SOLUTIONS TO JEE - 2008 PAPERS PAPER Œ II …1 SOLUTIONS TO JEE - 2008 PAPERS PAPER Œ II (CODE...

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1 SOLUTIONS TO JEE - 2008 PAPERS PAPER II (CODE 0) MATHS Section I (Straight Objective Type) This section contains 9 multiple choice questions. Each question has 4 choices (A) , (B) , (C) and (D) out of which ONLY ONE is correct. 1. A particle P starts from the point z 0 = 1 + 2 i , where i = 1 . It moves first horizontally away from origin by 5 units and then vertically away from origin by 3 units to reach a point z 1 . From z 1 the particle moves 2 units in the direction of the vector j i and then it moves through an angle 2 in anticlockwise direction on a circle with centre at origin , to reach a point z 2 . The point z 2 is given by (A) 6 + 7 i (B) 7 + 6 i (C) 7 + 6 i (D) 6 + 7 i Sol. (D) z 0 (1 + 2 i) z 1 (6 + 5 i) z 2 ( 6 + 7 i) 2. Let the function , g : ( , ) 2 , 2 be given by g (u) = 2 tan 1 (e u ) 2 . Then g is : (A) even and is strictly increasing in (0 , ) (B) odd and is strictly decreasing in ( , ) (C) odd and is strictly increasing in ( , ) (D) neither even nor odd , but is strictly increasing in ( , ) Sol. (C) g(u) = 2tan 1 (e u ) 2 g(u) = tan 1 (e u ) + tan 1 (e u ) 2 g(u) = tan 1 (e u ) cot 1 (e u ) g(u) = cot 1 (e u ) tan 1 (e u ) g(u) = g(u) odd function g(u) = u 2 e 1 2 > 0 (strictly increasing)

Transcript of SOLUTIONS TO JEE - 2008 PAPERS PAPER Œ II …1 SOLUTIONS TO JEE - 2008 PAPERS PAPER Œ II (CODE...

Page 1: SOLUTIONS TO JEE - 2008 PAPERS PAPER Œ II …1 SOLUTIONS TO JEE - 2008 PAPERS PAPER Œ II (CODE 0)MATHS Section I (Straight Objective Type) This section contains 9 multiple choice

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SOLUTIONS TO JEE - 2008 PAPERS

PAPER � II (CODE 0)

MATHS

Section I (Straight Objective Type)

This section contains 9 multiple choice questions. Each question has 4 choices (A) , (B) , (C) and (D)out of which ONLY ONE is correct.

1. A particle P starts from the point z0 = 1 + 2

i , where

i = 1 . It moves first horizontally away from

origin by 5 units and then vertically away from origin by 3 units to reach a point z1 . From z

1 the

particle moves 2 units in the direction of the vector j�i� and then it moves through an angle 2

in

anticlockwise direction on a circle with centre at origin , to reach a point z2 . The point z

2 is given by

(A) 6 + 7 i (B) � 7 + 6

i (C) 7 + 6

i (D) � 6 + 7

i

Sol. (D)z

0 (1 + 2

i)

z1 (6 + 5

i)

z2 (� 6 + 7

i)

2. Let the function ,

g : (�

, )

2,

2 be given by

g (u) = 2 tan�1 (eu) �

2

. Then �g� is :

(A) even and is strictly increasing in (0

, )

(B) odd and is strictly decreasing in (�

, )

(C) odd and is strictly increasing in (�

, )

(D) neither even nor odd , but is strictly increasing in (�

, )

Sol. (C) g(u) = 2tan�1(eu) � 2

g(u) = tan�1(eu) + tan�1 (eu) � 2

g(u) = tan�1(eu) � cot�1(eu)g(�u) = cot�1(eu) � tan�1(eu)g(u) = � g(�u)

odd function

g�(u) = u2e1

2

> 0 (strictly increasing)

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3. Consider a branch of the hyperbola ,

x2 � 2

y2 � 2 2 x � 4 2 y � 6 = 0 with vertex at the point A . .

Let B be one of the end points of its latus rectum . If C is the focus of the hyperbola nearest to thepoint A , then the area of the triangle ABC is :

(A) 1 � 32

(B) 23

� 1 (C) 1 + 32

(D) 23

+ 1

Sol. (B)

Hyperbola is

42x

2

22y

2

= 1

a = 2 , e = 23

b = 2

Area = 21

a (e �1) a

b2

23

� 1

4. The area of the region between the curves y = xcos

xsin1 and y = xcos

xsin1 bounded by the lines

x = 0 and x = 4 is :

(A) 12

022 t1t1

t

)(

d t (B)

12

022 t1t1

t4

)(

d t

(C) 12

022 t1t1

t4

)(

d t (D)

12

022 t1t1

t

)(

d t

Sol. (B)

dxxcosxsin1

xcosxsin1

4/

0

= dx

2x

tan1

2x

tan1

2x

tan1

2x

tan14/

0

= dx

2x

tan1

2x

tan12x

tan1

2

=

4/

0 2dx

2x

tan1

2x

tan2 =

12

022

dtt1)t1(

t4 as tan

2x

= t.

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5. Consider three points P = (� sin ( � ) , � cos ) , Q = (cos ( � ) , sin ) and

R = (cos ( � +) , sin ()) , where 0 < , , < 4

. Then :

(A) P lies on the line segment RQ (B) Q lies on the line segment PR(C) R lies on the line segment QP (D) P , Q , R are non-collinear

Sol. (D)P (�sin( � ), � cos) (x

1, y

1)

Q (cos( � ), sin) (x2, y

2)

and R (x2cos + x

1sin, y

2cos + y

1sin)

we see that T =

sincos

sinycosy,

sincos

sinxcosx 1212

and P, Q, T are collinear P , Q , R are non-collinear

6. An experiment has 10 equally likely outcomes . Let A and B be two non-empty events of theexperiment . If A consists of 4 outcomes

, the number of outcomes that B must have so that A and B

are independent is :(A) 2

, 4 or 8 (B) 3 , 6 or 9 (C) 4 or 8 (D) 5 or 10

Sol. (D)

105/p2

10p

104

)BA(P

5p2

is an integer p = 5 or 10.

