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AMITY INSTITUTE FOR COMPETITIVE EXAMINATIONS

PAPER-1: MATHEMATICS, PHYSICS & CHEMISTRYDo not open this Test booklet until you are asked to do so

Read carefully the Instructions on the Back Cover of this Test Booklet.

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1. Immediately fill in the particulars on this page of the Test Booklet with Blue/Black Ball Point Pen. Use ofpencil is strictly prohibited.

2.. The Answer Sheet is kept inside this Test Booklet. When you are directed to open the Test Booklet, take outthe Answer sheet and fill in the particulars carefully.

3. The Test is of 3 hours duration.4. The Test Booklet consists of 90 questions. The maximum marks are 432.5. There are three parts in the question paper. The distribution of marks subjectwise in each part is as under for

each correct response.Part-A - Mathematics (144 marks) - Question No. 1, 5 to 12 and 16 to 30 consist Four (4) Marks each

and Questions No. 2 to 4 and 13 to 15 consist Eight (8) marks each for each correct response.Part-B - Physics (144 marks) - Question No. 31 to 38, 43 to 49 and 52 to 60 consist Four (4) Marks

each and Questions No. 39 to 42 and 50 to 51 consist Eight (8) marks each for each correctresponse.

Part-C - Chemistry (144 marks) - Question No. 64 to 78 and 82 to 90 consist Four (4) Marks each andQuestions No. 61 to 63 and 79 to 81 consist Eight (8) marks each for each correct response.

6. Candidates will be awarded marks as in Instruction No. 5 for correct response of each question.1/4 (one fourth) marks will be deducted for indicated incorrect response of each question. No deductionfrom the total score will be made if no response is indicated for an item in the Answer Sheet.

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Test Booklet Code

CAMITY INSTITUTE FOR COMPETITIVE EXAMINATIONS



ANSWER FOR TEST CODE C

PART-A: MATHEMATICS

1. [1] 2. [1] 3. [1] 4. [4] 5. [4]

6. [2] 7. [3] 8. [4] 9. [4] 10. 4]

11. [3] 12. [4] 13. [3] 14. [3] 15. [2]

16. [2] 17. [2] 18. 1] 19. [3] 20. [1]

21. [1] 22. [2] 23. [3] 24. [3] 25. [3]

26. [4] 27. [2] 28. [3] 29. [4] 30. [4]

PART-B: PHYSICS

31. [4] 32. [4] 33. [1] 34. [2] 35. [4]

36. [3] 37. [2] 38. [2] 39. [2] 40. [1]

41. [2] 42. [1] 43. [2] 44. [3] 45. [4]

46. [3] 47. [4] 48. [3] 49. [3] 50. [1]

51. [4] 52. [1] 53. [2] 54. [1] 55. [2]

56. [4] 57. [1] 58. [3] 59. [2] 60. [4]

PART-C: CHEMISTRY

61. [3] 62. [4] 63. [1] 64. [4] 65. [4]

66. [2] 67. [1] 68. [3] 69. [4] 70. [1]

71. [3] 72. [2] 73. [2] 74. [4] 75. [2]

76. [4] 77. [1] 78. [1] 79. [1] 80. [4]

81. [2] 82. [4] 83. [2] 84. [1] 85. [4]

86. [4] 87. [3] 88. [3] 89. [3] 90. [3]




PART-A: MATHEMATICS

1. The line L given by 15x y

b passes through the point (13, 32). The line K is parallel to L and has

the equation 13

x yc . Then the distance between L and K is

(1)2317 (2)

2315 (3) 17 (4)

1715

Sol: Ans [1]

13 32 15 b b = –20

5 5bx y b

3 3x cy c are 11 lines

5

3b

c 15bc

34

c

Now given equation are

15 20x y

i.e., 4 20x y i.e., 4 20 0x y

4 13 3x y

i.e, 4 3x y 4 3 0x y

Distance between these 11 lines.

20 3 2316 1 17

2. The number of 3 × 3 non-singular matrices, with four entries as 1 and all other entries as 0, is(1) at least 7 (2) less than 4 (3) 5 (4) 6

Sol: Ans [1]3 rows contain 5 zeroes and no row can contain 3 zeroes

the rows will contain 2, 2 and 1 zeroesIt 1st row contains 2 zeroes and one 1,we will be left with a 2 × 2 determinant whose value is ab – cd 0

ab cdIt two of a, b, c, d are zeroes. Then either a = b = 0, c = d = 1 or a = b = 1, c = d = 0i.e., 2 ways. The remaining two elements of the matrix will be 1 and 0 which can be interchangedin 2 ways. Also in Its row, 2 zeroes and one 1 can be arranged in 3 ways.So total no. of such matrices = 3 × 2 × 2 = 12

Test Code-C AIEEE - 2010


Mathematics, Physics & Chemistry (Test Code: C) AIEEE-2010


3. Let f : R R be defined by

2 , 1( )

2 3, 1k x if x

f xx if x

If f has a local minimum at x = –1, then a possible value of k is

(1) –1 (2) 1 (3) 0 (4)12

Sol: Ans [1]

Hence K = –1

Directions : Questions number 4 to 8 are Assertion - Reason type questions. Each of these questionscontains two statements.

Statement - 1 : (Assertion) and

Statement - 2 : (Reason).

Each of these questions also has four alternative choices, only one of which is the correct answer.You have to select the correct choice.

4. Four numbers are chosen at random (without replacement) from the

set {1, 2, 3, ...., 20}.

Statement - 1 : The probability that the chosen numbers when arranged in some order will form

an AP is 185

Statement - 2 : If the four chosen numbers form an AP, then the set of all possible values ofcommon difference is {±1, ±2, ±3, ±4, ±5}.

(1) Statement-1 is false, Statement-2 is true.(2) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.(3) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1(4) Statement-1 is true, Statement-2 is false.


