B. 237°C C. 283°C D. 327°C 39. 100 ml of nitrogen at 0.30 atm pressure and 50 ml of carbon dioxide at 0.60 atm pressure are introduced into vessel of volume 150 ml without any change in temperature. What will be the ratio of the partial pressure of nitrogen to carbon dioxide in the vessel? A. 1 : 1 B. 1 : 2 C. 1 : 3 D. 2 : 1

OH

OH

OH

OH

NH2

Cl

CH 3

OCH3

I II A. I > II > III > IV C. IV > III > II > I

III IV B. II > IV > I > III D. III > I > II > IV

43. IUPAC name of CH 3 – CH– CH 2 – CH – CH 3 is: |

A. B. C. D.

OH 2-hydroxy-2-methyl pentanoic 2-hydroxy-4-methyl pentanoic 4-hydroxy-1-methyl pentanoic 4-hydroxy-2-methyl pentanoic

|

COOH acid acid acid acid

44. Blister copper obtained during extraction from cuprous oxide is called, because: A. It is the most impure form of copper B. It has blister like eruptions due to evolution of gas C. It is the most impure form D. It has a shining surface like blister 45. Which one of the following pairs of elements is called ‘chemical twins’ because of their very similar chemical properties? A. Hf and Zr B. Re and Fe C. Mn and W D. Mo and Tc

Biology 46. Which part of poppy plant is used to obtain the drug “Smack”? A. Flowers B. Leaves C. Roots D. Latex

48. All of the following are included in ‘Ex-situ conservation’ except: A. Wildlife safari parks B. Seed banks C. Botanical gardens D. Sacred groves

47. Which one of the following population interactions is widely used in medical science for the production of antibiotics? A. Commensalism B. Amensalism C. Parasitism D. Mutualism

49. A woman has an X-linked condition on one of her X chromosomes. This chromosome can be inherited by: A. Only daughters B. Both sons and daughters C. Only grandchildren D. Only sons

17 50. AGGTATCGCAT is a sequence from the coding strand of a gene. What will be the corresponding sequence of the transcribed mRNA? A. AGGUAUCGCAU B. UCCAUAGCGUA C. ACCUAUGCGAU D. UGGTUTCGCAT 51. According to Hugo de Vries, the mechanism of evolution is: A. Multiple step mutations B. Minor mutations C. Phenotypic variations D. Saltation 52. All of the following are part of an operon except: A. an operator B. a promoter C. an enhancer D. structural genes 53. Which of the following events does not occur in rough endoplasmic reticulum? A. Protein folding B. Phospholipid synthesis C. Cleavage of signal peptide D. Protein glycosylation 54. Which of these statements is incorrect? A. Enzymes of TCA cycle are present in mitochondrial matrix B. Oxidative phosphorylation takes place in outer mitochondrial membrane C. Glycolysis operates as long as it is supplied with NAD+ that can pick up hydrogen atoms D. Glycolysis occurs in cytosol 55. Select the incorrect match: A. Lampbrush Chromosome B. Polytene chromosomes C. Submetacentric chromosomes D. Allosomes -

Diplotene bivalents Oocytes of amphibians L-shaped chromosomes Sex chromosomes

56. Which of the following terms describe human dentition? A. Thecodont, Diphyodont, Homodont B. Pleurodont, Diphyodont, Heterodont C. Pleurodont, Monophyodont, Homodont D. Thecodont, Diphyodont, Heterodont 57. Nissl bodies are mainly composed of: A. Proteins and lipids B. Free ribosomes and RER C. Nucleic acids and SER D. DNA and RNA 58. Many ribosomes may associate with a single mRNA to form multiple copies of a polypeptide simultaneously. Such strings of ribosomes are termed as: A. Polysome B. Nucleosome C. Plastidome D. Polyhedral bodies 59. Which of the following options correctly represents the lung conditions in asthma and emphysema, respectively? A. Inflammation of bronchioles; Decreased respiratory surface B. Decreased respiratory surface; Inflammation of bronchioles C. Increased respiratory surface; Inflammation of bronchioles D. Increased number of bronchioles; Increased respiratory surface 60. The stage during which separation of the paired homologous chromosomes begins is: A. Pachytene B. Zygotene C. Diakinesis D. Diplotene

Mathematics 61. Find the coefficient of x10 in the binomial expansion 11

 2 3 of  2 x –  when x  0  x 11 A. C4 × 27 × 34 B. 11C7 × 27 × 33 C. 11C5 × 29 × 37 D. None of these 62. If y = tan x. cot 3x, x  R then: 1 1  y 1  y 1 A. B. 3 3 1  y3 C. D. None of these 3 63. Consider a circle with its centre lying on the focus of the parabola y2 = 2Px such that if touches the directrix of the parabola. Then a point of intersection of the circle and the parabola is:

 3P  P  B.  – , P  , P 2 2  P   P  C.  – , P D.  – , – P  2 2 64. A fair die is thrown twenty times. The probability that on the tenth throw the fourth six appears is: 20 120  57 C10  56 A. B. 610 6 20 6 84  5 C. D. None of these 610 A.

65. If R 

30 65 – 2965

30 64  2964 A. 0 < R  0.1 C. 0.5  R  1.0

, then: B. 0.1 < R  0.5 D. R > 1 (2015) KVPY—Practice Paper—3

18

Physics 66. A body is projected with a velocity of 40 ms–1. After 2s it crosses a vertical pole of height 20.4 m. The angle of projection be (Take, g = 9.8 ms–2): A. 90° B. 45° C. 30° D. 60° 67. A block of mass m is connected to another block of mass M by a spring (massless) of spring constant K. The blocks are kept on a smooth horizontal plane. Initially, the blocks are at rest and the spring is unstretched. Then a constant force F starts acting on the block of mass M to pull it. Then, the force on the block of mass m is: mF mF A. B. M mM MF  M  m F C.  D.   m  mM 68. In the circuit shown, the ammeter A reads a current of I1 amp. Now, the source of emf and the ammeter are physically interchanged i.e., the source is put between B and C and the ammeter between A and B. The ammeter now reads a current of I2 amp. Then: B A C R3

R2

A

R1

A. B. C. D.

I1 = I2 I1 > I2 I1 < I2 The relation between I1 and I2 will depend upon the relative values of resistances R1, R2 and R3

69. How much below the surface does the acceleration due to gravity become 70% of its value on the surface of the earth? (Radius of earth = 6.4 × 106 m) A. 1.02 × 105 m B. 1.92 × 106 m 7 C. 2.32 × 10 m D. 3.25 × 105 m 70. A glass tube of uniform internal radius r has a valve separating the two identical ends. Initially, the value is in a tightly closed position. End 1 has a hemispherical soap bubble of radius r. End 2 has sub-hemispherical soap bubble as shown in figure. Just after opening the valve.

2

1

A. Air from end 1 flows towards end 2. No change in the volume of the soap bubbles. B. Air from end 1 flows towards end 2, volume of the soap bubble at end 1 decreases. C. No change occurs. D. Air from end 2 flows towards end 1. Volume of the soap bubble at end 1 decreases.

Chemistry 71. The end product in the reaction is: CaCO3

Na 2 OH

Heat

CH 3 COOH   A   B  C A. B. C. D.

Methyl nitrate Acetoxime Acetaldehyde oxime Formaldehyde oxime

75. In the reaction: Heat

72. Which of the following cations will have minimum flocculation value for arsenic sulphide solution? A. Ca 2+ C.

B. Al3+

Na+

D. Mg2+

73. In the compound given below: +

74. In aqueous solution containing 6 g of urea in 500 mL of solution has density equal to 1.05. If the molar mass of urea is 60, then the molality of solution is: A. 0.19 B. 0.21 C. 0.23 D. 0.27 CH 3 – CH – CH 2 – O – CH 2 – CH 3  HI   | CH3 A. CH 3 – CH – CH 2 – I  CH 3CH 2 OH | CH3

B.

CH3 – CH – CH2 OH  CH 3 CH3 | CH3

C.

CH 3 – CH – CH 3  CH 3 CH 2 OH | CH3

D.

CH 3 – CH – CH 2 OH  CH 3CH 2 I | CH3

+

H 3N (Y)

N H3 (Z) COOH (X)

The correct order activity of the position X, Y and Z is: A. X > Y > Z B. Y > X > Z C. X > Y > Z D. Z > X > Y

(2015) KVPY—Practice Paper—3-II

19

Biology 76. Which of the following is true for nucleolus? A. Larger nucleoli are present in dividing cells B. It is a site for active ribosomal RNA synthesis C. It takes part in spindle formation D. It is a membrane-bound structure 77. Stomatal movement is not affected by: A. Temperature B. CO 2 concentration C. O2 concentration D. Light

78. Which among the following is not a prokaryote? A. Saccharomyces B. Oscillatoria C. Nostoc D. Mycobacterium 79. Which of the following is not a product of light reaction of photosynthesis? A. ATP B. Oxygen C. NADPH D. NADH 80. Stomata in grass leaf are: A. Dumb-bell shaped B. Barrel shaped C. Rectangular D. Kidney shaped

ANSWERS 1 C

2 A

3 D

4 C

5 A

6 C

7 A

8 A

9 A

10 B

11 B

12 D

13 C

14 B

15 A

16 A

17 C

18 C

19 B

20 A

21 B

22 A

23 A

24 C

25 B

26 C

27 B

28 C

29 A

30 D

31 B

32 A

33 C

34 B

35 C

36 D

37 A

38 D

39 A

40 A

41 B

42 A

43 D

44 B

45 A

46 D

47 B

48 D

49 B

50 A

51 D

52 C

53 B

54 B

55 B

56 D

57 B

58 A

59 A

60 D

61 A

62 D

63 A

64 C

65 D

66 C

67 B

68 A

69 B

70 B

71 B

72 B

73 C

74 A

75 D

76 B

77 C

78 A

79 D

80 A

EXPLANATORY ANSWERS 1. For log to be defined: x – 4 > 0 and x + 8 > 0  x > 4 ...(i) Further, according to the question, log10(x – 4)2 < log10(x + 8) As the base is greater than 1, log is an increasing function. or, (x – 4)2 < x + 8 2 or, x – 8x + 16 < x + 8 or, x2 – 9x + 8 < 0 or, (x – 1)(x – 8) < 0 Hence, 1 x> x< x< 4.

+ |y|  1 0, y > 0, 0, y < 0, 0, y > 0, 0, y < 0,

x x –x –x

+ – + –

y y y y

= = = =

1 1 1 1

P(A  B) = P(A) + P(B) – P(A  B) P(A) + P(B) = P(A  B) + P(A  B) Using the given information, 3 1 P(A) + P(B)   4 8 7  P(A) + P(B)  8 Hence, the correct option is (C).

20 5. In square ADCE, area = x2 So, side AC = x In Square AHIB, area = y2 So, side AB = y As, x2 > y2 so, x > y and In  ABC (ABC = 90°) AC2 – (AB)2 = (BC)2 Hence, area BCGF = (AC)2 – (AB)2 = x2 – y2 = (x – y) (x + y)

Let I =

=

n –2  x g( x)dx  

1



n2 (1  nx n )1/ n

x

= 2

2

 lim

x 0

(3  cos x )  4

1 tan 4 x lim x 0 4 x

a 3K 3K 1   . = b  2c  3d K  2K  3K 6K 2 10.

sin  tan    lim  1 and lim  1   0  0    

4 1 4

7.

Side of inner square =

Diameter of the circle = Radius of the circle =

Side 3

=

Side =

2 a 12.

3 a 2 2

13.

3

8. Given,

f (x) =



f f (x) =

and

f f f (x) = g(x) =

2

3  3  3 3 2  a  a 4 2 2  32 x

(1  x n )1/ n f ( x)



n times

1

x 4  x3  x2  x  1 (3)4  (3)3  (3)2  (3)  1 = 6 6 121 = (Remainder = 1). 6 C M

1 2 DL = DH + HL HL = OE =

14.

1 2 OB = AO = radius = 1.5 DO2 = OL2 + DL2

A D

E

O

H

L

B F

DL = DH +

x (1  2 x n )1/ n

2

G N

2

2

x (1  nx n )1/ n

2

1  3  1    =     DH   2 2 2

(1  3 x n )1/ n fo fo... of  ( x )   

3x 9  x  0, x = 2 4

 36   36  Number of 5s in 36! =       5   25  =7 +1 =8 Therefore, zeros in (36!)36! = 8 × 36!.

for n  2

[1  f ( x ) n ]1/ n x

x3/2 =

3x 2

| x | x negative number So, (| x | + x) > 0, In this case, we can only take positive values. So, the domain is (0, +) = 0 < x < +.

a

Side

3 (side)2 = 4

= x Case-I: When base x = 1 Case-II: When base x  1

11. In this function (| x | + x) should not be

a 2 2 Area of an equilateral triangle

=

3/2

Hence, two solutions.

2

a  a  a       2 2 2

2 2 This is circumcircle radius of equilateral triangle, R=

xx

then

= 2. 2

d (1  nx n ) dx  2 dx n (1  nx n )1/ n 1

a = 3K, b = K, c = 5K – 4K = K, d = 6K – 5K = K

2 sin 2 x (3  cos x ) (1 – cos 2 x )(3  cos x ) = lim tan 4 x x 0 x 0 x tan 4 x x  4x 4x x 0

dx

1– 1 (1  nx n ) n  C. n(n – 1)

9.

lim

2 sin 2 x

(1  nx n )1/ n

1

I =

6. We have,

= lim

n2 x n –1 dx

x n –1



1   DH   = 2  DH = 2

2–

1 2 2 –1  . 2 2

21 15. Given equation is sin (ex) = 5x + 5–x LHS = sin(ex) < 1,  x  R and RHS = 5x + 5–x  2  sin(ex) = 5x + 5–x has no solution.

1 1 1 =  RP R2 R3

24. We have,

R2 = 4  I1

16. Impulse = Change in linear momentum F × t = m(v – u)

I

I A R = 2 1

m(v – u) 150  10 (0 – 20)  t 0.1 = –30 N (Negative sign for opposing force) or

17.

F=

Work done = force × distance W = area enclosed by Fx curve From the given figure,

33 = 13.5 J. 2 18. According to Charle’s law, when P is constant T V As V is doubled, T becomes twice i.e., T = 2 × (27 + 273)K = 600 K = 600 – 273 = 327°C. x = sin t 2 y   = sin t cos  cos t sin 2 4 4

19. Given, and

x2   1   x  1   =    1 –    2   2   4   2  Squaring both sides, we get, a relation represents an ellipse. 20. Total distance travelled = 2XY XY XY  vu vd Total distance Av. velocity = Total time 2 vu vd 2XY vav = .  XY XY vu  vd  vu vd

Total time taken for round trip =

21. Voltmeter is connected in parallel, ammeter in series and the positive terminal of the battery is connected to the positive terminal of the measuring device and negative to negative. 22. A rectangular glass slab transmits a parallel beam of light as parallel beam with a lateral displacement.

q Magnetic dipole moment = 2M Angular momentum  Magnetic dipole moment (M)

23. We have,

M =

q  MR 2  1 R 4  .  = R 4   .  2M  2  4 4

1 1 1 =  RP 4 6 46 RP = = 2.4  46

 

2

2

A

W = 33



B

R3 = 6 

I2

–3

B 4.4  A

Now, 25. We have,

B

R S = 2.4 + 2 = 4.4 .

GM 2 (2R)2

GM 4R 3

= m2R R

=

2

M

m

GM



=



v = R = R 

4R 3 GM 4R

3



GM . 4R

26. Due to dust in atmosphere of the earth during the lunar eclipse red colour is seen. 27. Let F be the resistive force acting on each mass. Acceleration a1 in bigger mass M

Mg – F F  g– M M Acceleration a2 in smaller mass m, =

mg – F F  g– m m Since M > m, hence a1 > a2. Thus, bigger mass will reach first. =

28. According to work-energy theorem, 1 2 1 2 W = KE = mv2 – mv1 2 2 1 2 2 Now, s= t v= t 3 3 2 4 –1  v1 = 0, v2 =  2  ms 3 3 2 1 1 8  4 2 = W =  3   –  3  (0)  J.  3 2 2 3 29. If the magnetic field lines are parallel and equidistant then the strength of the magnetic field is uniform.

22 30. Given, f = –20 cm, v = –40 cm, u = ? 1 1 1  = As, f u v 1 1 1   = u –40 –20 1 1 1 –2  1 –1     = – u 20 40 40 40  u = –40 cm Hence, the object is placed at a distance of 40 cm from the pole of the mirror. 31. Number of W atoms per unit cell = 1 Number of O atoms per unit cell 1 = 12  (edge centre) = 3 4 Number of Na atoms per unit cell = 1 (cubic centre)  Formula = NaWO 3. 32. Given, PA0 = 70 mm, P = 84 mm, xA = 0.8  xB = 1 – 0.8 = 0.2 According to Raoult’s law P = PA 0xA + PB 0xB  84 = 0.8 × 70 + 0.2 × PB0  0.2 PB0 = 84 – 56 = 28 28  PB = = 140 mm. 0.2 33. Mass of 1 molecule in amu = 9 × 12 + 13 × 1 + 2.33 × 10–23 × 6.023 × 1023 = 108 + 13 + 14.0336 = 135.0336  135. 34. Since, r  n2, the radius of the second orbit will be 4a0 n We know that, mvr = , n = 2 and r = 4a0 2 h  v= 4 ma0 

Kinetic energy = =

 h  1 1 2 mv =  m   2  4 ma0  2

2

h2 32  2 ma02

35. Rate (activity) = Decay constant × Number of atoms The activity falls from 56 disintegration per minute (d pm) to 28 disintegration per minute (d pm) in 69.3 minutes.  The half-life period is 69.3 minutes.

0.693 0.693  = 10–2 min–1 T1/ 2 69.3 Number of atoms in the given sample = Activity/Decay-constant 20 atoms per min = = 2000 atoms. 10 –2 per min  Decay constant (k) =

36. When we go down a group in periodic table, the ionisation energy continuously decreases but in group 13, ionisation energy of Ga is greater than that of Al. This is an exceptional case and is explained on the basis of poor screening effect of d-electrons. 37. The melting points of LiCl, NaCl, KCl and RbCl are 601°C, 805°C, 770°C and 718°C respectively. Lattice energy decreases in order LiCl, NaCl, KCl and RbCl. On this basis alone, the melting points should also decrease in the same order, because of greater covalent character in LiCl, the melting point of LiCl is lower than that of NaCl and NaCl has the highest melting point among the given compounds. Hence, the highest melting point is of NaCl. 38. Given, T1 = 273 + 27 = 300 K and V1 = V/2 and V2 = V According to Charle’s Law, V = Constant T V1 V2 T1 = T2  

V V = T  T2 = 600 K 2  300 2 T2 = 600 – 273 = 327°C.

39. Let P1 be the pressure of nitrogen in the mixture P = 0.30 atm, P1 = ?, V = 100 ml and V1 = 150 ml According to Boyle’s law PV = P1V1  0.30 × 100 = P1 × 150

0.30  100 = 0.20 atm ...(i) 150 If P2 is the pressure of carbon dioxide (CO 2) in the mixture, then, 0.60 × 50 = 150 × P2 0.60  50  P2 = = 0.20 atm ...(ii) 150 From equations (i) and (ii), we get, P1 : P2 = 1 : 1. 

P1 =

40. For H 2SO 4

Weight of substance in one litre of solution Molecular weight X  10 M =  X = 980 98 Weight of substance in one litre of solution Normality = Equivalent weight 980  20 N= 49 10 ml of 20 N solution of H2SO 4 = 200 ml of N solution of H2SO 4 for NaOH 10 M =

23 Since equivalent and molecular weight are equal  1 M=1 N 100 ml of 1 N NaOH can be neutralized by 100 ml of 1 N solution of H 2SO 4. After neutralisation 100 ml of N solution of H2SO 4 will remain. So that the solution will acidic. 41. We have, CH3

CH3

CH3 Cl

Cl2/FeCl3

Cl

CH3

CCl3 Cl2 ; hv Trichloromethyl benzene

42. The correct order of acidic strength is: OH

> NH2

43.

> Cl

OH

> CH 3

OCH3

I II III Hence, I > II > III > IV.

IV

5 4 3 2 CH3 – CH – CH2 – CH – CH3 OH

y=



y=



(3 y – 1)( y – 3) 

>0 ( y – 3)2  (3y – 1)(y – 3) > 0 

1 or y > 3. 3



44. Solidified copper (Cu) obtained is called blister copper. Since it has blistered appearance due to the evolution of sulphur dioxide (SO 2) gas 2 Cu2O + Cu2S  6Cu + SO2. 45. Hafnium (Hf) and Zirconium (Zr) occur together in minerals and they exhibit similar properties due to similar atomic radii caused by lanthanide contraction. Therefore, the pair of elements is called chemical twins. 61. Suppose (r + 1)th term contains x10 in the binomial 11

r

 3 Cr (2 x 2 )11– r  –   x = (–1)r 11Cr 2(11 – r) 3r . x22 – 3r ...(i) 10 This term will contain x , if 22 – 3r = 10  r = 4 So, (4 + 1)th i.e., 5th term contains x10 Putting r = 4 in (i), we get T5 = (–1)4 11C4 211–4 . 34 . x10 = 11C4 27 . 34 . x10  Coefficient of x10 = 11C4 × 27 × 34. 11

y
0 y–3



COOH

3  expansion of  2 x 2 –   x

tan x tan x (1 – 3 tan 2 x )  tan 3 x 3 tan x – tan 3 x

3 – tan 2 x 3y – 1 tan2x = y–3 tan2x  0, Therefore,

But

o- and p-chlorotoluene (X)

OH



+

Friedel Crafts reaction

OH

y = tan x cot 3x

62.

3P 2 = 0, 4 Solving this equation with y2 = 2Px, or x 2  y 2 – Px –

3P 2 =0 4  (2x – P)(2x + 3P) = 0 P –3P  x= ,x= 2 2 P Putting x = on y2 = 2Px, we obtain y = ±P 2 –3P For x = , y is imaginary. Hence, the circle and 2 P  parabola intersect at  P , P and  , – P .   2 2  64. In the first 9 throws we should have three sixes and six non-sixes, and a six in the 10th throw and thereafter whatever face appears no matter. So required probability x 2  Px –

We obtain,

3

=

6

1 84  56  1  5 . C3        1  1  1.... 1   6  6 6 610 10 times

9

24 65. Applying componendo and dividendo 30 65 – 2965 30 64  2964

= = =

(30

65

65

– 29 )  (30

64

or

64

 29 )

(30 65 – 29 65 ) – (30 64  29 64 )

or

30 64 (30  1) – 2964 (29 – 1) 30 64 (30 – 1) – 2964 (29  1) 30 64  31 – 2964  28

30 64  29 – 2964  30  Numerator is greater than denominator  R > 1.

66. Given, u = 30 ms–1, BA = 20.4 m Let  be the angle of projection at O. Taking vertical motion of projectile from O to A, we have, u = u sin  = 40 sin  A u  sin  a = –9.8 ms–2 20.4 m s = 20.4 m, t = 2s As, s = ut 

O

 cos 

The pressure inside the sub-hemispherical bubble at the end 2 is S P2 = P0  4 r2

B

1  (–9.8  2  ) 2 20.4 = 80 sin  – 19.6 = 40

As, r2 > r1, So P2 < P1 Hence, air from end 1 flows towards end 2 and volume of bubble at end 1 decreases.

20.4 = (40 sin )  2 

 or

40 1  = sin 30° or  = 30°. 80 2 67. Acceleration of the system of two masses F a= Mm  Force on the block of mass m or

sin  =

f = ma =

mF (m  M)

R2 E R 2I  = R1  R 2  R 2  R1 R 2 R1   R1  R  R  2 3  ER 2 = ...(i) R1 R 2  R1 R 3  R 3 R 2

68. We have, I1 =

71. We have, CaCO3 Heat CH 3 COOH  (CH 3 COO)2 Ca 

CH3

| NH2OH CH 3 COCH 3   CH 3 – C  NHOH Acetoxime

72. As2 S3 is a negatively charged solution. According to Hardy Schulze rule, higher the charge on the active ion, more is its coagulating power. 73. The position (X) is most acidic because of –COOH group +

–N H3 group at position (Y) is more acidic than at Z because of the presence of electron with drawing –COOH group in close proximity. Hence, correct order of acidity is X > Y > Z.

After interchanging ammeter and source of emf, new current,

R 2I R2 E  = R1  R 2 R1  R 2  R1R 2   R3  R  R   1 2 ER 2  I1 = ...(ii) R1 R 2  R 2 R 3  R 3 R1

I2 =

Hence, 69. Given, Now, 

3R 3  6.4  10 6  10 10 = 1.92 × 106 m.

d=

70. Let r1 and r2 be the radius of hemispherical soap bubble and sub-hemispherical soap bubble at the ends 1 and 2 respectively. If P0 is the atmospheric pressure and S is the force of surface tension of the soap solution. The pressure inside the hemispherical bubble at the end 1 is S P1 = P0  4 r1



1 2 at 2

d 7 3  = 1– R 10 10

I1 = I2.

Mass of solution = 500 × 1.05 = 525 g Mass of urea = 6 g Mass of solvent = 525 – 6 = 519 g 6/66  1000 = 0.19 m.  Molality = 5/9 75. We have, 74.

Heat

g 70 7 = = and d = ? g 100 10 g = g(1 – d/R) 7 g d d = 1–  = 1– 10 g R R   

CH 3 – CH – CH 2 – O – CH 2 – CH 3  HI   | CH3 CH 3 – CH – CH 2 OH  CH 3CH 2 I | CH3 Halide ions attacks the smaller alkyl group.

25

Practice Paper (Solved)

3

Kishore Vaigyanik Protsahan Yojana (KVPY) STREAM – SA

Mathematics C

1

1. Let y 

1

2

1

3 2

R 6 cm

1 3  ...

What is the value of y? 11  3 A. B. 2 C.

15  3 2

D.

A

A. 10 : 3 C. 7 : 3

11 – 3 2

1

1 12



1 22

 1

1 22



1 32

 ...  1 

1 20072



A.

2008 –

1 2008

B.

2007 –

1 2007

C.

2008 –

1 2007

D.

2008 –

1 2009

1 20082

P

B

D

B. 2 : 1 D. 8 : 3

6. Four machines A, B, C and D are to be used for producing 10,000 units. Machine A alone can produce 50 units in a day. B is half as efficient as A. In how many days would the work be completed by all the four machines, start working together? A. 19 days B. 29 days C. 46 days D. None of these

15 – 3 2

2. Consider a triangle drawn on the X-Y plane with its three vertices of (41, 0), (0, 41) and (0, 0) each vertex being represented by its (X, Y) coordinates. The number of points with integer coordinates inside the triangle (excluding all the points on the boundary) is: A. 780 B. 800 C. 820 D. 741 3.

Q

S

 4 – x2  7. If f ( x )  sin log   , then the domain of f (x)  1 – x  is: A. (–1, 2) B. (–2, 1) C. (–2, 4) D. (–3, 6)



8. Area of the region bounded by the curve y = ex and lines x = 0 and y = e is: A. e – 1 unit2

B. e + 1 unit2

C. 1 unit2

D.

e

x2 y2 –  1 , | r | < 1 represents: 4. The equation 1– r 1 r A. an ellipse B. a hyperbola C. a circle D. None of the above

 ln y dy 1

9. The  PQR is inscribed in the circle, x2 + y2 = 25. If Q and R have coordinates (3, 4) and (–4, 3) respectively, then  QPR is equal to:   A. B. 3 2   C. D. 4 6

5. In the figure (not drawn to scale) given below, P is a point on AB such that AP : PB = 4 : 3. PQ is parallel to AC and QD is parallel to CP. In  ARC, ARC = 90° and in PQS, PSQ = 90°. The length of QS is 6 cm. What is the ratio of AP : PD? 25

(2015) KVPY—Practice Paper—4

26 10. The least positive integer n for which is: A. 6 C. 8

3

n 1 – 3 n 

1 12

B. 7 D. 9

11. The largest non-negative integer K such that 24 K divides 13! is: A. 2 B. 3 C. 4 D. 5 12. The diagram below represents three circular garbage cans, each of diameter 2 m. The three cans are touching as shown. Find, in metres, the perimeter of the rope encompassing the three cans.

A. 2 + 6 C. 4 + 6

B. 3 + 4 D. 6 + 6

13. A word of 6 letters is formed from a set of 16 different letters of the English alphabet (with replacement). Find out the probability that exactly 2 letters are repeated. 225  224  156 18080 A. B. 6 16 10 6 15  224  156 C. D. None of these 16 6  xx – x–x  14. If f ( x )  cot –1   , then f (1) equals: 2   A. –1 B. 1 C. log 2 D. –log 2

15. There are two candles of same length and same size. Both of them burn at uniform rate. The first one burns in 5 hours and the second one burns in 3 hours. Both the candles are left together. After how many minutes the length of the first candle is 3 times that of the other? A. 90 B. 120 C. 135 D. 150

Physics 16. A bus travelling the first one-third distance at a speed of 10 km h–1, the next one-third at 20 km h–1 and at last one-third at 60 km h–1. The average speed of the bus is: A. 12 km h–1 B. 18 km h–1 C. 22 km h–1 D. 26 km h–1 17. A satellite of the earth is revolving in a circular orbit with a uniform speed v. If the gravitational force suddenly disappears, the satellite will: A. move with a velocity v tangentially to the original orbit B. fall down with increasing velocity C. ultimately come to rest, somewhere on the original orbit D. continue to move with velocity v along the original orbit 18. A figure at point O, a mass m is performing vertical circular motion. The average velocity of the particle is increased. Then, at which point will the string break? A. A B. B C. C D. D

20. The potential difference between A and B in the figure is: 6

9

5

A

B 2A

12

2A

4V

A. 40 C. 48

B. 42 D. 52

21. Water flows through a frictionless tube with a varying cross-section as shown in Fig. Pressure p at points along the y-axis is represented by:

A. P

B. P

A

X

m D

R O

C

B

19. A truck and a car with the same kinetic energy are brought to rest by the application of the brakes which provide equal retarding force. Which of them will come to rest in a shorter distance? A. Truck B. Car C. Both truck and car would come to rest the same distance D. None of these

C. P

X

D.

X

P

X

22. The acceleration due to gravity on the surface of the moon is 1.7 ms–2. What is the time period of a simple pendulum on the moon, if its time period on the earth is 3.5 s? [g = 9.8 ms–2] A. 3.5 s B. 4.6 s C. 7.4 s D. 8.4 s 23. Atomic mass number of an element is 232 and its atomic number is 90. The end product of this

27 radioactive element is an isotope of lead (atomic mass 208 and atomic number 82). The number of  and particles emitted is: A.  = 3,  = 3 B.  = 6,  = 4 C.  = 6,  = 0 D.  = 4,  = 6 24. A ray of light passes from air to glass. The angle of incident is 45°. Find the angle of refraction, given refractive index of glass is 1.5.  2 A. sin –1    3  –1  1  C. sin    2

B.

 3 sin –1    5

–1 2 D. sin

25. A particle of mass m is projected with velocity v making an angle of 30° with the horizontal. The magnitude of angular momentum of the projectile about the point of projection, when the particle is at its maximum height h is: mv 2 A. 0 B. 2g C.

3 mv 2 2 g

D.

3mv 3 16 g

26. A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalence of this combination is R, then the ratio R/R is:

A. C.

1 5 1 25

B.

1 15

D. 15

27. A ball is thrown by one player reaches the other in 2s. The maximum height attained by the ball above the point of projection will be (g = 10 ms–2). A. 5 m B. 7 m C. 10 m D. 12 m 28. A car runs at a constant speed on a circular track of radius 100 m, taking 62.8 s for every circular lap. The average velocity and average speed for each circular lap respectively is: A. 0, 0 B. 10 ms–1, 10 ms–1 –1 C. 0, 10 ms D. 5 ms–1, 10 ms–1 29. The potential energy of a long spring, when stretched by 2 cm is U. If the spring is stretched by 8 cm, the potential energy stored in it is: A. 2 U B. 6 U C. 8 U D. 16 U 30. In a sample of helium (He) gas, its molecules have root mean square velocity 5/7 times the rms velocity of hydrogen gas molecules. If the temperature of the hydrogen gas is 0°C, then temperature of the gas will nearly be: A. 2°C B. 5.57°C C. 7.23°C D. 10.57°C

Chemistry 31. An ionic compound has a unit cell considering of A ions at the corners of a cube and B ions on the centres of faces of the cube. The empirical formula of the compound would be: A. AB 2 B. AB 3 C. A2B D. A3B

C. NH4Cl

D. Na2SO 4

35. Consider the following bromides: Me

Me

Me Br

Me Br (A)

(B)

Br (C)

The correct order of SN reactivity is: A. A > B > C B. C > A > B C. B > A > C D. A > C > B

32. At 80°C, the vapour pressure of pure liquid ‘A’ is 520 mm Hg and that of pure liquid ‘B’ is 1000 mm Hg. If a mixture of solution of ‘A’ and ‘B’ boils at 80°C and 1 atm pressure, the amount of ‘A’ in the mixture is (1 atm = 760 mm Hg). A. 32 mol per cent B. 42 mol per cent C. 50 mol per cent D. 52 mol per cent

36. The correct order of acidic strength is: A. CO 2 > N2O5 > SO3 B. Cl2O7 > SO2 > P4O10 C. K2O > CaO > MgO D. Na2O > Al2O3 > MgO

33. The activation energy for a reaction at the temperature T was found to be 2.303 RT mol–1. The ratio of the rate constant to Arrhenius factor is: A. 10–1 B. 10–2 C. 2 × 10–2 D. 2 × 10–3

37. For the four successive transition elements (Cr, Mn, Fe and Co) the stability of +2 oxidation states will be there in which of the following order? [At. no. of Cr = 24, Mn = 25, Fe = 26, Co = 27] A. Cr > Mn > Co > Fe B. Mn > Fe > Cr > Co C. Fe > Mn > Co > Cr D. Co > Mn > Fe > Cr

34. Among the electrolytes Na2SO 4, CaCl2, Al2(SO 4)3 and NH4Cl, then most effective coagulating agent for Sb 2S3 solution is: A. Al2(SO 4)3 B. CaCl2

38. The compound which gives off oxygen in moderate heating is: A. aluminium oxide B. cupric oxide C. zinc oxide D. mercuric oxide

28 39. The correct order of increasing bond angle in the following species is: A. Cl2O < ClO2– < ClO2 B. ClO2 < Cl2O < ClO2– C. ClO2– < Cl2O < ClO2 D. Cl2O < ClO2 < ClO2– 40. Ribose and 2-deoxyribose can be differentiated by: A. Barfoed’s reagent B. Osazone fermation C. Follen’s reagent D. Fehling’s reagent 41. The reaction of CH 3CH = CH A. CH3CH2CH Br OH B.