7. Let two non-collinear unit vectors b�anda� form an acute angle . A point P moves so that at any

time � t �

the position vector

OP (where O is the origin) is given by , tsinb�tcosa� . When P is

farthest from origin O , let M be the length of

OP and u� be the unit vector along

OP . Then :

(A) u� = b�a�

b�a�

and M = 2/1

b�.a�1 (B) u� = b�a�

b�a�

and M = 2/1

b�.a�1

(C) u� = b�a�

b�a�

and M = 2/1

b�.a�21 (D) u� = b�a�

b�a�

and M = 2/1

b�.a�21

Sol. (A)

tsinb�

tcosa�

u

P

O

�R� = cost2sin1 sin 2t = 1

t = 45º

cos1 ba1

and t = 45º so the unit vector along

OP is angle bisector of 2

a� and

2

b� is u� =

b�a�

b�a�

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8. Let I = 1ee

ex2x4

x

d

x , J =

1ee

ex2x4

x

d x

Then , for an arbitrary constant C , the value of

J � I equals :

(A) 21

log

1ee

1eex2x4

x2x4

+ C (B)

21

log

1ee

1eexx2

xx2

+ C

(C) 21

log

1ee

1eexx2

xx2

+ C (D)

21

log

1ee

1eex2x4

x2x4

+ C

Sol. (C)

J � I = 1ee

)1e(ex2x4

x2x

dx = 1zz

)1z(24

2

d

z where z = ex

1ee

1eenl

21

1z1

z

zdz

11

xx

xx

2

2 + C

J � I =

1ee

1eenl

21

xx

xx

+ C 21

log

1ee

1eexx2

xx2

+ C

9. Let g

(x) = log f

(x) , where f

(x) is a twice differentiable positive function on (0

, ) such that

f (x + 1) = x f

(x) . Then for N = 1 , 2 , 3 , ... , g

21

N � g

21

=

(A) � 4

2)1N2(

1...

251

91

1 (B) 4

2)1N2(

1...

251

91

1

(C) � 4

2)1N2(

1...

251

91

1 (D) 4

2)1N2(

1...

251

91

1

Sol. (A)g(x + 1) = log (f(x + 1) = log x + log(f(x))

= log x + g(x) g (x + 1) � g (x) = log x

g�(x + 1) � g�(x) = � 2x

1

g�

21

1 � g�

21

= � 4

g�

2

12 � g�

2

11 = �

94

......

.......

g�

21

N � g�

21

N = � 2)1N2(

4

Summing upto all term g�

2

1N � g�

21

= � 4

2)1N2(

1...

251

91

1

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Section II (Reasoning Type)

This section contains 4 reasoning type questions . Each question has 4 choices (A) , (B) , (C) and (D)out of which ONLY ONE is correct.

10. Suppose four distinct positive numbers a1 , a

2 , a

3 , a

4 are in G.P.

Let b1 = a

1 , b

2 = b

1 + a

2 , b

3 = b

2 + a

2 and b

4 = b

3 + a

4 .

Statement - 1 : The numbers b1 , b

2 , b

3 , b

4 are neither in A.P. nor in G.P.

Statement - 2 : The numbers b

1 , b

2 , b

3 , b

4 are in H.P.

(A) Statement - 1 is True , Statement - 2 is True ; Statement - 2 is a CORRECT explanation for

Statement - 1(B) Statement - 1 is True

, Statement - 2 is True ; Statement - 2 is a NOT CORRECT explanation

for Statement - 1(C) Statement - 1 is True

, Statement - 2 is False

(D) Statement - 1 is False , Statement - 2 is True

Sol. (C)b

1 = a

1 , b

2 = a

1 + a

2 , b

3 = a

1 + a

2 + a

3 , b

4 = a

1 + a

2 + a

3 + a

4

Hence b1 , b

2 , b

3 , b

4 are neither in A.P. nor in G.P. nor in H.P.

11. Let a , b

, c

, p

, q be real numbers . Suppose

, are the roots of the equation ,

x2 + 2 p x + q = 0 and ,

1 are the roots of the equation

a

x2 + 2

b

x + c = 0

, where 2 {� 1

, 0

, 1}

Statement - 1 : (p2 � q) (b2 � a c) 0

Statement - 2 : b p a or c q

a

(A) Statement - 1 is True , Statement - 2 is True ; Statement - 2 is a CORRECT explanation for

Statement - 1(B) Statement - 1 is True

, Statement - 2 is True ; Statement - 2 is a NOT CORRECT explanation

for Statement - 1(C) Statement - 1 is True

, Statement - 2 is False

(D) Statement - 1 is False , Statement - 2 is True

Sol. (A)Suppose roots are imaginary then =

and

1 = =

1 not possible

roots are real (p2 � q) (b2 � a c) 0

Statement -

1 is correct

ab2

= u and

u =

ac

, + = � 2 p , u

= q

If = 1 , then = q c = q a (not possible)

Also u + 1 = a

b2 � 2

p b = a

p (not possible)

12. Consider L1 : 2

x + 3

y + p � 3 = 0

L2 : 2

x + 3

y + p + 3 = 0

where �p� is a real number and C : x2 + y2 + 6 x � 10

y + 30 = 0

Statement - 1 : If line L1 is a chord of circle C

, then L

2 is not always a diameter of circle C

Statement - 2 : If line L1 is a diameter of circle C

, then line L

2 is not a chord of circle C

(A) Statement - 1 is True , Statement - 2 is True ; Statement - 2 is a CORRECT explanation for

Statement - 1(B) Statement - 1 is True

, Statement - 2 is True ; Statement - 2 is a NOT CORRECT explanation

for Statement - 1(C) Statement - 1 is True

, Statement - 2 is False

(D) Statement - 1 is False , Statement - 2 is True

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Sol. (C)Circle = (x + 3)2 + (y � 5)2 = 4

