SOLUTIONS TO JEE - 2008 PAPERS PAPER Œ I (CODE 3) MATHS - Rao IIT · 2011-12-02 · SOLUTIONS TO...

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SOLUTIONS TO JEE - 2008 PAPERS

PAPER � I (CODE 3)

MATHS

Section I (Straight Objective Type)

This section contains 6 multiple choice questions. Each question has 4 choices (A) , (B) , (C) and (D)out of which ONLY ONE is correct.

1. Let g (x) =

)1x(coslog

)1x(m

n

; 0 < x < 2 , �m� and �n� are integers

, m 0

,

n > 0 and let �p� be the

left hand derivative of x � 1 at x = 1 . If 1x

Lim g

(x) = p , then :

(A) n = 1 , m = 1 (B) n = 1

, m = � 1

(C) n = 2 , m = 2 (D) n > 2

, m = n

Sol. (C)

P = 0hlim

h)1(f)h1(f

=

h|h|

= � 1

1

LHD = � 1

)h1(xLim

)1x(coslog

)1x(m

n

= � 1

= 0hLim

hcoslog

hm

n

= � 1

Using L - Hosp. 0hLim

hsinhcosm

hcoshn1m

m1n

= � 1

0hLim �

mn

cos h hsin

h 1n

= � 1

n � 1 = 1 n = 2 n = m m = 2 n = m = 2

2. The total number of local maxima and local minima of the function ,

f (x) =

2x1,x

1�x3,)x2(3/2

3

is :

(A) 0 (B) 1 (C) 2 (D) 3

2

Sol. (C)One point of local maxima andone point of local minima � 3

� 1

3. Let �a� and �b� be non-zero real numbers . Then , the equation

,

(a x2 + b

y2 + c) (x2 � 5

x

y + 6

y2) = 0 represents :

(A) four straight lines ,

when c = 0 and a , b are of the same sign

(B) two straight lines and a circle , when a = b and c is of sign opposite to that of a

(C) two straight lines and a hyperbola , when a and b are of the same sign and c is of sign

opposite to that of a(D) a circle and an ellipse

, when a and b are of the same sign and c is of sign opposite to that

of aSol. (B)

x2 � 5 x y + 6

y2 = 0 represents position of straight line

a x2 + b

y2 + c = 0 represents circle if a = b 0 and c is of opposite sign

4. The edges

of

a

parallelopiped

are

of

unit length and are parallel to non-coplanar unit vectors c�,b�,a�

such that ,

b�.a� = c�.b� = a�.c� = 21

. Then , the volume of the parallelopiped is :

(A) 2

1(B)

22

1(C)

23

(D) 3

1

Sol. (A)

Volume of paralleopiped is[a b c]2 =

c.cb.ca.c

c.bb.ba.b

c.ab.aa.a

[a b c]2 =

21

[a b c] = 2

1

5. Consider the two curves , C

1 :

y2 = 4

x and C

2 :

x2 + y2 � 6

x + 1 = 0 . Then :

(A) C1 and C

2 touch each other only at one point

(B) C1 and C

2 touch each other exactly at two points

(C) C1 and C

2 intersect (but doe not touch) at exactly two points

(D) C1 and C

2 neither intersect nor touch each other .

Sol. (B)

01x6yx

x4y22

2

Solving x2 � 2 x + 1 = 0

(x � 1)2 = 0 D = 0So, touching each other at two points (1 , 2) and (1

, �

2)

3

6. If 0 < x < 1 ,

then 2x1

2/121�1� 1xcotsinxcotcosx

is equal to :

(A) 2x1

x

(B) x (C) x

2x1 (D) 2x1

Sol. (C)

0 < x < 1 Let cot�1 x =

2

1

x1

1sin and cot�1 x = cos�1

2x1

x

)1

2x1

x 2x1

2/12

22

21

x1

1

x1

x

2x1 2/12 xxx = x

2x1

Section II (Multiple Correct Answers Type)

This section contains 4 multiple correct answer(s) type questions . Each question has 4 choices(A) , (B) , (C) and (D) out of which ONE OR MORE is/are correct.