AIEEE-2010 Mathematics, Physics & Chemistry (Test Code: C)


Sol: Ans [4]

The common difference can be of the type

1, 2, 3, 4, 5, 7

Hence statement 2 is false5. Let A be a 2 × 2 matrix with non-zero entries and let A2 = I, where I is 2 × 2 identity matrix. Define

Tr(A) = sum of diagonal elements of A and |A| = determinant of matrix A.Statement - 1 : Tr(A) = 0.Statement - 2 : |A| = 1.(1) Statement-1 is false, Statement-2 is true.(2) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.(3) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1(4) Statement-1 is true, Statement-2 is false.

Sol: Ans [4]

Leta b

Ac d

where a, b, c, d are non-zero real numbers.

2 a b a bA

c d c d

=2

2

a bc ab bdac dc bc d

= 1 00 1

2 21a bc bc d

Also ( ) 0 0c a d a d as 0c

( ) 0 0b a d a d as 0b Now

2 2 2| | ( ) 1A ad bc a bc a bc a bc

0a d Trace A = 0

6. Let f : R R be a continuous function defined by

1( )2x xf x

e e

Statement - 1 : 1( )3

f c , for some cR

Statement - 2 : 10 ( )

2 2f x , for all xR





Sol: Ans [2]

A.M. > G.M.

2 22

x xe e

2 2 2x xe e

1( )

2 2f x

Also both ex and e–x are positive. So f(x) > 0

Since maximum value of f(x) is 1 0.353

2 2 and f(x) is continuous, f(x) will achieve the value

1 0.3333 . So statement 2 is correct explanation for statement - 1

7. Statement-1: The point A(3, 1, 6) is the mirror image of the point B(1, 3, 4) in the plane x – y + z = 5

Statement-2 : The plane x – y + z = 5 bisects the line segment joining A(3, 1, 6) and B(1, 3, 4).(1) Statement-1 is false, Statement-2 is true.(2) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.(3) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1(4) Statement-1 is true, Statement-2 is false.

Sol: Ans [3]

Mid point of AB is 3 1 6 3 6 4, ,

2 2 2

= (2, 2, 5) which lies in x – y + z = 5 and

Direction ratios of line AB are 1, –1, 1, which are D. Rs of normal of plane.

8. Let S1 = 10

10

1

( 1) jj

j j C

, S2 =

1010

1j

jj C

and S3 = 10

2 10

1j

jj C

.

Statement-1: S3 = 55 × 29

Statement-2 : S1 = 90 × 28 and S2 = 10 × 28





Sol: Ans [4]

S1 =10

10

1( 1) j

jj j C

=10

8 82

110 9 10 9 2j

jC

S2 =10 10

10 9 91

1 110 10 2j j

j jj C C

S3 =10 10 10

2 10 9 91 1

1 1 110 10( 1 1)j j j

j j jj C j C j C

= 8 91 110.9. 10j jC C

= 90.28 + 10.29 = 55 × 29

9. A line AB in three-dimensional space makes angles 45° and 120° with the positive x-axis and thepositive y-axis respectively. If AB makes an acute angle with the positive z-axis then equals(1) 75° (2) 30° (3) 45° (4) 60°

Sol: Ans [4]

1cos452

l

1cos1202

m n = cos

2 1 1 1cos 1

2 4 4

1cos2

= 60°

10. For a regular polygon, let r and R be the radii of the inscribed and the circumscribed circles. A falsestatement among the following is

(1) There is a regular polygon with 3

2rR (2) There is a regular polygon with

12

rR

(3) There is a regular polygon with 12

rR (4) There is a regular polygon with

23

rR

Sol: Ans [4]

sinxR n

tanxr n




cosrR n

If n = 3, 4, 6 then 1 1 3, &2 22

rR

23

rR is not possible

11. Let cos 4( )5

and let sin 5( )13

, where 0 < ,4

. Then tan 2 =

(1)207

(2)2516

(3)5633

(4)1912

Sol: Ans [3]

0 ,4

0 ,2

and 4 4

tan( ) tan( )tan 21 tan( ) tan( )

=

3 5564 12

3 5 3314 12

12. Let S be a non-empty subset of R. Consider the following statement :

P : There is a rational number x S such that 0x .

Which of the following statement is the negation of the statement P?

(1) x S and 0x x is not rational.

(2) There is a rational number x S such that 0x

(3) There is no rational number x S such that 0x

(4) Every rational number x S satisfies 0x

Sol: Ans [4]

Every rational number x S satisfies

0.x There is no positive rational number




13. Let p(x) be a function defined on R such that p´(x) = p´(1 – x), for all [0,1], (0) 1x p and

p(1) = 41. Then 1

0

( )p x dx equals

(1) 42 (2) 41 (3) 21 (4) 41

Sol: Ans [3]

Let ( ) ( ) (1 )h x p x p x

´( ) ´( ) (1 ) 0 0,1h x p x p x x

( )h x = constant 0,1

( ) ( ) (1 ) (0) (1) 42h x p x p x p p

1 1

0 0

( ( ) (1 )) 42 42p x p x dx dx

1 1

0 0

( ( ) (1 ) 42p x dx p x dx

1

0

2 ( ( ) 42p x dx

1

0

( ( ) 21p x dx

14. A person is to count 4500 currency notes. Let an denote the number of notes he counts in the nth

minute. If a1 = a2 = ... = a10 = 150 and a10, a11, ... are in an AP with common difference –2, then the

time taken by him to count all notes is

(1) 135 minutes (2) 24 minutes (3) 34 minutes (4) 125 minutes

Sol: Ans [3]

The currency notes conted in first 10 minutes = 1500 remaining 3000 are counted in the AP.

148 + 142 + .......... let it takes n minutes then

3000 = [2 148 ( 1) 2]2n n

3000 = 148n – n2 + n = 149 n – n2

n2 – 149n + 3000 = 0




(n – 125) (n – 24) = 0 n = 125 and n = 24.