CH3CH Br CH2

OH

with HBr gives:

OH

C. CH 3CH2CH Br

Br

D. CH 3CH Br CH2

Br

42. A solution of CuSO 4 is electrolysed for 10 minutes with a current of 1.5 amperes. What is the mass of copper deposited at the cathod? A. 0.135 g B. 0.193 g C. 0.296 g D. 0.315 g

43. Which of the following colligative properties can provide molar mass of proteins (or polymers or colloids) with greater precision? A. Depression in freezing point B. Osmotic pressure C. Relative lower of vapour pressure D. Elevation of boiling point 44. The conversion of molecules X to Y follows second order kinetic. If concentration of X is increased to three times, how will it affect the rate of formation of Y? A. 9 times B. 10 times C. 5 times D. 7 times 45. Substances which behave as normal electrolytes in solutions at low concentrations and exhibit colloidal properties at higher concentrations, are called: A. associated colloids B. emulsions C. lyophilic colloids D. macromolecular colloids

Biology 46. The Golgi complex participates in: A. Fatty acid breakdown B. Activation of amino acid C. Respiration in bacteria D. Formation of secretory vesicles 47. The two functional groups characteristic of sugars are: A. hydroxyl and methyl B. carbonyl and hydroxyl C. carbonyl and phosphate D. carbonyl and methyl 48. A ‘new’ variety of rice was patented by a foreign company, though such varieties have been present in India for a long time. This is related to: A. Co-667 B. Basmati C. Lerma Rojo D. Sharbati Sonora 49. Select the correct match: A. Ribozyme — Nucleic acid B. G. Mendel — Transformation C. T.H. Morgan — Transduction D. F2 × Recessive parent — Dihybrid cross 50. Which of the following is commonly used as a vector for introducing a DNA fragment in human lymphocytes? A. Retrovirus B. pBR 322 C.  phage D. Ti plasmid 51. In India, the organisation responsible for assessing the safety of introducing genetically modified organisms for public use is: A. Indian Council of Medical Research (ICMR) B. Genetic Engineering Appraisal Committee (GEAC) C. Research Committee on Genetic Manipulation (RCGM)

D. Council for Scientific and Industrial Research (CSIR) 52. The correct order of steps in Polymerase Chain Reaction (PCR) is: A. Extension, Denaturation, Annealing B. Denaturation, Annealing, Extension C. Denaturation, Extension, Annealing D. Annealing, Extension, Denaturation 53. Use of bioresources by multinational companies and organisations without authorisation from the concerned country and its people is called A. Bio-infringement B. Bioexploitation C. Biodegradation D. Biopiracy 54. Winged pollen grains are present in: A. Mustard B. Pinus C. Mango D. Cycas 55. After karyogamy followed by meiosis, spores are produced exogenously in: A. Neurospora B. Saccharomyces C. Agaricus D. Alternaria 56. Which one is wrongly matched? A. Uniflagellate gametes – Polysiphonia B. Unicellular organism – Chlorella C. Gemma cups – Marchantia D. Biflagellate zoospores – Brown algae 57. What is the role of NAD + in cellular respiration? A. It functions as an enzyme B. It is the final electron acceptor for anaerobic respiration C. It is a nucleotide source for ATP synthesis D. It functions as an electron carrier

29 58. Oxygen is not produced during photosynthesis by: A. Green sulphur bacteria B. Chara C. Cycas D. Nostoc

C. Fusion of two male gametes with one egg D. Fusion of one male gamete with two polar nuclei

59. Double fertilization is: A. Fusion of two male gametes of a pollen tube with two different eggs B. Syngamy and triple fusion

60. In which of the following forms is iron absorbed by plants? A. Ferric B. Both ferric and ferrous C. Free element D. Ferrous

Mathematics





61. The incentre of the triangle with vertices 1, 3 , (0, 0) and (2, 0) is:  3 2 1  A.  1, B.  ,  3 2   2  C.

 2 3 3, 2   

D.

 1   1,  3

62. If x, y and z are in AP and tan–1 x, tan–1 y and tan–1 z are also in AP then: A. x = y = z B. 2x = 3y = 6z C. 6x = 3y = 2z D. 6x = 4y = 3z

64. The sides of a triangle are in the ratio 1 : 3 : 2 , then the angles of the triangle are in the ratio: A. 1 : 3 : 5 B. 2 : 3 : 2 C. 3 : 2 : 1 D. 1 : 2 : 3 65. In the figure, find the area of the shaded portion, that is, the portion of the quadrant DOB which is not included in the 4 semicircles. Given that OA = AB = OC = CD = 2 and DOB = 90°. A.

63. The number of values of x in the interval [0, 5] satisfying the equation 3 sin2x – 7 sinx + 2 = 0 is: A. 0 B. 5 C. 6 D. 10

C.

1 2 5 1 – 2 2

5 –

D

C

O

B.

5 – 1

D.

5 –1 2

A

B

Physics 66. A ray of light is incident on the surface of the separation of a medium with the velocity of light at an angle 45° and is refracted in the medium at an angle 30°. What will be the velocity of light in the medium? A. 1.96 × 108 ms–1 B. 2.12 × 108 ms–1 C. 3.18 × 108 ms–1 D. 3.33 × 108 ms–1 67. A stone tied to the end of a string of 1 m long is whirled in a horizontal circle with a constant speed. If the stone makes 22 revolutions in 44 seconds, what is the magnitude and direction of the acceleration of the stone? A. 2 ms–2 and direction along the radius towards the centre B. 2 ms–2 and direction along the radius away from the centre C. 2 ms–2 and direction along the tangent to the circle D. 2/4 ms–2 and direction along the radius towards the centre 68. A player caught a cricket ball of mass 150 g moving at the rate of 20 ms–1. If the catching process be completed in 0.1 s, the force of the blow exerted by the ball on the hands of the player is: A. 10 N B. 20 N C. 30 N D. 300 N

69. Which of the diagrams in figure correctly shows the change in kinetic energy of an iron sphere falling freely in a lake having sufficient depth to impart it a terminal velocity: KE

KE

A.

B. Depth

Depth

KE

KE

C.

D. Depth

Depth

70. Two syrings of different cross-section (without needle) filled with water are connected with a tightly fitted rubber tube filled with water. Diameters of the smaller piston and larger piston are 1 cm and 3 cm respectively. If the force of 10 N applied to the smaller piston, then the force exerted on the larger piston is: A. 90 N B. 100 N C. 120 N D. 150 N

30

Chemistry A. since ionic product and solubility product are same precipitation will not occur B. since ionic product is greater than solubility product, no precipitate will be formed C. since ionic product is lesser than solubility product, precipitation will occur D. since ionic product is greater than solubility product, precipitation will occur

71. Energy of an electron in the ground state of hydrogen atom is –2.18 × 10–18 J. What is the ionisation enthalpy of atomic hydrogen in terms of J mol–1?

72.

A. 1.31 × 106

B. 2.35 × 105

C. 4.01 × 103

D. 5.76 × 104

MgBr (i) CO2 +

74. A solution has 1 : 4 mole ratio of pentane to hexane. The vapour pressures of pure hydrocarbons at 20°C are 400 mm Hg for pentane and 120 mm Hg for hexane. The mole fraction of pentane in the vapour phase would be: A. 0.286 B. 0.478 C. 0.527 D. 0.823

P

(ii) H3O

In the above reaction, product P is: CHO

COOH

A.

B.

2+

75. OH

Ph – C  C – CH3

+

A

is: OH

O

C.

Hg ; H

Ph

Ph OH

A.

D.

B.

H3C

H3C O

Ph

73. Predict if there will be any precipitate by mixing 50 mL of 0.01 M NaCl and 50 mL of 0.01 M AgNO 3 solution. The solubility product of AgCl is 1.5 × 10–10.

C.

O

D. H 3C

H3C

Biology 76. Which of the following elements is responsible for maintaining turgor in cells? A. Magnesium B. Calcium C. Potassium D. Sodium 77. Which one of the following plants shows a very close relationship with a species of moth, where none of the two can complete its life cycle without the other? A. Hydrilla B. Viola C. Banana D. Yucca 78. Pollen grains can be stored for several years in liquid nitrogen having a temperature of: A. –120°C B. –160°C C. –196°C D. –80°C

79. Niche is: A. all the biological factors in the organism’s environment B. the functional role played by the organism where it lives C. the range of temperature that the organism needs to live D. the physical space where an organism lives 80. Which of the following is a secondary pollutant? A. CO B. O3 C. SO 2 D. CO 2

ANSWERS 1 D 11 B 21 A 31 B

2 A 12 A 22 D 32 C

3 A 13 A 23 B 33 A

4 B 14 A 24 A 34 A

5 C 15 D 25 D 35 B

6 D 16 B 26 C 36 B

7 B 17 A 27 A 37 B

8 C 18 B 28 C 38 D

9 C 19 C 29 D 39 A

10 C 20 C 30 B 40 B

31 41 A 51 B 61 D 71 A

42 C 52 B 62 A 72 B

43 B 53 D 63 C 73 D

44 A 54 B 64 D 74 B

45 A 55 C 65 D 75 C

46 D 56 A 66 B 76 C

47 B 57 D 67 A 77 D

48 B 58 A 68 C 78 C

49 A 59 B 69 C 79 B

50 A 60 A 70 A 80 B

EXPLANATORY ANSWERS 1

y=

1.

2 

 y=

1 3 y

3 y 7  2y

3. Go through the method of induction 1

2y2 + 6y – 3 = 0

–6  36  24 y= 4



–6  60 –3  15  4 2 Since ‘y’ is a +ve number, Therefore, =

15 – 3 y= 2 Hence, option (D) is the answer. 2. Equation of the line x + y = 41. If the (x, y) coordinates of the points are integer, then their sum shall also be integers. So that x + y = K (1, a variable) as we have to exclude points lying on the boundry of the triangle. K can take all values from 1 to 40 only. K = 0 is also rejected because K = 0 will give the point A, which can not be taken.

1

1 2

1

1 2

 

1 2

1

1 2 Similarly, 1

1 12



2

2

1 22



 1

X

Now x + y = K (K = 1, 2, 3, ... 40) with K = 40, x + y = 40 by taking integral solutions. We get points (1, 39), (2, 38), (3, 37)....(39, 1) i.e., 39 points. x + y = 40 will be satisfied by 39 points, similarly x + y = 39 is satisfied by 38 points x + y = 38 is satisfied by 37 points x + y = 3 is satisfied by 2 points x + y = 2 is satisfied by 1 point x + y = 1 is satisfied by no point So, the total number of all such points is 39 + 38 + 37 + 36 + .... + 3 + 2 + 1 39  40 = = 780 points 2 Hence, option (A) is the answer.

2

2



1 3

2

1 20072

=



3 7 8 1    3– 2 6 3 3 1 20082

 2008 –

1 2008

Hence, option (A) is the answer. x2 y2 –  1 , where 1– r 1 r |r|< 1  1 – r is (+ve) and 1 + r is (+ve)

4. Given equation is

x2

5. PQ | | AC, 

CD AB 4  = QB PB 3

QD | | PC, 

CQ 4 PD  = QB 3 DB

–

y2

1 a2 b2 Hence, it represents a hyperbola when | r | < 1.  Given equation is of the form

(0, 41)

(41, 0)

1

 ...  1 

Y

A

9 3 1 =  2– 4 2 2

4 PD = 3 DB 4  PD = PB 7 AP 7 AP 7 4 7 AR    .  = =  4 PD 4 PB 4 3 3 PB 7 6. Machine A produces 50 units per day Machine B produces 25 units per day Machine C produces 50 units per day Machine D produces 100 units per day All four working together produce 225 units per day To produce 10,000 units, It will take

As,

10000 = 44.44 days 225 Hence, option (D) is the answer.

32

7. Given,

4 – x2 1– x 4 – x2  (1 – x) x  –2 Thus, domain  (–2,

For domain,

B

>0 >0

432  12  So, n = 8, only possible least positive integer. 1/ 3

11.

13.

Q (3, 4)

 2

24K  (23 × 3)K Exponent of 2 in 13! 13   13   13   2    2    3  = 10   2  2  Exponent of 3 in 13! 13   13  So, (23 × 3)3, So K = 3.  3  2 = 5   3 

E

Distance between 2 centres = 2 m  DC = DE = FA = 2 m Perimeter of the figure = BC + DE + FA + Circumference of sectors AOB, COB and FOE But three equal sectors of 120° = 1 full circle of same radius. Therefore, Perimeter of surface = 2r + BC + DE + FA = (2r + 6) m = (2 + 6) m [ r = 1]

(0, 1) 1

120°

F

y=e (1, e )

e

120° O'

A

Y

m1 =

C

120° O

>0

1  8. Shaded Area = e –   e x dx   1 unit2 0 

O

AOB = COD = FOE = 120°

12.

 4 – x2  f (x) = sin log    1 – x 

f (x) =

d  xx – x–x    2 2   x x – x – x  dx  1   2   1

–2

d x (x – x– x ) 4  ( x – x ) dx x

–x 2

 f (x) =

d x log x (e – e – x log x ) dx (x  x )

 f (x) =

–2 d d    ex log x (x log x) – e–x log x (–x log x) dx dx (xx  x– x )2  

 f (x) =  f (x) =

 f (1) =

–2

x

–x 2

–2 ( x x  x – x )2

–2(1  log x) x

(x  x





 x x (1  log x )  x – x (1  log x )

–x 2

)

–2  –1. (1  1)

(x x  x– x ) 

–2(1  log x) xx  x–x

33 15. Let V1 and V 2 are rates of burning for both candles respectively & L is the length of each candle.

L L , V2 = 5 3 Let after time ‘t’, their lengths are l1 & l2 V1 =

L t 5 L l2 = L – t & l1 = 3l2 (Given) 3 L   L L – t =  L – t 3  3  5 5–t =3 –t 5 5 – t = 15 – 5t 4t = 10  t = 2.5 hours = 150 minutes. l1 = L –

   

16. Let the total distance travelled = s

s /3 s /3 s /3   10 20 60 10 s s  = 180 18 s s  vav =  = 18 km h–1. t s /18 17. In the absence of gravitational force, the satellite will move with velocity v tangentially to the original orbit. Total time taken, t =

18. Since the tension in the string will be maximum when the particle is at the lowest point B

mv2 Then, TB = Weight of the particle + R Thus, the string will break at point B, when the average velocity v is increased. 19. By work-energy theorem, Loss of KE of the vehicle = Work done against retarding force = Retarding force × distance travelled

Loss in KE Retarding force As the both KE and retarding force are same, hence both truck and car would come to rest in the same distance.

21. Liquid flows from high pressure to low pressure. Hence, pressure of liquid in bigger diameter portion of tube is greater than in small diameter portion of tube. 22. Given, gm = 1.7 ms–2, ge = 9.8 ms–2, Tm = ?, Te = 3.5 s As,

Te = 2 

l ge

...(i)

and

Tm = 2 

l gm

...(ii)

From equation (i) & (ii),

Tm = Te

 

Tm = Te A ZX

23. We have, We have,

 

ge 9.8 = 3.5 = 8.4 s. gm 1.7

A – 4 Z – 2 + Y

A – 4 = 208 or, – 4 = 208 – 232

24 6 4 and 90 – 12 +  = 82 or =4 Hence,  = 6 and  = 4. or,  =

24. The refractive index of glass w.r.t air gn

a

=

 sin r =

sin i sin 45   sin r sin r sin 45  g

na

1

 sin r =

2





1/ 2 1.5

i = 45° Air Medium

r

 2 2 2   r = sin –1  . 3 3  3 

25. We have, L0 = Pr  L0 = mv cos  H = mg

 Distance travelled =

20. The arrangement shown in a series connection, whatever current enters at A to pass B. So, i = 2A, that resistance = 6 + 9 + 5 = 20  The effective potential drop across the resistances = 20 × 2 = 40 V The emf of two batteries (12 – 4 = 8V) is opposing the potential difference across AB. Therefore, the p.d. applied across AB = 40 + 8 = 48 V.

ge gm

=

3 v 2 sin 2 30  . 2 2g

v cos  H

3 mv 3 . 16 g

26. When resistance R is cut into five equal parts, each part will have a resistance of R/5. If these five resistances are connected in parallel the equivalent resistance.

1 1 1 1 1 1     = R eq R/5 R/5 R/5 R/5 R/5 5 5 5 5 5 25      R R R R R R R eq R 1 = or = . 25 25 R = Req

(2015) KVPY—Practice Paper—5

34

27. As,

T=

33. We have, 2.303 log K = 2.303 log A – Ea/RT Ea = 2.303 RT

2 u sin 2 u sin 2 = g g

u sin =1 g Now, Maximum height, or,

2.303 log K = 2.303 log A – log K = log A – 1

g  u 2 sin 2   10 2 u 2 sin 2    1 = 5 m. H = =  2  g 2  2 2g

Average velocity =

28.

 K K log   = –1  = 10–1.  A A

Displacement 0  0 Time taken 62.8

34. Since Sb 2S3 is a negatively charged solution, therefore, Al2(SO 4)3 is the most effective coagulating agent because higher the magnitude of opposite charge (Al 3+), higher is the coagulating power.

Distance 2 r  Time taken T 2   100 = = 10 ms–1. 62.8

Average speed =

1 K(2)2 2 1 2 and U = K(8) 2 Dividing equation (ii) by (i), we get,

29. We have,

U=

...(i)

2°

(vrms )He = (vrms )H 5 = 7

37. Mn2+ (3d5) has the highest stabilization due to half filled configuration. Fe2+ (3d6) is less stable because of one extra electron in half filled configuration. Cr2+ (d4) has one vacant subshell and Fe2+ is more stabilized as compared to Cr2+ because of exchange energy. Hence, correct order : Mn > Fe > Cr > Co.

5  5  2  273 = 278.57°C 77

31. Number of A ions per unit cell = 1 Number of B ions per unit cell =

1 6  3 2

 Empirical formula = AB3.

 

= = = =

PA 0 xA + PB 0 xB 520 xA + 1000 (1 – xA) 520 xA + 1000 – 1000 xA 240

240  0.5 480 Moles of A = 50%.

xA =

Br (B) (C) 1° carbocation Allyllic carbocation

(T) He  2 (273  0)  4

= 278.57 – 273 = 5.57°C.

Ptotal 760 760 480 xA

Br (A) 2° carbocation

Me Br

Me

36. Cl2O7, SO2 and P4O11 are the anhydrides of HClO 4, H2SO 3 and H 3PO 4 whose strengths are in the order HClO4 > H2SO 3 > H3PO 4.

2

32. We have,   

1°

Me

(T) He  (M) H (T) H  (M) He

(T) He  5   = 7 2  273 (T)He =

Me

It involves Order of stability of carbocation is Allyl > 3° > 2° > 1°  Order of reactivity is C > A > B.

U  8    = 16 U  2 U = 16 U.

30. We have,

35. Since formation of carbocation is the rate determining step in S N  reaction. Hence, the reactivity in SN  reaction depends upon the stability of the intermediate carbocations. The alkyl halide which gives more stable carbocation is more reactive towards SN reaction

...(ii)

2



2.303 RT RT

38. Hg2+ ion is the largest among Cu2+, Zn2+ and Al3+. Therefore, the bond between Hg and O is the weakest and most easily broken. 39. The structures of Cl2O–, Cl2O and ClO2 are as follows:  –O

– Cl  O

Cl – O – Cl

O  Cl  O 

The central atom in all these species is sp3 hybridised. In ClO 2, the repulsion between orbitals having lone pair of electrons and unpaired electron is less and hence bond angle is maximum in case of ClO 2. In ClO 2–, repulsion between bond pairs is greater than that in Cl2O. Hence, the bond angle in ClO2– is greater than that in Cl2O. The experimentally determined values of bond angles of Cl2O, ClO2– and ClO2 are 110.9°, 111.0° and 117.6° respectively. Hence, the correct order of increasing bond angle are : Cl2O < ClO2– < ClO2. (2015) KVPY—Practice Paper—5-II

35 40.

CHO

Therefore, it is an equilateral triangle So, the incentre coincides with centroid

CH = NNH C6H5

H – C – OH H – C – OH

C = NNH C6H5 C6H5 NHNH2

H – C – OH

H – C – OH

H – C – OH

CH2OH

CH2OH

Ribose

 0 1 2 0  0  3 I=  ,  3 3  

Osazone

In deoxyribose, one-OH group is missing, which will prevent the formation of osazone. 41. The addition of a proton at -carbon gives a carbocation (I) which is resonance stabilized because of electron donating effect of OH group. The addition of Br– ion to the carbocation gives the main product.   CH 3 – CH = CH

OH

+

[CH 3 – CH 2 – CH

CH 3 – CH 2 – CH

+

O H]

Br–

  CH 3 – CH 2 – CH Br

 2y   xz tan –1  = tan –1  2  1 – xz  1– y  xz xz = 1 – y2 1 – xz y2 = x2 Since, x, y and z are in an AP as well as in a G.P. x = y = z.

H+

OH (I)

 1  I =  1, .  3 62. Since, x, y and z are in an AP  2y = x + z Also, tan–1 x, tan–1 y and tan–1 z are in an AP 2 tan–1 y = tan–1 x + tan–1 z

OH Product

42. Given, current strength (I) = 1.5 A Time (t) = 10 × 60 = 600 s Quantity of electricity passed = I × t = 1.5 × 600 = 900 C Copper deposited as Cu2+ + 2e–  Cu (s) 2 mol of electrons or 2 × 96500 C of current deposit copper = 63.56 g  900 C of current will deposit copper

63.56  900 = 0.296 g. 2  96500 43. Osmotic pressure is preferred over the other colligative properties, because the polymers dissociate a high temperature and hence the use of other colligative properties will not give the precise value of the molar mass. =

44. The reaction, X  Y, follows second order Kinetic, hence the rate law equation will be Rate = KC2, where C = [X] If concentration of X increases three times, now [X] = 3C mol L–1  Rate = K(3C)2 = 9KC2 Hence, the rate of reaction will become 9 times, Thus, the rate of formation Y will increase 9 times. 45. Such substances, when dissolved in medium behave as colloidal particles at higher concentration due to formation of aggregated particles thus formed are called micelles. Micelles are aggregation of colloidal particles.





61. Let the vertices of triangle be A 1, 3 , B(0, 0) and C(2, 0). Here, AB = BC = CA = 2

63.    

3 sin2 x – 7 sin x + 2 3 sin2x – 6 sin x – sin x + 2 3 sin x (sin x – 2) – 1 (sin x – 2) (3 sin x – 1) (sin x – 2)

= = = =

0 0 0 0

1 [ sin x = 2 is rejected] 3

sin x =

 1 x = n  (–1) n sin –1   , n  I  3 For 0  n  5, x  [0, 5] There are six values of x  [0, 5] which satisfy the equation 3 sin2 x – 7 sin x + 2 = 0. 

64. Let a : b : c = 1 : 3 : 2  c2 = a2 + b2 B  Triangle is right angled at C or

C = 90° and

In  BAC, tan A =

a 1  b 3

a C

c A

b

a 1  b 3

 A = 30° and B = 60° [ A + B = 90°]  Ratio of angles A : B : C = 30° : 60° : 90° = 1 : 2 : 3. 65. Area of Quarter part of circle (DOB) =

(4)2  4 4

Area of 4 semi-circle = 4 

(1)2  2 2

1

36 Area of shaded region

71. Ionisation enthalpy of atomic hydrogen = E – E1 = 0 – (–2.18 × 10–18 J) = 2.18 × 10–18 J or 2.18 × 10–18 J atom –1 = 2.18 × 10–18 × 6.023 × 1023 = 1.31 × 106 J mol–1.

 (12 )  90 1   1  –  1  1 = 2   –   – 1 = 2 360  2  4 2 2   Required Area of shaded portion

= 4 – 2  66. We have, 

  5 – 1 = 2  – 1  – 1. 2 2 2

72. We have,

c1 sin i = c2 sin r

3  10 c2

8

=

O MgBr

sin 45 0.707  sin 30 0.5

+

COOH

c2 =

68. Given, m = 150 g, v = 20 ms–1, t = 0.1 s v 20 As, a=  = 200 ms–2 t 0.1 F = ma 

F=

150  200  30 N. 1000

69. When an iron sphere falls freely in a lake, its motion is accelerated due to gravity and retarted due to viscous force. The overall effect is increase in velocity and so, increase in kinetic energy till the sphere acquires terminal velocity which is constant. Hence, kinetic energy of the sphere, beyond this depth of lake becomes constant. 70. Since pressure is transmitted undiminished throughout the water F1 F2  A1 = A 2 Where F1 and F2 are the forces on the smaller and on the larger pistons respectively and A1 and A2 are the areas respectively.

+ Mg(OH) Br

73. We have, NaCl + AgNO3  AgCl + NaNO3 Given, Ksp = 1.5 × 10–10

1  10 –2 M = 0.5 × 10–2 M 2 1 –2 [Cl–] =  10 M = 0.5 × 10–2 M 2 Now, Kip = [Ag+] [Cl–] = (0.5 × 10–2) × (0.5 × 10–2) = 2.5 × 10–5 Kip > Ksp When K ip > Ksp, it results in precipitation. [Ag+] =

74. We have, po (pentane) = 440 mm Hg po (hexane) = 120 mm Hg

1  0.2 1 4 4  0.8 x(hexane) = 1 4 p(pentane) = 0.2 × 440 = 88 mm p(hexane) = 0.8 × 120 = 96 mm Total vap. pressure = 88 + 96 = 184 mm Mole fraction of pentane in vapour phase, x(pentane) =

y(pentane) =

 D2  A (D 2 / 2)2 F2 = 2 F1 = = F 1  D  F1 A1 (D1 / 2)2 1

Ph – C  C – CH3

(3  10 )

(1  10 –2 )2

Hg

2+

H2O, H+

Ph – C = CH CH3 OH

Tautomerises

–2 2

=

88  0.478. 184

75. We have,

2



H /H 2O

+ C=O

3  108  0.5 = 2.12 × 108 ms–1. 0.707 67. We have, a = r2,  = 2v Given, 22 revolutions = 44 sec 1 revolution = 44/22 = 2 s 1 v = Hz 2 42  a = r2 = 1  = 2 ms–2 4 It acts towards the centre, as the centripetal acceleration. 

C – OMg Br O

 10 N  90 N.

Ph – C – CH2 CH3 O

  

37

4

Practice Paper (Solved)

Kishore Vaigyanik Protsahan Yojana (KVPY) STREAM – SA

Mathematics 1. The polynomials ax3 + 3x2 – 13 and 2x3 – 5x + 9 are divided by x + 2, if the remainder in each case is the same. Then find the value of a: 5 5 A. B. – 9 9 2 2 C. D. – 7 7 2 2. If  and  are the roots of x + x + 2 = 0, then

10  10  –10   –10 A. 4096 C. 1024

6. If a sphere of the maximum volume is placed inside a hollow right circular cone with radius ‘r’ and slant height ‘l’ such that the base of the cone touches the sphere, then the volume of the sphere is:

is equal to: B. 2048 D. 512

A.

4  l  r   3  l –r

C.

4  l –r   3  l  r

3

3

B.

4 3  l –r2 r   l  r  3

D.

4 3  l  r 2 r   l – r  3

3

3

7. In the figure given below, a flower is inscribed in a circle of radius 1 cm, then find the Area of the flower.

3. If a, b, c and d are four positive numbers, then which of the following is not true? A. B. C. D.

a  a b  c d         4 b c d e e a  a b  b d         4. b c c e e a b c d e     5 b c d e a b c d e a 1      a b c d e 5

A.

 3 3 2   – 2  cm  

B.

2 – 3 3  cm

2

C.

 5 5 3  2  3 – 2  cm  

D.

4 – 3 3  cm

2

8. If e1 is the eccentricity of the conic 9x2 + 4y2 = 36 and e2 is the eccentricity of the conic 9x2 – 4y2 = 36, then:

4. The set of real values of x satisfying | x – 1 |  3 and | x – 1 |  1 is: A. [2, 4] B. (–, 2)  [4, ] C. [–2, 0]  [2, 4] D. None of these 5. There is a point inside a circle. What is the probability that this point is closer to the circumference than to the centre? 3 5 A. B. 4 7 2 1 C. D. 3 7

A.

e12 – e22  2

B.

2  e22 – e12  3

C.

e22 – e12  2

D.

e22 – e12  3

9. The mean deviation of the series a, a + d, a + 2d, .... a + 2n from its mean is: (n  1)d nd A. B. 2n  1 2n  1 (2 n  1)d n( n  1) d C. D. n(n  1) 2n  1 37

38 10. Three spheres are kept inside a cone, as given in the figure. Spheres are touching both the slant sides of the cone and the adjacent spheres. If the radius of the first sphere and the third sphere are 5 units and 20 units respectively, then find the radius of the second sphere. A. 10 units B. 12 units C. 15 units D. 18 units

2 1

12. In the figure given below CD, AE and BF are onethird of their respective sides. It follows that AN 2 : N2N1 : N1D = 3 : 3 : 1 and similarly, for lines BE and CE. Then, the area of triangle, N 1N2N3 is:

1  ABC 6

C.

1  ABC 9

D.

1  ABC 10

14. If you have 3 tickets of a lottery for which 10 tickets were sold and 5 prizes are to be given, then the probability that you will win at least one prize is: 7 9 A. B. 12 12 C.

N2 N1 C

D

1 12

D.

11 12

15. The set of real values of x satisfying | x – 1 |  3 and | x – 1 |  1 is: A. [2, 4] B. (–, 2)  [4, +] C. [–2, 0]  [2, 4] D. None of these

E

B

B.

13. A hollow right circular cylinder of radius r and height 4r is standing vertically on a plane. If a solid right circular cone of radius 2r and height 6r is placed with its vertex down in the cylinder, then the volume of the portion of the cone outside the cylinder is: 8 3 r A. B. 2r3 3 9 3 r C. D. 7r3 8

A

N3

1  ABC 7

3

11. Find the number of integral solutions of equation x + y + z + t = 25, x > 0, y > 1, z > 2 and t  0. A. 22C 3 B. 22C 5 C. 21C 3 D. 23C 7

F

A.

Physics 16. A ball is dropped vertically from a height d above the ground. It hits the ground and bounces up vertically to the height d/2. Neglecting subsequent motion and air resistance, its velocity v varies with the height above the ground as: v

v

A.

h

d

h

v

v

C.

d

B.

d h

D.

d

h

B. the mechanical efficiency of the machine increases C. both mechanical advantage and efficiency increase D. its efficiency increases, but its mechanical advantage decreases 18. If g is the acceleration due to gravity on the surface of the earth, the gain in the potential energy of an object of mass m raised from the surface of the earth to a height equal to the radius R of the earth is: 1 mgR A. mgR B. 2 1 mgR C. D. 2 mgR 4 19. What is the angle of projection for a projectile motion whose range R is n times the maximum height H?

 4 A.   tan –1    n 17. If a machine is lubricated with oil: A. the mechanical advantage of the machine increases

C.

 n   tan –1    4

B.  = tan–1 (n)

D.  = tan–1 (2n)

39 20. When a bicycle is in motion, the force of friction exerted by the ground on the two wheels is such that, it acts: A. in the forward direction on the front wheel and in the backward direction on the rear wheel B. in the backward direction on the front wheel and in the forward direction on the rear wheel C. in the forward direction on the both the front and the rear wheels D. in the backward direction on both the front and rear wheels 21. A mirror is inclinded at an angle of  with the horizontal. If the ray of light is incident at an angle of incidence , then the reflected ray makes the following angle with horizontal:  

A. 0 C. 2

H

B.  D. /2

22. In an electrical circuit three incandescent bulbs A, B and C of rating 40 W, 60 W and 100 W respectively are connected in parallel to an electric source. Which of the following is likely to happen regarding their brightness? A. Brightness of all bulbs will be the same B. Brightness of bulb A will be the maximum C. Brightness of bulb B will be more than that of A D. Brightness of bulb C will be less than that of B 23. No matter how far you stand from a spherical mirror, your image appears erect. The mirror may be: A. plane B. concave C. convex D. either plane or convex 24. A particle is projected from a point A with velocity u 2 at an angle of 45° with horizontal as shown in figure. It strikes the place BC at right angles. The velocity of the particle at the time of collision is: u 2

C 60°

45° A

B

A. u

B.

u 2 2u

3u D. 3 2 25. The effective resistance between points A and B is: C.

R

C.

C 2R

2R R

D

B

2R 3

B.

R 3

D.

3R 5

26. The minimum size of the mirror fixed on the wall of a room in which an observer at the centre of room may see the full image of the wall of height h behind him is: A.

h 3

B.

C.

2h 3

D. h

M

Incident ray

A

A. R

h 2

27. A light and heavy body have same kinetic energy, which of these will have higher momentum? A. light body B. heavier body C. both have equal momentum D. none of these 28. A point object O and a mirror M move with velocities of 3 cms–1 and 4 cms–1 respectively as shown in figure. O O  is the normal to the mirror. What is the velocity of the image?

–1

3 cms 60° O

O'

M

A. B. C. D.

5 7 8 9

cms–1 cms–1 cms–1 cms–1

29. A satellite is moving in a circular orbit at a height 100 km above the surface of the earth. A person inside the satellite feels weightless because: A. the centripetal force makes the satellite move in the circular orbit B. acceleration due to gravity is almost zero at such height C. the force due to earth and the moon are almost compensated at a such height D. the centripetal force makes the satellite move in a circular orbit 30. Which part of the solar cooker is responsible for greenhouse effect? A. Coating with black colour inside the box B. Mirror C. Glass sheet D. Outer cover of the solar cooker

40

Chemistry 31. Two elements X and Y from two compounds X 2Y3 and XY 2. If 0.3 mole of X 2Y3 weighs 31.8 g and 0.2 mole of XY 2 weighs 12.4 g. What are the atomic weights of X and Y? A. 18, 26 B. 26, 18 C. 12, 18 D. 16, 28

39. Identify the major products (P) and (Q) in the following reaction: + CH3CH2 CH2 Cl

An hyd. AlCl3

(P)

(i) O2 (ii) H3O+ , 

32. One gram of hydrogen is found to combine with 80 g of bromine. One gram of calcium (valency = 2) combines with 4 g of bromine. The equivalent weight of calcium is: A. 30 B. 40 C. 60 D. 20

A.

and CH3 CO CH3

B.

and CH3 CH2 CHO

33. If the molecule of HCl were totally polar, the expected value of dipole moment is 6.12 D (debye), but the experimental value of dipole moment is 1.03 D. The percentage ionic character is: A. 12 B. 17 C. 25 D. 30

C.

and CH3 COCH3

D.

and CH3 CH2 CHO

34. The increasing order of atomic radius for the elements Na, Rb, K and Mg is: A. Na < K < Mg < Rb B. K < Na < Mg < Rb C. Na < Mg < K < Rb D. Mg < Na < K < Rb 35. Dinitrogen and dioxygen are main constituents of air, but these do not react with each other to form oxides of nitrogen because: A. N2 and O2 are unreactive B. Oxides of nitrogen are unstable C. the reaction is endothermic and requires very high temperature D. the reaction can be initiated only in presence of a catalyst

(Q) + PhOH

40. Conversion of benzene to acetophenone can be brought by: A. Friedel Crafts acylation B. Friedel Crafts alkylation C. Wurtz reaction D. Wurtz-Fittig’s reaction LiAlH

Cu

dilute

4  X   Y  Z 41. CH 3 COOH  300°C NaOH

In above reaction Z is: A. Butanol C. Aldol

B. Acetal D. Ketol

42. In an experiment, 0.04 F was passed through 400 mL of a 1 M NaCl solution. What would be the pH of the solution after electrolysis? A. 13 B. 15 C. 8 D. 7

36. The decreasing order of acidic character among ethane (I), ethene (II), ethyne (III) and propyne (IV) is: A. III > IV > II > I B. I > II > III > IV C. IV > III > II > I D. II > III > I > IV

43. On addition of 1 mL solution of 10% NaCl to 10 mL sold solution in the presence of 0.0250 g of starch, the coagulation is just prevented. Starch has the gold number. A. 2.5 B. 25 C. 0.25 D. 0.025

37. For preparing an alkane, a saturated solution of sodium or potassium salt of a carboxylic acid is subjected to: A. hydration B. dehydration C. oxidation D. electrolysis

44. Oxidation states of the metal in the minerals haematite and magnetite, respectively are: A. II, III in haematite and II in magnetite B. III in haematite and II, III in magnetite C. II, III in haematite and III in magnetite D. II in haematite and II, III in magnetite

38. When neopentyl bromide is subjected to Wurtz reaction, the product formed is: A. 2, 2, 3, 3 -tetramethyl hexane B. 2, 2, 5, 5 -tetramethyl hexane C. 2, 2, 4, 4 -tetramethyl hexane D. 2, 2, 4, 4 -tetramethyl pentane

45. Bithional is added to soap as an additive to function as a/an: A. antiseptic B. dryer C. hardener D. softner

41

Biology 46. World Ozone Day is celebrated on: A. 5th June B. 22nd April C. 16th September D. 21st April 47. In stratosphere, which of the following elements acts as a catalyst in degradation of ozone and release of molecular oxygen? A. Carbon B. Oxygen C. Fe D. Cl 48. What type of ecological pyramid would be obtained with the following data? Secondary consumer : 120 g Primary consumer : 60 g Primary producer : 10 g A. Inverted pyramid of biomass B. Upright pyramid of biomass C. Upright pyramid of numbers D. Pyramid of energy 49. Natality refers to: A. Death rate B. Number of individuals entering a habitat C. Number of individuals leaving the habitat D. Birth rate 50. Which of the following has proved helpful in preserving pollen as fossils? A. Pollenkitt B. Sporopollenin C. Oil content D. Cellulosic intine 51. Which of the following pairs is A. Starch synthesis in pea B. T.H. Morgan C. XO type sex Determination D. ABO blood grouping 52. Select the correct match: A. Alec Jeffreys – B. Francois Jacob and – Jacques Monod C. Matthew Meselson – and F. Stahl D. Alfred Hershey – and Martha Chase

wrongly matched? : Multiple alleles : Linkage : Grasshopper : Co-dominance

Streptococcus pneumoniae Lac operon

53. Which of the following flowers only once in its lifetime? A. Bamboo species B. Papaya C. Mango D. Jackfruit 54. Select the correct statement: A. Franklin Stahl coined the term “linkage” B. Transduction was discovered by S. Altman C. Spliceosomes take part in translation D. Punnett square was developed by a British scientist 55. Offsets are produced by: A. Meiotic divisions B. Parthenogenesis C. Parthenocarpy D. Mitotic divisions 56. The experimental proof for semiconservative replication of DNA was first shown in a: A. Fungus B. Virus C. Plant D. Bacterium 57. Select the wrong statement: A. Cell wall is present in members of Fungi and Plantae B. Mitochondria are the powerhouse of the cell in all kingdoms except Monera C. Pseudopodia are locomotory and feeding structures in Sporozoans D. Mushrooms belong to Basidiomycetes 58. Casparian strips occur in: A. Epidermis B. Endodermis C. Cortex D. Pericycle 59. Which of the following statements is correct? A. Ovules are not enclosed by ovary wall in gymnosperms B. Stems are usually unbranched in both Cycas and Cedrus C. Horsetails are gymnosperms D. Selaginella is heterosporous, while Salvinia is homosporous 60. Pneumatophores occur in: A. Halophytes B. Submerged hydrophytes C. Carnivorous plants D. Free-floating hydrophytes

Pisum sativum TMV

Mathematics 61. The set of real values of x satisfying | | x – 1 | – 1 |  1 is: A. [–1, 3] B. [0, 2] C. [–1, 1] D. None of these

62. (x2 – 1) is a factor of f (x) = (x5 + ax4 + bx3 + cx2 + x + d). The graph of f (x) intersects y-axis at (0, –3), Find the value of (a + c). A. 0 B. 3 C. –3 D. –1 (2015) KVPY—Practice Paper—6

42 4 th of the 5 height from which it has fallen. What is the total distance that of travels before coming to rest if it is gently dropped from a height of 120 m? A. 1020 m B. 1080 m C. 1180 m D. None of these

63. After striking the floor a ball rebounds to

 3 – 6 4

B.