Distance between L1 and L

2

13

6 < radius

Statement -

2 is false , but Statement

-

1 is correct

13. Let a solution y = y (x) of the differential equation

,

x 1x2

d y � y

1y2

d x = 0

satisfy y (2) =

3

2

Statement - 1 : y (x) = sec

6xsec 1

Statement - 2 : y (x) is given by ,

y1

= x

32 �

2x

11

(A) Statement - 1 is True , Statement - 2 is True ; Statement - 2 is a CORRECT explanation for

Statement - 1(B) Statement - 1 is True

, Statement - 2 is True ; Statement - 2 is a NOT CORRECT explanation

for Statement - 1(C) Statement - 1 is True

, Statement - 2 is False

(D) Statement - 1 is False , Statement - 2 is True

Sol. (C)

1xx

xd2

= 1yy

yd2

sec�1 x = sec�1 y + C

sec�1 2 = sec�1

3

2 + C

C = 6

y = sec

6xsec 1�

cos�1 y1

= cos�1

x1

� cos�1

23

y2

= x3

� 2x

11

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Section III (Linked Comprehension Type)

This section contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions haveto be answered. Each question has 4 choices (A) , (B) , (C) and (D) , out of whoch ONLY ONE iscorrect.

Paragraph - I (Questions numbers 14 to 16)

Consider the function , f : (�

, ) (�

, ) defined by

f (x) =

1xax

1xax2

2

, 0 < a < 2 .

14. Which of the following is true ?(A) (2 + a)2

f (1) + (2 � a)2

f (�

1) = 0 (B) (2 � a)2

f (1) � (2 + a)2

f (�

1) = 0

(C) f (1) f

(�

1) = (2 � a)2 (D) f

(1) f

(�

1) = � (2 + a)2

Sol. (A)

F�(x) =

2

2 2

2a(x 1)

(x ax 1)

F�� (x) =

3

2 3

�4a(x 3x a)

(x ax 1)

F�� (1) = 3 2

�4a( 2 a) 4a

(2 a) (a 2)

F��� (�1) = 2

4a

(2 a)

(2 + a)2

f (1) + (2 � a)2

f (�

1) = 0

15. Which of the following is true ?

(A) f (x) is decreasing on (� 1

, 1)

and has a local minimum at x = 1

(B) f (x) is increasing on (� 1

, 1)

and has a local maximum at x = 1

(C) f (x) is increasing on (� 1

, 1)

but has neither a local maximum nor a local minimum at x = 1

(D) f (x) is decreasing on (� 1

, 1)

but has neither a local maximum nor a local minimum at x = 1

Sol. (A)

2

2 2

2a(x 1)0

(x ax 1)

2(x 1) 0 option (A)

f (x) is decreasing on (� 1

, 1)

and has a local minimum at x = 1

16. Let g (x) =

xe

02t1

)t(f

d

t . Which of the following is true ?

(A) g (x) is positive on (� , 0)

and negative on (0 , )

(B) g (x) is negative on (� , 0)

and positive on (0 , )

(C) g (x) changes sign on both (� , 0)

and (0 , )

(D) g (x) does not change sign on (� , )

Sol. (B)

g� (x) =

xx

2x

f '(e )e

1 e

g� (x) =

2xx

2x x 2 2x

2a(e 1)e

(e ae 1) (1 e )

e2x � 1 > 0 g (x) is negative on (� , 0)

and positive on (0 , )

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Paragraph - II (Questions numbers 17 to 19)

Consider the lines , L1 :

31x

= 1

2y =

21z

and L2 :

12x

= 2

2y = 3

3z

17. The unit vector perpendicular to both L1 and L

2 is :

(A) 99

k�7j�7i� (B)

35

k�5j�7i� (C)

35

k�5j�7i� (D)

99

k�j�7i�7

Sol. (B)

321

213

kji

= k�5j�7i�

Hence unit vector will be 35

k�5j�7i�

18. The shortest distance between L1 and L

2 is :

(A) 0 (B) 3

17(C)

35

41(D)

35

17

Sol. (D)Point (3 , 0 , 4)

Shortest distance = 35

)5(4)1(3

35

17

19. The distance of the point (1 , 1

, 1) from the plane passing through the point (�

1

, �

2

, �

1) and whose

normal is perpendicular to both the lines L1 and L

2 is :

(A) 75

2(B)

75

7(C)

75

13(D)

75

23

Sol. (C)Plane is given by � (x + 1) � 7 (y + 2) + 5 (z + 1) = 0

x + 7 y � 5

x + 10 = 0

Distance = 75

10571 =

75

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Section IV (Matrix Match Type)

This section contains 3 questions . Each question contains statementsgiven in two columns which have to be matched. Statements (A,B,C,D) inColumn I have to be matched with statements (p,q,r,s) in Column II. Theanswers to these questions have to be appropriately bubbled as illustratedin the following example.

p q r s

p q r s

p q r s

p q r s

p q r s

A

B

C

D

If the correct matches are A-p, B-q, B-r, C-p, C-q and D-s, then the correctlybubbled 4 × 4 matrix should be as follows.

[ For every correct answer + 6 marks will be awarded (if all options are marked correctly) and no negative marking for wrong answer ]

Match the statements/expression in Column I with the statements/expressions in column II .

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20. Consider the lines given by : L1 : x + 3

y � 5 = 0

L2 : 3

x � k

y � 1 = 0

L3 : 5

x + 2

y � 12 = 0

Column I Column II(A) L

1 ,

L

2 ,

L

3 are concurrent

, if (p) k = � 9

(B) One of L1 ,

L

2 ,

L

3 is parallel to atleast

one of the other two , if (q) k = � 6/5

(C) L1 ,

L

2 ,

L

3 form a triangle

, if (r) k = 5/6

(D) L1 ,

L

2 ,

L

3 do not form a triangle

, if (s) k = 5

Sol. A - s B - p , q C - r D - p , q , sx + 3

y � 5 = 0 and 5

x + 2

y � 12 = 0 intersect at (2 , 1)

Hence 6 � k � 1 = 0 , k = 5

for L1 , L

2 to be parallel

31

= k

3

k = � 9

For L1 , L

2 to be parallel

53

= 5k

k = 56

for k = 5 , � 9 , 56

they will form triangle

for k = 5 , k = � 9 , 56

they will not form triangle.