7. Let f (x) be a non-constant twice differentiable function defined on (�

, ) such that

f (x) = f (1 � x) and f

41

= 0 . Then :

(A) f (x) vanishes at least twice on [0

, 1] (B) f

21

= 0

(C)

2/1

2/1 f

21

x sin x d

x = 0 (D)

2/1

0

f (t) esin t d

x =

1

2/1 f (1 � t) esin t

d

t

Sol. (ABCD)f (x) = f

(1 � x)

So , f

(x) is symmetric about x =

21

Hence f

2

1 = 0So

, we can say f

(x) = 0 Atleast wice in [0 , 1]

Using Roll�s in

21

,41

and

43

,21

Now put x = x/2 + 1f(x/2 + 1) = f(1/2 �x) so f(x/2 + 1) is an even function and f(x/2 + 1)sin x is an odd function so

I =

2/1

2/1

f

21

x sin x d

x = 0

And 1

2/1 f (1 � t) esin t

d

t

2/1

0

f (x) esin x d

x By putting t = 1 � x

4

8. Let , S

n =

n

1k22 knkn

n

and T

n =

1n

0k22 knkn

n

for n = 1

, 2

, 3 , .... Then :

(A) Sn <

33

(B) S

n >

33

(C) T

n <

33

(D) T

n >

33

Sol. (AD)

Let Sn n

Lim Sn

Sn < n

Lim n1

n

1k

2

2

nk

nk1

1

< 1

02xx1

xd

=

1

0

1�

3

1x2tan

3

2

3

2

3

1tan3tan 1�1�

Sn <

33

similarly T

n >

33

9. Let P (x

1 , y

1) and Q (x

2 , y

2) , y

1 < 0

,

y

2 < 0 be the end points of the latus rectum of the ellipse

x2 + 4 y2 = 4 . The equations of parabolas with latus rectum PQ are :

(A) x2 + 2 3 y = 3 + 3 (B) x2 � 2 3 y = 3 + 3

(C) x2 + 2 3 y = 3 � 3 (D) x2 � 2 3 y = 3 � 3Sol. (BC)

Equation of Parabola whose vertices are

213

,0 and

213

,0 are

231

,0

21

,3

21

,3

213

,0

231

,0

21

,3

21

,3

213

,0

x2 + 2 3 y = 3 � 3

x2 � 2 3 y = 3 + 3

5

10. A straight line through the vertex P of a triangle PQR intersects the side QR at the point S and thecircumcircle of the triangle PQR at the point T . If

S is not the centre of the circumcircle

, then :

(A) SP1

+ TS

1 <

RSSQ

2

(B)

SP1

+ TS

1 >

RSSQ

2

(C) SP1

+ TS

1 <

RQ4

(D) SP1

+ TS

1 >

RQ4

Sol. (BD)PS . ST = QS . SRH.M. < G.M.

Q S

T

R

P

TS.SP >

TS1

SP1

2

SP1

+ TS

1 >

TS.SP

2

SP1

+ TS

1 >

RS.SQ

2

2RSSQ

= 2RQ

> SR.SQ QR > 2 SR.SQ

SP1

+ TS

1 >

RQ4

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.

11. Consider the system of equations , a

x + b

y = 0 , c

x + d

y = 0 , where a

, b

, c

, d {

0

, 1

]

Statement - 1 : The probability that the system of equations has a unique solution is 83

Statement - 2 : The probability that the system of equations has a solution is 1 .(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)System has unique solution if and only if a

d � b

c 0

For this a d = 0 and b = c = 1

or a = d = 1 and b c = 0 so total combination is 6

Total choice of a , b , c , d = 24

Probability = 42

6 =

83

6

12. Consider the system of equations , x � 2

y + 3

z = �

1

� x + y � 2 z = k

x � 3 y + 4

z = 1

Statement - 1 : The system of equations has no solution for k 3

Statement - 2 : The determinant

141

k21

131

0 , for k 3 .