3  – 4 12

D.

3  – 4 8

65. The piece of paper in the shape of a sector of a circle (see Fig. 1) is rolled up to form a right circular cone (see Fig. 2). The value of the angle  is: 

64. A semi-circle of diameter 1 unit sits at the top of a semi-circle of diameter 2 units. The shaded region inside the smaller semicircle, but outside the larger semicircle is lune. The area of lune is: A.

C.

12

1 unit 5 Fig. 2

Fig. 1

2 unit

3  – 4 24

A.

10  13

B.

9 13

C.

5 13

D.

6 13

Physics 66. A rocket is fired from the earth to the moon. The distance between the earth and the moon is r and the mass of the earth is 81 times the mass of the moon. The gravitation force on the rocket will be zero, when its distance from the moon is: r r A. B. 2 5 r r C. D. 10 15 67. A copper block of mass 2.5 kg is heated in a furnace to a temperature of 500°C and then placed on a large ice block. What is the maximum amount of ice that can melt? Specific heat of copper is 0.39 Jg–1C–1, Heat of fusion of water = 335 Jg–1. A. 1.0 kg B. 1.2 kg C. 1.5 kg D. 2.0 kg 68. A body of mass 2.0 kg makes an elastic collision with another body at rest and continues to move in the original direction with a speed equal to half its original speed. What is the mass of the second body?

A. 0.32 kg C. 0.50 kg

B. 0.67 kg D. 0.75 kg

69. A scooterist moving with a speed of 36 kmh–1, sees a child standing in the middle of the road. He applies brakes and brings the scooter to rest in 5 s, just in time to save the child. Then, the average retarding force on the vehicle, if mass of vehicle, and the driver is 300 kg. A. 100 N B. 200 N C. 300 N D. 600 N 70. Two charges ± 20 C are placed 10 mm apart. The electric field at point P, on the axis of the dipole 10 cm away from its centre O on the side of the positive charge is: A

O

–20  C

E

+20 C

NC–1

B. 2.4 × 106 NC–1

C. 3.6 × 106 NC–1

D. 5.2 × 105 NC–1

A. 1.2 ×

109

B

Chemistry 71. A 0.0020 m aqueous solution of an ionic compound Ca(NH 3)5(NO 2) Cl freezes at –0.00732°C. Number of moles of ions which 1 mol of ionic compound–produces on being dissolved in water will be (Kf = 1.86°C/m) A. 2 B. 3 C. 4 D. 5 72. An open flask contains air at 27°C. At what temperature should it be heated so that 1/3rd of air present in it goes out? A. 120°C B. 150°C C. 177°C D. 199°C

73. The rate of a gaseous reaction becomes half when volume of the vessel is doubled. What is the order of the reaction? A. 0 B. 1 C. 2 D. 3 74. Acetone is treated with excess of ethanol in the presence of hydrochloric acid. The product obtained is: A. (CH 3 )2 C

OH OC 2H 5

43 75. A compound with molecular mass 180 is acylated with C H 3COCl to get a compound with molecular mass 390. The number of amino groups present per molecule of the former compound is: A. 2 B. 5 C. 7 D. 9

O

B. CH 3CH2CH 2 – C – CH2 CH2 CH3 O

C. CH 3CH2CH 2 – C – CH3 D. (CH3)2 C

OC2H5 OC2H5

Biology 76. Sweet potato is a modified: A. Stem B. Rhizome C. Tap root D. Adventitious root

B. Cycads C. Conifers D. Deciduous angiosperms

77. Secondary xylem and phloem in dicot stem are produced by: A. Apical meristems B. Axillary meristems C. Phellogen D. Vascular cambium 78. Plants having little or no secondary growth are: A. Grasses

79. Annual migration does not occur in the case of: A. Arctic tern B. Salmon C. Siberian Crane D. Salamander 80. Which one of the following is not a living fossil? A. Peripatus B. King Crab C. Sphenodon D. Archaeopteryx

ANSWERS 1 A

2 C

3 D

4 C

5 A

6 B

7 B

8 B

9 C

10 A

11 A

12 A

13 D

14 D

15 C

16 B

17 B

18 B

19 A

20 B

21 A

22 C

23 D

24 D

25 A

26 A

27 B

28 B

29 A

30 C

31 B

32 D

33 B

34 D

35 C

36 A

37 D

38 B

39 A

40 A

41 C

42 A

43 B

44 B

45 A

46 C

47 D

48 A

49 D

50 B

51 A

52 B

53 A

54 D

55 D

56 D

57 C

58 B

59 A

60 A

61 A

62 B

63 B

64 B

65 A

66 C

67 C

68 B

69 D

70 C

71 A

72 C

73 B

74 D

75 B

76 D

77 D

78 A

79 D

80 D

EXPLANATORY ANSWERS P1(x) = ax3 + 3x2 – 13 P2(x) = 2x3 – 5x + a Since, P1(x) and P2(x) divided by (x + 2), then P1(–2) = P2(–2) a(–2)3 + 3(–2)2 – 13 = 2(–2)3 – 5(–2) + a –8a + 12 – 13 = –16 + 10 + a –8a – 1 = –6 + a 5 a= . 9

1. We have,

2. 

Now,

x2 + x + 2 = 0  +  = –1  = 2

10  10 

–10



–10

=

10  10 10  10 ()10

= ()10 = 210 = 1024. 3. Take any four positive numbers, then check the options.

44 4.

|x – 1|  3 –3  x – 1  3 –2  x  4

|x – 1| 1 x – 1  –1, x – 1  1 x  0, x  2

–2

0

2

4

[–2, 0]  [2, 4]. 5. Assume that the radius of the bigger circle is r, and the

r =

r  l2 – r 2 (l  r )

r =

r (l  r )(l – r ) r (l – r ) 1/ 2  (l  r ) (l  r )1/ 2

Area of sphere =

4 4 (l – r )3 / 2  (r )3 = r 3 . 3 3 (l  r )3 / 2

10. We can see in figure 1 that the AOAP, R

r , point will be closer to 2 the circumference than to the centre of the point is lying in the segment B. radius of the inner circle is

Area of segment B =

C

C'

Q B'

B P A'

A

3 2 r 4

O Fig. 1

BOBQ, and COCR will be similar

A

r /2

C B

B A

Hence, the probability of point being closer to the circumference

3 2 r 3 = 4 2  . 4 r

O

In fig-2,

r

O 2r

C

OA2 = AB2 – OB2 h 2 = l2 – r2 h=

l2 – r 2 .

We know that, In radius × Semi-perimeter = Area of the triangle r× s=

l  l  2r  lr 2 1 = × Base × height 2 1 2 2 =  (2r )  l – r 2 S=

r × (l + r) = r  l 2 – r 2

r2 – r1 r –r = 3 2 = K r2  r1 r3  r2

Hence, the three radii are in a GP

l

r

R

r3 r2 = r2 r1

r' B

Q

Using Componendo and Dividendo,

A

l

P Fig. 2

6. Consider the biggest cross-section of the cone as a isosceles triangle, therefore the circle inscribed in the triangle will be the biggest cross-section of the sphere.

l

r3 r2

r1

So, and hence,

r2 5 = 20 r2 r22 = 100 r2 =

100 = 10 units.

11. Given, x + y + z + t = 25 Where, x  1, y  2, z  3, t  0 Let, p = x – 1, q = y – 2, r = z – 3 and s = t Then, p + q + r + s = x + y + z + t – 6 = 25 – 6 = 19 Where, p, q, r, s  0  p + q + r + s = 19 p, q, r, s  0  Required number = Number of ways in which 19 identical things can be distributed among 4 persons when each person can get any number of things = n + r – 1Cr – 1 = 22C3.

45 13.  ECD is similar to  EAB, Hence, EC = 3r A

2r

Since, velocity is downwards (–ve). After collision, it reverses the direction with a smaller magnitude. But, the velocity is upwards (+ve). Graph (a) satisfies all these conditions. Moreover, at time t = 0, h = d From 0 to 1, v increases downwards At 1 to v, reverses its direction From 1 to 2, v decreases upwards After that, the next collision occurs.

B

C r

D

6r

4r E

17. When a machine is lubricated with oil, energy wastage against friction decreases. This increases the mechanical efficiency of the machine. Hence, (B) is correct answer.

r

18. On the surface of earth,

Then outside volume of Cone

1 1 (2r )2 (6r ) – (r 2 )  3r 3 3 3 3 = 8r – r = 7r3. =

14. Probability that you will win at least one prize = 1 – Probability that you will not win any prize 5

= 1–

C3

10

C3



11 . 12

|x – 1| 3 –3  x – 1  3 –2  x  4

15. or or

GMm ...(i) R At a height equal to radius of earth, GMm GMm – U2 = – ...(ii) Rr 2R  U = U2 – U1 GMm GMm GMm   = – 2R R 2R GM 1 But, g=  U = mgR . 2 R2 U1 = –

19. Given, R = nH or

–

–2

+

4

u 2 sin 2  u 2  2 sin  cos  = n. 2g g 4 or tan  = n

For |x – 1| 1 Solving this, we get, (x – 1)  –1 or (x – 1)  1 or x  0 or x  2 0

–

2

or

+



Combining both we get, –

–2

0

2

4

u 2 sin 2  u 2 sin 2 = n. 2g g

+

 4  = tan –1   .  n

20. When a bicycle is in motion, the force of friction acts in the backward direction on the front wheel and in the forward direction on the rear wheel. 21. Clearly, from the figure, the reflected ray is parallel to horizontal.

So, common set is [–2, 0]  [2, 4]. 16. For uniformly accelerated or decelerated motion, v2 = u2 + 2gh Hence, the v-h graph is a parabola At, t = 0, h = d

  

u 1

 For parallel, angle,  = 0. d 2 0

h

22. When bulbs are connected to their rated potential difference, greater the power, greater is the brightness. Hence, brightness of bulb B will be more than that of A.

46 23. In the both cases, the plane mirror or the convex mirror, the image formed is erect whatever may be the position of object from the mirror. Hence, mirror may be either plane or convex. 24. Let v be the velocity at the time of collision Then, u 2 cos 45 = v sin 60°

u 2  

1   = 2



3v 2 2

v=

u.

3 25. Here, points B and D are commons. So, 2R in arm DC and 2R in arm CB are in parallel between C and B. Their effective resistance

30. The glass sheet has a peculiar property that does not allows infrared radiations of large wavelength to pass through it. 31. Let the atomic weights be X and Y respectively.  0.3 × 2 × X + 0.3 × 3 × Y = 31.8 and 0.2 × 1 × X + 0.2 × 2 × Y = 12.4 On solving above equations, we get, X = 26 and Y = 18. 32.  One gram of hydrogen combines with 80 g of bromine  Equivalent weight of bromine = 80  4 g of bromine combines 1 gm of calcium  80 g of bromine combines

80  1 g of calcium 4 = 20 g  Equivalent weight of calcium = 20.

2R  2R  R. 2R  2R 26. From  OM 1M 2 and OAB,

=

=

A'

A

33. Percentage of ionic character

M1 h

y

O

x

x

O'

M2 B'

B

x M 1M 2 = 2x  y h hx M 1M 2 = (2 x  y)

Size of mirror, If,

x = y, then, M 1M 2 =

h . 3

27. Since, p  2 mE Then, kinetic energy in same. Hence, particle of heavier mass will have higher momentum. 28. Only velocity of the mirror along, OO = 2 cms–1 is relevant Relative velocity : velocity = 5 cms–1 4 cm s

O

–1

3 cm s –1 6 0°

experimental value of dipole moment  100 theoretical value of dipole moment 1.03  100  16.83%  17%. = 6.12 34. In alkali metal group, when we move down the group, atomic radius increases and while moving in the period from left to right atomic radius decreases due to increase in nuclear charge. So, the sequence is: Mg < Na < K < Rb. =

O'

M

 Velocity of the image relative to mirror = 5 cms–1 towards left Relative to laboratory frame : velocity of image = 7 cms–1. 29. When a satellite is moving in a circular orbit, the centripetal force makes the satellite to move in circular orbit. Now, the gravitational pull on a body providing centripetal force is balanced by centrifugal force. Due to which the person inside the satellite feels weightlessness.

35. Dinitrogen and dioxygen do not react with each other at a normal temperature. At high temperature (in an automobile engine), when fossil fuel is burnt, these gases combine to yield significant quantities of NO and NO2. 36. The correct order of decreasing acidity is: CH  CH > CH  C – CH 3 > CH 2 = CH2 > CH 3 – CH 3 III IV II I Hydrogen atoms attached to sp-hybridised carbon-atom are most acidic, followed by those attached to sp 2 hybridised carbon atom and those attached to sp 3 hybridised carbon are least acidic. 37. Alkanes are prepared from carboxylic acids by electrolysis. In Kolbe’s electrolysis, an aqueous solution of sodium or potassium salt of carboxylic acid is electrolysed and alkane is evolved at the anode. 38.

CH3

CH 3

CH3

2 – CH3 C – CH2 Br + 2 Na  CH3 – C – CH2 – CH2 – C – CH3 CH3 Neopentyl bromide

CH 3

CH3

2, 2, 5, 5-tetra methyl hexane

47 CH 3

61. | | x – 1 | – 1 |  1 It is possible when | x – 1 |  2 So, x could be  3 and  –1 So, set of real values of x is [–1, 3].

CH 3 CH

+ CH3CH2 CH2 Cl

39.

Anhyd. AlCl 3

Cumene P

62. x = 1 and x = –1 are two roots of f (x). Similarly, if we put x = 0, then f (x) = –3, Now solve the equation to get the value of a + c.

+ CH3 COCH3

63. So, the total distance covered

OH (i) O2 +

(ii) H3O , 

Acetone Q

120 4/5

40. We have, 120 m

COCH 3 + CH 3 COCl Acetyl Benzene chloride

Floor 120 ×

Anhyd.AlCl 3

Floor

= 120  2  120 

41. We have, Li Al H 4

CH 3 CH 2 OH

Cu 300°C

CH 3 CHO

dilute NaOH

CH3 – C –CH2 – CHO

4 4 4  2  120    ... 5 5 5

 4    = 120  240  5   1080 m. 4 1 –   5 

H Aldol

1 1 H2 + Cl2 2 2 1 F of electricity will produce 1 mole of OH – ions 0.04 F of electricity will produce = 0.04 mole of OH– ions. 0.04  1000  0.1 Molarity of solution formed = 400 – [OH ] = 0.1

42. NaCl (aq) + H 2O(l)  Na+ + OH–(aq) +

B

r2 1    .  = 2 2 3 6

A

1 1

1 O

 area of shaded region = – area of  OAB 6

10 –14  10 –13 0.1 pH = –log [H+] = –log (10)–13 = 13.

=

 3 – 6 4

Hence, area of lune = Area of semi-circle – area of shaded region

43. Gold number is the number of milligrams of protective colloids required to prevent the coagulation of 10 mL of gold sol., when 1 mL of a 10% NaCl is added to it.  Gold number = 0.025 × 1000 = 25.

45. An antiseptic is a substance which prevents the growth of micro-organisms as long as it remains in contact with them. Antiseptics do not harm the living tissues and can be applied on cuts and wounds. Bithional is used as antiseptic in medicated soap.

C

64. Area of sector OACB

[H +] =

44. In haematite (Fe2O3) oxidation number of Fe is = 2x + 3 × (–2) = 0  x=3 Magnetite (Fe3O4) is an equimolar mixture of FeO and Fe2O3  Oxidation number of iron is FeO is 2 and in Fe2O3 is 3.

Floor

4 4 4  = 120  2  120      ... 5 5 5 

OH



4 5

Acetophenone

This reaction is called Friedel Crafts acylation reaction.

CH 3 COOH

4 5

2  1  1 3    –  – =  2 2 6 4 

=

3    – = 4 8 6

65. 12



5

 

Slant Height = 13 by l = r 2 × 5 = 13 ×  =

10  . 13

3  – . 4 24

48 66. Let x be the distance of rocket from the moon where gravitational force on rocket is zero due to the moon and the earth. It will be so when the gravitational force on rocket due to the moon is equal and opposite to the gravitational force on rocket due to the earth. GMm G81M  m So, = x2 (r – x )2 81 1 or = 2 (r – x )2 x 

1 9 r = or x = . x r–x 10

mc T 2500  0.39  500  L 335 = 1500 g = 1.5 kg.

or  

(m1 – m2 )u1  2 m1u2 m1  m2

v (2 – m2 )v  2 m2  0 = 2 2  m2 2 + m2 = 2(2 – m2) = 4 – 2m2 3m2 = 4 – 2 = 2 m2 =

From,

 2| p|

=

2  2  10 –7  9  10 9

(10  10 –2 )3 4  0 r 2 = 3.6 × 106 NC–1.

71. The number of moles produced by 1 mole of ionic compound Tf = iKf m, i = Tf / Kfm Tf = 0 – (–0.00732) = 0.00732 °C Kf = 1.86 °C/m, m = 0.0020 i=

n

2 = 0.67 kg. 3 = 10 m/s, v = 0, t = 5s, f = ?,

69. Given, u = 36 kmh–1 m = 300 kg As, v = u + at 0 = 10 – a × 5 

 E =

73. For the gaseous reaction, A  products Suppose the order of reaction is n, so that rate = K [A]n or r = K [a]n ...(i) When volume of vessel becomes double, molar concentration becomes half. Since rate becomes half, therefore,

68. Given, m1 = 2 kg, u1 = v, m2 = ?, u2 = 0, v1 = v/2





0.00732  2. 1.86  0.0020 72. Let initial number of moles of air at 27°C (300 K) = n At temperature TK, the number of moles left n 2n = n–  3 3 At constant pressure and volume, n 1T 1 = n 2T 2 2n T  n × 300 = 3  T = 450 K or 177°C.

m =

v1 =

a W C. T = 0 D. T = W

23. If the resultant of all the external forces acting on a system of particles is zero, then from an inertial frame, one can surely say that: A. linear momentum of the system does not change in time B. kinetic energy of the system does not change in time C. angular momentum of the system does not change in time D. potential energy of the system does not change in time 24. If voltage across a bulb rated 200 V-100 W drops by 2.5% of its rated value, the percentage of the rated value by which the power would decrease is: A. 5% B. 10% C. 15% D. 20% 25. Which of the following statement is not a similarity between electrostatic and gravitational forces? A. both forces are central in nature B. both forces are conservative in nature C. both forces obey inverse square law D. both forces operate over very large distances 26. If Q, E and W denote respectively the heat added, change in internal energy and the work done in a closed cyclic process, then:

A. Q = 0 C. E = 0

B. W = 0 D. Q = W = 0

27. In torch, search light and headlight of vehicles, the bulb is placed: A. very near to the focus of the reflector B. between the pole and the focus of the reflector C. between the focus and centre of curvature of the reflector D. at the centre of curvature of the reflector 28. Fig. shows a ray of light as it travels from medium A to B. Refractive index of the medium B relative to medium A is: 45° 45°

Medium B 60°

30° Medium A

A. C.

2 3 1

2

B.

3 2

D.

2

29. A 600 W motor drives a pump which lifts 200 litres of water per minute to a height 6 m. The efficiency of the motor is nearly. (Take, g = 10 ms–2) A. 23% B. 33% C. 41% D. 25% 30. When a radioactive isotope 88Ra228 decay in series by the emission of 3  particles and -particle, the isotope finally formed: A. 83X216 B. 84X228 215 C. 83X D. 86X222

Chemistry 31. Three elements A, B and C crystallise into a cubic solid lattice. Atoms A occupy the corners, B occupy the cube centres and C occupy the edges. The formula of the compound is: A. AB 2C B. ABC 2 C. ABC 3 D. ABC 4 32. Concentrated aqueous sulphuric acid is 98% H 2SO 4 by mass and has a density of 1.80 gmL–1. Volume of acid required to make one litre of 0.1 M H 2SO 4 is: A. 2.55 mL B. 5.55 mL C. 3.55 mL D. 8.55 mL 33. For a first order reaction A  B, the reaction rate at reactant concentration of 0.01 M is found to be 2.0 × 10–5 mol L–1 s–1. The half-life period of the reaction is: A. 115 s B. 138 s C. 347 s D. 428 s

34. What would be the product formed when 1-bromo-3chloro cyclobutane reacts with two equivalents of metallic sodium is ether? A.

B.

C.

D.

Cl Br

35. In the reaction: OCH3

HBr

the products are:

A.

OH and CH3 Br

B.

Br and CH 3OH

C.

Br and CH 3 Br

D. Br

OCH3 and H2

52 36. Which of the following will be most readily dehydrated under acidic condition? O

O

A.

OH

B. OH O OH

C.

D. OH

37. The correct order of the bond energy is: A. I2 > Br2 > Fe > Cl2 B. Cl2 > Br2 > F2 > I2 C. I2 > Br2 > Cl2 > F2 D. Cl2 > F2 > Br2 > I2 38. A 4 : 1 mixture of helium (He) and methane (CH 4) is contained in a vessel at 10 bar pressure. Due to a hole in the vessel, the gas mixture leaks out. The composition mixture effusing out initially is: A. 1 : 2 B. 4 : 1 C. 8 : 1 D. 3 : 8 39. In which of the following, the oxidation number of oxygen has arranged in increasing order? A. OF2 < KO2 < BaO2 < O3 B. KO2 > OF2 < O3 < BaO2 C. O3 < OF2 < KO2 < BaO2 D. BaO 2 < KO2 < O3 < OF2

41. An electric current is passed through silver voltmeter connected to a water voltmeter. The cathode of silver voltmeter weighed 0.108 g more at the end of electrolysis. The volume of O2 at STP evolved is: A. 2.3 cm 3 B. 4.5 cm 3 3 C. 11.2 cm D. 5.6 cm 3 42. In a reaction, 2A + B  A 2B, the reactant B will disappear at: A. the same rate as A will decrease B. half the rate as A will decrease C. half the rate as A 2B will form D. twice the rate as A will decrease 43. Which of the following is the strongest acid? A. SO 2(OH)2 B. SO(OH)2 C. ClO3(OH) D. ClO2(OH) 44. The polymer containing strong intermolecular forces e.g., hydrogen bonding is: A. Nylon 6, 6 B. Teflon C. Natural rubber D. Polystyrene A 45. H 3 C – CH – CH  CH 2  HBr  | CH3

A (predominantly) is: A. CH 3 – CH – CH 2 – CH 2 Br

40. Consider the following aqueous solution and assume 100% ionization in electrolytes: (i) 0.1 m urea (ii) 0.04 m Al2(SO 4)3 (iii) 0.05 m CaCl2 (iv) 0.005 NaCl The correct statement regarding the above solution is: A. vapour pressure will be highest for solution (ii) B. freezing point will be lowest for solution (i) C. freezing point will be highest for solution (iv) D. boiling point will be highest for solution (iv)

B.

C. D.

| CH3 CH 3 – CH – CH – CH 3 | | CH3 Br Br | CH 3 – C – CH 2 CH 3 | CH3 CH 3 – CH – CH – CH 3 | | Br CH3

Biology 46. Which one of the following statements is correct with reference to enzymes? A. Apoenzyme = Holoenzyme + Coenzyme B. Holoenzyme = Apoenzyme + Coenzyme C. Coenzyme = Apoenzyme + Holoenzyme D. Holoenzyme = Coenzyme + Co-factor

49. Which among the following are the smallest living cells, known without a definite cell wall, pathogenic to plants as well as animals and can survive without oxygen? A. Bacillus B. Pseudomonas C. Mycoplasma D. Nostoc

47. Which cells of ‘Crypts of Lieberkuhn’ secrete antibacterial lysozyme? A. Argentaffin cells B. Paneth cells C. Zymogen cells D. Kupffer cells

50. Which one from those given below is the period for Mendel’s hybridization experiments? A. 1856 - 1863 B. 1840 - 1850 C. 1857 - 1869 D. 1870 - 1877

48. Phosphoenol pyruvate (PEP) is the primary CO 2 acceptor in: A. C3 plants B. C4 plants C. C2 plants D. C3 and C4 plants

51. Flowers which have single ovule in the ovary and are packed into inflorescence are usually pollinated by: A. Water B. Bee C. Wind D. Bat

53 52. Asymptote in a logistic growth curve is obtained when: A. The value of ‘r’ approaches zero B. K = N C. K > N D. K < N 53. Out of ‘X’ pairs of ribs in humans only ‘Y’ pairs are true ribs. Select the option that correctly represents values of X and Y and provides their explanation: A. X = 12, True ribs are attached dorsally to vertebral Y = 7 column and ventrally to the sternum. B. X = 12, True ribs are attached dorsally to verteY = 5 bral column and sternum on the two ends. C. X = 24, True ribs are dorsally attached to verteY = 7 bral column but are free on ventral side. D. X = 24, True ribs are dorsally attached to verteY = 12 bral column but are free on ventral side.

56. Among the following characters, which one was not considered by Mendel in his experiments on pea? A. Stem - Tall or Dwarf B. Trichomes - Glandular or Non-glandular C. Seed - Green or Yellow D. Pod - Inflated or Constricted 57. Which of the following cell organelles is responsible for extracting energy from carbohydrates to form ATP? A. Lysosome B. Ribosome C. Chloroplast D. Mitochondrion 58. If there are 999 bases in an RNA that codes for a protein with 333 amino acids, and the base at position 901 is deleted such that the length of the RNA becomes 998 bases, how many codons will be altered? A. 1 B. 11 C. 33 D. 333

54. MALT constitutes about ............... per cent of the lymphoid tissue in human body. A. 50% B. 20% C. 70% D. 10%

59. Which of the following are found in extreme saline conditions? A. Archaebacteria B. Eubacteria C. Cyanobacteria D. Mycobacteria

55. Homozygous purelines in cattle can be obtained by: A. mating of related individuals of same breed. B. mating of unrelated individuals of same breed. C. mating of individuals of different breed. D. mating of individuals of different species.

60. Receptor sites for neurotransmitters are present on: A. membranes of synaptic vesicles B. pre-synaptic membrane C. tips of axons D. post-synaptic membrane

Mathematics 61. If the sum of the distance perpendicular lines in a plane A. Square B. C. Straight D.

of a point from two is 1, then its locus is: Circle Two intersecting lines

62. The Area (in sq. units) bounded by the curves y = | x | – 1 and y = – | x | + 1 is: A. 1 B. 2 C. 2 2 D. 4

64. If in a PQR, sin P, sin Q, sin R are in AP then: A. the altitudes are in AP B. the altitudes are in HP C. the medians are in GP D. the medians are in AP 65. Seven white balls and three black balls are randomly placed in row. The probability that no two black balls are placed adjacently equals:

63. The number of orderd pairs (, ), where ,   (–, ) 1 satisfying cos ( – ) = 1 and cos ( + ) = is: e A. 0 B. 1 C. 2 D. 4

A.

1 2

B.

7 15

C.

2 15

D.

1 3

Physics 66. A uniform sphere of mass M and radius R exerts a force F on a small mass m situated at a distance of 2R from R m R P the centre O of the sphere. A spherical portion of diameter R 2R is cut from the sphere as shown

in Fig. The force of attraction between the remaining part of the sphere and the mass m will be: F 7 F A. B. 2 9 5 3 F F C. D. 7 4

54 67. A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8 kg. How much is the rise in temperature of the block in 2.5 minutes assuming 50% of power is used up in heating the machine itself or lost to the surroundings? Specific heat of aluminium = 0.91 J g–1°C–1. A. 103°C B. 110°C C. 115°C D. 120°C 68. A body of mass m falls from a height h and collides with another body of same mass at rest. After collision, the two bodies combine and move through distance d till they come to rest. What is the work done against the resistive force? h  A. mg (h + d) B. mg   2 d  2  d  1 mg(h  d ) C. mg  h   D. 2 2

69. A current of 3 A flows through the 2  resistor shown in the circuit. The power dissipated in the 5  resistor is: 2 2 2

A. 1 W C. 4 W

B. 2 W D. 5 W

70. The temperatures T 1 and T 2 of heat reservoirs in the ideal Carnot engine are 1500°C and 500°C respectively. If T1 increases by 100°C, what will be the efficiency of the engine? A. 59% B. 62% C. 90% D. 98%

Chemistry 71. Equimolar aqueous solutions of NaCl and BaCl2 are prepared. If the freezing point of NaCl is –2°C, the freezing point of BaCl2 solution is expected to be: A. –3°C B. –4°C C. –1.5°C D. –1.98°C Zn/dust

X 72. Phenol  The product Z is: A. Benzene C. Benzaldehyde

CH 2 Cl  Y Anhyd.AlCl 3

CHO CHO (i) NaOH/100°C

75.

+

major product

(ii) H /H 2O

is:

CHO OHC

COOH HOOC

Alkaline  Z KMnO 4

A. COOH HOOC

B. Benzoic acid D. Toluene

CH2OH HOH2C

B.

73. A mineral of titanium (perovakite) is formed to contain calcium ions at the corners, oxygen atoms at the face centres and titanium atoms at the centre of the cube. The oxidation state of titanium in the mineral is: A. +1 B. +2 C. +3 D. +4

CH2OH HOH2C OH

HOOC

C. COOH

O

74. Find the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO 4 in 2 litres of water at 25°C, assuming that it is completely dissociated. A. 3.27 × 10–2 atm B. 5.27 × 10–3 atm –4 C. 1.25 × 10 atm D. 2.37 × 10–2 atm

HO

O

D.

Biology 76. Artificial selection to obtain cows yielding higher milk output represents: A. stabilizing selection as it stabilizes this character in the population. B. directional as it pushes the mean of the character in one direction. C. disruptive as it splits the population into two, one yielding higher output and the other lower output. D. stabilizing followed by disruptive as it stabilizes the population to produce higher yielding cows.

77. The hepatic portal vein drains blood to liver from: A. Heart B. Stomach C. Kidneys D. Intestine 78. The water potential of pure water is: A. Zero B. Less than zero C. More than zero but less than one D. More than one

55 79. Which of the following represents order of ‘Horse’? A. Equidae B. Perissodactyla C. Caballus D. Ferus

80. Alexander Von Humbolt described for the first time: A. Ecological Biodiversity B. Laws of limiting factor C. Species are relationships D. Population Growth equation

ANSWERS 1 C

2 C

3 C

4 A

5 A

6 D

7 B

8 B

9 A

10 D

11 C

12 D

13 C

14 C

15 D

16 B

17 B

18 D

19 C

20 D

21 A

22 B

23 A

24 A

25 D

26 C

27 A

28 B

29 B

30 A

31 C

32 B

33 C

34 A

35 A

36 B

37 B

38 C

39 D

40 C

41 D

42 B

43 C

44 A

45 C

46 B

47 B

48 B

49 C

50 A

51 C

52 B

53 A

54 A

55 A

56 B

57 D

58 C

59 A

60 D

61 A

62 B

63 B

64 B

65 B

66 B

67 A

68 B

69 D

70 A

71 A

72 B

73 D

74 B

75 C

76 B

77 D

78 A

79 B

80 C

EXPLANATORY ANSWERS 1. OC = OD = R

AP = 4x

 OB = R 2 The diameter of the smaller circle

 



= R 2 –R  R Area of the semi-circle =

R 2

R 

Area of the circle =



2 –1



2





2 –1

22 Hence, the ratio of the area of the smaller circle to that of the semi-circle



=

2

or



2



2 – 1 : 2.

4z 12 x  3x  7z 7

PD =

2.

2



2 –1

C 4y R

Q 3y

A

4x

P

4z

D 3x

12 x  7 : 3. 7 3. Here angle covered by the large hand in 60 minutes = 360° and angle covered by the small hand in 60 360  15 minutes = 24 Required time when the hands are at right angles between 24 and 1 

3z

B

AP : PD = 4 x :

60 × 90 minutes past 24 (360 – 15) 60 = × 90 minutes past 24 345 360 = minutes past 24 23 15 = 15 minutes past 24 23 4. (1 + 2x + 3x2 + ... + 21x20) (21x20 + 20x19 + 19x18 + .... + 2x + 1) Coeff. of x30 is 11.21 + 12.20 + .... + 21.11 = 2[11.21 + 12.20 + 13.19 + 14.18 + 15.17] + 16.16 = 2[1225] + 256 = 2450 + 256 = 2706. =

56 5. x2 + 6x + 8 x R x2 – 2x – 8  0 (x – 4) (x + 2)  1 x  [–2, 4]

8. Since, a1, a2, a3 are in AP a2 – a1 = a3 – a2 = d

 1  1 1   ...    an –1 an –2   a1 a2 a2 a3  11 1 1 1 –  ...  –  d  a1 a2 a2 a3  9. For y = f (x) to be defined | x | – x > 0 or | x | > x It is true only if x < 0 So, the domain is – < x < 0. =

2

x + 6x + 8 –4

–3

–2

0

2

4

Clearly min value of expression is O at x = –2. 6.  OAC = 90° As AC is tangent and OA is radius as CAD = 45° So, OAD = 45° = AOD  OA = 2 Area of shaded region = 1 – (area of sector OAX – area of OAD)

O

then inequality  x2  x  log 6   > 1  x4  45° 45°

A

1

X 1

1

B

x2  x >6 x4

D

1

 1  1 3  1 = 1–   2  –  = 1–  –   – 4 2  4 2 2 4 2 7. Let the radius of the circle be r units D

x2  x > 0, As 0 < 0.8 < 1 x4

10.