21. Column I Column II

(A) The minimum value of 2x

4x2x2

is (p) 0

(B) Let A and B be 3 3 matrices of real numbers, whereA is symmetric

, B is skew-symmetric and

(A + B) (A � B) = (A � B) (A + B) . If (AB)t = (� 1)k AB ,

where (AB)t is the transpose of the matrix AB, then thepossible values of k are (q) 1

(C) Let a = log3 log

3 2 . An integer k satisfying

1 < a3k2

< 2 , must be less than (r) 2(D) If sin = cos , then the possible values of

1

2 are (s) 3

Sol. A - r B - q , s C - r , s D - p , r

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(A) y = 2x

4x2x2

xdyd

= 2

2

)2x(

4x2x)2x2()2x(

at x = 0 the minimum value is 2 r(B) (A + B) (A � B) = (A � B) (A + D)

(AB)t = (�1)k AB q, s

(C) (1) < 2�K , a32 < 2

k1 1 2

3 2 3

r, s

(D) sin = sin 2

= n+ (�1)n 2

In both the cases for n the value of

1

2 are even p,r

22. Consider all possible permutations of the letters of the word ENDEANOEL .

Column I Column II(A) The number of permutations containing the word ENDEA is (p) 5 !(B) The number of permutations in which the letter E occurs in

the first and the last positions is (q) 2 5 !(C) The number of permutations in which none of the letters

D , L , N occurs in the last five positions is (r) 7 5 !(D) The number of permutations in which the letters A , E , O

occur only in odd positions is (s) 21 5 !

Sol. A - p B - s C - q D - qTotal number of words are 9.(A) The number of permutaiotn containing (ENDEA) is 5 ! p

(B) If E occurs in the Ist and last position 2!7!

= 21 . 5 ! s

(C) D, L, N will not occur in last five position 2!4!

× 3!5!

= 2 . 5 ! q

(D) A, E, O occur only in odd position by 3!5!

× 2!4!

q

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JEE - 2008 PAPER PHYSICSPAPER � II (CODE 0)

Section I (Straight Objective Type)

This section contains 9 multiple choice questions. Each question has 4 choices (A) , (B) , (C) and (D)out of which ONLY ONE is correct.

23. Consider a system of three charges q q

,3 3

and 2q3

placed at points

A, B and C, respectively, as shown in the figure. Take O to be thecentre of the circle of radius R and angle CAB = 60º

Figure.

60º

B

C

A

Ox

y

(A) The electric field at point O is 2

0

q

8 R directed along

the negative x-axis(B) The potential energy of the system is zero

(C) The magnitude of the force between the charges at C and B is 2

20

q

54 R

(D) The potential at point O is 2

0

q12 R

Sol. (C)

60º

C

B

A

O

q3

q3

-2q3

3R

E

2E+E

2RR

Resultant field is not along negative x -axis

V = 0

14

q q 1 q 2q 1 q 2q 13 3 R 3 3 2R 3 3 3R

= 2

0

1 1 2q0

4 R 9 9 9 3

FCB

= 0

14 × 2

q 2q 13 3 3R

= 2

20

q

54 R

correct option is (C)

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24. A radioactive sample S1 having activity of 5Ci has twice the number of nuclei as another sample S2which has an activity of 10Ci. The half lives of S1 and S2 can be(A) 20 years and 5 years, respectively (B) 20 years and 10 years, respectively(C) 10 years each (D) 5 years each

Sol. (A)

A1 = 5 =

2/1t693.0

2 N

A2 = 10 =

2/1t693.0 N

105

= 2/1

2/1

tt

2

2/1

2/1

tt =

14

Correct option is (A)

25. A transverse sinusoidal wave moves along a string in the positivex-direction at a speed of 10cm/s. The wavelength of the wavesis 0.5 m and its amplitude is 10cm. At a particle time t, thesnap-shot of the wave is shown in figure. The velocity of point Pwhen its displacement is 5cm isFigure

xP

y

(A) 3 �j m / s

50

(B) 3 �j m / s

50

(C) 3 �i m / s

50

(D) 3 �i m / s

50

Sol. (A)

For wave going in positive x-direction, particle velocity dy dy

cdt dx

Here dy

vedx

, dy

vedt

Velocity is in positive y-direction.Correct answer is (A)

26. A block (B) is attached to two unstretched springs S1 and S2 with spring constants k and 4krespectively (see figure I). The other ends are attached to identical supports M1 and M2 not attachedto the walls. The springs and supports have negligible mass. There is no friction anywhere. Theblock displaced towards wall 1 by a small distance x (figure II) and released. The block returns andmoves a maximum distance y towards wall 2. Displacements x and y are measured with respect to

the equilibrium position of the block B. The ratio yx is

VVVVVVVVVV VVVVVVVVVVBM1S1S2M2

2 1

I

VVVVVVVVVV VVVVVVVVBM1S1S2M2

2 1

II

x

(A) 4 (B) 2 (C) 12

(D) 14

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Sol. (C)

12

kx2 = 214k y

2

x2 = 4y2

x = 2y

y 1x 2

Correct option is (C)

27. A bob of mass M is suspended by a massless string of length L.The horizontal velocity V at position A is just sufficient to make itreach the point B. The angle at which the speed of the bob is halfof that at A, satisfies

VA

L

B

(A) 4

(B) 4 2

(C) 3

2 4 (D)

34

Sol. (D)

1 1 5gLM(5gL)= M MgL(1 cos )

2 2 4

1 15mgL mgL(1 cos )

2 4

1 � cos = 158

cos = � 78

= 151º

Correct option is

28. A glass tube of uniform internal radius (r) has a valve separatingthe two identical ends. Initially, the valve is in a tightly closedposition. 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,

x

12

(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 increases

Sol. (B)R

2 > R

1

P2 < P

1

Air flows from end 1 to end 2 correct option is (B)