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






Sol. (A)

D = 431

211

321

= 6

D1 =

431

21k

321

= 3 � k

D2 =

411

2k1

311

= k � 3

D3 =

131

k11

121

= k � 3

System of the equation has no solution of k 3

13. Consider three planes , P

1 : x � y + z = 1

P2 : x + y � z = � 1

P3 : x � 3

y + 3

z = 2

Let L1 , L

2 , L

3 be the lines of intersection of the places

P

2 and P

3 ,

P

3 and P

1

and P

1 and P

2

respectivelyStatement - 1 : At least two of the lines L

1 , L

2 and L

3 are non-parallel

Statement - 2 : The three planes do not have a common point .(A) Statement - 1 is True







7

Sol. (D)

111

111

k�j�i�

ba

= k�2j�2

k�4j�4

331

111

k�j�i�

cb

k�2j�2

111

331

k�j�i�

ac

The dc�s of each of the line are proportional to (0 , 1 , 1)

So the three lines are respectively parallel.So statement

-

1 is wrong .

14. Let f and g

be real valued functions defined on interval (�

1

, 1) such that g

(x) is continuous

,

g (0) 0

, g

(0) = 0

, g

(0) 0 and

f (x) = g (x) sin x

Statement - 1 : 0xLim

[ g

(x) cot x � g

(0) cosec x

]

= f

(0)

Statement - 2 : f (0) = g

(0)







Sol. (B)f (x) = g (x) cos x + sin x . g

(x) f

(0) = g (0)

Now f (x) = 2 g

(x) cos x � g (x) sin x + sin x g

(x) f

(0) = 2 g

(0) = 0

But 0xLim

xeccos)0(gxcot)x(g = 0xLim xsin

)0(gxcos)x(g

= 0xLim xcos

xsin)x(gxcos)x(g = g

(0) = 0 = f

(0)

8

Section IV (Linked Comprehension Type)

This section contains 3 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 15 to 17)

A circle C of radius 1 is inscribed in an equilateral triangle PQR . The points of contact of C with thesides PQ

, QR

, RP

are

D

, E

, F respectively . The line PQ is given by the equation

,

3 x + y � 6 = 0 and the point D is

23

,233

. Further , it is given that the origin and the centre of

C are on the same side of the line PQ .

15. The equation of circle C is :

(A) 232x + (y � 1)2 = 1 (B) 232x + 2

2

1y

= 1

(C) 23x + (y + 1)2 = 1 (D) 23x + (y � 1)2 = 1

Sol. (D)

Equation of CD is

23

233x

= 21

23y

= � 1

C 1,3

Equation of the circle is 23x + (y � 1)2 = 1

16. Points E and F are given by :

(A)

23

,23

, 0,3 (B)

21

,23

, 0,3

(C)

23

,23

,

21

,23

(D)

23

,23

,

21

,23

Sol. (A)

Since the radius of the circle is 1 and C 1,3 , co-ordinates of F is 0,3

Equation of CE is

23

3x

=

21

1y = 1 E

23

,23

9

17. Equations of the sides QR , RP are :

(A) y = 3

2 x + 1

,

y = �

3

2 x � 1 (B) y =

3

1 x

,

y = 0

(C) y = 23

x + 1 ,

y = �

23

x � 1 (D) y = 3 x

,

y = 0

Sol. (D)

Equation of QR is y � 3 = 3 3x

y = x3Equation of RP is y = 0 .

Paragraph - II (Questions numbers 18 to 20)

Let A , B , C be three sets of complex numbers as defined below :

A = { z : Im z 1}

B = 3i2z:z

C = 2z)i1(Re:z

18. The number of elements in the set A B C is :

(A) 0 (B) 1 (C) 2 (D) Sol. (B)

A = Set of points on and above the line y = 1 in the Argand planeB = Set of points on the circle (x � 2)2 + (y � 1)2 = 32

C = Re (1 � i) z = Re (1 � i) (x + i y)

x + y = 2Hence (A B C) = has only one point of intersection .