Hence. Time taken = 24 + 10 = 34 minutes

15. Let f : R R be a positive increasing function with (3 )lim 1( )x

f xf x

Then (2 )lim( )x

f xf x

(1) 3 (2) 1 (3)23

(4)32

Sol: Ans [2]

Since (3 )lim 1( )x

f xf x

lim (3 ) lim ( ) lim (2 )x x x

f x f x f x

Limit = 1

16. Solution of the differential equation cos x dy = y (sin x – y) dx, 0 < x < 2

is

(1) tan x = (sec x + c) y (2) sec x = (tan x + c) y

(3) y sec x = tan x + c (4) y tan x = sec x + c

Sol: Ans [2]

2(tan ) secdy x y y xdx

2

1 1 (tan ) secdy x xy dx y

Put 1 , (tan ) secdtt t x xy dx

I.F. = tan secx dxe x

sec tant x x c

sec (tan )x y x c

17. The area bounded by the curves y = cos x and y = sin x between the ordinates x = 0 and x = 32

is

(1) 4 2 1 (2) 4 2 2 (3) 4 2 2 (4) 4 2 1

Sol: Ans [2]




The area is

4

0

2 (cos sin )x x dx

+ 5 4

/ 4

(sin cos )x x dx

= 2 2 2 2 2 4 2 2

18. The equation of the tangent to the curve 2

4y xx

, that the parallel to the x-axis, is

(1) y = 3 (2) y = 0 (3) y = 1 (4) y = 2

Sol: Ans [1]

3

81 0dydx x

x = 2

The point is (2, 3). Hence targent is y = 3.

19. Let f : (–1, 1) R be a differentiable function with f(0) = –1. Let f (0) = 1.

Let g(x) = [f(2f(x) + 2)]2. Then g (0) =

(1) –2 (2) 4 (3) –4 (4) 0

Sol: Ans [3]

´( ) 2 (2 ( ) 2) ´(2 ( ) 2) . 2 ´( )g x f f x f f x f x

´(0) 2 (2 (0) 2) ´(2 (0) 2) . 2 ´(0)g f f f f f

= 2 (0) ´(0) 2 ´(0)f f f

= 2 1 1 2 1 4x

20. There are two urns. Urn A has 3 distinct red balls and urn B has 9 distinct blue balls. From each urntwo balls are taken out at random and then transferred to the other. The number of ways in which thiscan be done is(1) 108 (2) 3 (3) 36 (4) 66

Sol: Ans [1]

The no of ways = 2 2

3 9C C

= 108




21. Consider the system of linear equations :

1 2 3

1 2 3

1 2 3

2 32 3 33 5 2 1

x x xx x xx x x

The System has(1) no solution (2) infinite number of solutions(3) exactly 3 solutions (4) a unique solution

Sol: Ans [1]

Since D = 0 and D1 0So no solution.

22. Let ˆâ j k and ˆˆ ˆc i j k . Then the vector b

satisfying 0a b c and . 3a b

is

(1) ˆˆ ˆ 2i j k (2) ˆˆ ˆ 2i j k (3) ˆˆ ˆ2 2i j k (4) ˆˆ ˆ 2i j k

Sol: Ans [2]

c (a b)

a c a (a b) a b a a a b 3a 2b

ˆ ˆ ˆ ˆ â c 3a 2i j k 3j 3kb

2 2

= ˆ ˆ î j 2k

23. For two data sets, each of size 5, the variances are given to be 4 and 5 and the corresponding meansare given to be 2 and 4, respectively. The variance of the combined data set is

(1)132

(2)52

(3)112

(4) 6

Sol: Ans [3]

Let 2ix 2

45

and 2iy 4

55

xi2 + 5 × 4 – 4xi = 20

xi2 + 20 – 40 = 20

xi2 = 40

and yi2 = 105

Variance = 2 2 2 2

1 5 1 5(x 3) ...(x 3) (y 3) ...(y 3)10

= 40 6 10 9 5 105 6 20 9 5

10

= 55 1110 2




24. The circle x2 + y2 = 4x + 8y + 5 intersects the line 3x – 4y = m at two distinct points if(1) 35 < m < 85 (2) –85 < m < –35 (3) –35 < m < 15 (4) 15 < m < 65

Sol: Ans [3]

6 16 m 55

–25 < –10 – m < 25

–15 < –m < 35

–35 < m < 15

Hence Ans. is 3

25. An urn contains nine balls of which three are red, four are blue and two are green. Three balls aredrawn at random without replacement from the urn. The probability that the three balls have differentcolours is

(1)223

(2)13

(3)27

(4)121

Sol: Ans [3]

Probability 1 1 1

3

3 4 2 29 7

C C C

C

26. If two tangents drawn from a point P to the parabola y2 = 4x are at right angles, then the locus of P is(1) 2x – 1 = 0 (2) x = 1 (3) 2x + 1 = 0 (4) x = –1

Sol: Ans [4]

Clearly locus is directrix which is x = –1.

27. If the vectors ˆˆ ˆ 2a i j k , ˆˆ ˆ2 4b i j k

and ˆˆ ˆc i j k are mutually orthogonal, then

( , ) =(1) (3, –2) (2) (–3, 2) (3) (2, –3) (4) (–2, 3)

Sol: Ans [2]

b c 0 2 4 0a c 0 4 2 0

= –3, = 2

28. The number of complex numbers z such that | 1| | 1| | |z z z i equals

(1) (2) 0 (3) 1 (4) 2Sol: Ans [3]

Z = 0 satisfies the relation. So Ans is 3

29. If and are the roots of the equation x2 – x + 1 = 0, then 2009 + 2009 =(1) 2 (2) –2 (3) –1 (4) 1




Sol: Ans [4]

Roots are – and –2

So, (–)2009 + (–2)2009 = –( + 2) = 1

30. Consider the following relations :

R = {(x, y) | x, y are real numbers and x = wy for some rational number w};

S = {( ,m pn q

) | m, n, p and q are integers such that n, q 0 and qm = pn}. Then

(1) R and S both are equivalence relations

(2) R is an equivalence relation but S is not an equivalence relation

(3) neither R nor S is an equivalence relation

(4) S is an equivalence relation but R is not an equivalence relation

Sol: Ans [4]

Since S is equivalence but R is not so answer is 4.