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29. A vibrating string of certain length under a tension T resonates with a mode corresponding to thefirst overtone (third harmonic) of an air column of length 75 cm inside a tube closed at one end. Thestring also generates 4 beats per second when excited along with a tuning fork of frequency n. Nowwhen the tension of the string is slightly increased the number of beats reduces to 2 per second .Assuming the velocity of sound in air to be 340 m/s, the frequency n of the tuning fork in Hz is(A) 344 (B) 336 (C) 117.3 (D) 109.3

Sol. (A)

Frequency of air column/string , f = c

34L

= 3 × 340

4 0.75

= 340 Hzfrequency of tuning fork = 336 Hz or 344 HzAs increase in tension in string decreases the beat frequency, the unknown frequency, n = 344 Hz Correct option is (A)

30. A parallel plate capacitor C with plates of unit area and separation d is

filled with a liquid of dielectric constant K = 2. The level of liquid is d3

initially. Suppose the liquid level decreases at a constant speed V, thetime constant as a function of time t is

Rd3

d

C

(A) 06 R

5d 3Vt

(B)

02 2 2

(15d 9Vt) R

2d 3dVt 9V t

(C) 06 R

5d 3Vt

(D)

02 2 2

(15d 9Vt) R

2d 3dVt 9V t

Sol. (A)

d 3

-vt )(

2d 3

+vt)(

1 2

1 1 1C C C =

0 0

2d dvt vt

3 31 1 2

= 0

1 2d d vtvt

3 6 2

= 0

14d 6vt d 3vt

6

= 0

15d 3vt

6

c = 06

5d 3vt

Time constant = RC = 06 R

5d 3vt

correct option is (A)

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31. A light beam is traveling from Region I to Region IV (Referfigure). The refractive index in Regions I, II, III and IV are

n0,

0 0n n,

2 6 and 0n

8, respectively. The angle of incidence

for which the beam just misses entering Region IV is

n0

Region I Region II Region III Region IV

n2

0 n6

0 n8

0

0 0.2m 0.6m

(A) 1 3sin

4

(B) 1 1sin

8

(C) 1 1sin

4

(D) 1 1

sin3

Sol. (B)

n0sin = 0n

8 sin 90º

sin= 18

= sin�1 18

correct option is (B)

Section III (Reasoning Type)This section contains 4 reasoning type questions . Each question has 4 choices (A) , (B) , (C)and (D) out of which ONLY ONE is correct.

32. Statement - 1 : For an observer looking out through the window of a fast moving train, the nearbyobjects appear to move in the opposite direction to the train, while the distant objectsappear to be stationary

Statement - 2 : If the observer and the object are moving at velocities 1V

and 2V

respectively with

reference to a laboratory frame, the velocity of the object with respect to the observer

is 2 1V V

(A) Statement - 1 is True , Statement - 2 is True ; Statement - 2 is a CORRECT explanation for

Statement - 1(B) Statement - 1 is True

, Statement - 2 is True ; Statement - 2 is a NOT CORRECT explanation

for Statement - 1(C) Statement - 1 is True

, Statement - 2 is False

(D) Statement - 1 is False , Statement - 2 is True

Sol. (B)

33. Statement - 1 : It is easier to pull a heavy object than to push it on a level ground.Statement - 2 : The magnitude of frictional force depends on the nature of the two surfaces in contact.(A) Statement - 1 is True

, Statement - 2 is True ; Statement - 2 is a CORRECT explanation for

Statement - 1(B) Statement - 1 is True

, Statement - 2 is True ; Statement - 2 is a NOT CORRECT explanation

for Statement - 1(C) Statement - 1 is True

, Statement - 2 is False

(D) Statement - 1 is False , Statement - 2 is True

Sol. (B)

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34. Statement - 1 : For practical purposes, the earth is used as a reference at zero potential in electricalcircuits.

Statement - 2 : The electrical potential of a sphere of radius R with charge Q uniformly distributed

on the surface is given by 0

Q4 R

(A) Statement - 1 is True , Statement - 2 is True ; Statement - 2 is a CORRECT explanation for

Statement - 1(B) Statement - 1 is True

, Statement - 2 is True ; Statement - 2 is a NOT CORRECT explanation

for Statement - 1(C) Statement - 1 is True

, Statement - 2 is False

(D) Statement - 1 is False , Statement - 2 is True

Sol. (B)

35. Statement - 1 : The sensitivity of a moving coil galvanometer is increased by placing a suitablemagnetic material as a core inside the coil.

Statement - 2 : Soft iron has a high magnetic permeability and cannot be easily magnetized ordemagnetized.

(A) Statement - 1 is True , Statement - 2 is True ; Statement - 2 is a CORRECT explanation for

Statement - 1(B) Statement - 1 is True

, Statement - 2 is True ; Statement - 2 is a NOT CORRECT explanation

for Statement - 1(C) Statement - 1 is True

, Statement - 2 is False

(D) Statement - 1 is False , Statement - 2 is True

Sol. (C)

Paragraph - I (Questions numbers 36 to 38)

The nuclear charge (Ze) is non-uniformly distributed within a nucleus ofradius R. The charge density (r) [charge per unit volume] is dependentonly on the radial distance r from the centre of the nucleus as shown infigure. The electric field is only along the radial direction.Figure.

Rr

d

(r)

a36. The electric field at r = R is

(A) independent of a (B) directly proportional to a(C) directly proportional to a2 (D) inversely proportional to a

Sol. (A)Electric field at r = R,

E = 20

1 Ze4 R

correct option is (A)

37. For a = 0, the value of d (maximum value of as shown in the figure) is

(A) 33Ze

4 R(B) 3

3Ze

R(C) 3

4Ze

3 R(D) 3

Ze

3 R

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Sol. (B)

Rr

d

dr d

R

ze = R 20

d4 r dr r d

R

=

4 3d R R4 d

R 4 3

=

3 3dR dR4

4 3

3dR4

12

d = 3

3Ze

Rcorrect option is (B)

38. The electric field within the nucleus is generally observed to be linearly dependent on r. This implies

(A) a = 0 (B) a = R2

(C) a = R (D) 2R

a3

Sol. (C)Electric field is linearly dependent on r when density is uniform inside nucleus.