C O

C



x2  x –6 x4 x 2  x – 6 x – 24  x4  (x – 8)(x + 3) (x – 4)  –4 (–4, –3) 

P

OP = (1 OB = (1 In OPB : OP2 + BP2 = (1 – r)2 + (1)2 = (1 1  r= 4

1

>0 > 0 x  –4 < x < –3 or x > 8  (8, ).

11. We have tn = Sn – Sn – 1  n  2 tn =

A

>0

1 1 n(2n2  9n  13) – (n –1) 2(n –1)2  9(n –1)  13 6 6





=

1 2 n3 – (n – 1)3  9 n2 – (n – 1)2  13(n – n  1)  6

=

1 2 6 n – 6 n  2  9(2 n – 1)  13 6



 



B

– r) + r) OB2 + r)2 units

1 Area of smaller circle =   square units 16 Sum of the Area of quarter circles    =   square units 4 4 2 Area of shaded region

   9 = 2 –    = 2 –  3.14  16 2  16 = 0.23 square units Hence, option (B) is the closest.

1 2 6 n  12 n  6  = (n + 1)2 6 Also, t1 = S1 = 4 = (1 + 1)2 =

n





n

tr =

r 1

1

 (r  1)  2 (n  1)(n  2) – 1 r 1

=

1 n(n  3). 2

12.  is the angle between the tangent and the line circle with centre (2, –1) and r = 3 perpendicular from centre on 3x – 4y = 5 is P=

64–5 1 5

57 17. Clearly, from the figure, that medium B is rarer and medium A is denser. The refractive index of a rarer medium to the denser medium is less than unity. (2, –1) line  P=1  – 90° 2 2



9 0°

3x – 4y = 5

r i

–

Medium A T



sin (90 – ) =



cos  =

13.

1 3 1 2 2  So, sin  = . 3 3

1 1 3 3 m(v cos30)2 = mv 2   E. 2 2 4 4

Q 3kA(100 – ) 2 k .A( – 25)  = t x x

–2 X –2

2

or  

Y'

 –x

 e(II) ( x



0

 2 x ) dx  4 (I)

I.B.P. 14. 3 – x  0  x3 2– 3– x  0  43 –x  x  –1

–1 0 2 3

1  2 – 3 – x

 3– x  1  3– x1x 2 Hence, x  [–1, 2]. n 2 cos ( n  1) , n = 17,  =  15. S = 9  2 sin 2 17 sin 18 .cos   –1. =  sin 18 sin

16. As, or

=

19. For slabs in series, same amount of heat passes through each slab.

X'

and

1 2 mv 2 KE at the highest point

18. Initial KE, E =

Y

Required area =

Medium B

v2 – u2 = 2as v2 = 2as

[ u = 0]

 v = 2 as  v  s or v2  s Which is equation of parabola. Thus, the graph is parabolic path.

x

100 °C

300 – 3 = 2 – 50 5 = 350  = 70°C.

x

1 3k

25 °C

2k

20. According to work-energy theorem, the kinetic energy remains constant, when no work is done by the external force, so,

1 2 mv = constant 2 or | v | = constant Hence, change in its speed is zero. 21. If the acceleration due to gravity is g, on the earth’s surface of radius R, then the change value of acceleration due to gravity, at the distance ‘r’ from the centre of the earth is given as

4 1 4 g = G   r 3     Gr 3  r2 3  g  r. 22. Let the angle subtended by the wire at the point, where the sparrow sits be 2. Then 2T cos  = W, sin ,   90°, as the wire will say by a small amount. Hence, T >> W. 23. Since Fext = 0 So, P = constant Fext = 0, does not mean that  = 0  L may or may not be constant. As internal energy may change, then kinetic and potential energy may change. Hence, linear momentum of the systems does not change in time. (2015) KVPY—Practice Paper—8

58 V2 R As the resistance of the bulb is constant

33. We have,

[A] = 0.01 M Rate = 2.0 × 10–5 mol L–1s–1 Rate = k[A]

24. Power, P =

P 2 V = P V P 2 V  100 =  100 % decrease in power = P V = 2 × 2.5% = 5%.



25. Electrostatic force operates over distance which are not large, while gravitational forces operate over very large distances.

Rate 2.0  10 –5  = 2.0 × 10–3 [A] 0.01 0.693 0.693 t1   = = 347 s. 2 k 2.0  10 –3 34. Since alkyl bromide is more reactive than alkyl chloride, therefore, intramolecular Wurtz reaction occurs on the side of Br atom forming bicyclo product as

k=

Br

26. Internal energy depends only on the initial and final states of temperature and not on the path. In a cyclic process, an initial and final states are the same, the change in internal energy is zero. 27. This is to convert a diverging beam of light coming from the source into a parallel beam of light. Hence, option (A) is correct. sin 60 3 /2 3    sin 45 1 / 2 2 29. We have, mgh = 200 × 10 × 6

28. As, nBA =

3 . 2

mgh 200  10  6  P1 = = 200 W t 60 P0 = 600 W

As, Since,

200  100 = 33.33%  33%. 600 30. Decrease in mass number due to emission of 3particles and a -particle = 3 × 4 = 12 Decreases in charge number in the process =3 ×2–1 =5  For the resulting element, A = 228 – 12 = 216 Z = 88 – 5 = 83.  Efficiency,

31.

=

Number of A atoms = Number of B atoms = Number of C atoms = 

Formula =

1 8 1 8 1 ×1 =1 1  12  3 4 ABC3.

32. 98% H2SO 4 by mass means 98 g of H2SO 4 are present in 100 g of solution. 100 Volume of solution = = 55.56 mL 1.80 98 / 98  100 = 18.0 M Molarity of solution = 55.56 Now, M 1V 1 = M 2V 2  18.0 × V1 = 0.1 × 1000 0.1  1000  V1 = = 5.55 mL. 18

+

–Na + 2Na

Dry ether

Cl 1-Bromo-3-Cyclobutane

Cl

HBr

OCH3

35.

–NaCl

–NaBr

Bicyclo compound

OH + CH3Br

36. -hydroxy aldehydes or -hydroxy ketones readily under dehydration to give -unsaturated aldehyde or ketone. +

O

O

OH

H

OH2

+

O +

H

–H –H2O

37. Bond energy of halogens decreases down the group as the size of atom increases. The bond energy of F2, is however, lower than that of Cl2 and Br2, because of inter-electronic repulsion present in the small atom of F. Thus, the bond energy decreases in the order Cl2 > Br2 > F2 > I2. 38. We know that, rHe = rCH 4

M CH 4 M He



16 2 4

[ M He = 4, M CH = 16] 4 i.e., the rate of effusion of He is two times faster than CH 4. Since 4 : 1 is initial composition, then the composition of mixture effusing out initial is 8 : 1. 39. The oxidation number of oxygen in O 3 = 0 Oxidation number of oxygen in OF2 is x + (–2) = 0  x = +2 Oxidation number of oxygen in KO 2 is 1 + 2x = 0 –1  x= 2 and oxidation number of oxygen in BaO 2 is 2 + 2x = 0  x = –1  Correct increasing order is BaO 2 < KO2 < O3 < OF2.

59 Y

40. The freezing point of a substance is the temperature at which vapour pressure of the substance becomes equal to vapour pressure of its liquid. More the number of particles of solute more will be the depression in freezing point and more the freezing point will decrease. Therefore, lesser the number of particles more will be the freezing point of substance. Hence, the highest freezing point of option (iv).

y = –|x| + 1 (0, 1)

Y'

63. Since, cos ( – )  – But, –2 –

108 0.108 = 8 Wt. of O2



Wt. of O2 =



Volume of O2 =

0.108  8  8  10 –3 108 22400  8  10 –3 32

–

The reactant B will disappear at half the rate as A will decrease. 43. In ClO3(OH) or HClO 4 is the strongest acid because of larger electronegativity and higher oxidation state of Cl atom. Hence, ClO3 (OH) is the strongest acid. 44. Nylon 6, 6 has strong intermolecular forces. It has amide linkages and therefore, has strong hydrogen bonding between two polyamide chains. 45. We have, –

CH3 – CH – CH = CH2 + HBr

–Br

CH3

+ CH3 – CH – CH – CH3 CH3 (2° - Carbocation)

Br + +Br– CH3 – C – CH2 – CH3 CH3 – C – CH2 – CH3 CH3

CH3

1 2n  –  < 2 [as, , (–, )] 0

1 1  cos 2 = < 1, e e which is true for four values of [as –2 < 2 < 2] 64. By the law of sine Rule a b c   k [say] = sin P sin  sin R

Also,

d[A 2 B] 1 d[A] d[B] – = – 2 dt dt dt

= = < =

2  2 = 2 square units.

Given, cos ( + ) =

= 5.6 cm 3. 42. We have, 2A + B  A2B

1  a  P1 =  2

P

2 c b a P P 2 P  P1 = R a k sin P Q 2 2 Similarly, P2 = and P3 = k sin  k sin R Since, sin P, sin Q and sin R are in AP, hence P1, P2, P3 are in H.P. 

P1 =

2

62. The region is clearly square with vertices at the points (1, 0), (0, 1), (–1, 0) and (0, –1)

3

1

65. The number of ways of placing 3 black balls without any restriction is 10C3. Since, we have total 10 places of putting 10 balls in a row. Now the number of ways in which no two black balls put together is equal to the number of ways of choosing 3 places marked ‘....’ out of eight places. _W_W_W_W_W_W_W_ This can be done in 8C3 ways 8

 Required Probability =

C3

10

3° - Carbocation (more stable)

61. By the given conditions, we can take two perpendicular lines as x and y-axis. If (h, k) is any point on the locus, then | h | + | k | = 1, Therefore, the locus is | x | + | y | = 1. This consist of a square of side 1. Hence, the required locus is a square.

X

(0, –1)

Area of square =



1, 2 hydride shift

(1, 0)

X'

41. We have,

108 Weight of Ag deposited = 8 Weight of O 2 produced

y = |x| – 1

(–1, 0)

C3



8 76 7  . 10  9  8 15

66. The gravitational force of attraction on mass m at P due to solid sphere is F= or

GMm

GMm (2R)

2



GMm 4R 2

= 4F ...(i) R2 Mass of the spherical portion removed from sphere,

60

M =

M

3



4  R M      3 2 8

4 3 R 3 Gravitational force of attraction on mass m at P, if mass of the spherical portion removed is present there is

F =

G(M / 8) m (3R / 2) 2



GM m R2

1 4 1 4 2F   = 4F    8 9 8 9 9

 Gravitational force of attraction on mass m at P due to removing part of the sphere is F = F – F = F –

2F 7F .  9 9

67. Given, P = 10 kW = 104 W, mass, m = 8 kg = 8 × 103 g Rise in temperature T = ? time, t = 2.5 min = 2.5 × 60 s = 150 s Specific heat, C = 0.91 Jg–1C–1 Total energy = P × t = 104 × 450 = 15 × 105 J As 50% of energy is lost,  Energy available, 1 5 Q =  15  10 = 7.5 × 105 J 2 As, Q = mc T or

Q 7.5  10 5  T = = 103°C mc 8  10 3  0.91

where V  Pd across 5  = 5 V 25  5 W.  Power = 5 70. We have, T1 = 1500 + 100 = 1600°C = 1600 + 273 = 1873 K and T2 = 500°C = 500 + 273 = 773 K

T2 773 1100  1–  T1 1873 1873 1100  100%  59%. = 1873 71. As, Tf = 2, i = 2, Tf = i Kf m For NaCl, 2 = 2 × Kf × m For BaCl2, Tf = ?, i = 3 Tf = 3 × Kf × m Since, K f and m are same equimolal solution T f 3 = or Tf = 3 2 2  Freezing point of BaCl2 solution = 0 – 3 = –3°C. 72. We have, As,

 

m 2 gh  0 = 2 mv v=

...(i)

gh / 2

Work done against the friction,

1 2 = (2 m) gd  (2 m) v 2  gh  h  = (2 m) gd  m   = mg   2 d  . 2 2 69. 2 , 4  and (1  + 5) are in parallel. Their potential 3A difference are the same i 2 A B v = 2 × i1 4 i C D = 4 × i2 = 6 × i3 1 5 i  i3 = 1 A E F Total potential difference = 5 × 1 + 1 × 1 = 6V 2

CH 3

OH Zn dust

CH 3Cl

Alkaline

Anhyd. AlCl3

KMnO 4 Benzoic acid

= 8 (corners) ×

1 = 1 8

O atoms per unit cell = 6(face centres) ×

1 = 3 2

Ti atoms per unit cell = 1  Formula of mineral = CaTiO 3 Suppose oxidation state of Ti be x +2 + x + 3(–2) = 0 or x = +4. 74. If K 2SO 4 is completely dissociated, K2SO 4  2K+ + SO42–, i = 3 Molecular mass of K2SO 4 = 2 × 39 + 32 + 4 × 16 = 174 As,  = i cRT

WB  RT 3  25  10 –3  0.082  298 = MB  V 174  2.0

= i

= 5.27 × 10–3 atm. 75.

CHO

CHO NaOH, 100°C Intermolecular cannizzaro reaction

3

 Power dissipated in 5  resistance =

COOH

73. Number of Ca atoms per unit cell

68. Given, m1 = m, v1 =

2gh m2 = m, v2 = 0 If v is the combined velocity after collision, using the principle of conservation of linear momentum, m1v1 + m2v2 = (m1 + m2)v

 = 1–

–

COO

HOH2C

COO– HOH2C

COOH HOH2C +

H , H2O

V2 R

–

CH2OH

  

OOC

CH2OH HOOC

61

Practice Paper (Solved)

6

Kishore Vaigyanik Protsahan Yojana (KVPY) STREAM – SA

Mathematics 1. What is the remainder when (1! + 2! + 3! + .... + 1000!) is divided by 5? A. 1 B. 2 C. 3 D. 4

A. 6 C. 14

B. 10 D. 18

9. D, E and F are the mid points of the sides AB, BC and CA, respectively, and X, Y and Z are the mid-points of DE, EF and FD respectively. It is given that the circum radius of triangle ABC is 8 3 cm and triangle XYZ is inscribed by a circle (in cm 2) what is the ratio of shaded area to that of area of ABC?

2. How many zeroes will be there at the end of the expression N = 7 × 14 × 21 × ... × 777? A. 24 B. 25 C. 26 D. None of these

A

3. How many times does the digit 6 appears when we count from 11 to 400? A. 34 B. 74 C. 39 D. 79

B

6. Let x  4  4 – 4  4 – .... , Then x equals:

C.

 13 – 1   2   

B.

3

7. The least positive integer n for which is: A. 6 C. 8

3

n 1 – 3 n 

C

E

A.

3 16

B.

5 16

C.

7 16

D.

9 16

10. Corners are sliced off a unit cube, so that the six faces each become regular octagons. What is the total volume of removed tetrahedron?

 13  1  2   

D.

F Y

X

5. Find the value of (a + b) if the polynomial x3 – ax2 – 13x + b has (x – 1) and (x + 3) as factors. A. 2 B. 15 C. 18 D. None of these

A. 3

Z

D

1  1 1   4. If  x    3 , then the value of  x 6  6   x 4  4    x x  x  is: A. 15134 B. 15132 C. 15144 D. None of these

1 12

B. 7 D. 9

A.

5 2 –7 3

B.

10 – 7 2 3

C.

3–2 2 3

D.

8 2 – 11 3

11. The 288th term of the series a, b, b, c, c, c, d, d, d, d, e, e, e, e, e, ..... is:

8. The number of all 3-digit numbers abc (in base 10) for which (a × b × c) + (a × b) + (b × c) + (c × a) + a + b + c = 29 is:

A. u C. w 61

B. v D. x

62 12. If x, y and z are distinct positive real numbers; the 2

2

2

x ( y  z )  y ( x  z )  z ( x  y) would be: xyx A. Greater than 4 B. Greater than 5 C. Greater than 6 D. None of the above

A. B. C. D.

13. How many positive real numbers x satisfy the equation x3 – 3| x | + 2 = 0? A. 1 B. 3 C. 4 D. 6

1 1 1  a, x 2  3  b, then x 3  2 is: x x x a3 + a2 – 3a – 2 – b a3 – a2 – 3a + 4 – b a3 – a2 + 3a – 6 – b a3 + a2 + 3a – 16 – b

14. If x 

15. If x, y are real numbers such that 3 then the value of (x + y) | (x – y) is: A. 0 B. 1 C. 2 D. 3

x 1 y

–3

x –1 y

 24 ,

Physics 16. A boat which has a speed of 5 kmh–1 in still water crosses a river of width 1 km along the shortest possible path in 15 min. The velocity of the river water in kmh–1 is: A. 2 B. 3 C. 4 D. 5 17. A body is moved along a straight line by a machine delivering constant power. The distance moved by the body in time t is proportional to: A. t B. t1/2 3/4 C. t D. t3/2 18. A satellite S is moving in an elliptical orbit around the earth. The mass of the satellite is very small compared to the mass of the earth is: A. the acceleration of S is always directed towards the centre of the earth B. the angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant C. the total mechanical energy of S varies periodically with time D. the linear momentum of S remains constant in magnitude 19. A ray of light from a denser medium strikes a rarer medium at an angle of incidence (i). The reflected and refracted rays make an angle of 90° with each other. The angles of reflection and refraction are r and r. The critical angle is: i

A.

B. Time

Time

Temperature

C.

Temperature

Time

D.

Time

21. A wooden block, with a coin placed on its top, floats in water as shown in figure. The distance l and h are shown there. After sometime, the coin falls into the water. Then: A. l increases and h decreases B. l decreases and h increases C. both l and h increase D. both l and h decrease

Coin l h

22. The three resistances of equal value are arranged in the different combinations shown below. Arrange them in increasing order of power dissipation. I.

III.

i

II. i

IV.

i

i

r

A. I < II < III < IV C. IV < II < I < III

r'

A. sin–1 (tan r) C. sin–1 (tan i)

Temperature

Temperature

B. sin–1 (cos i) D. tan–1 (sin i)

20. Liquid oxygen at 50 K is heated to 300 K at constant pressure of 1 atm. The rate of heating is constant. Which of the following graphs represents the variation of temperature with time?

B. III < II < IV < I D. II < III < IV < I

23. A ray of light incident at an angle of 60° on one face of a rectangular glass slab of thickness 2 3 mm and refractive index is: A. 2 mm C. 4 mm

3 . The lateral shift produced in mm

B. 3 mm D. 5 mm

63 24. In the given figure, a ball rolls without sliding in a horizontal surface. It ascends a curved track upto height h and returns value of h is h1 for the sufficiently rough curved track to avoid sliding and h2 for smooth curved track, then: A. h1 = h2 B. h1 < h2 C. h1 > h2 D. h2 = 2h

I

I

A.

B.

h

t

t

I

I

C.

D. t

25. A clock hung on a wall has marks instead of numerals on its dial. On the adjoining wall, there is a plane mirror and the image of the clock in the mirror indicates the time 7 : 10. Then the time on the clock is: A. 3 : 50 B. 4 : 50 C. 5 : 40 D. 7 : 10 26. A fuse wire is a wire of: A. low resistance and low melting point B. high resistance and low melting point C. low resistance and high melting point D. high resistance and high melting point 27. When an electric heater is switched on, the current flowing through it (I) is plotted against time (t). Taking into account the variation of resistance with temperature, which of the following best represents the resulting curve?

t

28. An astronaut accidently gets separated out of his/her small spaceship accelerating in interstellar space at a constant rate of 100 ms–2. What is the acceleration of the astronaut the instant after he/she is outside the spaceship? (Assume that there is no nearby stars to exert gravitational force on him/her) A. 0 ms–2 B. 10 ms–2 –2 C. 100 ms D. 4.9 ms–2 29. A juggler maintains four balls in motion, making each of them to rise a height of 20 m from his hand. What time interval should he maintain for the proper distance between them? [Take, g = 10 ms–2] A. 2 s B. 4 s C. 1 s D. 5 s 30. The activity of a radioactive sample diminishes from 1024 to 128 in 2 minutes. In 6 minutes, the activity diminishes to: A. 2 B. 3 C. 4 D. 8

Chemistry 31. The Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron from Bohr orbits in an atom of hydrogen? A. 2  5 B. 3  2 C. 4  1 D. 5  2 32. The first element of a group in the periodic table of elements often shows the resemblance to the second element of the neighbouring group on the right, which is: A. ionisation energy B. electronegativity C. diagonal relationship D. parallel relationship 33. Which of the following will give yellow precipitate on shaking with an aqueous solution of NaOH followed by acidification with dil. HNO 3 and addition of AgNO 3 solution? CH 3

A.

I Br

B.

Cl

C.

CH 2 – Cl

D.

Br

Cl

O

A.

O C H 2I

OCH3

C.

D. OH

ONa

35. In the given transformation which of the following is the most appropriate reagent? CH = CHCOCH3

Reagent

HO

CH = CHCH2CH3

HO

A. NH2 – NH2/OH– C. Zn-Hg/HCl

B. NaBH4 D. Na, liq. NH3

36. The correct order of increasing basicity of the given conjugate bases (R = CH 3) is: A. R– < HC = C– < RCOO– < NH2– B. RCOO – < HC = C– < NH2– < R– C. RCOO – < NH2– < HC = C– < R– D. RCOO – < HC  C– < R– < NH2– 37.

F

NO2

(i) (CH 3) 2NH DMF, 

A

H2/Pd-C

B

The product B is: CH3

OH + CH2I2 + NaOH

34.

B.

O

A.

O2 N

B.

O2 N

N(CH3 )2

D.

H 2N

NH2

NH2

OH

The product is:

N

C.

H 2N

N(CH3 )2

64 38. Helium (He) atom is two times heavier than a hydrogen molecule. At 298 K, the average kinetic energy of a helium atom is: A. half that of a hydrogen molecule B. same as that of a hydrogen molecule C. two times that of a hydrogen molecule D. four times that of a hydrogen molecule 39. 0.001 mole of strong electrolyte Zn(OH)2 is present in 200 mL of an aqueous solution. The pH of this solution is: A. 12 B. 15 C. 18 D. 20 40. The oxidation state of S in H 2S2O8 is: A. 0 B. +6 C. +5 D. +7 41. Which of the following factors is of no significance for roasting sulphide ores to the oxides and not subjecting the sulphide ores to carbon reduction directly? A. Metal sulphides are less stable than the corresponding oxides B. CO 2 is more volatile than CS2

C. Metal sulphides are thermodynamically more stable than CS2 D. CO 2 is thermodynamically more stable than CS2 42. A hydrocarbon contains 10.5 g carbon per gram of hydrogen. The emiprical formula of the hydrocarbon is: A. C2H 3 B. C3H 7 C. C7H 8 D. C3H 9 43. What weight of glycerol should be added to 600 g of water in order to lower its freezing point by 10°C? A. 182 g B. 209 g C. 297 g D. 315 g 44. Assuming that Hund’s rule is violated, the bond order and magnetic nature of the diatomic molecule B2 is: A. 1 and diamagnetic B. 0 and diamagnetic C. 1 and paramagnetic D. 0 and paramagnetic 45. Equal masses of methane and oxygen are mixed in an empty container at 25°C. The fraction of the total pressure exerted by oxygen is: A. 1/2 B. 1/3 C. 2/3 D. 3/4

Biology 46. DNA fragments are: A. Positively charged B. Negatively charged C. Neutral D. Either positively or negatively charged depending on their size 47. A baby boy aged two years is admitted to play school and passes through a dental check-up. The dentist observed that the boy had twenty teeth. Which teeth were absent? A. Incisors B. Canines C. Pre-molars D. Molars 48. Anaphase Promoting Complex (APC) is a protein degradation machinery necessary for proper mitosis of animal cells. If APC is defective in a human cell, which of the following is expected to occur? A. Chromosomes will not condense B. Chromosomes will be fragmented C. Chromosomes will not segregate D. Recombination of chromosome arms will occur 49. An important characteristic that Hemichordates share with Chordates is: A. Absence of notochord B. Ventral tubular nerve cord C. Pharynx with gill slits D. Pharynx without gill slits 50. The genotypes of a Husband and Wife are IAIB and IAi. Among the blood types of their children, how many different genotypes and phenotypes are possible?

A. B. C. D.

3 3 4 4

genotypes; genotypes; genotypes; genotypes;

3 4 3 4

phenotypes phenotypes phenotyeps phenotypes

51. Transplantation of tissues/organs fails often due to nonacceptance by the patient’s body. Which type of immune-response is responsible for such rejection? A. Auto-immune response B. Cell-mediated immune response C. Hormonal immune response D. Physiological immune response 52. Adult human RBCs are enucleate. Which of the following statement(s) is/are not most appropriate explanation for this feature? (a) They do not need to reproduce (b) They are somatic cells (c) They do not metabolize (d) All their internal space is available for oxygen transport A. Only (d) B. Only (a) C. (a), (c) and (d) D. (b) and (c) 53. Lungs are made up of air-filled sacs, the alveoli. They do not collapse even after forceful expiration, because of: A. Residual Volume B. Inspiratory Reserve Volume C. Tidal Volume D. Expiratory Reserve Volume 54. Zygotic meiosis is characteristic of: A. Marchantia B. Fucus C. Funaria D. Chlamydomonas

65 55. Select the correct route for the passage of sperms in male frogs: A. Testes  Bidder’s canal  Kidney  Vasa efferentia  Urinogenital duct  Cloaca B. Testes  Vasa efferentia  Kidney  Seminal Vesicle  Urinogenital duct  Cloaca C. Testes  Vasa efferentia  Bidder’s canal  Ureter  Cloaca D. Testes  Vasa efferentia  Kidney  Bidder’s canal  Urinogenital duct  Cloaca 56. Which one of the following statements is not valid for aerosols? A. They are harmful to human health B. They alter rainfall and monsoon patterns C. They cause increased agricultural productivity D. They have negative impact on agricultural land 57. Viroids differ from viruses in having: A. DNA molecules with protein coat B. DNA molecules without protein coat

C. RNA molecules with protein coat D. RNA molecules without protein coat 58. During DNA replication, Okazaki fragments are used to elongate: A. The leading strand towards replication fork B. The lagging strand towards replication fork C. The leading strand away from replication fork D. The lagging strand away from the replication fork 59. Plants which produce characteristic pneumatophores and show vivipary belong to: A. Mesophytes B. Halophytes C. Psammophytes D. Hydrophytes 60. The process of separation and purification of expressed protein before marketing is called: A. Upstream processing B. Downstream processing C. Bioprocessing D. Post-production processing

Mathematics  1 1 61. If 3 f ( x )  5 f    – 3 for all non-zero x, then  x x f (x) = 1 3 1  3   A. B.   5 x – 6  –  5 x – 6 14  x 16  x C.

1  3   –  5 x  6 14 x

D. None of these

62. If A = cos2 x + sin4 x, then: 3 3  A 1 A2 A. B. 4 4 3 3 C. –  A  D. None of these 4 4 63. The given figure show s a set of concentric squares of the diagonal of the innermost square is 2 units and if the distance between the corresponding corners of any two successive squares is 1 unit. Find the difference between the area of eight and seventh square, counting from the innermost square:

A. 10 2 square units C. 35 2 square units

B. 30 square units D. None of these

D Q 64. In the parallelogram P, Q, R and S are midpoints of the sides AB, CD, R DA and BC respectively. AS, BQ, CR and DP are joined. Find the A B P ratio of the area of the shaded region to the area of the parallelogram ABCD: 1 1 A. B. 5 4 4 1 C. D. 15 6

C

S

65. In n positive integers are taken at random and multiplied together, then the probability that the last digit of the product be 2, 4, 6 or 8 is: 4n  2n 4n  2n A. B. 5n 5n C.

4n – 2n 5n

D.

4 n  5n 2n

Physics 66. A ball of mass 0.2 kg is thrown vertically upwards by applying force by hand. If the hand moves 0.2 m, while applying the force and the ball goes upto 2 m height further, find the magnitude of force. [Take g = 10 ms–2] A. 10 N B. 15 N C. 20 N D. 25 N

67. A 3 kg ball strikes a heavy rigid wall with a speed of 10 ms–1 at an angle of 60° with the wall. It gets reflected with the same speed at 60° as shown in figure. If the ball is in contact with the wall for 0.2 s, what is the average force exerted on the ball by the wall?

I

6 0°

6 0°

W a ll

(2015) KVPY—Practice Paper—9

66 A. 100 N C. 200 N

B. 150 N D. 350 N

68. A vessel of height 2d is half-filled with a liquid of refractive index 2 and the other half with a liquid of refractive index n. (The given liquids are immiscible). Then the apperent depth of the inner surface of the bottom of the vessel (neglecting the thickness of the bottom of the vessel) will be: A.

n

nd

B.

d  2n 2n

C.



d n 2



C.

E 2

D.

2 E 3

70. A cylindrical vessel is filled with water upto height H. A hole is bored in the wall at a depth h from the free surface of water. For maximum range h is equal to:





A. H





C.

d n 2 d n 2

D.

69. A satellite with kinetic energy E is revolving round the earth in a circular orbit. The minimum additional kinetic energy required for it to escape into lunar space is: A. E B. 2E

n 2

2 H 3

B.

H 2

D.

3 H 4

Chemistry 71. When a bottle of dry ammonia and a bottle of dry HCl connected through a long tube are opened simultaneously at both ends, at first: A. a white ring is formed at the centre of the tube B. a white ring is formed near the ammonia bottle C. entire length of tube turns white D. a white ring is formed near HCl bottle 72. When bauxite powder is mixed with coke and reacted with nitrogen at 2075 K, carbon monooxide and X are formed. What is the gas formed, when X is reacted with water? A. O2 B. N2 C. N2O D. NH3

B. Neopentane C. Tertiary butyle chloride D. Neohexane 74. Heptane and octane form ideal solution. At 373K, the vapour pressures of the two ligaments components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane? A. 73.43 kPa B. 83.21 kPa C. 92.03 kPa D. 103.41 kPa 75. In the following species, the one which is likely to be the intermediate during benzoin condensation of benzaldehyde is:

73. Which branched chain isomer of the hydrocarbon with molecular mass 72 U gives only one isomer of mono substituted alkyl halide? A. Isohexane

A.

Ph — C = O

C.

Ph — C

OH OH

B.

Ph — C  O

D.

Ph — C

OH OH

Biology 76. Identify the wrong statement in context of heartwood. A. Organic compounds are deposited in it B. It is highly durable C. It conducts water and minerals efficiently D. It comprises of dead elements with highly lignified walls 77. Spliceosomes are not found in cells of: A. Plants B. Fungi C. Animals D. Bacteria 78. Which of the following statements is correct? A. The ascending limb of loop of H enle is impermeable to water B. The descending limb of loop of Henle is impermeable to water

C. The ascending limb of loop of Henle is permeable to water D. The descending limb of loop of Henle is permeable to electrolytes 79. Which ecosystem has the maximum biomass? A. Forest ecosystem B. Grassland ecosystem C. Pond ecosystem D. Lake ecosystem 80. The final proof for DNA as the genetic material came from the experiments of: A. Griffith B. Hershey and Chase C. Avery, Mcleod and McCarty D. Hargobind Khorana (2015) KVPY—Practice Paper—9-II

67

ANSWERS 1 C 11 D 21 D 31 D 41 B 51 B 61 B 71 D

2 A 12 C 22 B 32 C 42 C 52 C 62 A 72 D

3 D 13 A 23 A 33 A 43 C 53 A 63 B 73 B

4 A 14 A 24 C 34 B 44 A 54 D 64 A 74 A

5 D 15 D 25 B 35 A 45 B 55 D 65 C 75 C

6 B 16 B 26 A 36 B 46 B 56 C 66 C 76 C

7 C 17 D 27 B 37 C 47 C 57 D 67 B 77 D

8 D 18 A 28 A 38 B 48 C 58 D 68 D 78 A

9 B 19 A 29 C 39 A 49 C 59 B 69 A 79 A

10 B 20 B 30 A 40 B 50 C 60 B 70 B 80 B

EXPLANATORY ANSWERS 1. By Remainder Theorem (1!  2!  3!  4!  5!  ...1000!) = 5 = 1 + 2 + 1 + (–1) + 0 + 0 + .... + 0 = 3. 2.

N = 7 × 14 × 21 × .... × 777 = 7(1 × 2 × 3 × ... × 111) = 7 × 111! Number of zeroes = 20 + 4 = 24.

x4 

5

111

5

20 4

3. We know that, in every consecutive 100 numbers, every digits comes 10 times at unit place and 10 times at tens place. Then from 11 to 100, 6 will appear for 19 times and from 100 to 400, 6 will appear for 3 × 20 = 60 times. Hence, answer = 19 + 60 = 79 times. 1 x = 3 4. ...(i) x Do square of both sides of equation (i) 2 1  x    = (3)2 x 1 x2  2 = 9 – 2 = 7  ...(ii) x Do cube of both sides of equation (ii) 3  2 1 x    = (7)3 x2  1 1   x 6  6  3  x 2  2  = 343 x x   1 6 x  6 = 343 – 3 × 7 x = 343 – 21 = 322 Do square of both sides of equation (ii) 2

 2 1  x  2  = (7)2 x

1

 2 = 49  x 4 

1

= 47 x x4  6 1  4 1  Then,  x  6   x  4  = 322 × 47 = 15134. x x 5. As, (x – 1) and (x + 3) are the factors of P(x) then P(1) = 0, P(–3) = 0 (1)3 – a(1)2 – 13(1) + b = 0 1 – a – 13 + b = 0 –a – 12 + b = 0 a – b = –12 ...(i) P(–3) = 0 (–3)3 – a(–3)2 – 13(–3) + b = 0 –27 – 9a + 39 + b = 0 9a – b = 12 ...(ii) equation (i) – equation (ii), –a = 0 a = 0, b = 12  a + b = 12 + 0 = 12. 6.

4

x=

4 4– x x2 = 4  4 – x  x2 – 4 = 4 – x Now, Put the values from options Only option (B) satisfies the conditions.



7.

 n  11/ 3 –  n1/ 3  n  11/ 3 –  n1/ 3

1 12 1 < 12
 432 12  So, n = 8, only possible least positive integer.

68 8. abc + ab + bc + ca + a + b + c = 29 ab(c + 1) + b(c + 1) + a(c + 1) + c + 1 = 30 (a + 1)(b + 1)(c + 1) = 30 9a 1 0  b, c  9 a, b, c  I, Then, the number of solution = 18. 1 1 1 of ABC  of of ABC 1 1 5 4 4  9. 4 =  . 4 16 16 ar (ABC)

10. Assume side of Regular polygon = x x  AB = 2 A 45° B x AB + BC + CD = 1 45° x x x x =1 2 2 2x  x = 1 x





2 1 = 1  x =

V= Height =

x 2

 1–

1 2

Then,

1 1 2  a and x  3  b x x 2 1  1 2  x   = a2  x  2  2 = a2 x x

...(i)

3

C

D

1 × Base Area × Height 2 1 2

2

1 1  3–2 2 1–    2 4 2 Area of one of the tetrahedron is

Area of Base =

1 × Base Area × Height 3 10 – 7 2 1 3–2 2  1   1 – =  =   3  4   24 2 Since, there are 8 removed corners, we get an answer

=

10 – 7 2 . 3 11. The number of terms of the series forms the sum of first n natural numbers i.e., n(n +1). Thus, the first 23 letters will account for the first 23 × 24 = 276 terms of the series. The 288th term will be the 24th letter, which is x. Hence, option (D) is the answer. of

12. Here, x, y and z are distinct positive real numbers So,

x>0  x3 – 3x + 2 = 0  (x – 1)(x2 + x – 2) = 0  (x – 1)(x + 2)(x – 1) = 0  x = 1, –2 When x = 1  x > 0 Only one positive real root. 14. Given, x 

2 –1

, Base = 1 –

13. x3 – 3| x | + 2 = 0

x 2 ( y  z )  y 2 ( x  z )  z 2 ( x  y) xyz

x x y y z z =      (We know that) y z x z x y a b  x y  y z   z x =            that   2 b a  y x  z y  x z  If a and b are distinct numbers > 2 + 2 + 2  6 Hence, option (C) is the answer.