PART-B: PHYSICS

31. A ball is made of a material of density where oil < < water with oil and water representing thedensities of oil and water, respectively. The oil and water are immiscible. If the above ball is inequilibrium in a mixture of this oil and water, which of the following pictures represents its equilibriumposition?

(1)

Water

Oil (2)

Oil

Water (3)

Water

Oil (4)Oil

Water

Sol: Ans [4]

Since, water is denser than ball and oil, ball is denser than oil.

Directions: Questions number 32 – 33 are based on the following paragraph.

A nucleus of mass M + m is at rest and decays into two daughter nuclei of equal mass M/2 each.Speed of light is c.

32. The speed of daughter nuclei is

(1)mc

M

(2)mc

M m

(3)mc

M m

(4)2 mcM




Sol: Ans [4]

2 2122 2

M v mc

2 mv cM

33. The binding energy per nucleon for the parent nucleus is E1 and that for the daughter nuclei is E2.Then

(1) E2 > E1 (2) E1 = 2E2 (3) E2 = 2E1 (4) E1 > E2

Sol: Ans [1]

Radioactive decay happens in such a way that binding energy per nucleon increases.

34. In a series LCR circuit R = 200and the voltage and the frequency of the main supply is 220 V and50 Hz respectively. On taking out the capacitance from the circuit the current lags behind the voltageby 30°. On taking out the inductor from the circuit the current leads the voltage by 30°. The powerdissipated in the LCR circuit is

(1) zero W (2) 242 W (3) 305 W (4) 210 W

Sol: Ans [2]The circuit is in resonance.Power = I2R

= 2220 200

200

= 242 W

35. Let there be a spherically symmetric charge distribution with charge density varying as

5( )4o

rrR

upto r = R, and (r) = 0 for r > R, where r is the distance from the origin. The

electric field at a distance r (r < R) from the origin is given by

(1)4 53 4

o

o

r rR

(2)

53 4

o

o

r rR

(3)

4 53 3

o

o

r rR

(4)

54 3

o

o

r rR

Sol: Ans [4]

From Gauss theorem E.4r2 = 1( )en

o

q

qen = 2

0

5 44

r

or r drR

54 3

o

o

r rER




Directions: Questions number 36 – 38 are based on the following paragraph

An initially parallel cylindrical beam travels in a medium of refractive index (I) = o + 2I, where o

and 2 are positive constants and I is the intensity of the light beam. The intensity of the beam isdecreasing with increasing radius.

36. The speed of light in the medium is

(1) directly proportional to the intensity I (2) maximum on the axis of the beam

(3) minimum on the axis of the beam (4) the same everywhere in the beam

Sol: Ans [3]

2oc Iv

2o

cvI

37. As the beam enters the medium, it will

(1) diverge near the axis and converge near the periphery

(2) travel as a cylindrical beam

(3) diverge

(4) converge

Sol: Ans [2]

38. The initial shape of the wave front of the beam is

(1) convex near the axis and concave near the periphery

(2) planar

(3) convex

(4) concave

Sol: Ans [2]

For cylindrical beam wavefront is planar and perpendicular to beam.

39. A point P moves in counter-clockwise direction on a circular path as shownin the figure. The movement of ‘P’ is such that it sweeps out a lengths = t3 + 5, where s is in metres and t is in seconds. The radius of thepath is 20 m. The acceleration of ‘P’ when t = 2s is nearly

(1) 7.2 m/s2 (2) 14 m/s2

y

x

B

O A

P( , )x y

20 m

(3) 13 m/s2 (4) 12 m/s2




Sol: Ans [2]

2

2 6td sa tdt

2 2 2(3 )/20c

ds ta Rdt

at t = 2, at = 12 , ac = 7.2

2 212 (7.2)a = 14 m/s2

40. Two identical charged spheres are suspended by strings of equal lengths. The strings make an angle of30° with each other. When suspended in a liquid of density 0.8 g cm–3, the angle remains the same. Ifdensity of the material of the sphere is 1.6 g cm–3, the dielectric constant of the liquid is

(1) 2 (2) 1 (3) 4 (4) 3

Sol: Ans [1]Initially Tcos 15° = mg

T sin15° = 2

21

4 o

qr

Finally T cos 15° = mg 1 l

T sin 15° 2

21

4 o

qk r

r

T15°

mg

Fe

Solving we get k = 2.

41. Two conductors have the same resistance at 0°C but their temperature coefficients of resistance are1 and 2. The respective temperature coefficients of their series and parallel combinations are nearly

(1)1 2

1 21 2

,

(2) 1 2 1 2,2 2

(3) 1 2

1 2,2

(4) 1 2

1 2 ,2

Sol: Ans [2]

In series Re = Ro (1 + 1t) + Ro (1 +2t)

= 2Ro 1 2( )1

2t

1 2

2e

In parallel 1 2

1 1 1(1 ) (1 )e o oR R t R t




1

oR [1–1t + 1– 2t]

1

oR [2 –(1 + 2)t ]

1 2( )2 12o

tR

1 2

1 2

( )12 22 1

2

o oe

R RR tt

1 2

2e

42. The potential energy function for the force between two atoms in a diatomic molecule is approximately

given by U(x) = 12 6a b

x x where a and b are constants and x is the distance between the atoms. If the

dissociation energy of the molecule is D = [U(x = ) – Uat equilibrium], D is

(1)2

4b

a(2)

2

6ba

(3)2

2b

a(4)

2

12b

a

Sol: Ans [1]

dyFdx

, F = 0

x = xo = 1/ 62a

b

at x = xo, u = 4ba

x = , u = 0

2 2

04 4b bDa a

Directions: Question number 43 – 44 contain Statement-1 and Statement -2. Of the four choicesgiven after the statements, choose the one that best describes the two statements.