R r

d

correct option is (C)

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Paragraph - II (Questions numbers 39 to 41)

A uniform thin cylindrical disk of mass M and radius R isattached to two identical massless springs of springconstant k which are fixed to the wall as shown in thefigure. The springs are attached to the axle of the disksymmetrically on either side at a distance d from its centre.The axle is massless and both the springs and the axleare in a horizontal plane. The unstretched length of eachspring is L. The disk is initially at its equilibrium positionwith its centre of mass (CM) at a distance L from the wall.

The disk rolls without slipping with velocity 0 0�V V i

. The

coefficient of friction is .

39. The net external force acting on the disk when its centre of mass is atdisplacement x with respect to its equilibrium position is

(A) �kx (B) � 2kx (C) � 2kx3

(D) � 3

xk4

Sol. (D)

2kx

a

f

2kx × R = 23 a

MR2 R

a = 4 kx3 M

Net force = Ma = 4

Kx3

leftside

correct option is (D)

40. The centre of mass of the disk undergoes simple harmonic motion with angular frequency equal to

(A) kM

(B) 2kM

(C) 2k3M

(D) 4k3M

Sol. (D)

2M

RT 22K

I

=

MM

222K

= 3M

24K

2K 4K3MM

R

2

I

correct option is (D)

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41. The maximum value of V0 for which the disk will roll without slipping is

(A) M

gk

(B) M

g2k

(C) 3M

gk

(D) 5M

g2k

Sol. (C)

22 20

2

V1 3 1MR Kx 2

2 2 2R ...(1)

2Kx � mg = m × 4 kx3 m

mg = kx × 4

23

= kx × 23

x = 3 mg

2K

.....(2)

From (1) & (2)

V0 = g

3MK

correct option is (C)

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Section IV (Matrix Match Type)

This section contains 3 questions . Each question contains statementsgiven in two columns which have to be matched. Statements (A,B,C,D) inColumn I have to be matched with statements (p,q,r,s) in Column II. Theanswers to these questions have to be appropriately bubbled as illustratedin the following example.

p q r s

p q r s

p q r s

p q r s

p q r s

A

B

C

D

If the correct matches are A-p, B-q, B-r, C-p, C-q and D-s, then the correctlybubbled 4 × 4 matrix should be as follows.

[ For every correct answer + 6 marks will be awarded (if all options are marked correctly) and no negative marking for wrong answer ]

42. Column I gives a list of possible set of parameters measured in some experiments. The variationsof the parameters in the form of graphs are shown in column II. Match the set of parameters given inColumn I with the graphs given in Column II . Indicate your answer by darkening the appropriatebubbles of the 4 × 4 matrix given in the ORS.

Column I Column II(A) Potential energy of a simple pendulum (y axis)

as a function of displacement (x) axis (p)

x

y

O

(B) Displacement (y axis) as a function of time (x axis) fora one dimensional motion at zero or constant acceleration

when the body is moving along the positive x-direction (q)

x

y

O

(C) Range of a projectile (y axis) as a function of its velocity

(x axis) when projected at a fixed angle (r)

x

y

O

(D) The square of the time period (y axis) of a simple

pendulum as a function of its length (x axis) (s)

x

y

O

Sol. A � p B � q, s C � s D � q

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43. An optical component and an object S placed along its optic axis are given in Column I . Thedistance between the object and the component can be varied. The properties of images are given inColumn II. Match all the properties of images from Column II with the appropriate componentsgiven in Column I. Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix

given in the ORS.Column I Column II

(A)S

(p) Real image

(B)S

(q) Virtual image

(C)S

(r) Magnified image

(D)S

(s) Image at infinity

Sol. A - p, q, r, s B - q C - p, q, r, s D - p, q, r, s

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44. Column I contains a list of processes involving expansion of an ideal gas. Match this with columnII describing the thermodynamic change during this process. Indicate your answer by darkening theappropriate bubbles of the 4 × 4 matrix given in the ORS.

Column I Column II(A) An insulated container has two chambers (p) The temperature of the

separated by a valve. Chamber I contains an gas decreasesideal gas and the chamber. II has vacuum .The valve is opened.

vacuum

III

ideal gas

(B) An ideal monatomic gas expands to twice its (q) The temperature of the gas

original volume such that its pressure P 2

1

V , increases or remains

where V is the volume of the gas constant(C) An ideal monoatomic gas expands to twice its (r) The gas loses heat

original volume such that its pressure P 4 / 3

1

V ,

where V is its volume(D) An ideal monoatomic gas expands such that its (s) The gas gains heat

pressure P and volume V follows the behaviorshown in the graph

2V1V1 V

P

Sol. A - q B - p, r C - p, s D - q, s

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JEE - 2008 PAPER CHEMISTRYPAPER � II (CODE 0)

Section I (Straight Objective Type)

This section contains 9 multiple choice questions. Each question has 4 choices (A) , (B) , (C) and (D)out of which ONLY ONE is correct.

45. The correct stability order for the following species is

(I) O

(II) (III) O

(IV)

(A) (II) > (IV) > (I) > (III) (B) (I) > (II) > (III) > (IV)(C) (II) > (I) > (IV) > (III) (D) (I) > (III) > (II) > (IV)

Sol. (D)

(I) H C 3

C

CH3

OCH2

CH3H C 3

C

CH3

OCH2

CH3

Stabilizes by resonance and have six -hydrogen atom (hyperconjugation)

(II) H C�CH�O�HC3

CH3

CH3

H C�CH==O�HC3

CH3

CH3

Stabilizes by resonance and have only three -hydrogen atoms

(III) H C�CH�CH �HC3 2

CH3

CH3

have five a-hydrogen atoms

(IV) H C�CH �CH �HC2 2 2

CH3

CH3

have only two -hydrogen atoms I > III > II > IV

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46. Cellulose upon acetylation with excess acetic anhydride/H2SO4 (catalytic) gives cellulose triacetate whosestructure is:

(A)

AcO

O

HOAc H

O

H OAc

H

H

O

AcO

O

HOAc H

O

H OAc

H

H

AcO

O

HOAc H

H OAc

O

H

H

(B)