1   x   = a3 x 1 1  x 3  3  3  x   = a3  x x Add equation (i) and (ii)

...(ii)

1  2 1  3 1   x  3    x  2   2  3  x   = a2 + a3 x x x 1  b   x 3  2   2  3a = a 2 + a 3  x  1 x 3  2 = a3 + a2 – 3a – b – 2. x 15. Let, 3x/y = t t 3t – = 24 3  8t = 3 × 24  t = 9 So, 3x/y = 9 x 1 x xy 2 1 y   3. = 2  = y x x–y 2 –1 –1 y 16. Shortest possible path comes, when absolute velocity of boatman comes perpendicular to river current as shown in figure. vr

W  t= vb

W

2 W v v br – vr2 (where W  width of the river) 1 1  = 2 4 5 – vr2

Given, W = 1 km, t = 15 min = and vbr = 5 km/h 1 = 4

 

vb

br

15 1 h h 60 4

1 25 – vr2

25 – vr2 = 16  vr2 = 9 vr = 3 kmh–1.

River Current

69 17. We have, P = constant  Work done in time t, W = Pt From work-energy theorem, net work done is change in kinetic energy,

1 2 mv = Pt 2 v  t1/2 By integrating, we get, s  t3/2. 

 D sin i =  R sin r 

3 and t = 2 3 m sin 60 1 sin i sin i  As,  = or sin r = = 2 sin r  3  r = 30° Now, lateral shift,

2 3 1 t sin(i – r ) =  = 2 mm. cos r 3 2 2 24. When the track (case A) only translation KE changes to the gravitational PE, but when the track is rough (case B) both the translational and rotational KE changes to PE. Thus, PE = mgh will move in the case B than in the case A. Hence, h1 > h2. d=

18. The force on the satellite always towards the earth, therefore, acceleration of the satellites S is always directed towards the centre of the earth is zero. Hence, the angular momentum of S about the centre of the earth is constant throughout. Since, force (F) is conservative in nature, so the mechanical energy of the satellite remains constant. Speed of the satellite (S) is maximum, when it is the nearest to the earth and minimum when it is the farthest. 19. We have, r + r + 90° = 180°  r = 90° – r Further, i = r Applying Snell’s law,

23. Given, i = 60°,  =

i

r

Denser 90° Rarer

r i=r

or

 D sin r =  R sin (90 – r )   R cos r



R –1    = tan r or C = sin  R  D  D  = sin–1 (tan r).

20. The temperature of liquid oxygen will first increase in the same phase. Then, phase change (liquid to gas) will take place. During which temperature will remains constant. After that, the temperature of oxygen in gaseous state will further increase. 21. l will decrease because the wooden block moves up and h will decrease because the coin will displace the volume of water (V 1) equal to its own volume, when it is in the water whereas, when it is on the block it will displace the volume of water (V2) whose weight is equal to weight of the coin and since density of coin is greater than the density of water, V 1 < V2. 22. As, P = i2 R Current is same, So P  R In the first part, it is 3r, in the second part, it is 2/3 r r in the 3rd part, it is and in the fourth part, the net 3 3r resistance is . 2 Since, RIII < RII < RIV < RI  PIII < PII < PIV < PI or III < II < IV < I.

25. The image in a plane mirror undergoes depth inversion, hence the time on the image is a much ahead of 12 hours zero minutes as the time on actual clock is behind 12 hours zero minutes. Time on the actual clock = 7 hours 10 minutes Time on the image clock = (12 hours and zero min.) – (7 hours and 10 mins.) = 12:00 – 7:10 = 4:50. 26. We know that, the fuse wire is a device to prevent a high current from passing through the circuit. It consists of piece of wire made from a metal of low resistance and low melting point. 27. When we move in the direction of the current in a uniform conductor, the potential decreases linearly. When we pass through the cell, from its negative to positive terminal, the potential increases by an amount equal to its potential difference. This is less than its emf, as there is some potential drop across its internal resistance, when the cell is deriving current. 28. The moment of the astronaut is out of spaceship, net force acting on the astronaut is zero. This is because there is no nearby stars to exert gravitational force on him/her and the small spaceship exerts negligible gravitational attraction on him/her. As net force acting on the astronaut is zero, the acceleration of the astronaut is zero. 29. Given, S = 20 m, Using, v2 = u2 + 2aS For upward motion, we have, 0 = u2 + 2 × (–10) × 20 or u = 20 ms–1 If t is the total time of flight of the ball in going up and coming back, then total displacement in time t is zero. i.e., S = 0 1  Using the relation, S = ut  at 2 1  0 = 20t  (–10)t 2 = 20t – 5t2 or t = 4S 2 4  Time interval of each ball =  1S. 4

70 39. We have, Zn(OH)2  Zn2+ + 2OH–

3

1  1 128 30. We have, =   8  2 1024 3 half-life period, s = 2 minutes 6 minutes means 9 half-life periods,

[OH –] = pOH = = pOH = pH + 2 =

9

 1   1  2. N = N 0   = 1024   512   2 31. The electron has minimum energy in the first orbit and its energy increases as n increases. Here, n represents the number of orbit, i.e., 1st, 2nd, 3rd, .... . The third line from the red end corresponds to yellow region i.e., 5. In order to obtain less energy electron tends to come in 1st or 2nd orbit. So the jump involved may be either 5  1 or 5  2. 

32. Certain elements of 2nd period show similarity with the diagonal elements in the next period as shown below: Group 1

Group 2

Group 13

Group 14

2nd Period

Li

Be

B

C

3rd Period

Na

Mg

Al

Si

Hence, Li resembles Mg, Be resembles Al and B resembles Si is called the diagonal relationship and due to the reason that these pairs of elements have almost identical radii and polarising power (i.e., charge/ size ratio). 33. Alkyl halides are more reactive than aryl halides and therefore, only (A) and (D) will react. Since precipitate has yellow colour, it must be of AgI. 34. We have, –

OH

+

ONa

–

OH

O

CH 2I2

NaOH +

–2NaOH

O

ONa

CH = CHCOCH3 NH 2 NH 2 ,OH

35.

Wolf Kishner reduction

HO

CH = CHCH2 CH3 –

HO

36. Stronger the acid, weaker is the conjugate base. Base + Proton  Conjugate acid Conjugate acid formed will be: RCOOH, NH3, HC  CH, CH4 Acidic strength follows order RCOOH > HC  CH > NH3 > CH4  Increasing basicity will be RCOO – < HC  C– < NH2– < R–. CH3 F

NO2

DMF, 

N CH 3

NO2

N

40. We have, H 2S2O8 i.e., O

O

H–O–S–O–O–S–O–H O

O

Two oxygen atoms are involved in peroxide linkage, hence oxidation number will be –1 (each).  H2S2O8 = 2(+1) + 2(x) + 2(–1) + 8(–2) = 0  2 + 2x – 14 = 0  2x – 12 = 0  x = +6. 41. Oxidising roasting is a very common type of roasting in metallurgy and is carried out to remove sulphur and arsenic in the form of their volatile oxides. CS2 is more volatile than CO 2. Hence, option (B) is of no significance for roasting sulphide ores to their oxides. The reduction process depends on the thermodynamic stability of the products and not on their volatility. 42. Number of moles of atoms of carbon is 11.5 g of the 10.5  0.875 hydrocarbon = 12 Number of moles of atoms of hydrogen is 11.5 g of 1.0 1 the hydrocarbon = 1.0 Ratio of number of moles of C and H = 0.875 : 1 = (0.875 × 8) : (1 × 8) = 7 : 8 Hence, the empirical formula of the hydrocarbon is C7H 8 .



CH3 H2 /Pd



43. We have, Tf = Kf × m =

37. We have, (i) (CH3 )2 NH

1000 200 1 × 10–2 mol L–1 –log [OH –] = –log (1 × 10–2) 2 log10 2, pH + pOH = 14 14  pH = 12.

[OH –] = 0.001  2 

NH2

CH3

38. As, KE  T, Kinetic energy depends on temperature of the gas and is independent of the nature of the gas. Hence, at 298 K, KE of both hydrogen and helium is same.

W2 =

K f  W2  1000 M 2  W1

10  92  600 = 296.77 g  297 g. 1.86  1000

44. Assuming that Hund’s rule is violated, electronic configuration of B2 is 1s2, *1s2, 2s2, *2s2, 2px2  Bond order = (6 – 4)/2 = 2/1 = 1 and the molecule is diatomic.

x x 3x   16 32 32 P(O 2) = P(Total) × x O2

45. Total moles of mixture =

71 64. Q and P are mid-points of DC and AB respectively  area of ADP = area of DPQ = Area of PQB

P(O2 ) x / 32 1  . = xO2  P(Total) 3 x / 32 3

 61. We have,

In ADH,

1 (area of parallelogram ABCD) 4 DR = RA, RE | | AH  DE = EH

In ABG,

AP = PB =

= area of QBC =

 1 1 3 f ( x)  5 f   = – 3  x x 1 put x = , we get, x  1 3 f    5 f ( x) = x – 3  x equation (i) × 3 – equation (ii) × 5, we get,

...(i)

...(ii)

1  9f (x) – 25f (x) = 3  – 3 – 5( x – 3) x  3 –16f (x) = – 9 – 5 x  15 x  3  –16f (x) = –  –  5 x – 6  x  f (x) =

1 BG 2 But, BG = DE (by symmetry)  DE = EH = 2HP 

[For 0 < x < 1, xn < x  n  N – {1}]  cos2 x + (sin2 x)2  cos2 x + sin2 x for all x

= 1 – sin2 x + (sin2 x)2

3 1 1 1  2 = 1 –   – sin 2 x  (sin 2 x) 2  =   – sin  4 2 4 4  2 1 2  Now,  – sin x   0 for all x 2

3 1   – sin 2 4 2

S

B

area of shaded region 1 1 = 4  . parallelogram ABCD 20 5

65. Last digit of the product will be 1, 2, 3, 4, 6, 7, 8 or 9 if and only if each of the n positive integers ends in any of these digits. Now the probability of an integer ending 1, 2, 3, 4, 6, 7, 8 or 9 is 8/10. Therefore, the probability that the last digit of the n

 x 

2

3  x  for all x  4 3  A for all x 4 3 From (i) and (ii), we obtain  A  1 for all x. 4 63. The diagonal of the innermost square is 2 units  Diagonal of the 7th square = 14 and Diagonal of the 8th square = 16 

C

1 area of parallelogram ABCD 20

Hence,

(sin2 x)2  sin2 x

A = cos2 x + sin4 x

G H

A

=

Q F

DE =

0  sin2 x  1 for all x

Again,

E R

P 2 DP 5 1 2 In ADP, DR = DA, DE = DP 2 5 1 2 = Area of RDE =   area of ADP 2 5 1 2 1 =    area of parallelogram ABCD 2 5 4

62. We have, A = cos2 x + sin4 x = cos2 x + (sin2 x)2 Now, –1  sin x  1 for all x

A  1 for all x

D

PH =



1  3   –  5 x – 6 . 16  x



1 AB, PH | | BG 2

1 2  Area of 7th square =  (14)  98 2 1 2 Area of 8th square =  (16)  128 2  Their difference = 128 – 98 = 30 square units.

2

 4 product of n integers is 1, 2, 3, 4, 6, 7, 8 or 9 is   .  5 The probability for an integer to end in 1, 3, 7 or 9 is 4 2  . Therefore, the probability for the product of 10 5 n  12  n positive integers to end in 1, 3, 7 or 9 is    5 Hence, the required probability n

n

4n – 2n  4  2 . =   –    5  5 5n 66. Let v be the velocity given by hand, when the ball reaches a height h. 

1 2 mv = mgh 2

...(i)

If F is the force applied by the hand, then as per question,

1 2 mv  F  0.2 = mg (h + 2) 2

72 Using (i), mgh + F × 0.2 = mgh + mg × 2

 The horizontal range will be maximum. 1 2 –1/ 2  (H – 2 h)  0 If dR/dh = 0 i.e, 2  (hH – h ) 2 or H = 2h  h = H/2.

mg  2  10 mg 0.2 F = 10 × 0.2 × 10 = 20 N.



F=



67. Clearly, from the figure, components of momentum perpendicular to wall cancel out, and components along the wall add. As, F × t = Change in momentum = 2 mv cos 

2 mv cos  2  3  10 cos 60  t 0.2 60 10   150 N. = 2 2



=

 Due to first liquid,

d

d

d d  x2 = x2 n  Total apparent depth

rNH3 36.5  = 1.46 17 rHCl rHCl So, the rate of diffusion of NH 3 is greater than the rate of diffusion of HCl. Hence, a white bottle ring of NH 4Cl will be formed near HCl bottle.





GM r

E=

1 2 1 2GM GMm mve  m   2E 2 2 r r  Additional energy needed = E – E = 2E – E = E.

26.0  0.26 100 35  0.31 Mole of octane = 114 0.26  0.456 x(heptane) = 0.26  0.31

and 

Total pressure = 105.2 × 0.456 + 46.8 × 0.544 = 47.97 + 25.46 = 73.43 kPa.

R = ut =

75. The mechanism of benzoin condensation is O

Ph — C + : CN

Ph — C — CN

OH PhCHO

H O

CN

Ph — C — CN

H

–

O

Ph — C — C — Ph

2 gh  2(H – h) / g

–

O

OH

2(H – h) / g

0.31  0.544 0.26  0.31

x(octane) =

70. Horizontal velocity of water flowing out of the hole,

1 2 (H – h) = gt or t = 2  Horizontal range,

CH 3 Only one monochloro product is possible

Mole of heptane =

...(ii)

2gh Height of the hole from the ground level = (H – h) The time taken by water to cover vertical distance (H – h) will be

CH 3 — C — CH 2Cl

CH 3

...(i)

u=

CH 3 Chlorination

Neopentane

74.

E =

M HCl  M NH 3

CH 3 — C — CH 3

d d n 2   . 2 n n 2

d

1 2 1 GMm mv0  2 2 r Kinetic energy for escaping body, Kinetic energy,

=

CH 3

69. Orbital velocity of satellite, v0 =

; M  molecular mass

73. The molecular mass indicates the compound is pentane. Since it forms only one mono substituted alkyl halide, the compound is neopentane in which all the H atoms are equivalent.

2

n=

= x1 + x2 =

M

72. Bauxite powder on mixing with coke followed by reaction with N2 produces aluminium nitrite which on reaction with water produces ammonia. Al2O3 + 3C + N2  2AlN + 3CO Bauxite (X) HN + 3H2O  Al(OH)3 + NH3 (X)

Real depth d  Apparent depth x

2 = x or x1 = 1 Due to the second liquid,

rNH3



F=

68. As,

1

71. Rate of diffusion 

H

–

OH

Ph — C — C — Ph CN

O

Ph — C — C — Ph + : CN

H

OH

= 2 h(H – h)

 Ph — C — CN is the intermediate.   

OH

H

–

73

Practice Paper (Solved)

7

Kishore Vaigyanik Protsahan Yojana (KVPY) STREAM – SA

Mathematics 1. In the following figure AE = EB, BD = 2DC, what is the ratio of the areas of PED and ABC? A

E

A. C.

B. D.

2. The sum of all non-integer x5 – 6x4 + 11x3 – 5x2 – 3x + A. 6 B. C. –5 D.

D. 8 : 1

4 2 :1

2 2 :1

7. If p and q are the roots of the quadratic equation 2x2 + 6x + N = 0 (N < 0), then what is the maximum p q  ? value of q p A. –2 B. –3 C. –4 D. None of these

roots of the equation 2 = 0 is: –11 3

 n – 1  n – 1  n  n n! is:  5    6    7 where,     r  (n – r )! r ! A. 12 B. 13 C. 14 D. 15

n  1 8. Let f (n)     where [x] denotes the integral  2 100  100

part of x. Then the value of

 f ( n)

is:

n 1

Directions (Qs. No. 4 and 5): In the given below figure a square ABCD is inscribed in a circle of radius 5 cm. The square ABCD is inscribed by a circle which is inscribed by a triangle EFG. The triangle EFG is inscribed by a circle, which is again inscribed by a triangle XYZ. E

64 : 3

D.

6. What is the number of distinct triangles with integral valued sides and perimeter as 14? A. 6 B. 5 C. 4 D. 3

1 6 1 12

3. The least value of a natural number n such that

A

C.

C. 16 : 3

C

D

1 4 1 9

B.

5. Find the ratio of the area of square ABCD and the area of triangle XYZ. A. 16 : 3 3 B. 64 : 3 3

P

B

A. 4 : 1

A. 50 C. 1

B. 51 D. None of these

9. ABCD is a circle and circles are drawn with AO, CO, DO and OB as diameters, areas E and F are shaded. E/F is equal to:

B

C

X Y F D

Z

A

G C

4. Find the ratio of the radius of the outermost circle to the radius of the circle inscribed in triangle XYZ.

E

G

G

E

G G

F

F

F

F

G G

E

G

G

D

E

B

73

(2015) KVPY—Practice Paper—10

74 A. 1 C.

B.

 4

12. The number of pairs of real (x, y) such that x = x2 + y2 and y = 2xy is: A. 4 B. 3 C. 2 D. 1

1 2

D. None of these

10. In the figure ABC is a right angle triangle with B = 90°, BC = 21 cm and AB = 28 cm. With AC as diameter of a semicircle and with BC as radius, a quarter circle is drawn. Find the area of the shaded portion correct to two decimal place.

1947

13. The value of

1



2  21947

n0

A.

A

C.

28 cm

is equal to:

n

487

1946

B.

21945

1947

21947 1948

D.

1947

21947

2

B

21 cm

A. 428.75 cm 2 C. 214.37 cm 2

14. Let f (x) be a quadratic polynomial with f (2) = 10 and f (–2) = –2, Then the coefficient of x in f (x) is: A. 1 B. 2 C. 3 D. 4

C

B. 857.50 cm 2 D. 371.56 cm 2

11. At what time between 10 o’clock and 11 o’clock are the two hands of clock symmetric with respect to the vertical line (give the answer to the nearest second)? A. 10 h 9 m 13 s B. 10 h 9 m 14 s C. 10 h 9 m 22 s D. 10 h 9 m 50 s

15. In a triangle ABC, it is known that AB = AC, suppose D is the mid-point of AC and BD = BC = 2. Then the area of the triangle ABC is: A. 2 B. 2 2 C. D. 2 7 7

Physics 16. Two particles A and B are projected with same speed so that the ratio of their maximum heights reaches is 3 : 1. If the speed of A is doubles without altering other parameters, the ratio of the horizontal ranges attained by A and B is: A. 1 : 2 B. 2 : 1 C. 1 : 4 D. 4 : 1 17. A block of mass 0.1 kg is held against a wall applying a horizontal force of 5 N on the block. If the coefficient of friction between the block and the wall is 0.5, the magnitude of the frictional force acting on the block is: A. 1.5 N B. 2.5 N C. 3.5 N D. 3.0 N 18. Which of the following figure represents variation of total mechanical energy of a pendulum oscillating in air as function of time? E

A.

E t

B.

E

C.

t E

t

D.

t

B. ivory ball will not rise C. ivory ball will rise to a greater height than wet clay ball D. both balls will rise to the same height 20. A given system under goes a change in which work done by the system equals the decrease in its internal energy. The system must have undergone: A. isothermally B. adiabatically C. isobatically D. isochorically 21. The resistance between A and B in the given figure will be: 10  10 

A. B. C. D.

50 

20  30  90  more than 10  but less than 20 

22. The stress versus strain graphs for wires of two materials A and B are shown in figure. If the Y A and YB are the Young’s modulus of materials, then:

Stress

Y

19. Two identical solid balls, one of ivory and the other of wet clay are dropped from the same height on the floor. After striking the floor: A. ivory ball will rise to the lesser height than wet clay ball

20 

O

A B 60° 30° Strain

X

75 A. Y B = 2 YA

B. YA = YB

C. 2YA = YB

D. YA = 3YB

23. A 10 mm long awlpin is placed vertically in front of a concave mirror. A 5 mm long image of the awlpin is formed at 30 cm in front of the mirror. The focal length of this mirror is: A. –10 cm B. –20 cm C. –25 cm D. –30 cm 24. A spherical conductor with a radius of 0.20 m has an excess charge of 10 C. What is the electric field at the surface of the sphere? A. 2.25 × 106 NC–1 B. 3.25 × 105 NC–1 C. 4.25 × 108 NC–1 D. 5.25 × 107 NC–1 25. Which of the following graph represents the variation of magnetic flux density B with distance r for a straight long wire carrying an electric current? B

27. The work function of sodium is 2.3 eV. Find the maximum wavelength for the light that will cause photoelectrons to be emitted from the sodium. A. 5100 Å B. 5000 Å C. 5400 Å D. 5800 Å 28. The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object? A. between principal focus and centre of curvature B. at the centre of curvature C. beyond the centre of curvature D. between the pole of the mirror and its principal focus 29. With what velocity should a particle be projected so that its height becomes equal to radius of the earth?

B

A.

B. r

1

1

A.

 GM  2   R 

B.

 8GM  2   R 

1

1

C.

 2GM  2   R 

D.

 4GM  2   R 

r

B

B

C.

D. r

r

26. A convex lens of 10 cm focal length is combined with a concave lens of 6 cm fixed focal length. What is the focal length of the combination? A. 10 cm B. –15 cm C. 15 cm D. –20 cm

30. A player caught a cricket ball of mass 150 g moving at the rate of 20 ms–1. If the catching process be completed in 0.1 s, the force of the blow exerted by the ball on the hands of the player is: A. 0.2 N B. 10 N C. 20 N D. 30 N

Chemistry 31. If energy of the electron in hydrogen atom in some excited state is –3.4 eV, then what will be its angular momentum? A. 1.2 × 10–34 kg m 2 s–1 B. 2.5 × 10–31 kg m 2 s–1 C. 3.1 × 10–33 kg m 2 s–1 D. 6.3 × 10–32 kg m 2 s–1 32. Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidising property is: A. O > F > N > Cl B. Cl > F > N > O C. F > O > Cl > N D. F > Cl > O > H 33. Which of the following graphs correctly describes the behaviours of a fixed mass of an ideal gas? (T is measured in K)

P

P

C.

D. T

O

34. The reagents for the following conversions is/are: Br

A. B. C. D.

Br

A.

B. O

T

O

P

H––H

alcoholic KOH followed by NaNH 2 alcoholic KOH Zn/CH3OH

35. KI in acetone, undergoes SN2 reaction with each of A, B, C and D. The rates of the reaction vary as: O

A

PV

?

aq. K O H follow ed by N aN H 2

H 3C – Cl

V

V

O

Cl B

A. A > C > B > D C. C > A > D > B

Cl

Cl C

D

B. A > B > C > D D. D > A > C > B

76 A. AB 3 C. A3B 4

36. Find the product of the given reaction: CH 3 +

H 

CH 3 OH

CH 3

40. A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. What is the mass percentage of the resulting solution? A. 65.5% B. 40.5% C. 75.1% D. 95.5%

CH3

A.

B. CH 3

CH3

CH3

C.

D. CH 3

CH3

CH3

37. One mole of alkene on ozonolysis gives 2 moles of butanone. The alkene is: A. 3, 4-dimethylhex-3 ene B. 2, 3-dimethylhex-2 ene C. 2, 3-dimethylhex-3 ene D. 3, 4-dimethylhex-2 ene 38. Arrange the following acides in order of their increasing acidity: COOH

COOH

COOH

NO 2 B

A. D < B < A < C C. B < C < A < D

OH C

41. The rate of reaction is given by rate, r = k[H +]n. If the rate becomes 100 times, when the pH changes from 2 to 1, the order of the reaction is: A. 1 B. 2 C. 3 D. 4 42. Which of the following is the correct order of increasing oxidising character of oxoacids of chlorine? A. HClO < HClO2 < HClO3 < HClO4 B. HClO < HClO4 < HClO3 < HClO2 C. HClO4 < HClO3 < HClO2 < HClO D. HClO3 < HClO4 < HClO2 < HClO 43. Which one of the following metals is required as cofactor by all enzymes utilizing ATP in phosphate transfer? A. Na B. Mg C. Ca D. K

COOH

NO 2

A

B. A2B D. A2B 3

D

B. C < D < B < A D. B < C < D < A

39. In a solid ‘AB’ having NaCl structure ‘A’ atoms occupy the corners of the cubic unit cell. If all the face-centred atoms along one of the axes are removed, the resultant stoichiometry of the solid is:

44. When 2.91 g of a mercuric compound is vapourised in a 1 litre bulb at 680 K, the pressure is 458 mm. The molecular weight of the compound is: A. 82.18 B. 112.35 C. 271.6 D. 312.4 45. Identify the least stable ion amongst the following: A. Li–1 B. Be – – C. B D. C –

Biology 46. The function of copper ions in copper releasing IUD’s is: A. They suppress sperm motility and fertilizing capacity of sperms B. They inhibit gametogenesis C. They make uterus unsuitable for implantation D. They inhibit ovulation 47. An example of colonial algae is: A. Chlorella B. Volvox C. Ulothrix D. Spirogyra 48. Root hairs develop from the region of: A. Maturation B. Elongation C. Root cap D. Meristematic activity 49. Hypersecretion of Growth Hormone in adults does not cause further increase in height, because: A. Growth Hormone becomes inactive in adults. B. Epiphyseal plates close after adolescence.

C. Bones loose their sensitivity to Growth Hormone in adults. D. Muscle fibres do not grow in size after birth. 50. Which of the following in sewage treatment removes suspended solids? A. Tertiary treatment B. Secondary treatment C. Primary treatment D. Sludge treatment 51. Select the mismatch: A. Pinus B. Cycas C. Salvinia D. Equisetum

-

Dioecious Dioecious Heterosporous Homosporous

52. What is the criterion for DNA fragments movement on agarose gel during gel electrophoresis? A. The larger the fragment size, the farther it moves B. The smaller the fragment size, the farther it moves C. Positively charged fragments move to farther end D. Negatively charged fragments do not move

77 53. In Bougainvillea thorns are the modifications of: A. Stipules B. Adventitious root C. Stem D. Leaf 54. The association of histone H1 with a nucleosome indicates: A. Transcription is occurring B. DNA replication is occurring C. The DNA is condensed into a Chromatin Fibre D. The DNA double helix is exposed 55. A temporary endocrine gland in the human body is: A. Pineal gland B. Corpus cardiacum C. Corpus luteum D. Corpus allatum 56. Select the mismatch: A. Frankia B. Rhodospirillum C. Anabaena D. Rhizobium

-

A. anterior pituitary gland and of LH and oxytocin. B. anterior pituitary gland and of LH and FSH. C. posterior pituitary gland and of oxytocin and FSH. D. posterior pituitary gland and of LH and relaxin.

stimulates secretion stimulates secretion stimulates secretion stimulates secretion

58. A gene whose expression helps to identify transformed cell is known as: A. Selectable marker B. Vector C. Plasmid D. Structural gene 59. Presence of plants arranged into well defined vertical layers depending on their height can be seen best in: A. Tropical Savannah B. Tropical Rain Forest C. Grassland D. Temperate Forest

Alnus Mycorrhiza Nitrogen fixer Alfalfa

57. GnRH, a hypothalam ic horm one, needed in reproduction, acts on:

60. Functional megaspore in an angiosperm develops into: A. Ovule B. Endosperm C. Embryo sac D. Embryo

Mathematics A 61. Suppose S1 and S2 are two R B unequal circles; AB and CD Q are the direct comm on P D tangents to these circles. A C S traverse common tangent PQ cuts AB on R and CD on S. If AB = 10, then RS is: A. 8 B. 9 C. 10 D. 11

62. In the given circle, AC is the diameter of the circle, ED is parallel to AC. CBE = 65°, find DEC. B

O

E

A. 35° C. 45°

A. 156 407

B.

78 407

C.

D.

234 407

312 407

24

h

13

64. Let r be a root of the equation x2 + 2x + 6 = 0, The value of (r + 2) (r + 3) (r + 4) (r + 5) is equal to: A. 51 B. –51 C. –126 D. 126 65. Let a, b, c be non zero real numbers such that a + b + c = 0. Let q = a2 + b2 + c2 and r = a4 + b4 + c4. Then, A. q2 < 2r always B. q2 = 2r always C. q2 > 2r always D. q2 – 2r can take both positive and negative value

65° A

63. The base of a pyramid is a rectangle of side 18 m × 26 m and its slant height to the shorter side of the base is 24 m. Find its volume.

C D

B. 55° D. 25°

Physics 66. Two conducting wires of same material, and of equal lengths and areas are first connected in series and then parallel in a circuit. The ratio of heat produced in series and parallel combinations would be: A. 1 : 4 B. 4 : 1 C. 2 : 1 D. 1 : 2

67. Water drops fall at regular intervals from a tap which is 5 m above the ground. The third drop is leaving the tap at the instant the first drop touches the ground. How far above the ground is second drop at that instant? [Take g = 10 ms–2]

78 A. 1.25 m C. 3.75 m

B. 2.50 m D. 4.50 m

68. A dish of mass 20 gram is kept floating horizontally in air by firing 300 bullets per minutes which bound from the dish with the same speed in opposite direction. What is the velocity of the bullet at the time of impact, if each bullet weighs 5 grams? A. 1.29 ms–1 B. 3.92 ms–1 C. 5.27 ms–1 D. 6.54 ms–1 69. A mass M suspended from a spring of neglible mass. The spring is pulled a little and then released so that the mass executes SHM of time period T. If the mass is increased by m, the time period becomes 5T/3.

Then the ratio of m/M is: A. 11/9 B. 16/9 C. 7/5 D. 25/9 70. A plane mirror having square shape is mounted parallel to a vertical wall at some distance from it. A point light source is fixed on the wall. Light from it gets reflected from the mirror and forms a patch on the wall. When mirror is moved parallel to itself towards the wall: A. the path does not change remains stationary B. area of patch remains the same C. centre of path does not remain stationary D. none of these

Chemistry 71. Al2O3 is reduced by electrolysis at low potentials and high currents. If 4.0 × 104 A of current is passed through molten Al2O 3 for 6 hours, what mass of aluminium is produced? Assume 100% current efficiency, atomic mass of Al = 27 g/mol. A. 2.3 × 103 g B. 8.1 × 104 g C. 4.5 × 105 g D. 6.3 × 102 g 72. The activation energies of two reactions are E 1 and E2 (E1 > E2). If the temperature of the system increased from T1 to T2, the rate constant of the reaction changes from K1 to K1 in the first reaction and K 2 to K2 in the second reaction, predict which of the following expression is correct. K1 K 2 K1 K 2 A. K  K B. K  K 1 2 1 2 C.

K1 K 2  0 K1 K 2

D.

K1 K 22  K1 K 2

73. An alkyl halide with molecular formula C6H13Br on dehydrohalogenation gave two isomeric alkenes X and

Y with molecular formula C 6 H 12 on reductive ozonolysis, X and Y gave four compounds CH 3COCH3, CH3CHO, CH3CH2CHO and (CH3)2CHCHO. The alkyl halide is: A. 2-bromo-2, 3-dimethyl butane B. 3-bromo-2-methyl pentane C. 2-bromo hexane D. 2, 2-dimethyl-1–bromobutane 74. Isopropyl benzene is oxidised in the presence of air to compound ‘A’. When compound ‘A’ is treated with dilute mineral acid, the aromatic product formed is: A. benzene B. phenol C. tolune D. benzaldehyde 75. In the following equilibrium reaction, 2A  B + C the equilibrium concentration of A, B and C are 1 × 10–3 M, 2 × 10–3 M and 3 × 10–3 M respectively at 300 K. The value of K C for this equilibrium at the same temperature is: A. 3 B. 5 C. 6 D. 7

Biology 76. DNA replication in bacteria occurs: A. During S phase B. Within nucleolus C. Prior to fission D. Just before transcription 77. Which among these is the correct combination of aquatic mammals? A. Seals, Dolphins, Sharks B. Dolphins, Seals, Trygon C. Whales, Dolphins, Seals D. Trygon, Whales, Seals

78. Coconut fruit is a: A. Drupe C. Nut

B. Berry D. Capsule

79. Double fertilization is exhibited by: A. Gymnosperms B. Algae C. Fungi D. Angiosperms 80. Which of the following components provides sticky character to the bacterial cell? A. Cell wall B. Nuclear membrane C. Plasma membrane D. Glycocalyx

79

ANSWERS 1 D

2 D

3 C

4 B

5 B

6 C

7 A

8 B

9 A

10 A

11 B

12 A

13 A

14 C

15 C

16 D

17 B

18 B

19 C

20 A

21 D

22 D

23 B

24 A

25 D

26 B

27 C

28 D

29 A

30 D

31 A

32 C

33 A

34 B

35 D

36 A

37 A

38 B

39 C

40 A

41 B

42 C

43 B

44 C

45 B

46 A

47 B

48 A

49 B

50 C

51 A

52 B

53 C

54 C

55 C

56 B

57 B

58 A

59 B

60 C

61 C

62 D

63 A

64 C

65 B

66 A

67 C

68 B

69 B

70 B

71 B

72 A

73 B

74 B

75 C

76 C

77 C

78 A

79 D

80 D

EXPLANATORY ANSWERS 6.  a + b > c So, maximum length of any particular side can be 6 units Now, if a = 6, then b + c = 8, then the possible sets are (6, 6, 2), (6, 5, 3) and (6, 4, 4) If a = 5, then b + c = 9, so the possible set is (5, 5, 4) So, the number of distinct triangle = 4.

2. (x – 1) (x – 2) (x3 – 3x + 1) = 0  Required sum of roots = 3 3.

n–1C

5

+

n–1C

< nC7 nC < nC 6 7 6

n! 6!(n – 6)! n– 6 n nmin

n! 7!( n – 7)! >7 > 13 = 14.


0 So, p and q are real

7. p + q = –3 and pq =

4. Let the radius of outermost circle = R Then, the radius of circle EFG =

Then, the radius of circle XYZ =

R 2

R / 2  

p q ( p  q)2 18  = –2 –2 q p pq N p q Therefore, maximum value of  = –2. q p Again,

R

2 2 2 So, ratio of the radius of the outermost circle to the radius of circle inscribed in XYZ 1  2 2 : 1. = 1: 2 2 5. Let the side of ABCD = x x Then, the radius of circle EFG = 2 3 x Then, side of EFG = 2 3 x So, side of XYZ = 4 3 3 2 x . Then, area of XYZ = 64

8. For all the value of n < 50, f (n) = 0 and for all the n < 50, f (n) = 1, Hence, 51 such values are there. 9. AO = CO = DO = BO = radius of bigger circle = r (let) Then, Area of (G + F) =

r 2 2

Area of 2(G + F) = r2 Also, Area of 2G + F + E = r2 i.e., 2G + F + F = 2G + F + E F =E So, the ratio of E and F = 1 : 1.

80 A

10.

y

x 28 cm D

E

B

 AC =

y= 

1  Solutions  , 2

1  1 1   , –  . 2 2 2

z

w C

21 cm 2

1947

2

(28)  (21)  784  441  1225 = 35 cm

13.

 n0

Area of  ABC (x + w) =

1

Total terms = 1948



2  21947

1  21  28 = 294 cm 2 2

T1 =

Area of sector w + z

11.  Let at x minute past 10 o’clock they become symmetric We know that speed of hour hand and minute hand is 1 : 12  When minute hand moves x minute distance then x hour hand moves minute distance 12 x 12 x x–1 11  = 10 – x 12 10 – x 2 13 x 10  = 10 12 120  x= 13  x = 9 minute 13.8 second therefore required time is 10 h 9 m 13.8 sec  10 h 9 m 14 sec. 12. x = x2 + y2 and y = 2xy I. When y = 0  x = x2  x = 0, x = 1 Solutions are (0, 0) (1, 0) 1 II. When x = 2 1 1 1 – = y2   = y2 2 4 4

1



1947

1 2

1 1947 21947  2 2



1 1947 2 2

Similarly,

693 2 r 2 22 (21)2 1386    = cm 2 4 7 4 4 Area of semicricle y + z =

r 2 22 35 35 1 1995      = cm 2 2 7 2 2 2 4 693 1995  2w + x + y + 2z = 294  2 4 = 294 + 346.5 + 481.25 = 1121.75 ...(i) w + x + y + z = Area of right angle triangle + Area of semicircle = 294 + 481.25 = 775.25 cm 2 ...(ii) 2 × equation (ii) – equation (i) Shaded Area, x + y = 1550.5 – 1121.75 = 428.75 cm 2.