43. Statement-1: When ultraviolet light is incident on a photocell, its stopping potential is Vo and themaximum kinetic energy of the photoelectrons is Kmax. When the ultraviolet light is replaced by X-rays, both Vo and Kmax increase.

Statement-2: Photoelectrons are emitted with speeds ranging from zero to a maximum value becauseof the range of frequencies present in the incident light.




(1) Statement-1 is false, Statement-2 is true

(2) Statement-1 is true, Statement-2 is false

(3) Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation of Statement-1

(4) Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation of Statement-1.

Sol: Ans [2]

Since eVo = Kmax and X-ray > v

Since photoelectrons have different energies due to losses in metal during interaction with other atoms.

44. Statements-1: Two particles moving in the same direction do not lose all their energy in a completelyinelastic collision.

Statement-2: Principle of conservation of momentum holds true for all kinds of collisions.

(1) Statement-1 is false, Statement-2 is true

(2) Statement-1 is true, Statement-2 is false

(3) Statement-1 is true, Statement-2 is true, Statement-2 is the correct explanation of Statement-1

(4) Statement-1 is true, Statement-2 is true, Statement-2 is not the correct explanation of Statement-1.

Sol: Ans [3]

45. A radioactive nucleus (initial mass number A and atomic number Z) emits 3-particles and 2 positrons.The ratio of number of neutrons to that of protons in the final nucleus will be

(1)12

4A Z

Z

(2)4

2A Z

Z

(3)8

4A Z

Z

(4)4

8A Z

Z

Sol: Ans [4]

+3 2eZ Z 6 Z 6 2A A 12 A 12X Y Z

No. of protons = Z – 8

No. of Neutrons = (A–12) – (Z – 8)

= A – Z – 4.

46. If a source of power 4 kW produces 1020 photons/second, the radiation belongs to a part of the spectrumcalled

(1) microwaves (2) -rays (3) X-rays (4) ultraviolet rays

Sol: Ans [3]

34 83 20 6.6 10 3 104 10 10

= 49.5Å

X-rays.




47. The figure shows the position-time (x – t) graph of one-dimensional motion of a body of mass 0.4 kg.The magnitude of each impulse is

2

2 40 6 8 10 12 14 16

x(m)

t(s)(1) 1.6 Ns (2) 0.2 Ns (3) 0.4 Ns (4) 0.8 Ns

Sol: Ans [4]Impulse = mV= 0.4 ( – 1 – 1)= – 0.8 Ns

48. The combination of gates shown below yields

(1) XOR gate (2) NAND gate

(3) OR gate (4) NOT gate

A

B

X

Sol: Ans [3]

49. Two long parallel wires are at a distance 2d apart. They carry steady equal currents flowing out of theplane of the paper as shown. The variation of the magnetic field B along the line XX is given by

(1) X

B

d d

(2) X

B

d d

(3) X

B

d d

(4) X

B

d d

Sol: Ans [3]

50. Let C be the capacitance of a capacitor discharging through a resistor R. Suppose t1 is the time takenfor the energy stored in the capacitor to reduce to half its initial value and t2 is the time taken for thecharge to reduce to one-fourth its initial value. Then the ratio t1/t2 will be

(1)14

(2) 2 (3) 1 (4)12




Sol: Ans [1]

/toq q e

2 /

4to

oq q e ...(i)

since 2

2qu

c ;

2t

ou u e

12

2

to

ou

u e ...(ii)

Solving (i) and (ii)

1

2

14

tt

51. A rectangular loop has a sliding connector PQ of length l and resistance R and it is moving with aspeed v as shown. The set-up is placed in a uniform magnetic field going into the plane of the paper.The three currents I1, I2 and I are

(1) I1 = I2 = I = B v

Rl

(2) I1 = I2 B v6 R

l, I =

B v3R

lP

v

I2I

QI1

l

(3) I1 = – I2 = B v

Rl

, I = 2 B v

Rl

.

(4) I1 = I2 = B v3R

l, I =

2 B v3R

l

Sol: Ans [4]

2. 3

Bvl BvlIR R RR

R R

I1 = I2 = I/2

52. For a particle in uniform circular motion, the acceleration a at a point P (R, ) on the circle of radiusR is (Here is measured from the x-axis)

(1)2v

R cos i –

2vR

sin j (2)2 2v vˆ ˆ

R Ri j

(3)2 2v vˆ ˆcos sin

R Ri j (4)

2 2v vˆ ˆsin cosR R

i j




Sol: Ans [1]

2 2ˆ ˆcos sinv va i j

R R X

Y

R

a v R= /2

P

53. Two fixed frictionless inclined planes making an angle 30° and 60° with the vertical are shown in thefigure. Two blocks A and B are placed on the two planes. What is the relative vertical acceleration ofA with respect to B?(1) zero(2) 4.9 ms–2 in vertical direction

30°60°

A

B(3) 4.9 ms–2 in horizontal direction(4) 9.8 ms–2 in vertical direction

Sol: Ans [2]

. . .A B A g B ga a a

in vertical direction

(g sin 60°). cos30° – (gsin 30°) cos60°= g sin (60° – 30°)= g/2

54. A small particle of mass m is projected at an angle with the x-axis with an initial velocity v0 in the

x – y plane as shown in the figure. At a time t < 0v sin ,g

the angular momentum of the particle is

(1)12

mgv0t2 cos k

(2)12

mgv0t2 cos i

(3) – mg v0t2 cos j O x

yv0

(4) mgv0t cos k

where ˆ ˆ,i j and k are unit vectors along x, y and z-axis respectively..