AcO

O

HOH H

O

H OH

H

H

O

AcO

O

HOH H

O

H OH

H

H

AcO

O

HOH H

H OH

O

H

H

(C)

AcO

O

HOAc H O

H OAc

H

O

AcO

O

HOAc H

H OAc

O

AcO

O

HOAc H

H OAc

O

H

H HH H

(D)

AcO

O

HH H O

OAc OAc

H

O

AcO

O

HH H

OAc OAc

O

AcO

O

HH H

OAc OAc

O

H

H HH H

Sol. (A)As in cellulose 1-4 glycosidic linkage is present

47. In the following reaction sequence, the correct structure of E, F and G are:

O O

OHPh *

Heat [E]

NaOH

I2 [F] + [G]

(* implies 13C labelled carbon)

(A) E =

O

CH3Ph *F =

O

O NaPh *G = CHI3

(B) E =

O

CH3Ph * F =

O

O NaPhG = CHI3

(C) E =

O

CH3Ph * F =

O

O NaPhG = *CHI3

(D) E =

O

CH3Ph * F =

O

O NaPhG = *CH I3

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Sol. (C)

Ph�C�CH �C�OH2*

OO

Ph�C�CH +CO3 2*

O

(E)

Ph�C�CH3*

O

Ph�C�COONa + CHI3*

O

(F)

I2

NaOH (G)

48. Among the following, the coloured compound is:(A) CuCl (B) K3[Cu(CN)4] (C) CuF2 (D) [Cu(CH3CN)4]BF4

Sol. (C)The crystalline form of CuF2 is blue coloured.

49. Both [Ni(CO)4] and [Ni(CN)4]2� are diamagnetic. The hybridisations of nickel in these complexes, respec-

tively, are(A) sp3, sp3 (B) sp3, dsp2 (C) dsp2, sp3 (D) dsp2, dsp2

Sol. (B)Ni(CO)4 = sp3 [Ni(CN)4]

2� = dsp2

50. The IUPAC name of [Ni(NH3)4] [NiCl4] is(A) Tetrachloronickel (II) - tetraamminenickel (II)(B) Tetraamminenickel (II)-tetrachloronickel (II)(C) Tetraammnenickel (II)-tetrachloronickelate (II)(D) Tetrachloronickel (II)-tetraamminenickelate (0)

Sol. (C)The IUPAC name of [Ni(NH3)4] [NiCl4] is Tetraammnenickel (II)-tetrachloronickelate (II).

51. Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The timerequired to liberate 0.01 mol of H2 gas at the cathode is (1 Faraday = 96500 C mol�1](A) 9.65 × 104 sec (B) 19.3 × 104 sec (C) 28.95 × 104 sec (D) 38.6 × 104 sec

Sol. (B)

No. of eq. = 96500

it = 2 × 0.01 =

96500t1010 3

t = 1.93 × 104

52. Among the following, the surfactant that will form micelles in aqueous solution at the lowest molar con-centration at ambient condition is(A) CH3(CH2)15N

+ (CH3)3 Br � (B) CH3(CH2)11OSO3�Na+

(C) CH3CH2)6COO � Na+ (D) CH3(CH2)11N+(CH3)3Br �

Sol. (B)

53. Solubility product constants (Ksp) of salts of types MX, MX2 and M3X at temperature �T� are 4.0 × 10�8,3.2 × 10�14 and 2.7 × 10�15, respectively. Solubilities (mol dm�3) of the salts at temperature �T� are in the

order(A) MX > MX2 > M3X (B) M3X > MX2 > MX (C) MX2 > M3X > MX (D) MX > M3X > MX2

Sol. (D)

For MX, S = spK = 8104 = 2 × 10�4

MX2 = S = 4

K3 sp

= 4102.3

314

= 2 × 10�5

M3XTS = 27

K4 sp

= 27107.2

415

= 10�4

MX > M3X > MX2

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Section II (Reasoning Type)

This section contains 4 reasoning type questions . Each question has 4 choices (A) , (B) , (C) and (D)out of which ONLY ONE is correct.

54. Statement - 1 : Aniline on reaction with NaNO2/HCl at 0ºC followed by coupling with b-naphthaol gives a

dark blue coloured precipitate.Statement - 2 : The colour of the compound formed in the reaction of aniline with NaNO2/HCl at 0ºC

followed by coupling with b-naphthol is due to the extended conjugation.(A) Statement - 1 is True

, Statement - 2 is True ; Statement - 2 is a CORRECT explanation for

Statement - 1(B) Statement - 1 is True

, Statement - 2 is True ; Statement - 2 is a NOT CORRECT explanation

for Statement - 1(C) Statement - 1 is True

, Statement - 2 is False

(D) Statement - 1 is False , Statement - 2 is True

Sol. (D)

Benzene diazonium chloride (C H N Cl6 5 2 )

55. Statement - 1 : [Fe(H2O)5NO]SO4 is paramagneticStatement - 2 : The Fe in [Fe(H2O)5NO]SO4 has three unpaired electrons.(A) Statement - 1 is True

, Statement - 2 is True ; Statement - 2 is a CORRECT explanation for

Statement - 1(B) Statement - 1 is True

, Statement - 2 is True ; Statement - 2 is a NOT CORRECT explanation

for Statement - 1(C) Statement - 1 is True

, Statement - 2 is False

(D) Statement - 1 is False , Statement - 2 is True

Sol. (A)Magnetic moment of [Fe(H2O)5NO]SO4 is 3.9 BM which confirms the presence of three unpaired electonso it is a paramagnetic.

56. Statement - 1 : The geometrical isomers of the complex [M(NH3)4Cl2] are optically inactive.Statement - 2 : Both geometrical isomers of the complex [M(NH3)4Cl2] possess axis of symmetry.(A) Statement - 1 is True

, Statement - 2 is True ; Statement - 2 is a CORRECT explanation for

Statement - 1(B) Statement - 1 is True

, Statement - 2 is True ; Statement - 2 is a NOT CORRECT explanation

for Statement - 1(C) Statement - 1 is True

, Statement - 2 is False

(D) Statement - 1 is False , Statement - 2 is True

Sol. (A)

M

Cl

Cl

NH3

NH3H3N

NH3

trans

M

NH3

NH3

Cl

ClH3N

NH3

cisThe molecule should not posses alternate axis of symmetry to be optically active.