1 2



1947

 n0

1 4

=

1947

2  2

f (x) 10 –2 12

14.

974

487



21947 21945 2 = ax + bx + c = 4a + 2b + c = 4a – 2b + c = 4b  b = 3.

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

AB = AC ABC ~ BDC

15.

AB BC AC  = BD DC BC

 x 22  22 –    2 x 2  x 2 – 22 cos A =  2. x. x 2.2.2

x2

x2 4 4

8–

2

2x – 4

2

=

8x2 – 16 = 8 x 2 –

x4 4

x4 = 64 x= 2 2 S= Area of  ABC = 16. We have,

2 2 2 2 2  2 2 1 2

2

H=



u 2 sin 2  2g

For A,

H1 =

u 2 sin 2 1 , 2g

and for B,

H2 =

u 2 sin 2  2 2g

Given, 



2  1 2 2 – 1 (1)(1)  7.

3 H1 sin 2 1 H =  1 = 1 H2 H2 sin 2 2 sin 2 1 3 = 1 sin 2 2

81

i.e.,

1 3 and sin 2 = 2 2 2 u12 sin 1 cos 1 R1 = (For A) g

sin 1 =

Now, range,

2 u12

= and

R2 =

3 1 . 2 2 g

2 u22 sin  2 cos 2 (For B) g

1 3 2 u22 . . 2 2 = g

  17. We have,

R1 u 2 4u 2 = 12  2 R2 u2 u R1 : R2 = 4 : 1.

23. Given, h0 = 10 mm, h = –5 mm, u = –30 cm As,  Now,



N

W = 0.98 N

For vertical equilibrium of block, f = mg = 0.98 N < (f)max. 18. When the pendulum is oscillating in air, its total mechanical energy decreases exponentially with time. Thus, the correct figure is (B). 19. As the ivory ball is more elastic than wet clay ball, hence it will tend to regain its shape instantaneously after the collision. Therefore, there will be a large energy and momentum transfer compared to the wet clay. Thus, the ivory ball will rise higher after the collision. 20. When a gas expands suddenly, (adiabic change) work is done by the gas. Hence, internal energy of the system decreases.

1 1 1 1 10  5  2    21. We have, = RP 10 20 50 100 100 17 =  RP = 17 100 100 270  RAB = 10   . 17 17 Stress 22. As, Y= = Slope of stress-strain curve Strain = tan  YA tan 60 3   3  YA = 3YB.  = 1 YB tan 30 3

u = –60 cm

1 1 1 1 1  =   f v u –60 –30 (1  2) –1  = – 60 20 f = –20 cm.

24. We have, surface charge density

 10  10 –6  cm –2 A 4 (0.2)2 The electric field just outside of the surface of a charged sphere is given by: =

 10  10 –6 1   2 0 4 (0.2) 8.85  10 –12 6 –1 = 2.25 × 10 NC .

E=

(f )max = N = 0.5 × 5 = 2.5 5N

h1 –v –5 (–30) h2 = u  10 = – u

25. The magnetic field induction at a point due to a long current carrying wire is related with distance r by relation

1 r Hence, graph (D) is correct. B

26. Given, f1 = 10 cm, f2 = –6 cm, f = ? As,  27. As,   

1 1 1 1 1 1 = f  f  10 – 6  – 15 f 1 2 f = –15 cm. KEmax = h  – 0 

hc – 0 

hc – 0 > 0 (KE is always –ve)  ch <  0

ch 3  108  6.62  10 –34  0 2.3  1.6  10 –19 = 5400 Å.

max =

28. In case of concave mirror, there is only one case, when the formed is virtual, erect and enlarge. This is when the object is placed between pole and focus of the mirror. 2 gh 29. Here, v2 = , given h = R h 1 R 

v2 = gR  v =

gR 

GM . R (2015) KVPY—Practice Paper—11

82 30. Given, m = 150 g, v = 20 ms–1, t = 0.1 s v 20 a=  = 200 ms–2 t 0.1 As, F = ma 150  200  30 N . F= 1000 31. Energy of the electrons is –3.4 eV. This indicates that the electron is in second orbit.  Angular momentum, (mvr)

acidity is more pronounced at p-position than that at m-position. Moreover, o-substituted benzoic acids are generally stronger acids than benzoic acid known as ortho effect. Therefore, the correct order is: COOH

nh 2  6.6  10  2 2  3.14 = 2.1 × 10–34 kg m 2 s–1. 32. Electronegativity character  non-metallic character  oxidising power.  The order of electronegativity is F > O > Cl > N. 33. The graph (A) is an isobar drawn on the basis of Charles’s law, at constant pressure i.e., V  T. 34. Simple alkyl halides are dehydrohalogenated by using a strong base such as alc. KOH but vinyl halides require much stronger base such as NaNH 2 for dehydrohalogenation. CH 2 – CH 2 Br

Alc. KOH –RBr, –H 2O

CH 2 = CH

Br

NaNH 2 HC  CH –NaBr, –NH 3

Br

35. The bulky groups result into stearic hindrance in the formation of transition state. Hence, B is less reactive than A. In the compound (D), the transition state is highly stabilized by Ph – C– group and hence it is || O most reactive. Therefore, the order of reactivity is D > A > C > B. 36. We have, CH 3 CH 3

CH3

H+/

OH

OH CH 3

+

–H

CH 3 –H2O

CH3

CH 3

Methyl Shift

+

+ 2

CH3

+

CH 3

CH3

OH C

CH3 CH2 C == C – CH2 – CH2

O3

O

CH3 CH2 – C

C – CH2 CH3 O

CH3 CH3 3, 4-Dimethyl hex-3-ene

O 2CH3CH2 C CH3 O Butanone

38. Electron donating groups tend to decrease acid character while electrons withdrawing groups tend to increase acidic character. The effect of substituents on


HClO3 > HClO4. (2015) KVPY—Practice Paper—11-II

83 43. Mg binds to the phosphate group in ATP, hence making a complex that acids in the transfer of ATP phosphate e.g., Mg acts as a cofactor to the enzyme creative kinase which converts creating to creative phosphate.

Creative Creative + ATP   Kinase, Mg2  creative phosphate + ADP. 44. Volume of vapours at STP

458  1  273  0.24 L 680  760 Molecular weight is the weight of 22.4 L of vapours at STP. =

2.91  22.4  271.6.  Molecular weight = 0.24 45. The electron affinity of Be is minimum among Li, Be, B and C. Since minimum energy is liberated in formation of Be–, therefore Be– is the least stable. 61.  

RP = RA = 10 –  RS = 10 –  +  A

10 – 

R  Q



P

...(i)

B

...(ii)

65° O 65° 90°  E

V2 1  V2  t    t 2R 2 R 

A 24 C

13

1 × Area of Base × Height 3

1  (18  26)  407 = 156 407 . 3

...(i) R R V

...(ii)

67. Time taken by one drop to fall through 5 m is,

(13)2

407

V

H1 1  . = 2 H2 4   1 V  t  2 R 

= – h2 = 576 – 169 h2 = 407

=

=

V2 t R eq.

1  V2   t 2  R 

D

(24)2

H1 =

C

 AC | | ED,  65° + 90° +  = 180°  = 180° – 155°  = 25°.

Volume of pyramid =

 l The resistance of both wires    will be equal.  A R R Let this resistance be R.

 V2  H2 = 2   t  R  From equation (i) and (ii), we get,

B

h=

q2 . 2 66. When the conducting wires are of same material same length (l) and same areas (A). r=

V2 V2 In parallel, H2 = t  t R eq. R/2

62.  CBE = CAE = 65° [Angle by same chord in same sector]

63.

 q2  2 q – 2 r=  4  

In series,

Also, SQ = SD = 10 –   RS = 10 –  +  (i) and (ii)   = , Hence, RS = 10.

h2

65. a + b + c = 0, abc  R  0 a2 + b2 + c2 + 2(ab + bc + a) = 0 q = a 2 + b 2 + c2, r = a 4 + b 4 + c4 r = q2 – 2(a2b2 + b2c2 + c2a2) r = q2 – 2[(ab + bc + ca)2 – 2abc(a + b + c)]

D

C S

A

64. r be a root  r2 + 2r + 6 = 0 ...(i) Now, (r + 2) (r + 3) (r + 4) (r + 5) = (r2 + 5r + 6) (r2 + 9r + 20) = 3r(7r + 14) using (i) = 21(r2 + 2r) using (i), r2 + 2r = –6 = 21(–6) = –126.

t=

2s  g

25  1s 10

h

 Time interval for successive two drops is

B

The distance travelled by falling drop in 2

1 second. 2

1 second is, 2

1  1 s =  10    = 1.25 m  2 2  Height of the second drop from the ground = 5 – 1.25 = 3.75 m.

84 68. Let v be the velocity at the time of impact, m = 5 g = 5 × 10–3 kg

3 × 96500 C of electric deposit Al = 27 g  8.64 × 108 C of electricity will deposit Al

300 5 60 Change in linear momentum of each bullet = 2 mv  Change in linear momentum/second = 2 mv × 5 which is the force exerted by the impact. The dish will keep floating in air, if this force balances the weight of the dish = (Mg) i.e., 2mv × 5 = Mg 10 × (5 × 10–3) × v = 20 × 10–3 × 9.8

27  8.64  10 8 3  96500 = 8.16 × 104 g.

Number of bullets fired/s =



T = 2

M K

T = 2 

K 2 E2 = K1 2.303 R

 T2 – T1     T1T2 

E1 > E2

K1 K1 >1 K 2 log K2



K1 K > 2 . K1 K2

or 1

73.

CH3 – CH – CH – CH2 – CH3 CH3

2

–HBr

CH 3 – C = CH – CH 2 – CH3 (X) CH 3 + H 3C – CH – CH = CH – CH 3

Br

3-Bromo-2-methyl pentane

m 25 16  T   5 –1  . =   –1    –1      M T 3 9 9

or

log

...(ii)

1

2

 T2 – T1     T1T2 

log

T M  m 2  m2 =   1    T  M   M



E1 K1 = 2.303 R K1

Since,

...(i)

Mm K From equation (i) & (ii), we get and

log

72. We have,

20  9.8 = 3.92 ms–1. 50

v=

69. We have,

=

Ozonolysis CH 3 (Y) CH3 – C = O + CH3 CH2 CHO

70. Consider the mirror is mounted such that the source lies on the normal drawn from the centre of the mirror. Then side view will be as shown in figure.

Ozonolysis

CH3 – CH – CHO + CH3 CHO

CH3

CH3



74. We have,



CH 3

V

S

CH 3 – CH CH 3

HOO – C – CH 3 O2

From this particular case centre of the patch coincides with the source and linear dimensions of the patch are twice the linear dimensions of the mirror. Hence, the patch will be square in shape. If the mirror is moved parallel to itself with velocity v along the line, normal to the wall and passing through the source as shown in figure, then the patch will remain unchanged in shape and size. Hence, option (B) is correct. Al3+

3e–

71. We have, +  Al Quantity of electricity passed =I ×t = 4 × 104 × 6 × 3600 = 8.64 × 108 C

OH H 2 O/H

+

+ (CH 3)2 CO

Air

Acetone Isopropyl benzene

Phenol

75. For the reaction, 2A  B + C Equilibrium constant [KC] =

[B][C] [A]2

[A] = 1 × 10–3 M, [B] = 2 × 10–3 M [C] = 3 × 10–3 M Putting the values of [A], [B] and [C] KC =

  

2  10 –3  3  10 –3 (1  10 –3 )2

 6.

85

Practice Paper (Solved)

8

Kishore Vaigyanik Protsahan Yojana (KVPY) STREAM – SA

Mathematics 1. A small bird is taking rest at the centre of the base of a hemispherical cage, suddenly it stands up, flies to the topmost point in the cage, then in a straight line to the cage door at the intersection of the curved surface and the base. In the process, it covers a total distance of 482 cm. What is the radius of the hemisphere? A. 100 cm B. 150 cm C. 200 cm D. 50 cm

7. In triangle ABC, with A = 90°, the bisectors of the angles B and C meet at P. The distance from P to the hypotenuse is 4 2. The distance AP is: A. 8 B. 4 C.

3. Find the last non zero digit of 96! A. 2 B. 4 C. 6 D. 8 4. The number (1024)1024 is obtained by raising (16)16 to the power n. What is the value of n? A. 64 B. 642 C. 6464 D. 160

9. A regular octagon is formed by cutting congruent isosceles right angled triangles from the corners of a square. If the square has side length 1, the side-length of the actagon is:

5. The sum of the infinite series: 1 2 3 n    ...  n  ... 10 10 2 10 3 10

1 9

B.

10 81

C.

1 8

D.

17 22

C.

A 1 5A 2

A.

3 –1 2

B.

2 –1

C.

5 –1 4

D.

5 –1 3

10. Let A 1, A2, A3, ... An be a nine sided regular polygon with side length 2 units. The difference between the lengths of the diagonals A1 A5 and A2 A4 equals:

6. In a Rhombus one of the diagonal is twice the other diagonal. Let A be the area of the Rhombus in square units. Then each side of the Rhombus is: A.

4 2

8. A circle ‘A’ of radius 10 cm is inscribed in an equilateral XYZ. Another circle B of radius 6 cm is drawn such that it is concentric with circle A. If a third circle C is drawn in such a way that it touches circle A internally and B touches C internally, then find the radius of circle C. A. 4 cm B. 5 cm C. 3.5 cm D. 4.5 cm

2. How many zeroes will be there at the end of 1003 × 1001 × 999 × .... × 123? A. 224 B. 217 C. 0 D. None of these

A.

D.

8 2

A. 2  12 C. 6

12 – 2 B. D. 2

11. The number of solutions are in [0, 2] such that

B.

1 2A 2

sin(2 x ) 4 

D.

1 4A 4

A. 2 C. 8 85

1 is: 8

B. 4 D. 16

86 12. Let S1 be the sum of areas of the squares whose sides are parallel coordinate axes. Let S2 be the sum of areas of the slanted squares as shown in the figure. Then S1/S2 is:

13. The real numbers x satisfying those which satisfy. A. x < 1 C. 5 < x < 1

a

a 2

A. 2

B.

C. 1

D.

x5  1 are precisely 1– x

B. 0 < x < 1 D. –1 < x < 1

14. The number of pairs of real (x, y) such that x = x2 + y2 and y = 2xy is: A. 4 B. 3 C. 2 D. 1 15. How many positive real number x satisfy the equation x3 – 3| x | + 2 = 0? A. 1 B. 3 C. 4 D. 6

2 1

2

Physics 16. A stone weighing 3 kg falls from the top of a tower 100 m high and buries itself 2 m deep in the sand. The time of penetration is: A. 0.3 s B. 0.09 s C. 0.9 s D. 1.8 s

21. A steel wire of length 0.72 m has a mass of 5 × 10–3 kg. If the wire is under tension of 60 N, the speed of transverse wave on the wire is: A. 51 ms–1 B. 63 ms–1 –1 C. 93 ms D. 103 ms–1

17. The potential difference between the points A and B of the following circuit is (consider internal resistance of the all negligible):

22. Which of the following can make a parallel beam of light, when light from a point source is incident on it? A. concave mirror as well as convex lens B. convex mirror as well as concave lens C. two plane mirrors placed at 90° to each other D. concave mirror as well as concave lens

10  A

5V 20 

B

2V

A. 4 V C. 9 V

B. 6 V D. 12 V

18. A simple pendulum is oscillating without damping. When the displacement of the bob is less than maximum, its acceleration vector is a correctly shown in:  a

B.

A.

 a

C.

 a

D.

 a

19. A marble block of mass 2 kg lying on ice, when given a velocity of 5 ms–1, is stopped by friction in 10 s. Determine the coefficient of friction. A. 0.2 B. 0.05 C. 0.01 D. 0.50 20. Find the wavelength associated with an electron subjected to potential difference of 1.25 kV. A. 0.347 Å B. 0.257 Å C. 0.457 Å D. 0.211 Å

23. In order to produce solar energy during sunlight, where energy is stored in the batteries? A. nickel-cadmium B. nickel-zinc C. zinc-cadmium D. nickel-sulphur 24. A positively charged particle (alpha particle) projected towards west is deflected towards north by a magnetic field. The direction of magnetic field is: A. towards east B. towards south C. upwards D. downwards 25. A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located? A. The image is located at 30 cm infront of the mirror B. The image is located at 30 cm behind the mirror C. The image is located on focus D. None of these 26. A body of mass M 1 collides elastically with another mass M 2 at rest. There is maximum transfer of energy, when: A. M 1 > M 2 B. M 1 < M 2 C. M 1 = M 2 D. Same for all values of M 1 and M 2

87 27. If W 1, W 2 and W 3 represent the work done in moving a particle from A to B along three different paths 1, 2 and 3 respectively (as shown) in the gravitation field of a point mass m, find the correct relation between W 1, W 2 and W 3. m 1

2 3

A

A. W 1 > W 2 > W 3 C. W 1 < W 2 < W 3

B. W 1 = W 2 = W 3 D. W 2 > W 1 > W 3

28. You are given water, mustard oil, glycerine and kerosene. In which of these media a ray of light incident obliquely at same angle would bend the most?

A. Kerosene C. Mustard oil

B. Water D. Glycerine

29. Acceleration due to gravity on moon is 16 ms–2. An inflated balloon is released on the moon. It will: A. move down with acceleration 1.6 ms–2 B. move up with acceleration 1.6 ms–2 C. move up with acceleration 9.8 ms–2 D. move down with acceleration 9.8 ms–2 30. A man slides down on a telegraphic pole with an acceleration equal to one-fourth of acceleration due to gravity. The frictional force between man and pole is equal to in terms of man’s weight W is: W W A. B. 2 4 3 W C. W D. 4

Chemistry 31. Irrespective of the source, pure sample of water always yields 88.89% mass of oxygen and 11.11% mass of hydrogen. This is explained by the law of: A. constant volume B. constant composition C. conservation of mass D. multiple proportion

36. The temperature at which 10% aqueous solution (W/V) of glucose will exhibit the osmotic pressure of 16.4 atm is (R = 0.082 dm 3 atm K–1 mol–1). A. 110 K B. 230 K C. 360 K D. 410 K

32. The vapour pressure of two liquids X and Y are 80 and 60 Torr respectively. The total vapour pressure of the ideal solution obtained by mixing 3 moles of X and 2 moles of Y would be: A. 72 Torr B. 80 Torr C. 120 Torr D. 150 Torr

37. End product of the following reaction is:

33. The sum of pH and pKb for a basic buffer solution is 13. The ratio of the concentration of the base to that of the salt is: A. 0.5 B. 0.1 C. 1 D. 10 34. If three elements X, Y and Z crystallize in a cubic solid with X atoms at the corners, Y atoms at the cube centres and Z atoms at the faces of the cube, then the formula of the compound is: A. XYZ2 B. X2YZ C. X2Y3Z D. XYZ3 O NH2OH

35.

H+

X

Na/C2H 5OH

Y

Identify Y in the following sequence. O

N – CH3

A.

H

H

C – NHCH3

B.

H NH2

C.

H

O CH3

D.

H N

O

O + HBr

O

O

A.

HO

OH

B. Br Br

Br OH

C.

Br

OH

D. HO

Br

38. In the reaction p-chlorotoluene with KNH 2 in liq. NH3, the major product is: A. o-toluidine B. m-toluidine C. p-toluidine D. p-chloroaniline 39. 4.5 g of aluminium (at. mass = 27 amu) is deposited at cathode from Al3+ solution by certain quantity of electric charge. The volume of hydrogen produced at STP from H + ions in solutions by the same quantity of electric charge will be: A. 5.6 L B. 11.2 L C. 22.4 L D. 44.8 L 40. The correct order of increasing acid strength of the compounds: 1. CH 3COOH 2. MeOCH2COOH Me COOH 3. CH 3COOH 4. Me A. 1 < 2 < 3 < 4 B. 1 < 4 < 2 < 3 C. 4 < 2 < 3 < 1 D. 2 < 4 < 1 < 3

88 41. The reaction of chloroform with alcoholic KOH and p-toluidine form: A.

CH 3

NC

B.

CH 3

C.

CH 3

NHCl 2

D.

CH 3

CN2

N2Cl

42. Which is the monomer of neoprene in the following? A.

CH 2  C – CH  CH 2 | CH3

B. CH 2 = CH – CH = CH2 C.

43. For the reaction A + 2B  2C, the rate of forward reaction is 2000 times the rate of the backward reaction when the concentration of each reactant and product is 2.0 m. What is the value of equilibrium constant? A. 0 B. 1 C. 2 D. 13 44. Which one of the following sets gives the correct arrangement of the compounds involved based on their bond strengths? A. HF < HCl < HBr < HI B. HF > HCl > HBr > HI C. HI > HBr > HCl > HF D. HF > HBr > HCl > HI 45. Potassium permanganate acts as on oxidant in alkaline and acidic medium. The final products formed from KMnO4 in the two conditions are respectively. A. Mn2+ and Mn3+ B. MnO2 and Mn2+ C. Mn3+ and Mn2+ D. MnO2– and Mn3+

CH 2  C – CH  CH 2 | Cl

D. CH 2 = CH – C  CH

Biology 46. Life cycle of Ectocarpus and Fucus respectively are: A. Haplontic, Diplontic B. Diplontic, Haplodiplontic C. Haplodiplontic, Diplontic D. Haplodiplontic, Haplontic 47. Which one of the following is related to Ex-situ conservation of threatened animals and plants? A. Wildlife Safari parks B. Biodiversity hot spots C. Amazon rainforest D. Himalayan region 48. Good vision depends on adequate intake of carotene rich food. Select the best option from the following statements. (a) Vitamin A derivatives are formed from carotene. (b) The photopigments are embedded in the membrane discs of the inner segment. (c) Retinal is a derivative of Vitamin A. (d) Retinal is a light absorbing part of all the visual photopigments. Options: A. (a) and (b) B. (a), (c) and (d) C. (a) and (c) D. (b), (c) and (d) 49. Thalassemia and sickle cell anemia are caused due to a problem in globin molecule synthesis. Select the correct statement. A. Both are due to a qualitative defect in globin chain synthesis. B. Both are due to a quantitative defect in globin chain synthesis. C. Thalassemia is due to less synthesis of globin molecules. D. Sickle cell anemia is due to a quantitative problem of globin molecules.

50. Which of the following are not polymeric? A. Nucleic acids B. Proteins C. Polysaccharides D. Lipids 51. A disease caused by an autosomal primary nondisjunction is: A. Down’s Syndrome B. Klinefelter’s Syndrome C. Turner’s Syndrome D. Sickle Cell Anemia 52. With reference to factors affecting the rate of photosynthesis, which of the following statements is not correct? A. L ight saturation for C O 2 fixation occurs at 10% of full sunlight. B. Increasing atmospheric CO 2 concentration up to 0.05% can enhance CO 2 fixation rate C. C3 plants respond to higher temperatures with enhanced photosynthesis while C 4 plants have much lower temperature optimum. D. Tomato is a greenhouse crop which can be grown in CO2 - enriched atmosphere for higher yield. 53. Fruit and leaf drop at early stages can be prevented by the application of: A. Cytokinins B. Ethylene C. Auxins D. Gibberellic acid 54. The region of Biosphere Reserve which is legally protected and where no human activity is allowed is known as: A. Core zone B. Buffer zone C. Transition zone D. Restoration zone 55. In case of poriferans, the spongocoel is lined with flagellated cells called: A. ostia B. oscula C. choanocytes D. mesenchymal cells

89 56. A decrease in blood pressure/volume will not cause the release of: A. Renin B. Atrial Natriuretic Factor C. Aldosterone D. ADH 57. A dioecious flowering plant prevents both: A. Autogamy and xenogamy B. Autogamy and geitonogamy C. Geitonogamy and xenogamy D. Cleistogamy and xenogamy 58. Which of the following facilitates opening of stomatal aperture? A. Contraction of outer wall of guard cells B. Decrease in turgidity of guard cells C. Radial orientation of cellulose microfibrils in the cell wall of guard cells

D. Longitudinal orientation of cellulose microfibrils in the cell wall of guard cells 59. The DNA fragments separated on an agarose gel can be visualised after staining with: A. Bromophenol blue B. Acetocarmine C. Aniline blue D. Ethidium bromide 60. Which statement is wrong for Krebs’ cycle? A. There are three points in the cycle where NAD + is reduced to NADH + H + B. There is one point in the cycle where FAD + is reduced to FADH 2 C. During conversion of succinyl CoA to succinic acid, a molecule of GTP is synthesised D. The cycle starts with condensation of acetyl group (acetyl CoA) with pyruvic acid to yield citric acid

Mathematics 61. ABCD is a square. A circle is inscribed in the square. Also taking A, B, C, D (the vertices of square) as the centres of four quadrants drawn inside the circle, which are touching each other on the mid-points of the sides of square. Area of square is 16 cm 2. What is the area of shaded region?

63. In the adjoining figure O is the centre of circle. The radius OP bisects a rectangle ABCD at right angle. DM = NC = 2 cm and AR = SB = 1 cm and KS = 4 cm and OP = 5 cm. What is the area of rectangle? D A

M R

P L K

N

C

S

B

O

A. 8 cm 2 C. 12 cm 2 A. 4 – 8 C. 4 – 10 62. The

value

| x – 1 |log3 x A. 3 C. 3

2

B. 2 – 8 D. None of these of

–2 log x

9

x

satisfying

the

equation

 ( x – 1) 7 :

B. 10 cm 2 D. None of these

64. The number of distinct prime divisor of the number 5123 – 2533 – 2593 is: A. 4 B. 5 C. 6 D. 7 65. Let loga b = 4, logc d = 2 where a, b, c, d are natural numbers. Given that b – d = 7, then value of c – a is: A. 1 B. –1 C. 2 D. –2

B. 34 D. log43

Physics 66. A projectile is projected in the upward direction making an angle of 60° with horizontal direction with velocity of 147 ms–1. Then, the time after which its inclination with the horizontal is 45° is: A. 5.49 s B. 7.12 s C. 9.57 s D. 12.5 s

67. A 0.5 kg ball moving with a speed of 12 m/s strikes a hard wall at an angle of 30° with the wall. It is reflected with the same speed and at the same angle. If the ball is in contact with the wall for 0.25 s, the average force action on the wall is: A. 15 N B. 20 N C. 30 N D. 24 N

30° 30°

(2015) KVPY—Practice Paper—12

90 68. The tension in a string holdings a solid block below the surface of a liquid (of density greater than that of solid) as shown in Fig. is T0 T0, when the system is at rest. What will be the tension in the string, if the system has upward acceleration a? T0 a A. T  B. T = T0 [g – a]/g g C. T = T0 [(g + a)/g] D. T = T0 (g + a)

69. The position of an object which when placed in front of a concave mirror of a focal length 20 cm produces a virtual image which is thrice the size of the object is: A. 13.33 cm B. 15.45 cm C. 18.54 cm D. 11.23 cm 70. A bullet fired into a fixed target loses half of its velocity penetrating 3 cm. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion? A. 5 cm B. 4 cm C. 1 cm D. 2 cm

Chemistry 71. The packing efficiency of the two-dimensional square unit cell shown is:

COOH

SOCl2

B

NH3

C

NaOH Br2

Br

A

COOH

SO2 NH2

A. 78.54% C. 38.69%

A.

B. 72.45% D. 81.92%

74. In a set of reactions m-bromobenzoic acid gave a product D. Identify the product D.

B.

NH 2

72. The number of photons emitted per second by a 60 W source of monochromatic light of wavelength 663 nm is (h = 6.63 × 10–34 Js) A. 2 × 1020 B. 3 × 10–20 C. 4 × 10–20 D. 1.5 × 1020 73. The coagulating power of an electrolyte for arsenious sulphide decreases in the order: A. Ba2+ > Al3+ > Na+ B. Al3+ > Ba2+ > Na+ C. PO 43– > SO42– > Cl– D. Cl– > SO42– > PO43–

D

NH2

C.

CONH2

D.

Br

Br

75. The compound that does not liberate CO 2, on treatment with aqueous sodium bicarbonate solution is: A. Salicylic acid B. Benzoic acid C. Carbonic acid (phenol) D. Benzene sulphonic acid

Biology 76. Mycorrhizae are the example of: A. Fungistasis B. Amensalism C. Antibiosis D. Mutualism 77. The pivot joint between atlas and axis is a type of: A. Fibrous joint B. Cartilaginous joint C. Synovial joint D. Saddle joint 78. Which of the following is correctly matched for the product produced by them? A. Acetobacter aceti : Antibiotics B. Methanobacterium : Lactic acid C. Penicillium notatum : Acetic acid D. Sacchromyces cerevisiae : Ethanol

79. Frog’s heart when taken out of the body continues to beat for sometime. Select the best option from the following statements. (a) Frog is a poikilotherm. (b) Frog does not have any coronary circulation. (c) Heart is “myogenic” in nature. (d) Heat is autoexcitable. A. Only (c) B. Only (d) C. (a) and (b) D. (c) and (d) 80. Myelin sheath is produced by: A. Schwann Cells and Oligodendrocytes B. Astrocytes and Schwann Cells C. Oligodendrocytes and Osteoclasts D. Osteoclasts and Astrocytes

91

ANSWERS 1 C

2 C

3 C

4 D

5 B

6 C

7 A

8 A

9 B

10 D

11 C

12 A

13 D

14 A

15 A

16 B

17 A

18 D

19 B

20 A

21 C

22 A

23 A

24 C

25 A

26 C

27 B

28 D

29 A

30 D

31 B

32 A

33 B

34 D

35 C

36 C

37 B

38 B

39 A

40 B

41 A

42 C

43 B

44 B

45 B

46 C

47 A

48 B

49 C

50 D

51 A

52 C

53 C

54 A

55 C

56 B

57 B

58 C

59 D

60 D

61 A

62 B

63 B

64 C

65 A

66 A

67 D

68 C

69 A

70 C

71 A

72 A

73 B

74 C

75 C

76 D

77 C

78 D

79 D

80 A

EXPLANATORY ANSWERS 1. Let the radius of the cage = r Using the figure, CB = r 2 Then, OC + CB = 482

9S = 10

r

A B r O r  r 2 = 482 Hence, r = 199.66 = 200 cm (approximately) 1

2. Since all the numbers in the expressions are odd. Therefore the product of all odd numbers would also be odd. Hence, number of zeroes is zero.

1 9S 10 . = ,S = 9 10 81 Area of Rhombus =

6.

Each diagonal, d1 =

A d2 = 2 A

(1024)1024 = (16)16n 10 1024

2 

16 n

 

= 24

We know that,

a=

1 d12  d 22 2

a=

1 1 A  4A  a = 5A 2 2

10 × 1024 = 4 × 16n

10  1024 n= 4  16 n = 160. 5.

1 2 3 n  2  3  ....  n  ... 10 10 10 10 1 2 S   .... = 10 10 2 10 3 Subtracting 1 1 1 9S  2  3  ... = 10 10 10 10

1  d1  d2 2

1  d  (2 d ) = A 2 d= A

3. 96! = 292 × 346 × 522 × .... Now 522 and 222 can be eliminated since these will result in zeroes, then the unit digit of the remaining = 270 × 346 = 4 × 9 = 6. 4.

1 10 1 1– 10

C

7.

C

D

S=

4 2 F

4 2 P 4 2

A

E

B

92 x2 cos 40° = x2 – 2 x2(1 – cos 40°) = 2 1 x= sin 20

 Incircle is formed whose radius = 4 2 PE = r = 4 2 In

PF = r = 4 2 also PF = AE APE, (AP)2 = (AE)2 + (PE)2 2

2

   4 2 

= 4 2 

Now,

 64

AP = 8.

   (A 1A5)2 = 2 x 2  1 – cos   9 = 2x2 (1 – cos 160°) = 4x2 sin2 80° A1A5 = 2x sin 80° Similarly, In A2OA4 A2A4 = 2x sin 40° (ii) – (iii) A1A5 – A2A4 = 2x (sin 80° – sin 40°) = 2 (using (i))

X

A P O'

B O M

C Y

Z

MN = NP + PO + OM = 2 + 3 + 3 = 8 cm Hence, the radius of circle C = 4 cm.





9.

2





11.

=1

2

x 2  x 2 – (A1 A 5 )2 8 = 9 2x2

In A1OA5,

8. O is the centre of triangle, circle A and circle B where as O is the centre of circle C Since diameter of circle O is

N

cos

...(i)

1 8 Range  x + [0, 2] Let 2x = y given Range 0  x  2 0  2x  4  0  y  4

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

sin(2 x)4 =

O



2

3

4

1 8 From above graph, we can say total eight solution.

sin(2 x)4 =

45° 45°  2



12.

  2 = 1 





2 1 = 1



1

=

2 1

 2 –1.

A9

10.

A8

A1 2

x

2 9



 2



A7

x

O

13. A2 2

A6

A3 2 A5

A1OA 2 =

cos

2 = 9

A4

2  40 9 x2  x2 – 4 2x

2

a2 a2   ... = 4 16

a2 4a2  1 3 1– 4 a2 a2 a2 a2 4a2    ... = 2  S2 = 1 2 8 32 6 1– 4 S1  = 2. S2 S1 = a2 



x5  0 1– x0 x 1 – x x + 5 > 1 + x2 – 2x x2 – 3x – 4 < 0 (x – 4) (x + 1) < 0 x  (–1, 4) Using eq. (i) and (ii), x  (–1, 1).

...(i)

...(ii)

93 14. x = x2 + y2 and y = 2xy I. When y = 0  x = x2  x = 0, x = 1 Solution are (0, 0) (1, 0)

1 1 1 II. When x =  – = y2 2 2 4 1 1 = y2  y =  4 2 1  1 1  1  Solutions  ,   , –  .  2 2  2 2 15. x3 – 3| x | + 2 = 0 Let, x x3 – 3x + 2 (x – 1) (x2 + x – 2) x

> = = =

0 0 0 1

–1  9 1 2 Now, Let x HBr > HCl > HF, hence the order of bond strength is HF > HCl > HBr > HI.

44. We have, bond strength 

45. In alkaline medium, KMnO 4 is first reduced to manganate and then to insoluble manganese dioxide 2MnO4– + H2O  2MnO2 + 2OH– + 3[O] In acidic medium, manganous sulphate is formed MnO4– + 8H+ + 5e–  Mn2+ + 4H2O. 61.

6 = 2 25 = OL = KL = Area of rectangle =

[ ML = LN =

 r 2 1  –  AB  BC  sin 90 Area of leaf = 2  2  4   (2)2 1  = 2 –  2  2  1 = 2( – 2) 4 2   C 2 A

2

B

4

Area of shaded region = 2 × 2( – 2) = 4 – 8. 62. Consider option (B) |34 – 1| log3(34)2 –2 log819 = 8 | 80 |log 33 – log 8181 = log338 – log8181 = 8– 1= 7= Hence, option (B) is correct.

(34

– (80)7 7 7 7

D 2 cm

M

A 1 cm R

66. Let the body be projected from O with velocity u making an angle 60° with the horizontal and it reaches at A after time t with velocity v making an angle 45° with the horizontal direction. Resolving u and v into horizontal and vertical components. We have, u cos 60° = v cos 45°

u cos 60 cos 45 v sin 45° = u sin 60° – gt

or N

L 4 cm

65. logab = 4 and logcd = 2 b = a4 and d = c2 4 a – c2 = 7 (a2 – c)(a2 + c) = 7 a2 – c = 7 and a2 + c = 1 not possible or a2 – c = 1 and a2 + c = 7 2a2 = 8 a2 = 4 a = ±2 a = 2, c = 3  c – a = 1.

or

P

63.