Sol: Ans [1]2ˆ ˆcos ( sin ½ )o or V t i V t gt j

ˆ ˆcos ( sin )o ov V i V gt j

L r mv

21 ˆcos2 oL mgV t k




55. The equation of a wave on a string of linear mass density 0.04 kg m–1 is given by y = 0.02(m)

sin 20.04( ) 0.50( )

t xs m

. The tension in the string is

(1) 0.5 N (2) 6.25 N (3) 4.0 N (4) 12.5 N

Sol: Ans [2]

since sin 2 t Xy AT

T = 0.04 = 0.50 m

12.5m/svT

T = v2

2250.042

= 6.25 N

56. A diatomic ideal gas is used in a Carnot engine as the working substance. If during the adiabatic expansionpart of the cycle the volume of the gas increases from V to 32 V, the efficiency of the engine is(1) 0.99 (2) 0.25 (3) 0.5 (4) 0.75

Sol: Ans [4]

For adiabatic extension 1 11 1 2 2TV T V

215

for diatomic gas

2/ 51

2

32T VT V

2

1

14

TT

2

11 0.75T

T

57. A thin semi-circular ring of radius r has a positive charge q distributed uniformly over it. The net field

E

at the centre O is

(1) 2 20

ˆ2

q jr

(2) 2 2

0

ˆ2

q jr

O i

j

(3) 2 20

ˆ4

q jr (4) 2 2

0

ˆ4

q jr




Sol: Ans [1]

At centre 2KE

R

( / )2 2o o

q RR R

2 22 o

qR

58. The respective number of significant figures for the numbers 23.023, 0.0003 and 2.1 × 10–3 are(1) 5, 5, 2 (2) 4, 4, 2 (3) 5, 1, 2 (4) 5, 1, 5

Sol: Ans [3]

59. A particle is moving with velocity ˆ ˆv ( ),K yi xj where K is a constant. The general equation for its

path is(1) xy = constant (2) y2 = x2 + constant (3) y = x2 + constant (4) y2 = x + constant

Sol: Ans [2]

xdxv kydt

; ydyv kxdt

dy kxdt

;dx kydt

dy xdx y

y dy = x dx by integrating y2 = x2 + const.

60. In the circuit shown below, the key K is closed at t = 0. The current through the battery is

(1)2

VR at t = 0 and 1 2

2 21 2

VR RR + R

at t =

KV

L R1

R2(2)

1 2

1 2

V(R R )R R

at t = 0 and 2

VR

at t =

(3)1 2

2 21 2

VR RR + R at t = 0 and

2

VR at t =

(4)2

VR at t = 0 and 1 2

1 2

V(R R )R R

at t =

Sol: Ans [4]

At t = 0, inductor behaves as open circuit. In steady state it does not offer any opposition to flow ofdirect current.




PART-C: CHEMISTRY

61. A solution containing 2.675g of CoCl3.6NH3 (molar mass = 267.5 g mol–1) is passed through a cationexchanger. The chloride ions obtained in solution were treated with excess of AgNO3 to give 4.78g ofAgCl (molar mass = 143.5 g mol–1). The formula of the complex is (Atomic mass of Ag = 108 u)(1) [CoCl3(NH3)3] (2) [CoCl(NH3|)5]Cl2 (3) [Co(NH3)6]Cl3 (4) [CoCl2(NH3)4]Cl

Sol: Ans [3]

No. of moles of CoCl3.6NH3 = 2.675 0.01267.5

No. of moles AgCl = 4.78 0.03

143.5

No. of Cl– combine Ag+ = 0.03 = No. of Cl– produced by complex. [Co(NH3)6]Cl3 [Co(NH3)6] + 3Cl–

moles 0.01 produced 0.01 0.03

62. The standard enthalpy of formation of NH3 is –46.0 kJ mol–1. If the enthalpy of formation of H2 fromits atoms is –436 kJ mol–1 and that of N2 is –712 kJ mol–1, the average bond enthalpy of N – H bondin NH3 is(1) 1056 kJ mol–1 (2) –1102 kJ mol–1 (3) –964 kJ mol–1 (4) +352 kJ mol–1

Sol: Ans [4]

2 2 3N 3H 3NH 0H = 2 × –46.0 kJ mole–1

26H 3H 0H = 3 × –436 kJ mole–1

22N N 0H = –712 kJ mole–1

–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

36H 2N 2NH 0H = –2212 kJ mole–1

Averave bond energy N – H = 2112

6 = +352 kJ mol–1

63. The time for half life period of a certain reaction A Products is 1 hour. When the initial concentrationof the reactant ‘A’, is 2.0 mol L–, how much time does it take for its concentration to come from 0.50to 0.25 mol L–1 if it is zero order reaction?(1) 0.25 h (2) 1 h (3) 4 h (4) 0.5 H

Sol: Ans [1]

Rate = k[A]0 = k 0

1/ 2

[A ]k2t

0t[A ] [A ] [2.0]k k

t 2 1

= 1.0 mol lit–1 hr–1

0.5 0.25k 1.0t

, t = 0.25 h




64. If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution,the change in freezing point of water (Tf), when 0.01 mol of sodium sulphate is dissolved in1 kg ofwater, is (Kf = 1.86 K kg mol–1).(1) 0.0744 K (2) 0.0186 K (3) 0.0372 K (4) 0.0558 K

Sol: Ans [4]

22 4 4Na SO 2Na SO i = 3

Af f

A

n 0.01T ik 3 1.86w (kg) 1

Tf = 0.0558 k

65. From amongst the following alcohols the one that would react fastest with conc. HCl and anhydrousZnCl2 is(1) 2-Methylpropanol (2) 1-Butanol(3) 2-Butanol (4) 2-Methylpropan-2-ol