57. STATEMENT-1 : There is a natural asymmetry between converting work to heat and converting heat towork.

Statement - 2 : No process is possible in which the sole result is the absorption of heat from a reserviorand its complete conversion into work.

(A) Statement - 1 is True , Statement - 2 is True ; Statement - 2 is a CORRECT explanation for

Statement - 1(B) Statement - 1 is True

, Statement - 2 is True ; Statement - 2 is a NOT CORRECT explanation

for Statement - 1(C) Statement - 1 is True

, Statement - 2 is False

(D) Statement - 1 is False , Statement - 2 is True

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27

Sol. (A)Statement 2 is IInd law of thermodynamics which concludes that total heat can never be converted intoequivalent amount of work.

Section III (Linked Comprehension Type)

This section contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions haveto be answered. Each question has 4 choices (A) , (B) , (C) and (D) , out of whoch ONLY ONE iscorrect.

Paragraph for Question Nos. 58 to 60.

A tertiary alcohol H upon acid catalysed dehydration gives a product I. Ozonolysis of I leads to compoundsJ and K Compound J upon reaction with KOH gives benzyl alcohol and a compound L, whereas K onreaction with KOH gives only M.

M = Ph

O

HPh

H C3

58. Compound H is formed by the reaction of

(A)

O

CH3Ph + PhMgBr

(B)

O

CH3Ph + PhCH2MgBr

(C)

O

HPh + PhCH2MgBr

(D)

O

HPh + MgBrPh

Me

Sol. (B)

Ph�C�CH3

O

+ PhCH MgBr2 Ph�C�CH3

OH

H C�Ph2

59. The structure of compound I is

(A) PhH

Ph CH3

(B) PhH

H C3 Ph

(C) CH Ph2H

Ph CH3

(D) HPh

H C3 CH3

Sol. (A)

Ph�C�CH3

OH

H2C�Ph

H /+

� H O2

Ph�C=CH�Ph

CH3

(I)

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28

60. The structures of compound J, K and L, respectively, are(A) PhCOCH3, PhCH2COCH3 and PHCH2COO�K+ (B) PhCHO, PhCH2CHO and PhCOO�K+

(C) PhCOCH3, PhCH2CHO and CH3COO�K+ (D) PhCHO, PhCOCH3 and PhCOO�K+

Sol. (D)

Ph�C=CH

CH3

Ph

O3

H O/Zn2Ph�C=O + Ph�CHO

CH3

(K) (J)

Hence (D) is correct

Paragraph for Question Nos. 61 to 63.

In hexagonal systems of crystals, a frequently encountered arrangement of atoms is described as ahexagonal prism. Here, the top and bottom of the cell are regular hexagons and three atoms are sandwichedin between them. A space-filling model of this structure, called hexagonal close-packed (HCP), is constitutedof a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible.Three spheres are then placed over the first layer so that they touch each other and represent the secondlayer. Each one of these three spheres touches three spheres of the bottom layer. Finally, the secondlayer is covered with a third layer that is identical to the bottom layer in relative position. Assume radius ofevery sphere to be �r�.

61. The number of atoms in this HCP unit cell is(A) 4 (B) 6 (C) 12 (D) 17

Sol. (B)

Total effective number of atoms = 12 × 61

+ 2 × 21

+ 3 = 6

62. The volume of this HCP unit cell is

(A) 24 2 r3 (B) 15 2 r3 (C) 12 2 r3 (D) 33

64r3

Sol. (A)

Height of unit cell = 32

r4

Base area = 6 × 43

(2r)2

Volume = height × base area

= 24 3r2

63. The empty space in this HCP unit cell is(A) 74% (B) 47.6% (C) 32 % (D) 26%

Sol. (D)Packing fraction = 74%Empty space = 26%

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29

Section IV (Matrix Match Type)

This section contains 3 questions . Each question contains statementsgiven in two columns which have to be matched. Statements (A,B,C,D) inColumn I have to be matched with statements (p,q,r,s) in Column II. Theanswers to these questions have to be appropriately bubbled as illustratedin the following example.

p q r s

p q r s

p q r s

p q r s

p q r s

A

B

C

D

If the correct matches are A-p, B-q, B-r, C-p, C-q and D-s, then the correctlybubbled 4 × 4 matrix should be as follows.

[ For every correct answer + 6 marks will be awarded (if all options are marked correctly) and no negative marking for wrong answer ]

64. Match the compounds in Column I with their characteristic test(s)/reaction(s) given in Column II. Indicateyour answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS.

Column I Column II

(A) H N � NH Cl 2 3 (p) sodium fusion extract of the compoundgives Prussian blue colour with FeSO4

(B) HONH3I

COOH(q) gives positive FeCl3 test

(C) HO NH Cl 3(r) gives white precipitate with AgNO3

(D) O N2 NH � NH Br 3

NO2

(s) reacts with aldehydes to form thecorresponding hydrazone derivative

[ A � r , s B � p, q C � p, q, r D � p, s ]

65. Match the conversions in Column I with the type(s) of reaction(s) give in Column II. Indicate youranswer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS

Column I Column II(A) PbS PbO (p) roasting(B) CaCO3 CaO (q) Calcination(C) ZnS Zn (r) carbon reduction(D) Cu2S Cu (s) self reduction

[ A � p B � q C � p, r D � p, s ]

66. Match the entries in Column I with the correctly related quantum number(s) in Column II. Indicate youranswer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS

Column I Column II(A) Orbital angular momentum of the (p) Principal quantum number

electron in a hydrogen-like atomic orbital(B) A hydrogen-like one-electron wave (q) Azimuthal quantum number

function obeying Pauli principle(C) Shape, size and orientation of hydrogen (r) Magnetic quantum number

like atomic orbitals(D) Probability density of electron at the (s) Electron spin quantum number

nucleus in hydrogen-like atom[ A � q B � s C � p, q, r D � p, q, r ]