4 cm K

2 cm

C

t =

u sin 60 – v sin 45 g

 u cos 60  u sin 60 –  sin 45  cos 45  t = g 147  3 1  u –  [sin 60 – cos 60] =  9.8  2 2  g = 5.49 s.

S 1 cm B

(OS)2 = (OR)2 + (KS)2 25 = (OK)2 + 16 OK = 3 and ON 2 = OL2 + LN2

v =

=

5 cm

O

(OL)2 + 9 4 cm OL – OK = 1 cm 1 × 10 = 10 cm 2.

64. 5123 – (2533 + 2593) = 5123 – [(512) (2532 + 2592 – 253 × 259)] = 512 (5122 – (512)2 – 3(253) (259)) = 512 (3.253 – 259) = 29.3.253.7.37 = 29.3.(11).(23).7.37 So, number of distinct prime divisors are 6.

or

1)7

3 cm]

67. Given, m = 0.5 kg, v = 12 m/s,  = 30°, t = 0.25 s, F= ? As, Impulse = Change in linear momentum  F × t = 2 mv sin 30° (perpendicular to the wall)  F=

2mv sin 30 2  0.5  12 1  = 24 N. = t 0.25 2

96 68. Let V be the volume of block of density . Let  be the density of liquid. Mass of the block, m = V Initially, for the equilibrium of block, we have, Upward thrust = Vg = T0 + Vg or T0 = V( – )g ...(i) When lift is accelerated upwards, then g = g + a, so, T = V( – ) (g + a) ...(ii)

 ( g  a)  T = T0  .  g  69. Given, m = 3, u = ?, f = –20 cm

v –v 3 =  v – 3u u u According to mirror formula, We know that,



1 1  v u 1 1  –3u u –1  3 3u 2 3u u

=

4r 2

1 –20 –1 = 20 –1 = 20 = –13.33 cm.

 Packing efficiency =

2  r 2

 2 2r 

2



 4

As,

E=



60 =



n=



n = 2000 × 1014 = 2 × 1020.

n  6.63  10 –34  3  10  663  10 –9 60  663  10 –9 6.63  10 –34  3  10 8

73. According to Hardy-Schulze rule, greater is the valency of the oppositively charged ion of the electrolyte being added, faster is the coagulation. Therefore, for a negatively charged solution like As2S3, the order is Al3+ > Ba2+ > Na+.

1  u 1 2 mu = m    F  3 2  2 2

1 2 mv = F × S 2 From equation (i) and (ii), we get, 3 3 = 4 S or S = 4 cm  Further distance travelled = 4 – 3 = 1 cm.

B

1 1  2 4 Area occupied by particles = 2 × r2

=

Case-II,

A

No. of particles = 4 

1 f

3  1 2  mu  = F × 3 42

 2 2r

nhc  n  no. of photons emitted per second

2

or

a=

m= –

70. According to law of conservation of energy, Case-I,

2a = 4r

C

72. Power = Energy per second  60 W = 60 J/s Given, E = 60 J  = 663 nm = 663 × 10–9 m and c = 3 × 108 ms–1







D

= 0.7857 or 78.57%.

ga = g



AC = 4r

V( – )( g  a) T = V( – ) g T0





But,

...(i)

74.

COOH

COCl

SOCl2

...(ii)

NH 3

Br CONH 2

Br NH 2

NaOH, Br2

Area covered by particles Total area If a is length of unit cell. Face diagonal, AC = 2a

Br

71. Packing efficiency =

Br

75. Phenol is weaker acid than carbonic acid (H 2CO 3) and does not liberate CO 2 on treatment with aqueous sodium bicarbonate solution.   

97

Practice Paper (Solved)

9

Kishore Vaigyanik Protsahan Yojana (KVPY) STREAM – SA

Mathematics 1. ABCD is a rectangle of dimension 6 cm × 8 cm. DE and BF are the perpendiculars drawn on the diagonal of the rectangle. What is the ratio of the shaded to that of unshaded region? D

6. If a, b and c are distinct real numbers such that a : (b + c) = b : (c + a) then: A. a, b, c are all positive B. a, b, c are all negative C. a + b + c = 0 D. ab + bc + ca + 1 = 0

C

R 7. In the adjoining figure D ABCD, P and R are the mid-points of the sides AB S and CD. ABCD is a parallelogram. What is the A P ratio of the shaded to the unshaded region? 1 1 A. B. 2 3 1 C. D. None of these 4 8. In the adjoining figure ABCD is a D P rectangle. Find the m axim um number of rectangle including the E largest possible rectangle: A Q A. 16 B. 7 C. 18 D. 24

F E A

B

A. 7 : 3

B. 16 : 9

C. 4 : 3 2

D. Data insufficient

2. Total number of digits in the product of (4)1111 × (5)2222 is: A. 3333 B. 2223 C. 2222 D. Can’t be determined 3. If (x – 5) (y + 6) (z – 8) = 1331, then the minimum value of x + y + z is: A. 40 B. 33 C. 19 D. Not unique 33

4. Which of the following is correct if A  33 , 3

C

Q

B

R

C

F

S

B

9. In the given triangle ABC, the length of sides AB and AC is same (i.e., b = c) and 60° < A < 90° then the possible length of BC is:

33

B  333 , C  33

and D  3333 ? A. A > B = C > D B. C > A > B > D C. A > C > D > B D. C > B > D > A

A c

5. Let AB be a line segment of length 2. Construct a semicircle S with AB as diameter. Let C be the midpoint of the arc AB. Construct 1 another semicircle T external to the triangle ABC with chord AC as diameter. The area of the region inside the semicircle T but outside S is:  1 A. B. 2 2  1 C. D. 2 2

B

b

C

a

A. b < a < 2b

B.

C.

D.

bab 3

c  a  3a 3 cac 2

10. The number of integers ‘a’ such that 1  a  100 and aa is a perfect square is: A. 50 B. 53 C. 55 D. 56 97

(2015) KVPY—Practice Paper—13

98 11. The least positive integer n for which n 1 – n –1  0.2 is: A. 24 B. 25 C. 26 D. 27 12. How many natural numbers n are there such that n! + 10 is a perfect square? A. 1 B. 2 C. 4 D. Infinitely many 13. If x, y are real numbers such that 3 then the value of (x + y)/(x – y) is: A. 0 B. 1 C. 2 D. 3

x 1 y

–3

x –1 y

 24

14. The number of positive integers n in the set {1, 2, 3, ..... 100} for which the number A. B. C. D.

12  2 2  32  ....  n 2 1  2  3  ...  n

33 34 50 100

15. Three circles of radius 1, 2 and 3 units respectively touch each other externally in the plane. The circumradius of the triangle formed by joining the centres of the circle is: A. 1.5 B. 2 C. 2.5 D. 3

Physics 16. A solid sphere, a hollow sphere and a ring are released from the top of an inclinded plane (frictionless) so that they slide down the plane. Then maximum acceleration down the plane is (for no rolling): A. ring B. hollow sphere C. solid sphere D. all same

3.6

–1

v (ms )

17. An elevator is going up, the variation in the velocity of the elevator with time (t) is given in the graph. What is the height to which elevator takes the passengers?

O

2

10

12

Time (s)

A. 36 m C. 60 m

B. 48 m D. 72 m

18. When light travels from the glass to air, the incident angle is 1 and the refracted angle is 2. The true relation is: A. 1 = 2 B. 1 < 2 C. 1 > 2 D. None of these 19. U234 has 92 protons and 234 total nucleons in its nucleus. It decays by emitting an -particle. After the decay it becomes: A. Th230 B. U232 C. Ra230 D. Pa232 20. A proton travelling 23° w.r.t the direction of a magnetic field of a strength 2.6 mT experience a magnetic force of 6.5 × 10–17 N. The speed of the proton is: A. 3 × 104 m/s B. 4 × 105 m/s C. 7 × 106 m/s D. 9 × 105 m/s 21. One side of a glass slab is silvered as shown in figure. A ray of light incident on the other side at angle of incidence, i = 45°. Refractive index of glass is given as 1.5. The deviation of ray of light from the initial path, when it comes out of the slab is:

45°  = 1.5

A. 45° C. 120°

B. 90° D. 180°

22. A boy can throw a stone up to a maximum height of 10 m. The maximum horizontal distance that the boy can throw the same stone up to will be: A. 10 m B. 15 m C. 20 m D. 25 m 23. Wind is blowing West to East along two parallel tracks. Two trains moving with same speed in opposite directions have the steam of one double that other. The speed of each train is: A. equal to that of wind B. double that of wind C. three times that of wind D. half that of wind 24. The variation of net downward force F on a body on a rough inclined plane versus sine of angle of inclination () is shown by the following four graphs. The correct one is:

F

F

A.

B. sin 

sin 

F

F

C.

D. sin 

sin  (2015) KVPY—Practice Paper—13-II

99 25. Consider two observers moving with respect to each other at a speed v along a straight line. They observe a block of mass m moving a distance l on a rough surface. The following quantities will be same as observed by the two observers. A. acceleration of the block B. work done by friction C. kinetic energy of the block at time t D. total work done on the block 26. A satellite is moving around the earth with speed v in a circular orbit of radius r. If the orbit radius is decreased by 1% its speed will be: A. increase by 1% B. decrease by 1% C. increase by 0.5% D. decrease by 0.5% 27. A bird sitting on the floor of a wire cage which is being carried by a boy starts flying. The boy will feel that the box is now: A. lighter B. heavier C. no change in weight D. lighter in the beginning and heavier later 28. At what temperature is the rms velocity of a hydrogen molecule equal to that of an oxygen molecule at 47°C? A. 5 K B. 10 K C. 20 K D. 40 K

29. An uniform wire of resistance 20  having resistance 1 /m is bent in the form of a circle as shown in figure. If the equivalent resistance between A and B is 1.8 , then the length of the shorter section is:

A

B

A. 0.1 m C. 4 m

B. 2 m D. 5.2 m

30. Two balloons are blown into spherical shape of unequal sizes and are connected through a narrow tube as shown in figure. Then: B A

A. the smaller balloon becomes bigger B. the bigger balloon becomes smaller C. the smaller becomes smaller and the bigger becomes bigger D. there is no change in their sizes.

Chemistry 31. A metal crystallises with The edge of the unit cell the metal atom is: A. 124 pm C. 288 pm

a face-centred cubic lattice. is 408 pm. The diameter of

33. If the equilibrium constant for N 2(g) + O2(g)  2NO(g) 1 1 is K, the equilibrium constant for N 2 ( g)  O 2 ( g)  2 2 NO(g) will be:

C.

1

A.

OH

1 K 2

D. K2

B.

50% KOH Cl

+ Cl

Cl CH 2COO –

CH 2 OH +

C. Cl

Cl COO –

CH 2 OH

D.

+ OH

OH

35. Write the IUPAC name of the following: CH 3 – CH – CH – C – CH 2 – CH 3

34. Predict the product of given reaction: CHO

COO –

CH 2 OH

B. K

K2

OH +

B. 184 pm D. 402 pm

32. Dissolving 120 g of urea (mol. wt. 60) in 1000 g of water gave a solution of density 1.15 g/m 3. The molarity of the solution is: A. 2.05 M B. 2.22 M C. 3.15 M D. 3.75 M

A.

CH 2 OH

NH 2 OH O

A. B. C. D.

5-Amino-4-hydroxy 3-Amino-2-hydroxy 4-Amino-5-hydroxy 5-Amino-1-hydroxy

hexan-3-one hexan-1-one hexan-2-one hexan-4-one

100 36. Concentrated nitric acid, upon long standing turns yellow-brown due to the formation of: A. N2O B. N2O4 C. NO2 D. NO

Br

39. Phenol when it first reacts with concentrated sulphuric acid and then with concentrated nitric acid gives: A. o-nitrophenol B. p-nitrophenol C. nitrobenzene D. 2, 4, 6-trinitrobenzene 40. The correct order of decreasing acid strength of trichloro acetic acid (A), trifluoroacetic acid (B), acetic acid (C) and formic acid (D) is: A. A > C > B > D B. B > A > D > C C. B > D > C > A D. A > B > C > D 41. The final product in the following sequence of reaction is: NH2 Br2/OH

–

NaNO2

HBF4

HCl

F

Br

A.

37. When H 2O2 is shaken with an acidified solution of K2Cr2O7 in presence of ether, the ethereal layer turns blue due to the formation of: A. CrO5 B. Cr2O 3 C. CrO 42– D. Cr2(SO 4)3 38. Chlorobenzene can be obtained from benzene diazonium chloride by: A. Friedel Crafts reaction B. Wurtz reaction C. Fittig reaction D. Gattermann’s reaction

NH2

F

F

B. Br

Br NH 2

NH – Br

Br

C.

F

D. F

NO2

42. Among the following, the molecule of highest dipole moment is: A. H2O B. NH3 C. CHCl3 D. CCl4 43. The volumes of gases H 2, CH 4, CO 2 and NH 3 adsorbed by 1 g of activated charcoal at 298 K are in the order: A. NH3 > CO2 > CH4 > H2 B. CO 2 > NH3 > H2 > CH4 C. CH 4 > CO2 > NH3 > H2 D. H2 > CH4 > CO2 > NH3 44. The vapour pressure of water is (2.3 kPa at 300 K). What is the vapour pressure of 1 molal solution of a non-volatile solute in it is: A. 8.7 kPa B. 10.05 kPa C. 12.08 kPa D. 15.03 kPa 45. The half-life of a reaction is inversely proportional to the square of the initial concentration of the reactant. Then, the order of the reaction is: A. 1 B. 2 C. 3 D. 0

Biology 46. A foreign DNA and plasmid cut by the same restriction endonuclease can be joined to form a recombinant plasmid using: A. Ligase B. Eco RI C. Taq polymerase D. Polymerase III 47. Which of the following is not a component of downstream processing? A. Expression B. Separation C. Purification D. Preservation 48. Which of the following restriction enzymes produces blunt ends? A. Hind III B. Sal I C. Eco RV D. Xho I 49. Which kind of therapy was given in 1990 to a fouryear-old girl with adenosine deaminase (ADA) deficiency? A. Radiation therapy B. Gene therapy C. Chemotherapy D. Immunotherapy

50. How many hot spots of biodiversity in the world have been identified till date by Norman Myers? A. 43 B. 17 C. 25 D. 34 51. The primary producers of the deep-sea hydrothermal vent ecosystem are: A. coral reefs B. green algae C. chemosynthetic bacteria D. blue-green-algae 52. Which of the following is correct for r-selected species? A. Small number of progeny with large size B. Large number of progeny with small size C. Large number of progeny with large size D. Small number of progeny with small size 53. If ‘+’ sign is assigned to beneficial interaction, ‘–’ sign to detrimental and ‘0’ sign to neutral interaction, then the population interaction represented by ‘+’ ‘–’ refers to:

101 A. parasitism C. amensalism

B. mutualism D. commensalism

57. Methanogens belong to: A. Slime moulds B. Eubacteria C. Archaebacteria D. Dinoflagellates

54. Which of the following is correctly matched? A. Stratification – Population B. Aerenchyma – Opuntia C. Age pyramid – Biome D. Parthenium – Threat to hysterophorus biodiversity 55. Red List contains data or information on: A. marine vertebrates only B. all economically important plants C. plants whose products are in international trade D. threatened species 56. Which one of the following is wrong for fungi? A. They are both unicellular and multicellular B. They are eukaryotic C. All fungi possess a purely cellulosic cell wall D. They are heterotrophic

58. Select the wrong statement: A. Diatoms are microscopic and float passively in water B. The walls of diatoms are easily destructible C. ‘Diatomaceous earth’ is formed by the cell walls of diatoms D. Diatoms are chief producers in the oceans 59. The label of a herbarium sheet does not carry information on: A. height of the plant B. date of collection C. name of collector D. local names 60. Conifers are adapted to tolerate extreme environmental conditions because of: A. presence of vessels B. broad hardy leaves C. superficial stomata D. thick cuticle

Mathematics 61. In the adjoining figure ABC is an equilateral triangle inscribing a square of maxium possible area. Again in the square there is an equilateral triangle whose side is same as that of the square. Further the smallest equilateral triangle inscribes a square of maximum possible area. What is the area of the innermost square if the each side of the outermost triangle be 0.01 m?

E, Q and C, F, R lie in the same straight lines respectively. What is the ratio of perimeters of ABC : DEF : PQR? A. 3 2 : 2 2 :1 B. C.

C

D.

A

A. C.

873 – 504 3  cm 873 – 405 2  cm

2

2

63. pqr is a three-digit natural number such that pqr = p3 + q3 + r3. What is the value of r? A. 0 B. 1 C. 3 D. Can not be determined

B

0.01

B.

738 – 504 3  cm

2

64. How many divisors of 105 will have at least one zero at its end? A. 9 B. 12 C. 15 D. 25

D. None of these

62. In the adjoining figure three congruent circles are touching each other. Triangle ABC circumscribes all the three circles. Triangle PQR is formed by joining the centres of the circle. There is a third triangle DEF. Points A, D, P and B,

C

65. Let x, y, z be non zero real number such that

F

x y z y z x x3 y3 z3    7 and    9 , then 3  3  3  3 y z x x y z y z x

R

A

D

P

   2 1  3  :  2  3  : 2 2 1  3  : 2 3 : 3

2 4 3 : 2 3 : 3

Q

E

B

is equal to: A. 152 C. 154

B. 153 D. 155

102

Physics 66. If the acceleration due to gravity at earth is ‘g’ and mass of the earth is 80 times that of the moon and radius of the earth is 4 times that of the moon, the value of ‘g’ at the surface of the moon will be: g g A. B. 2 5 g g C. D. 10 15 67. A ball falls vertically on to a floor, with momentum p, and then bounces repeatedly, the coefficient of restitution is e. The total momentum imparted by the ball to the floor is: (1  e)  1 A. p  1 –  B. p (1 – e) e C. p(1 + e)

D.

 1 p 1    e

68. A projectile can have the same range R for two angles of projection. If t1 and t2 be the times of flights in the

two cases, then the product of the two time of flights is proportional to: 1 A. R B. R 1 C. D. R 2 R2 69. A small block of mass 100 g moves with uniform speed in a horizontal circular groove, with vertical side walls of radius 25 cm. If the block takes 0.2 s to complete one round, the normal contact force by the side wall of the groove is: A. 12.25 N B. 18.45 N C. 24.65 N D. 32.35 N 70. A string of mass 2.50 kg is under a tension of 200 N. The length of the stretched string is 20 m. If the transverse jerk is struck at one end of the string, how long does the disturbance take to reach the other end? A. 0.5 s B. 1.0 s C. 1.5 s D. 2.0 s

Chemistry 71. How many grams of concentrated nitric acid solution should be used to prepare 250 mL of 2.0 M of HNO3? The concentrated acid is 70% HNO 3 is: A. 45.0 g conc. HNO3 B. 52.0 g conc. HNO3 C. 60.0 g conc. HNO3 D. 72.0 g conc. HNO3

A. B. C. D.

74. Methyl benzoate can be prepared by:

72. In the dichromate dianion, the nature of bonds are: A. six equivalent Cr-O bonds and one O-O bond B. six equivalent Cr-O bonds and one Cr-Cr bond C. six non equivalent Cr-O bonds D. six equivalent Cr-O bonds and one Cr-O-Cr bonds

A. C6H5 COOH + CH2N2   

H B. C6H5 COOH + CH3OH  Pyridine C. C6H5 COCl + CH3OH    D. All the above methods

73. Identify ‘C’ in the following: + CH3 – CH – CH3

Anhyd. AlCl3 HCl

A

O3 130°C

B

Ethanol Propanone Cumene hydroperoxide Alcohol

75. Anilinium hydrogensulphate on heating with sulphuric acid at 453-473 K produces: A. sulphanilic acid B. aniline C. anthranilic acid D. benzene sulphonic acid

dil H2 SO4 100°C

Cl Phenol + C

Biology 76. Which one of the following statements is wrong? A. Laminaria and Sargassum are used as food B. Algae increase the level of dissolved oxygen in the immediate environment C. Algin is obtained from red algae, and carrageenan from brown algae D. Agar-agar is obtained from Gelidium and Gracilaria

78. How many plants among Indigofera, Sesbania, Salvia, Allium, Aloe, mustard, groundnut, radish, gram and turnip have stamens with different lengths in their flowers? A. Six B. Three C. Four D. Five

77. The term ‘polyadelphous’ is related to: A. calyx B. gynoecium C. androecium D. corolla

80. Free-central placentation is found in: A. Citrus B. Dianthus C. Argemone D. Brassica

79. Radial symmetry is found in the flowers of: A. Cassia B. Brassica C. Trifolium D. Pisum

103

ANSWERS 1 B

2 B

3 A

4 B

5 B

6 C

7 B

8 C

9 D

10 C

11 C

12 A

13 D

14 B

15 C

16 D

17 A

18 B

19 C

20 B

21 B

22 C

23 C

24 B

25 A

26 C

27 A

28 C

29 B

30 C

31 C

32 A

33 A

34 C

35 A

36 C

37 A

38 D

39 A

40 B

41 A

42 A

43 A

44 C

45 C

46 A

47 A

48 C

49 B

50 D

51 C

52 B

53 A

54 D

55 D

56 C

57 C

58 B

59 A

60 D

61 A

62 C

63 D

64 D

65 C

66 B

67 B

68 A

69 C

70 A

71 A

72 D

73 B

74 D

75 A

76 C

77 C

78 C

79 B

80 B

EXPLANATORY ANSWERS

1.

5. ACB = 90°

1  DE  AE Area of DAE = 2 1 Area of DEC  DE  CE 2

AC =

2 C

2

AE (AD)2  6  9     = CE (DC)2  8  16

1

 The area of shaded to unshaded region = 2. (4)1111 × (5)2222 = 22222 × 52222 = 1 × 102222 Hence, there will be 2223 number of digit. 3. (x – 5) = (y + 6) = (z – 8) = 11  x = 16, y = 5, z = 19 Thus, (x + y + z) = 16 + 5 + 19 = 40. 3

33

A= 3

327

16 . 9

2 1  2   =  2   2  

6.

a : (b + c) = b : (c + a) add 1 both sides, abc abc = bc ca a + b + c = 0.

7. Let Area of | | gm ABCD = 16 unit2

3

R

D

C 2

2

C= 3 C>A

2

S

2

2

2

27

A = 33 , D = 3333 A > D ( 327 > 333) Thus, the correct relation is C > A > B > D. Now,

B

  1 1   1 – –  = –   .  4 2 4 4 2 2

333

Hence,

1

O 2

Required Area = Area of semicircle having AC as diameter = Area under Arc OAC but outside triangle AOC

Area of BCF 9 = Area of BFA 16

4.

1

A

Similarly, in ABC,

A

Required Ratio =

Q

2

2 P

4 1  . 12 3

B

104 8. Number of total rectangle = 4C2 × 3C2 =6 ×3 = 18.

15. Formed triangle will be right angle whose sides are 3, 4, 5 1

O1

9. At A = 60°,

O2 2

1

BC = b = c A = 90°

and at

2

3

3 O3

BC =

2b  2c 60° < A < 90°



BC = c  a  c 2 10. All even values of a i.e., 50 and 1, 9, 25, 41, 81 total 55. 12. If n = 1, 2, 4, 5, n! + 10 is not a perfect square If n = 3, n! + 10 is a perfect square If n > 5 n! + 10 = 10

[3 × 4 × 6 ... × n + 1]

length of hypotenuse 2 5 =  2.5. 2 16. Since inclined plane is frictionless, then there will be no rolling and the mass will only slide down. Hence, acceleration will be same for all the bodies. So, Circum radius =

17. The height due to which an elevator takes the passengers will be equal to the area under the v-t graph.

then exponent of 2 is one so it is not a perfect square. 3x/y = t,

13.

t = 24 3 8t = 3 × 24 t=9 x/y 3 =t 3x/y = 9 x =2 y

1 (12  8)  3.6 2 1 =  20  3.6  36 m. 2



H=

3t –

  So,

18. We have, Incident ray

1 (glass)

1

2

x = 2y 

14.

3y xy  3. = y x–y

n(n  1)(2n  1) 2 6n(n  1) 



2n  1 =K 3 n=

3K – 1 2

1

3K – 1  100 2

Refracted ray  2 (Air)

Since, 2 < hence, 2 > or 1 < (According to Snell’s

19. -(2He4) particle has mass number 4 and atomic number 2. Hence, after emission of an -particle, the mass number of the U 234 will be reduced by 4 and after decay it becomes Ra230. 92 U

20. As,

234

  particle

  90Ra 230 . v=

3  3K  201

201 1K  3

=

1  K  67  Number of odd integers = 34.

1 1 2 law)

=

F qBsin  6.5  10 –17 1.6  10 –19  2.6  10 –3 sin 23 

6.5  10 5 = 4 × 105 ms–1 1.6  2.6  0.39

105 21. We have, Incident ray

45°

45°

r

Emergent ray 45° 45°

r



dv 1 GM GM 1 dr dr /  = 2 r3/ 2 r 2 r v dv  100 v

 % increase in velocity =

r

= From the figure, clearly, the angle between the incident ray and the emergent ray is 90°. 22. The maximum speed with which the boy can throw the stone is v=

= 10 2 m/s Maximum range, when projectile is thrown at an angle of 45°. 

Rmax



10 2 v2 =  g 10

2



24. When a body of mass m is held on a rough inclined plane of inclination , the net downward force on the body is, F = mg sin  – f Where f is constant force of friction. As sin  increases, f increases. Hence, (B) is correct answer. 25. When two observers are moving w.r.t. each other at a speed v along a straight line, acceleration of the block, if any, will be same. Distance moved may be different. Therefore, workdone/kinetic energy of the block may appear different. 26. Orbital speed of satellite, GM r

we get,

=

GM 2r 3 / 2

or

TH = TOXY

dr (in magnitude) ...(ii)

MH M OXY

= (40  273)

2 320  = 20 K. 32 16

29. Let the resistance of the shorter part AB be x. Total resistance is 20 . Hence, the resistance of longer AB part will be (20 – x). With respect to A and B, the two portions are connected in parallel. So,

Req =

(20 – x ) x  1.8 (20 – x )  x

On solving, we get, x = 2  The resistance per unit length is 1 /m. Thus, the length of the shorter part = 2 m. 30. Excess pressure inside the balloon is given by P = 4T/r  P × 1/r Hence, pressure of air in balloon A > Pressure of air in balloon B.  air will go from A to B, thus, making A smaller and B bigger. 31. For fcc lattice, radius (r) is related to edge length (a) a  r= , a = 408 pm 2 2

408 = 144 pm 2  1.414 Diameter = 2 × 144 pm = 288 pm. r=



1 GM r –3 / 2 dr dv = – 2

3RTOXY M OXY

3RTH  MH

TH T = OXY MH M OXY

...(i)

If r decreases, v increases; differentiating (i),

Vrm s =



 20 m.

23. Let u and v be the speed of train and wind respectively. The speed of steam track of train moving in the direction of wind = u – v The speed of stream track of train moving in the opposite direction of wind = u + v As per question, u + v = 2(v – u) or u = 3v Hence, the speed of each train is three times that of wind.

v=

27. The air pushed down by the wings of the bird will go out of the wire cage. 28. As,

2 gh  2  10  10

1 dr 1  100   1  0.5%. 2 r 2

32. Total mass of solution = 1000 + 120 = 1120 g Volume of solution =

1120  973.9 1.15



120 / 60  1000 = 2.05 M. 973.9

Molarity =

(2015) KVPY—Practice Paper—14

106 [NO]2 K= [N 2 ][O2 ]

33. As, and

...(i)

1/ 2

[N 2 ]

Br 2/OH

41.

[NO]

K =

NH 2 –

Br

+

NH 2

Br

Br

Br

HCl (0-5°C)

[O 2 ]

Br

Br

From (i) & (ii), we get, K =

–

NaNO 2

...(ii)

1/ 2

N 2 Cl

F Br

K.

Br

HBF 4

34. Cannizzoro reaction, O CH = O

Br

C–O

CH 2OH

50%, KOH

42. We have,

+

Cl

Cl

Cl

1.

35. We have,

H

O

Cl

5-Amino-4-hydroxyhexan-3-one

3.

36. Nitric acid upon long standing turns yellow brown due to decomposition by sunlight into NO2 Sunlight

4HNO3   4NO 2  2H 2O  O 2 37. Acidified K 2 Cr 2O 7 solution reacts with hydrogen peroxide (H 2O2) to give a deep blue solution due to formation of peroxo compound CrO 5 K2Cr2O7 + H2SO 4 + 4H2O2  2CrO 5 + K2SO 4 + 5H2O. 38. Chlorobenzene can be obtained from benzene diazonium chloride by Sandmeyer’s or Gattermann’s reaction: –

OH

39.

HNO 3 –SO 3

=

1  0.0177 1  1000 /18

= 0.0177

123 – PA = 0.0177 123 PA = 12.08 kPa.

45. We have, t1/2  [A]01–n where n  order of reaction 1

o-Nitrophenol

40. As I-effect increases, –COOH group becomes more electrons deficient and tendency to lose H + increases and therefore, acidic strength increases. Thus, CF3COOH > CCl3COOH > CH 3COOH. As, +I effect increases, acidic strength decreases i.e., CH 3COOH < HCOOH. Hence, correct order is CF3COOH > CCl3COOH > HCOOH > CH3COOH.

Cl

Cl Chloroform  = 1.01 D

44. Mole fraction of solute

 NO 2

4.

43. Gases which are more easily-liquefiable are more easily adsorbed, because the vander Waals or intermolecular forces which are involved in adsorption are more predominent in such gases. Hence, the correct order is NH 3 > CO2 > CH4 > H2.

 OH

Cl

Hence, the highest dipole moment of water molecule is 1.84 D.

+ N2

SO 3 H Conc. H 2SO 4

Cl

P0

OH

H

H Ammonia  = 1.46 D

C

C

P 0 – PA

i.e., Gattermann’s reaction

N

H

Cl Carbon tetrachloride  = 0, Symmetrical

Cl

Cu, HCl

2.

Cl

NH 2 OH O

+

H

Water  = 1.84 D

6 5 4 3 2 1 CH 3 – CH – CH – C – CH 2 – CH 3

N2 Cl

H

Since,   

t1/2 

[A 0 ]2

t1/2  [A0]–2 1 – n = –2 n = 3.

61. Let the side of square = a In DBF,

sin 60° =

a DB

107 A 60°

a

B

a

60°

D 2a 3

DM = DM + MN + NE =

60°

PAM = QBN = 30°

60°

60°

b a

F

G

1 PM = tan 30° = 3 AM

Again,

C

2a A

3 2a 3

2r

Q

M

2r

N

30°

= 0.01 3



= r 3  2r  r 3  2r 1  3



3–2





AB = BC = AC = 2r 1  3





= 0.01 3 – 2 3







2









 (0.01) 21 – 12 3 m

= 21 – 12 3 cm [ 1 m = 100 cm]



Area of innermost square = b2 = 21 – 12 3



= (21)2  12 3

2



2



– 2  21  12 3

= 441  432 – 504 3 = 873 – 504 3 cm 2. 62. Let the radius of each circle be r unit then, PQ = QR = PR = 2r P

2r

64. Total number of factors of 105 = 36 Total number of factor of 10 5 ending with no zero = 11 Hence, the total number of factors having at least one zero at its end = 25. 65. a3 + b3 + c3 – 3abc = (a + b + c) [(a + b + c)2 – 3(ab + bc + ca)] = 7[(7)2 – 3(9)] = 7(49 – 27) = 7 × 22 = 154. 66. Let M and R be the mass and radius of the earth, M and R are mass and radius of the moon. Then, R =

M

2r

N

E

PDM = QEN = 30°

DM = cos 30° DP DM = DP × DM =

r 3 2



63. pqr can be 370 or 371. Therefore, it is not possible to arrive at a unique answer.

Q

30°











= 2 1  3 : 2  3 : 2.

b = 0.01 3 – 2 3 3 – 2 3 m = 0.01 3 – 2 3



Ratio of perimeter of ABC : DEF : PQR

Similarly let side of smaller square = ‘b’

D

B

0.01 3 32

then,

P

AM = r 3  BN AB = AM + MN + NB

= 0.01

a=



DE = DF = EF = 2  3 r

3 a = 2 DB

 a







E

a

DB =

r 3 r 3  2r   2 3 r 2 2

3 [DP = QE = (r)] 2

R M and M = 4 80

Let g and g, the acceleration due to gravity on the surface of the earth and moon respectively. Then,

g=

and

g =

=

GM R2

GM  R

2

g . 5

G

M / 80 (R / 4)

2



1 GM  5 R2

108 67. When a particle undergoes normal collision with a floor or a wall, with the coefficient of restitution (e), the speed after collision is e times the speed before collision. Therefore, change in momentum after first impact = ep – (–p) = p(1 + e). After the second impact, change in momentum would be e(ep) – (–ep) = ep(1 + e) and so on.  Total change in momentum of ball = momentum imparted to floor = p(1 + e) [1 + e + e2 + ...]

As,

71. As,

= p(1 + e)/(1 – e).



Mass of HNO3 =

68. The horizontal range is same, when angle of projection is  or (90° – ). Therefore, t1 = 2 u sin/g



t1 t2 =

= 

g

2

sin 2 

2  250  63 = 31.5 g 1000

31.5  100 = 45.0 g. 70

=

72. In Cr2O72– ion, there are six equivalent Cr-O bonds and one Cr-O-Cr bond O O – O

O Cr

Cr

CH3CHCH3

73.

+ CH3 – CH – CH3

2

CH3C – O – O – H

v (7.85)  r 0.25

O + CH3 – C – CH3

100°C

2

Propanone

 C6H5COOCH 3 + N2 74. C6H5 COOH + CH2N2  H

C6H5COOH + CH 3OH  C6H5COOCH 3 + H2O Pyridine  C6H5COOCH3 + HCl C6H5COCl + CH3OH     Hence, all the above methods used to prepare methyl benzoate.

75. We have,

M = 2.50 kg,

+

–

NH3 HSO4

T = 200 N, l = 20 m Mass per unit length,

+

NH3

–

SO3

m –1

NH2 Re-arrangement

180 – 200°C

M 2.5  l 20

= 0.125 kg

OH

dil.H2SO4

= 246.49 ms–2 This acceleration is towards the centre of the circular groove. The only force in this direction is normal contact force due to side walls.  R = ma = 0.1 × 246.49 = 24.65 N.

m=

O2 130°C

CH3

2 r 22 0.25  2  t 7 0.2

= 7.85 ms–1

70. Given,

Anhyd. AlCl3

Cl

69. Speed of the block is,

a=

O O – O

2 R g

t1t2  R

v=

Mass / 63  1000 250

 Mass of HNO3 required for 70% solution

2u sin  2u cos   g g 2u 2

Moles of HNO3  1000 Volume of solution

Molarity =

2=

2u sin(90 – ) 2u cos   g g

l 20  = 0.5 s. v 40

t=



t2 =

200 = 40 ms–1 0.125

 Time taken by disturbance to reach the other end,

 1  = p(1  e)   1 – e 

and

T  m

v=

SO3H Sulphanilic acid

  

109

Practice Paper (Solved)

10

Kishore Vaigyanik Protsahan Yojana (KVPY) STREAM – SA

Mathematics 1. What is the unit digit of the expression: 8889235! + 2229235! + 6662359! + 9999999! A. 9 B. 6 C. 7 D. 1

7. Suppose we have two circles of radius 2 each in the plane such that the distance between their centres is

2 3 . The area of the region to both circle lies between: A. 0.5 and 0.6 B. 0.65 and 0.7 C. 0.7 and 0.75 D. 0.8 and 0.9

2. The last digit of the following expression is: (1!)1 + (2!)2 + (3!)3 + (4!)4 + ... (10!)10 A. 4 B. 5 C. 6 D. 7

8. The sum of all non-integer x5 – 6x4 + 11x3 – 5x2 – 3x + A. 6 B. C. –5 D.

3. What is the highest power of 40 which can exactly divide 4000? A. 888 B. 999 C. 754 D. None of these

9. If a, b, c, d are positive real numbers, such that a a ab abc abcd then,    b  2c  3d 3 4 5 6 is:

4. If x satisfies | x – 1 | + | x – 2 | + | x – 3 |  6 then: A. 0  x  4 B. x  0 or x  4 C. x  –2 or x  4 D. x  –2 or x  4

1 2 C. 2 A.