Sol: Ans [4]Lucas Test for alcoholsTertiary alcohols are most reactive

C OH

CH3

CH3

CH3 2-methylpropan-2-ol

66. If 10–4 dm3 of water is introduced into a 1.0 dm3 flask at 300 K, how many moles of water are in thevapour phase when equilibrium is established? (Given: Vapour pressure of H2O at 300 K is 3170 Pa;R = 8.314 JK–1 mol–1)(1) 4.46 × 10–2 mol (2) 1.27 × 10–3 mol (3) 5.56 × 10–3 mol (4) 1.53 × 10–2 mol

Sol: Ans [2]

Volume occupied by water vapours = 1.0 – 10–4 V = 1.0 dm3

PV = nRT 35

PV 3170 1.0n 1.27 10RT 0.0821 10 300

mol

67. In the chemical reactions,

NH2

2 4NaNO HBFHCl,278K A B

the compounds ‘A’ and ‘B’ respectively are

(1) benzene diazonium chloride and fluorobenzene

(2) nitrobenzene and chlorobenzene

(3) nitrobenzene and fluorobenzene

(4) phenol and benzene




Sol: Ans [1]

NH2

2NaNOHCl, 278K

N2Cl4HBF

F

2 3N BF

Benzenediazonium chloride

68. Consider the following bromides:

Me BrMe

Br

MeMe

BrA B C

The correct order of SN1 reactivity is(1) C B A (2) A B C (3) B C A (4) B A C

Sol: Ans [3]

Reactivity of SN1 based on stability of carbocation intermediate under identical condition.

MeMe

Me Me

CH2+

(B) > (C) > (A)

69. Which one of the following has an optical isomer?(1) [Co(H2O)4(en)]3+ (2) [Zn(en)2]2+ (3) [Zn(en)(NH3)2]2+ (4) [Co(en)3]3+

(en = ethylenediamine)

Sol: Ans [4]

[Co(en)3]3+ its mirror image is Non-super imposable.

70. The Gibbs energy for the decomposition of Al2O3 at 500°C is as follows:

12 3 2 r

2 4Al O Al O , G 966kJmol3 2

The potential difference needed for electrolytic reduction of Al2O3 at 500°C is at least(1) 2.5V (2) 5.0V (3) 4.5V (4) 3.0V

Sol: Ans [1]

G = –nF Ecell value for ‘n’ = No. of electron gain

2 3 22 4Al O Al O3 3

4 e- lose

4 e- gain

G = + 966 kJ mol–1 966 × 103 = 4 × 96500 × Ecell

Ecell = 2.5V




71. On mixing, heptane and octane form an ideal solution. At 373 K, the vapour pressures of the twoliquid components (heptane and octane) are 105 kPa and 45 kPa respectively. Vapour pressure of thesolution obtained by mixing 25.0g of heptane and 35g of octane will be (molar mass of heptane= 100g mol–1 and of octane = 114 g mol–1)(1) 96.2 kPa (2) 144.5 kPa (3) 72.0 kPa (4) 36.1 kPa

Sol: Ans [3]0 0hep OctP 105kPa, P 45kPa

hep

25100x 0.44825 35

100 114

, xoct = 0.552

0 0T hep hep oct octP P x P x (105 0.448) (45 0.552)

PT = 72.0kPa

72. The edge length of a face centered cubic cell of an ionic substance is 508 pm. If the radius of thecation is 110 pm, the radius of the anion is(1) 618 pm (2) 144pm (3) 288 pm (4) 398 pm

Sol: Ans [2]

Distance between cation and anion = 508 254

2

r+ + r– = 254 r– = 254 – 110 = 144 pm

73. The correct order of increasing basicity of the given conjugate bases (R = CH3) is

(1) 2RCOO NH HC C R (2) 2RCOO HC C NH R

(3) 2RCOO HC C R NH (4) 2R HC C RCOO NH

Sol: Ans [2]

Based on strength of their conjugate acids.

74. For a particular reversible reaction at temperature T, H and S were found to be both +ve. If Te isthe temperature at equilibrium, the reaction would be spontaneous when(1) Te is 5 times T (2) T = Te (3) Te T (4) T Te

Sol: Ans [4]G = H – TS for G to be negative, T > Te

75. Consider the reaction:

2 2Cl (aq) H S(aq) S(s) 2H (aq) 2Cl (aq)

The rate equation for this reaction is

rate = k[Cl2][H2S]

Which of these mechanisms is/are consistent with this rate equation?




A. 2 2Cl H S H Cl Cl HS (slow)

Cl HS H Cl S(fast)

B. 2H S H HS (fast equilibrium)

2Cl HS 2Cl H S(slow) (1) Neither A nor B (2) A only (3) B only (4) Both A and B

Sol: Ans [2]

Slow step is the rate determining step.

76. Percentages of free space in cubic close packed structure and in body centered packed structure arerespectively(1) 32% and 48% (2) 48% and 26% (3) 30% and 26% (4) 26% and 32%

Sol: Ans [4]P.F. for CCP is 74%hence empty space is 26%P.F for BCC is 68%hence empty space is = 32%

77. Out of the following, the alkene that exhibits optical isomerism is(1) 3-methyl-1-pentene (2) 2-methyl-2-pentene(3) 3-methyl-2-pentene (4) 4-methyl-1-pentene

Sol: Ans [1]

C CH

CH3

H

CH2

CH3

CH2

3-methylpent-1-eneChiral centre

78. One mole of a symmetrical alkene on ozonolysis gives two moles of an aldehyde having a molecularmass of 44 u. The alkene is(1) 2-butene (2) ethene (3) propene (4) 1-butene

Sol: Ans [1]3 2O / H O/ H

3 3 3CH CH CH CH 2CH CHO

CH3CHO has molecular mass = 44u

79. 29.5 mg of an organic compound containing nitrogen was digested according to Kjeldahl’s method andthe evolved ammonia was absorbed in 20 mL of 0.1 M HCl solution. The excess of the acid required15 mL of 0.1 M NaOH solution for complete neutralization. The percentage of nitrogen in the compoundis(1) 23.7 (2) 29.5 (3) 59.0 (4) 47.4




Sol: Ans [1]

Total meq. of HCl = N × V = 20 × 0.1 = 2 meq.