5. In the adjoining figure AB = CD = 2BC = 2BP = 2CQ. In the middle, a circle with radius 1 cm is drawn. In the rest figure all are the semicircular arcs. What is the perimeter of the whole figure? A

1 cm

P

B

O

A. 4  C. 10 

C

Q

D

O3

O6

A. 136.83 C. 139.83

O4

O

D. Non determinable

11. Three taps A, B, C fill up a tank independently in 10 hours, 20 hours, 30 hours respectively. Initially the tank is empty and exactly one pair of taps is open during each hour and every pair of taps is open at least for one hour. What is the minimum number of hours required to fill the tank? A. 8 B. 9 C. 10 D. 11

6. If the radius of each circle is 6 cm, then find the area of shaded region.

O1

B. 1

10. If  and  are acute angles such that cos2 + cos2 = 3 1 and sin  . sin  = then  +  equals: 2 4 A. 30° B. 45° C. 60° D. 90°

B. 8  D. None of these

O2

roots of the equation 2 = 0 is: 11 3

12. Observe that, at any instant, the minute and hour hands of a clock make two angles whose sum is 360°. At 6 : 15 the difference between these two angles is: A. 165° B. 170° C. 175° D. 180°

O5

B. 129.63 D. 138.83 109

110 13. The number of all 2-digit numbers n such that n is equal to the sum of the square of digit in its tens place and the cube of the digit in units place is: A. 0 B. 1 C. 2 D. 4

A. 10 C. 5 15. Let a 

14. In the given diagram rope is wound round the outside of a circular drum whose diameter is 70 cm and a bucket is tied to the other end of the rope. Find the number of revolutions made by the drum of the bucket is raised by 11 m:

B. 2.5 D. 5.5

12 2 32 10012    ...  1 3 5 2001

and b 

12 2  3 5

32 10012  ...  . The integer closet to a – b is: 7 2003 A. 500 B. 501 C. 1000 D. 1001 

Physics 16. The frequency of a light wave in a material is 2 × 1014 Hz and wavelength is 5000 Å. The refractive index of material will be: A. 1.33 B. 1.30 C. 1.40 D. 3.00 17. A rifle shoots a bullet with a muzzle velocity of 500 ms–1 at the small target 50 m away. To hit the target the rifle must be aimed (Take g = 10 ms–2): A. 5 cm above the target B. 5 cm below the target C. 10 cm below the target D. exactly at the target 18. A balloon with mass m is descending down with an acceleration a (where a < g). How much mass should be removed from it so that it starts moving up with an acceleration a? ma ma A. B. ga g–a 2ma 2 ma C. D. ga g–a 19. In the case of a sphere falling through a viscous medium, it attains terminal velocity, when: A. viscous force plus buoyant force becomes equal to force of gravity B. viscous force is zero C. buoyant force becomes equal to force of gravity D. viscous force plus force of gravity becomes equal to buoyant force 20. A rise of temperature of 4°C is observed in a conductor by passing a current. If the current is tripled, the rise of temperature will be: A. 12°C B. 28°C C. 36°C D. 48°C 21. A person sitting in the train moving with constant horizontal velocity drops a ball vertically downwards. The path observed by the man in the train moving in the opposite direction to the first train on a parallel track is a/an: A. parabola

B. circular path C. ellipse D. straight line vertically downwards 22. Three forces acting on a body are shown in Fig. To have the resultant force only along the y-direction, the magnitude of the minimum additional force needed along OX is: Y

4N 30°

1N 60°

X

O 2N

A. 1.5 N C. 2.5 N

B. 0.5 N D. 3.5 N

23. An ammeter and voltmeter are joined in series to a cell. The readings are x and y respectively. If a resistance is joined in parallel with voltmeter: A. x will decrease and y will increase B. x will increase and y will decrease C. both x and y will increase D. both x and y will decrease 24. In the given reactions, which of the following nuclear fusion reaction is not possible? A.

13 1 6 C  1H

 146 C  4.3 MeV

B.

12 1 6 C  1H

 137 N  2 MeV

C.

14 1 15 7 N  1H  8 O  7.3MeV

D.

235 1 92 C  0 n

 140 54 Xe 

94 1 38Sr  0 n  200 MeV

25. At what height above surface of earth value of g is same as in a mine 100 km deep? A. 20 km B. 35 km C. 50 km D. 65 km 26. The excess of pressure inside the first soap bubble is three times that inside the second bubble. The ratio of volume of the first to that of the second bubble is: A. 1 : 9 B. 1 : 27 C. 9 : 1 D. 27 : 1

111 27. Two absolute scales A and B have triple points of water defined to be at 200 A and 350 B. The relation between TA and TB is: 2 4 A. TA  TB B. TA  TB 7 7 4 2 C. TB  TA D. TB  TA 7 7

29. To convert a galvanometer into a voltmeter one should connect a: A. high resistance in series with galvanometer B. low resistance in series with galvanometer C. high resistance in parallel with galvanometer D. low resistance in parallel with galvanometer 30. A ray of light strikes a transparent rectangular slab (of refractive index 2 ) at an angle of incidence of 45°. The angle between the reflected and refracted rays is: A. 80° B. 95° C. 105° D. 120°

28. A simple pendulum of frequency v falls freely under gravity from certain height from the ground level. Its frequency of oscillation will: A. remain unchanged B. be greater than v C. be less than v D. becomes zero

Chemistry 31. A solid is formed by two elements A and B. The element B forms cubic close packing and atoms of A occupy two-thirds of tetrahedral voids. The formula of the compound is: A. A2B 3 B. AB 2 C. A3B 2 D. AB 3 32. A 5.2 molal aqueous solution of methyl alcohol CH 2OH is supplied. What is the mole fraction of methyl alcohol in the solution? A. 0.032 B. 0.052 C. 0.086 D. 0.092 33. 20.0 g of a magnesium carbonate sample decomposes on heating to give carbon dioxide and 8.0 g magnesium oxide. What will be the percentage purity of magnesium carbonate in sample? A. 72% B. 84% C. 94% D. 62%

36. A graph is plotted between pressure and volume at different tempera- P 150 K tures. On the basis of the graph, what 100 K changes will you observe in volume, O if the: V I. Pressure is increased at constant temperature? II. Temperature is decreased at constant pressure? A. Volume increases in (I) and decreases in (II) B. Volume decreases in (I) and increases in (II) C. Volume increases in both cases D. Volume decreases in both cases 37. The structure of the major product formed in the following reaction: CH2Cl NaCN DMF CH 2CN

34. In the following compounds: I.

H

A. I > III > II > IV C. IV > I > III > II

NC

CN

I CH 2 Cl

CH 2CN

C.

D. CN

IV. N

B.

N

H

III.

CH2Cl

A.

II. N

is:

CH 3

N H

B. II > I > III > IV D. III > I > IV > II

35. The electronic configurations of four elements are given below. Arrange these elements in the correct order of the magnitude (without sign) of their electron affinity? I. 2s2 2p5 II. 3s2 3p5 2 4 III. 2s 2p IV. 3s2 3p4 Select the correct answer using the codes given below: Codes: A. III < IV < I < II B. I < II < IV < III C. II < I < IV < III D. I < III < IV < II

I

38.

H 3C

OH

+

H –H2O

[F]

Br2/CCl4

C4H8Br2 5 such products are possible

How many structures of F are possible? A. 3 B. 4 C. 5 D. 6 39. Write the IUPAC name of the following: CH 3 – CH – C – CH – O CH2CH3 OCH3 O

A. B. C. D.

CH3

3-Ethoxy-2-methoxypentan-3-one 2-Ethoxy-4-methoxypentan-3-one 4-Ethoxy-2-methoxypentan-1-one 3-Ethoxy-3-methoxypentan-2-one

112 40. In the chemical reactions: NH 2 NaNO 2 HCl/278K

X

HBF 4

Y

The compound ‘X’ and ‘Y’ respectively are: A. phenol and benzene B. nitrobenzene and flourobenzene C. benzene diazoniumchloride and fluorobenzene D. nitrobenzene and flourobenzene 41. In an alkaline medium, glycine predominantly exists as in a/an: A. anion B. cation C. covalent form D. zwittor ion 42. The first order reaction is 50% complete in 69.3 minutes. Time required for 90% completion for this reaction is: A. 110 minutes B. 230 minutes C. 260 minutes D. 360 minutes

43. Which one of the following statements is incorrect about enzyme catalysis? A. enzymes are least reactive at optimum temperature B. enzymes are mostly protein in nature C. enzymes action is specific D. enzymes are denatured by ultraviolet rays and at high temperature 44. In the cyanide extraction process of silver from argentite ore, the oxidizing and reducing agents used are: A. HNO3 and CO respectively B. HNO3 and Zn dust respectively C. O2 and Zn dust respectively D. O2 and CO respectively 45. Nitric acid can be obtained from ammonia via the formation of intermediate compounds: A. nitrogen and nitrous oxide B. nitric oxide and nitrogen dioxide C. nitric oxide and dinitrogen pentaoxide D. nitrogen and nitric oxide

Biology 46. Cortex is the region found between: A. endodermis and vascular bundle B. epidermis and stele C. pericycle and endodermis D. endodermis and pith 47. The balloon-shaped structures called tyloses: A. are linked to the ascent of sap through xylem vessels B. originate in the lumen of vessels C. characterize the sapwood D. are extensions of xylem parenchyma cells into vessels 48. A non-proteinaceous enzyme is: A. Deoxyribonuclease B. Lysozyme C. Ribozyme D. Ligase 49. Select the mismatch. A. Methanogens – Prokaryotes B. Gas vacuoles – Green bacteria C. Large central vacuoles – Animal cells D. Protists – Eukaryotes

52. During cell growth, DNA synthesis takes place in: A. M phase B. S phase C. G1 phase D. G2 phase 53. Which of the following biomolecules is common to respiration-mediated breakdown of fats, carbohydrates and proteins? A. Acetyl CoA B. Glucose-6-phosphate C. Fructose 1,6-bisphosphate D. Pyruvic acid 54. A few drops of sap were collected by cutting across a plant stem by a suitable method. The sap was tested chemically. Which one of the following test results indicates that it is phloem sap? A. Absence of sugar B. Acidic C. Alkaline D. Low refractive index

50. Select the wrong statement. A. Mycoplasma is a wall-less microorganism B. Bacterial cell wall is made up of peptidoglycan C. Pili and fimbriae are mainly involved in motility of bacterial cells D. Cyanobacteria lack flagellated cells

55. You are given a tissue with its potential for differentiation in an artificial culture. Which of the following pairs of hormones would you add to the medium to secure shoots as well as roots? A. Gibberellin and abscisic acid B. IAA and gibberellin C. Auxin and cytokinin D. Auxin and abscisic acid 56. Phytochrome is a: A. chromoprotein B. flavoprotein C. glycoprotein D. lipoprotein

51. A cell organelle containing hydrolytic enzymes is: A. mesosome B. lysosome C. microsome D. ribosome

57. Which is essential for the growth of root tip? A. Mn B. Zn C. Fe D. Ca

113 58. The process which makes major difference between C3 and C4 plants is: A. respiration B. glycolysis C. Calvin cycle D. photorespiration

C. Microscopic, motile asexual reproductive structures are called zoospores D. In potato, banana and ginger, the plantlets arise from the internodes present in the modified stem

59. Which one of the following statements is not correct? A. Water hyacinth, growing in the standing water, drains oxygen from water that leads to the death of fishes B. Offspring produced by the asexual reproduction are called clone

60. Which one of the following generates new genetic combinations leading to variation? A. Nucellar polyembryony B. Vegetative reproduction C. Parthenogenesis D. Sexual reproduction

Mathematics 61. Find the length of the string wound on a cylinder of

E

X

1 height 48 cm and a base diameter of 5 cm. The 11 string makes exactly four complete turns round the cylinder while its two ends touch the cylinders top and bottom. A. 192 cm B. 80 cm C. 64 cm D. cannot be determined 62. Find the remainder when 2222 5555 + 55552222 is divided by 7? A. 1 B. 3 C. 0 D. 5

F

A H

B

D

K

C

A.

1 4

B.

1 5

C.

1 6

D.

1 8

65. Five real numbers x1, x2, x3, x4, x5 are such that: x1 – 1  2 x2 – 4  3 x3 – 9  4 x 4 – 16  5 x5 – 25

63. If K is any natural number, such that 100 < K < 200, how many values of K exist such that K has ‘z’ zeroes at its end and (K + 2)! has ‘z + 2’ zeroes at its end? A. 2 B. 4 C. 6 D. None of these

x1  x2  x3  x 4  x5 2 x1  x2  x3  x 4  x5 The value of is: 2 A. Not uniquely determined B. 55 C. 110 D. 210 

64. In the figure, AKHF, FKDE and HBCK are unit squares; AD and BF intersect in X. Then the ratio of the areas of triangles AXF and ABF is:

Physics 66. A train of 150 m length is going towards north direction at a speed of 10 ms–1. A bird flies at the speed of 5 ms–1 towards south direction parallel to the railways track. The time taken by the bird to cross the train is: A. 10 s B. 15 s C. 18 s D. 5 s 67. A vessel contains oil (density 0.8 g cm –3) over mercury (density = 13.6 g cm –3). A homogeneous sphere floats with half its volume immersed in mercury and the other half in oil. What is the density of material of sphere? A. 2.3 g cm –3 B. 4.5 g cm –3 –3 C. 7.2 g cm D. 11.2 g cm –3

68. Two spherical bodies of mass m and 5 M and radii R and 2 R are released in free space with initial separation between their centres equal to 12 R. If they attract each other due to gravitation force only, then the distance covered by the smaller body before collision is: A. 1.5 R B. 2.5 R C. 4.5 R D. 7.5 R 69. A block of mass 3 kg in contact with a second block of mass 2 kg rests on a horizontal frictionless surface. A horizontal force of 10 N is applied to push the first block. The force with which the first block pushes the second block is: (2015) KVPY—Practice Paper—15

114 A. 0 C. 8 N

B. 4 N D. 12 N

A.

70. Two beams of light are incident normally on water B.

4   RI   . If the beam 1 passes through a glass 3 3   Refractive index   slab of height h as shown in 2 figure, the time difference for both the beams for reaching the bottom is:

C. D.

2

h 6c

Water

h 2c 3h 4c 6h c

h2

Glass h

Chemistry 71. In the reaction, A + B  C + D, the |B|0 time taken for 75% reaction of A is twice the time taken for 50% reaction [B] of A. The concentration of B varies O with reaction time as shown in the figure. The overall order of the reaction is: A. 1 B. 2 C. 3 D. 0

2. 3. 4. A. C.

Time

72. The freezing point of an aqueous solution is –0.186°C. If the values of K b and K f of water are respectively 0.52 K Kg mol–1 and 1.86 K Kg mol–1, then the elevation of boiling point of the solution in K is: A. 1.02 B. 0.052 C. 1.32 D. 2.34

Benzaldehyde into benzyl alcohol Cyclohexanone into cyclohexane Cyclohexanone into cyclohexanol 1, 2 B. 1, 3 1, 3, 4 D. 2, 3, 4

74. The compound formed when calcium–acetate and calcium formate is dry distilled: A. Acetophenone B. Acetone C. Benzaldehyde D. None of these 75. An ion with mass number 56 contains 3 units of positive charge and 30.4% more neutrons than electrons. Assign the symbol of this ion:

73. Under Wolff-Kishner reduction conditions, the conversion’s which may be brought about are: 1. Benzophenone into diphenyl methane

A.

56 3 26 Fe

B.

C.

24 Mg2  12

D.

27 13

Al3

51 Cr 2  24

Biology 76. In majority of angiosperms: A. a small central cell is present in the embryo sac B. egg has a filiform apparatus C. there are numerous antipodal cells D. reduction division occurs in the megaspore mother cells 77. Pollination in water hyacinth and water lily is brought about by the agency of: A. bats B. water C. insects or wind D. birds 78. The ovule of an angiosperm is technically equivalent to:

A. megaspore C. megasporophyll

B. megasporangium D. megaspore mother cell

79. Taylor conducted the experiments to prove semiconservative mode of chromosome replication on: A. E. coli B. Vinca rosea C. Vicia faba D. Drosophila melanogaster 80. The mechanism that causes a gene to move from one linkage group to another is called: A. crossing-over B. inversion C. duplication D. translocation

ANSWERS 1 A 11 A 21 A

2 D 12 A 22 B

3 B 13 C 23 B

4 B 14 C 24 A

5 C 15 B 25 C

6 D 16 D 26 B

7 C 17 A 27 B

8 D 18 C 28 D

9 A 19 A 29 A

10 C 20 C 30 C

(2015) KVPY—Practice Paper—15-II

115 31 A

32 C

33 B

34 A

35 A

36 D

37 D

38 A

39 B

40 C

41 A

42 B

43 A

44 C

45 B

46 B

47 D

48 C

49 C

50 C

51 B

52 B

53 A

54 C

55 C

56 A

57 D

58 D

59 D

60 D

61 B

62 C

63 C

64 B

65 B

66 A

67 C

68 D

69 B

70 A

71 A

72 B

73 B

74 A

75 A

76 D

77 C

78 B

79 C

80 D

EXPLANATORY ANSWERS 1. First of all we find the unit digit individually of all the four term. The unit digit of 8889235! is equal to the unit digit of 89235!. Now unit digit of 89235! is equal to the unit digit of 84 (since 9235! is divisible by 4) which is 6. Similarly, (222)9235! = (2)4 = 6 (666)2359! = (6)4 = 6 (999)9999! = (9)4 = 1  6 + 6 + 6 + 1 = 19  unit digit = 9. 2. The unit digit of (1!)2 = 1 The unit digit of (2!)2 = 4 The unit digit of (3!)3 = 6 The unit digit of (4!)4 = 6 The unit digit of (5!)5 = 0 The unit digit of (6!)6 = 0 Thus the last digit of (7!)7, (8!)8, (9!)9, (10!)10 will be zero  Required last digit = 1 + 4 + 6 + 6 + 0 + 0 + ... + 0 = 17  unit digit = 7.

4. Since | x | = x or –x  |x – 1| + |x – 2| + |x – 3| 3x – 6 or 6 – 3x x 5.

Then, 6.

   

6 6 6 4 or x  0.

AB = 2 × 2 = 4 cm CD = 4 cm AP = BP = BC = CQ = QD = 2 cm = (2 +  + ) + 2 + (2 +  + ) = 10. D

BC = O1O2 = 12 cm In ABO 1, tan 60° = AB =

6 3

C

6 AB

B 60° A

O2 6 cm O1

2 3

 AB = CD = 2 3  Side of Hexagon = 2 3  12  2 3 = 4 3  12

3. 40 = 8 × 5 = 23 × 5 2

4000

5

4000

2

2000

5

800

2

1000

5

160

2

500

5

32

2

250

5

6

2

125

2

62

2

31

2

15

2

7

2

3

3 2 (a ) 4

Area of Hexagon = 6  999

1 3994

= 6

3 4 3  12 4

= 6

3 48  144  96 3 4

= 6

3 192  96 3 4







23m × 5n = 8m × 5n 28994 × 5999 = (23)1331 × 2 × 5999 = 2 × (8)1331 × (5)999 = 2 × (8)332 × (8 × 5)999

2





= 6 3 48  24 3

1

Now, since,

Thus, the highest power of 40 is 999 that can exactly divide 4000!.



= 144 3 2  3













= 144 2 3  3 = 930.82 cm 2 (6)2

Area of 7 circle = 7 × = 22 × 36 = 792 cm 2 Area of shaded region = 930.82 – 792 = 138.83 cm 2.

116 7. Let two circles are x2 + y2 = 4 and 2

x – 2 3



In 3 hours = (A + B) + (B + C) + (C + A) = (6 + 3) + (3 + 2) + (2 + 6) = 9 + 5 + 8 = 22 unit Rest Tank = 60 – 22 = 38 unit For the minimum time the rest tank must be filled with A and B taps (6 + 3 = 9 units) So, the rest (38 unit) part of tank will take 5 hours more So, the tank will be filled in 8th hours.

 y2 = 4

Equation of the common chord is x  3  A



 

3,1 , B



3, –1

A

So, AC 1B = 60° AB = 2

( 3)

x C1

C2 B

2MC1 = 3 Required Area = 2[area of sector C1AB – area C1AB]

a b c d

= = = =

 = 360 –

3K K 5K – 4K = K 6K – 5K = K

a 3K 3K 1   . = b  2c  3d K  2K  3K 6K 2 10.

3 2 1  cos 2 1  cos 2 3  = 2 2 2 cos 2 + cos 2 = 1 2 cos ( + ). cos ( – ) = 1

cos2  + cos2  =

...(i)

1 4 1 2 sin . sin  = 2 1 cos ( – ) – cos ( + ) = ...(ii) 2 From (i) and (ii), cos ( + ). cos ( – ) = cos ( – ) – cos ( + ) and

sin  sin  =

1 cos ( + ) = and cos ( – ) = 1 2  += . 3

  11.

A 10 h

B 20 h

C 30 h

3 unit/h 6 unit/h



195 720 – 195 525    2 2 2 525 – 195 330   165 . Difference = 2 2 13. n = ab ab = a2 + b2  10a + b = a2 + b2  a(10 – a) + b(1 – b) (1 + b) = 0  a(10 – a) = (b – 1) (b) (b + 1) If b = 2; a(10 – a) = 6  no value of ‘a’ b = 3; a(10 – a) = 24  a  (4, 6) Numbers are 43 and 63 b = 4; a(10 – a) = 60, no value of a b = 5; a(10 – a) = 120, no need to check further  Numbers are 43 and 63.

 1 1  = 2   22  –  2  3   0.723. 2 3 2   8. (x – 1) (x – 2) (x3 – 3x + 1) = 0  Required sum of roots = 3. 9.



195  1 12.  = 90  15      2 2

14. Distance Covered in one revolution = 2r n × d = 11 m

22 70  = 11 7 100 11  7  100 5 n= 22  70 Number of revolution = 5. n

15. a – b =

10012 – 1000 2 10012 – 2001 2003 = 1 for all K  0 it follows that, a – b = ...  1001 – 500.25  500.75 a – b  501. .... 

Since, Hence, 16. As,  =

2 unit/h

Let, Tank capacity = 60 unit Exactly one pair of taps is open during each hour and every pair of tap is open at least for one hour

12 22 – 12 32 – 22 42 – 32     ... 1 3 5 7

Velocity of light in vacuum c  Velocity of light in medium v v

 In medium,

v = v = 2 × 1014 × 5000 × 10–10 v = 108 ms–1



=

vvacuum 3  108  3. vmedium 108

117

17. We have,  

23. When a resistance is joined in parallel with voltmeter, the total resistance of the circuit decreases. The current will increase, x will increase. The equivalent resistance of voltmeter decreases and hence, the voltmeter reading y decreases.

Distance 50   0.1s Velocity 500 1 2 1 2 h = 0  gt   10  (0.1) 2 2 = 0.05 m = 5 cm h = 5 cm above target. t =

18. Let F be the upthrust of the air. As, the balloon is descending down with an acceleration a;  ma – F = ma ...(i) Let mass m0 be removed from the balloon so that it starts moving up with an acceleration a. Then, F – (m – m0)g = (m – m0)a  F – mg + m0g = ma – m0a ...(ii) Adding equations (i) and (ii), we get m0g = 2ma – m0a F F  m0g + m0a = 2ma a m0(g + a) = 2ma a

2ma . mg (m – m )g ga 19. When a sphere falling through a viscous medium it accelerates initially due to gravity. As a result of this its velocity increases, hence, viscous force plus buoyant, force becomes equal to force of gravity, the net force becomes zero and hence does the acceleration. Then the sphere descends with a constant velocity. This constant velocity is called terminal velocity. 

m0 =

2



2

T2  I2   3I    1  9 =   T1  I1   I1 

22. The resultant force will be along y-directions, when the resultant of components of various forces along x-direction is zero. If F is the force required along OX, then, F + 1 cos 60° + 2 cos 60° = 4 cos 60° F = cos 60°(4 – 1 – 2) =

1  0.5 N. 2

d  2h   g = g  1 –   g  1 –  R R



2h d = R R d 100  h= = 50 km. 2 2

or or 26. As,  or



P=

4S 1 or P  R R

P1 R = 2 P2 R1 R 3P2 R 1 = 2 or 1 = R2 P2 R1 3 4 R13   3 3 R V1 1  1 3  1 =    . = 4  3 R2  V2 3  27 R 2 3

27. Given, the triple point of water on scale A = 200 A Triple point of water on scale B = 350 B 200 A = 350 B = 273.16 K (Given)

273.16 273.16 K and 1B = K 200 350 If TA and TB represent the triple point of water on scales A and B, then or

1A =

273.16 273.16 TA = TB 200 350

T2 = T1 × 9 = 4°C × 9 = 36°C.

21. Since, ball has two component velocities perpendicular to each other as observed by a man on the train moving in the opposite direction wrt, first train, hence the path of ball will be parabolic path.



25. Let acceleration due to gravity at height h be the same as that at a depth d deep in to the earth, where d = 100 km

0

20. Heat produced in wire of resistance R in time t is, H = i2Rt Heat required to raise the temperature of a wire by T is  Q = msT Also, Q=H  msT = I2Rt or T  l2 

24. During nuclear fusion, two or more light nuclei combine to form a heavier nucleus. As 6 C 14 is radioactive, so, (A) is not possible.

200 TA 4 = or TA = TB . 350 TB 7

or

28. Frequency of oscillation of simple pendulum is

1 g 2 l where l  the length of the pendulum when this pendulum falls freely under gravity, then frequency of oscillation will becomes v=

v =

1 2

g 1  l 2

g–a l

118

1 g–g [ a = g for freely fall] 2 l

= 

2

1 0  0. 2 1

v =

32. 5.2 molal solution means that 5.2 moles of solute (CH 3OH) are present in 1000 g of water. Moles of CH3OH, nCH OH = 5.2

1000  55.56 18 5.2  0.086. nCH OH = 2 5.2  55.56 33. We have, MgCO3  MgO + CO2 84 g 40 g Moles of water, nwater =

29. Voltmeter is used to measure the potential difference across the resistance and it is connected in parallel with the circuit. A high resistance is connected to the galvanometer in series, so that only a small fraction (Ig) of the main circuit current (I) pass through it. If a considerable amount of current is allowed to pass through the voltmeter, then the reading obtained by this voltmeter will not be close to the actual potential difference between the same two points.

84 g 5 84 100   84%.  Purity percentage = 5 20  8g MgO will be form

Rb R

high resistor

34.

voltmeter

Ia

N

N

H

R' I E, r

I. Ip belongs to sp3 orbital

II. Ip is in sp2 orbital, hence not a part of aromatic sextet

N

N

H

H

30. Given, i = 45° According to Snells’ law at air glass surface, we get a sin i = g sin r sin i = 

sin r = =

2 sin r 

1 2 1 2



Incidence ray

Reflected ray i

sin i

O

r

Air (a = 1)  Glass (g = 2) Refracted ray

sin 45

1 sin r = 2

 1 r = sin –1    30  2 From Fig., we see that, r +  + r = 180°  r +  + 30° = 180°  45° +  + 30° = 180°   = 180° – 75° = 105° Thus, the angle between reflected ray and refracted ray is 105°. 31. Let the number of B atoms = n Number of tetrahedral voids = 2n  Number of A atoms = 2 n 

1 2n  3 3

III. IP belongs to sp3 but dispersed to some extent towards o

IV. Although N is sp2 hybridised, IP on N is in p-orbital, hence, forms a part of aromatic sextet Hence, order of basicity is I > III > II > IV. 35. Given electronic configurations i.e., I. 1s2 2p5 II. 3s2 3p5 2 4 III. 2s 2p and IV. 3s2 3p4 belong to flourine, chlorine, oxygen and sulphur respectively. Their electron affinity is in the following order i.e., O < S < F < Cl Hence, correct order is III < IV < I < II. 36. Volume decreases, when pressure is increased at constant temperature. When temperature is decreased at constant pressure, the volume decreases. 37. Alkyl halides are more reactive than aryl halides and hence, only the halogen is the side chain, i.e., Cl is substituted. CH2Cl NaCN DMF

 Formula A : B

2n : n or 2 : 3 3  Hence, formula is A2B3.

I

CH2CN

I

119

CH 3

H3C

H

H3C

CH 3

H3C +

H

62. 35555 + 42222

H3C +

H

But-1-ene Cis - But-2-ene

trans - But-2-ene

OCH 3 O 1 CH 3 2-Ethoxy – 4 – Methoxypentan-3-one –

N2+ Cl NaNO2, HCl 278 K

N2+ BF4 HBF 4 –HCl

1851

 9   64 

–

F

 –BF3

tan  =

64.

tan  =

alkaline

 H3 N – CH 2 – COO –   H 2 N – CH 2 – COO – medium



k =

sin  =

0.693  0.01 69.3

2 5

2.303 100 2.303 log t = = = 230.3 min. 0.01 10 0.01

43. The rate of enzyme activity rises rapidly with temperature and becomes maximum at a definite temperature called the optimum temperature on either side of the optimum temperature, the enzyme activity decreases. 44. Silver ore is oxidised by using oxygen from air 4Ag + 8NaCN + 2H2O + O2 (air)  4Na [Ag (CN)2] + 4NaOH In this Ag(O) is oxidised to Ag (+1) Silver is precipitated from the solution by addition of zinc powder, then 2 Na [Ag(CN)2] + Zn  Na2 [Zn(CN)4] + 2 Ag  Silver gets reduced from Ag(+1) to Ag(O). 45. In Ostwald’s process, the formation of HNO3 occurs as, Pt

4NH3  5O2    4NO  6H2 O  Energy 750.908 C

61.

12 16

22 56  = 16 cm 7 11

The length of one complete turn =

740

 42

 16 (By Fermat Theorem)

AB , X = 90° AF 2 1

A

90 – 

F

E

 X

H

D

K

AX AF

1

= AX; XF =

B

C

5

1 2 1 1    (AX)  (XF) Area of AXF 2 5 5 1 = 2 =  . 1 Area of ABF 1 5  (AB)  (AF)  2 1 2 2 65.

       x – 8 x – 16    x – 8 x – 25   0   x – 1 – 1   x – 4 – 2    x – 9 – 3  x – 16 – 4   x – 25 – 5 x

1

– 2 x1 – 1  x 2 – 4 x 2 – 4  x3 – 6 x3 – 9 4

4

2

1

5

5

2

2

2

3

2

4

Now,

2

5

x1 – 1 – 1 = 0 x2 – 4 – 2 = 0 x3 – 9 – 3 = 0 x 4 – 16 – 4 = 0 x5 – 25 – 5 = 0

2NO + O2  2NO2 3NO2 + H2O  2HNO3 + NO

The base circumference = d 

740

63. There are 6 values of K for which (K + 2)! has 2 more zeroes than K! The values are 148, 149, 173, 174, 198, 199.

41. In alkaline medium, amino acids exist as negatively charged particles,

42. We have,

 

2

= (–1)1831 + 9 + (1)740 × 16 = –9 + 16 = 7 Remainder = 0.

5 4 3 2 CH 3 – CH – C – CH – O C H 2 CH 3

NH 2

  3  4 3

=  27

39. We have,

40.

1851

 

= 33

CH 2

CH 3

–

38.

H

+

+H –H2 O

90

OH

16 2  12 2 = 20 cm Hence, total length = 4 × 20 = 80 cm.

x1 = 2, x2 = 8, x3 = 18, x4 = 32, x5 = 50 x1  x2  x3  x 4  x5  = 55. 2 66. Let the positive direction of x-axis to be from south to north. Then, Velocity of train VT = + 10 ms–1 Velocity of bird, VB = –5 ms–1 Relative velocity of the bird wrt train, = VB – VT = (–5 ms–1) – (10 ms–1) = –15 ms–1 So, the bird appears to move with a speed of 15 ms–1 from north to south. 150  Time taken by bird to cross the train = = 10 s. 15

120 67. Suppose V be the volume of sphere and  its density. Let 0 and m be the densities of oil and mercury respectively. Since the sphere is floating, therefore, weight of sphere = weight of oil displaced + weight of mercury displaced

V V  og  m g 2 2 o  m 0.8  13.6  or = = 7.2 gcm –3. 2 2 68. Let two spheres collide at point P, after time t, when the distance covered by smaller sphere is x1 and the bigger sphere is x2. Gravitational force on each sphere, or

GM  5 M (12R)2

x1

x2 P

R

2R

9R

Acceleration of smaller sphere, F G5M  a1 = M (12R)2 Acceleration of bigger sphere, F GM  5M (12R) 2 Using equation of motion,

a2 =

...(i)

=

h h h – = ( n g – nw ) ( c / n g ) ( c / nw ) c

=

h  3 4  h  9 – 8 h  –    = c  2 3 c  6  6c

1 1 a1 t 2 and x2 = a2 t 2 2 2 x1 a1 5  = x2 a2 or x1 = 5x2 ...(iii) Given, x1 + x2 = 9R ...(iv) From equations (iii) & (iv), we get, x1 = 7.5 R. F m1  m2 Given, m1 = 3 kg and m2 = 2 kg, F = 10 N

a=

10 10  = 2 ms–2 3 2 5  Force on the second block pushed by the first, f1 = m2a = 2 × 2 = 4 N. a=

70. The refractive index (RI) of glass is greater than that of water. Therefore, the speed of light in glass is less than that of water. It is given as, v=

71. For A, if t1/ 2  x, then t75% = 2x. It means the order wrt A is one. From the given graph, the concentration of B decreases linearly with time. So the rate wrt B remains constant. Hence, the order of reaction wrt B is zero.  Rate = k[A]1[B]0  order = 1 + 0 = 1. T f

72. As,



=

Kf

Tb Kb Tf = 0 – (–0.186) = 0.186°C

1.86 0.186 0.186  0.52  0.052. = or Tb = 0.52 Tb 1.86

73. Wolff Kishner reduction is

x1 =

69. We have,

Hence, the time difference for the both beams for h reaching the bottom is . 6c

or ...(ii)

h h – v g vw

t = t1 – t2 =

Vg =

F=



 Time difference for the rays,

c , where, c = 3 × 108 ms–1 n

NH – NH / Ethylene glycol / KOH

2 2  C  O     CH 2

Carbonyl compound Alkane i.e., 1, 3 are the Wolff Kishner reduction. 74. We have, CH 3 COO CH 3 COO

O OCH Ca + Ca



O OCH

75. Let the number of protons is x. So, the number of electrons will be x – 3. 130.4  Number of neutrons = ( x – 3)  100  Mass number = Number of protons + number of neutrons 130.4  56 = x  ( x – 3)  100 On solving, we get, x = 26 Hence, the number of protons or the atomic number is 26. The element having the atomic number 26 is iron (Fe). Thus, the symbol of the iron is

  

2CH3 CHO Acetaldehyde + 2CaCO 3

56 Fe3 26

.

R

R

KVPY STREAM–SA

Practice Test Papers (Solved)

Purple Patches of the Book: The book comprises numerous questions in several Practice Test Papers which will prove extremely useful for aspirants to be familiar with the current exam pattern, the type of questions asked, and their answers.

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All the questions incorporated in Practice Test Papers in the book have been solved by respective subject-experts with due diligence.

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KVPY STREAM–SA

The present book of Practice Test Papers has specially been published for the aspirants of Kishore Vaigyanik Protsahan Yojana (KVPY–Stream-SA) (Class XI) Fellowship Exam. The book is highly recommended for the aspirants to sharpen their problem solving skills, speed and accuracy, and help them prepare well by practising through these papers to face the prestigious exam with Confidence, Successfully.

KVPY Kishore Vaigyanik Protsahan Yojana

STREAM–SA Practice Test Papers (Solved) For Class XI

HSN Code : 49011010

All Questions Solved by Experts with Selected Explanatory Answers

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