Meq of NaOH required = N × V = 15 × 0.1 = 1.5 meq.

Meq. of HCl required to neutralise released NH3 = 2 – 1.5 = 0.5 meq.

N% = 3

1.4 NV 1.4 0.5 23.7%w 29.5 10

N is equal to molarity because valency factor (n) for both HCl and NaOH is 1

80. Ionisation energy of He+ is 19.6 × 10–18J atom–1. The energy of the first stationary state (n = 1) of Li2+

is(1) –2.2 × 10–15 J atom–1 (2) 8.82 × 10–17 J atom–1

(3) 4.41 × 10–16 J atom–1 (4) –4.41 × 10–17 J atom–1

Sol: Ans [4]2 2 18

1 1 22 2

2 2 1 2

E Z n 19.6 10 4 1E Z n E 9 1

18

29 19.6 104E

4

= –4.41 × 10–17J atom–1

81. The energy required to break one mole of Cl – Cl bonds in Cl2 is 242 kJ mol–1. The longest wavelengthof light capable of breaking a single Cl – Cl bond is (c = 3 × 108 ms–1 and NA = 6.02 × 1023 mol–1)(1) 700 nm (2) 494 nm (3) 594 nm (4) 640 nm

Sol: Ans [2]hcE

3 34 8

23

242 10 6.6 10 3 106.022 10

, = 494 × 10–9 = 494 nm

82. Solubility product of silver bromide is 5.0 × 10–13. The quantity of potassium bromide (molar masstaken as 120 g mol–1) to be added to 1 litre of 0.05 M solution of silver nitrate to start the precipitationof AgBr is(1) 6.2 × 10–5 g (2) 5.0 × 10–8 g (3) 1.2 × 10–10 g (4) 1.2 × 10–9 g

Sol: Ans [4]

Ksp = 5× 10–13 = [Ag+] [Br–]

5 × 10–13 = [0.05] [Br–]

[Br–] = 1 × 10–11 = [KBr]

[KBr] = 11 B KBr

B

W W1 10M V 120 1

WKBr = 1.2 × 10–9g




83. The correct sequence which shows decreasing order of the ionic radii of the elements is

(1) Na+ F– Mg2+ O2– Al3+ (2) O2– F– Na+ Mg2+ Al3+

(3) Al3+ Mg2+ Na+ F– O2– (4) Na+ Mg2+ Al3+ O2– F–

Sol: Ans [2]

Based on effective nuclear charge.

84. In aqueous solution the ionization constants for carbonic acid are

K1 = 4.2 × 10–7 and K2 = 4.8 × 10–11

Select the correct statement for a saturated 0.034 M solution of the carbonic acid.

(1) The concentrations of H+ and 3HCO are approximately equal

(2) The concentration of H+ is double that of 23CO

(3) The concentration of 23CO is 0.034 M.

(4) The concentration of 23CO is greater than that of 3HCO

Sol: Ans [1]

Value of K1 >> K2

3[H ] [HCO ] based on K1

85. At 25°C, the solubility product of Mg(OH)2 is 1.0 × 10–11. At which pH, will Mg2+ ions start precipitatingin the form of Mg(OH)2 from a solution of 0.001 M Mg2+ ions?

(1) 11 (2) 8 (3) 9 (4) 10

Sol: Ans [4]

Ksp = [Mg2+] [OH–]2

1 × 10–11 = [1 × 10–3] [OH– ]2

[OH–1] = 10–4 pOH = 4

pH = 10

86. The polymer containing strong intermolecular forces e.g. hydrogen bonding, is

(1) polystyrene (2) natural rubber (3) teflon (4) nylon 6,6

Sol: Ans [4]

In nylong 6,6 due to n

C

O

NH group. H bond is here.




87. The main product of the following reaction2 4conc.H SO

6 5 2 3 2C H CH CHOHCH(CH ) ?

(1) C C

H5C6

H H

CH CH3

CH3

(2) C CH2

CH2

CH3

CH2

H5C6

(3)

H5C6

H CH

H

CH3

CH3 (4) C C

CH2

H CH3

CH3H5C6

Sol: Ans [3]

CH CH

CH3

CH3CH2

OH

H5C6

2 4H (conc.H SO )

CH CH

CH3

CH3CH2

OH2+

H5C62H O CH

+CH

CH3

CH3CH2H5C6

H- shift (1,2)

CH2 CH

CH3

CH3CH+

H5C6HCH CH

CH3

CH3CHH5C6

Among cis and trans isomers, trans is more stbale hence (3)88. Biuret test is not given by

(1) urea (2) proteins (3) carbohydrates (4) polypeptidesSol: Ans [3]

Because it does not contain nitrogen.

89. The correct order of 20M / M

E values with negative sign for the four successive elements Cr, Mn, Fe

and Co is(1) Fe Mn Cr Co (2) Cr Mn Fe Co(3) Mn Cr Fe Co (4) Cr Fe Mn Co

Sol: Ans [3]According to electrochemical series.

90. Three reactions involving 2 4H PO are given below:

i) 3 4 2 3 2 4H PO H O H O H PO ii) 2

2 4 2 4 3H PO H O HPO H O iii) 2

2 4 3 4H PO OH H PO O In which of the above does 2 4H PO act as an acid?(1) (iii) only (2) (i) only (3) (ii) only (4) (i) and (ii)

Sol: Ans [3]

ii) 22 4 2 4 3H PO H O HPO H O in this step, 2 4H PO is proton donor..

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