19. Let �z� be any point in A B C . Then

,

2i1z +

2i5z lies between :

(A) 25 and 29 (B) 30 and 34 (C) 35 and 39 (D) 40 and 44Sol. (C)

The points (� 1

, 1) and (5 , 1) are the extremities of a diameter of the given circle

Hence 2i1z +

2i5z = 36

20. Let �z�

be any point in A B C and let �w� be any any point satisfying w � 2 � i < 3 .

Then z � w+ 3 lies between :(A) � 6 and 3 (B) � 3 and 6 (C) � 6 and 6 (D) � 3 and 9

Sol. (D)

wz < wz and wz = Distance between z and w

z is fixed. Hence distance between z and w would be maximum for diametrically opposite points.

wz < 6

� 6 < z � w< 6 � 3 < z � w+ 3 < 9

10

Paragraph - III (Questions numbers 21 to 23)

Consider the functions defined implicitly by the equation ,

y3 � 3

y + x = 0 on various intervals in

the real line .If x (�

, �

2) (2

, )

, the equation implicitly defines a unique real valued differentiable function

y = f (x) .

If x (� 2 , 2)

, the equation implicitly defines a unique real valued differentiable function y = g

(x)

satisfying g (0) = 0 .

21. If f 210 = 2 2 , then f

210 is equal to :

(A) 23 37

24(B) � 23 37

24(C)

37

243 (D) �

37

243

Sol. (B)

ydxd

= � 3 y2 + 3

2

2

yd

xd = � 6

y

2

2

xd

yd = �

3

22

ydxd

ydxd � 23 37

24

22. The area of the region bounded by the curve ,

y = f (x) , the x-axis and the lines x = a and x = b ,

where � < a < b < � 2

, is :

(A) b

a 1)x(f3

x2

d x + b

f (b) � a

f (a) (B) �

b

a 1)x(f3

x2

d x + b

f (b) � a

f (a)

(C) b

a 1)x(f3

x2

d x � b

f (b) + a

f (a) (D) �

b

a 1)x(f3

x2

d x � b

f (b) + a

f (a)

Sol. (A)

y = 2)x(f13

1

The required area = b

a

f (x) d

x =

ba

)x(fx � b

a

x f (x) d

x

= b f (b) � a f (a) + b

a 1)x(f3

x2

d x

23.

1

1

g (x) d

x =

(A) 2 g (�

1) (B) 0 (C) � 2

g (1) (D) 2

g (1)

Sol. (D)

We have y = 2)x(f13

1

g(x) is an odd

Hence

1

1

g (x) = g (1) - g (� 1) = 2

g (1)

11

JEE - 2008 PAPER PHYSICSPAPER � I (CODE 3)


This section contains 6 multiple choice questions. Each question has 4 choices (A) , (B) , (C)and (D) out of which ONLY ONE is correct.

24. An ideal gas is expanding such that PT2 = constant. The coefficient of volume expansion of the gasis

(A) 1T

(B) 2T

(C) 3T

(D) 4T

Sol. (C)Coefficient of volume expansion is defined as

= VdTdV

Here, PT2 = K

or

VnRT

T2 = K .....(1)

Differentiation of (1) givesnR × 3T2dT = KdV .....(2)Dividing (2) by (1)

VdTdV

= T3

or = T3

Correct option is (C)

25. Two beams of red and violet colours are made to pass separately through a prism (angle of the prismis 60º). In the position of minimum deviation, the angle of refraction will be

(A) 30º for both the colours (B) greater for the violet colour(C) greater for the red colour (D) equal but not 30º for both the colours

Sol. (A)Under minimum deviation ,

r1 = r

2 =

2A

= 30 º for all colours of light

Correct option is (A)

26. A spherically symmetric gravitational system of particles has a mass density 0 for r R

0 for r R

where 0 is a constant. A test mass can undergo circular motion under the influence of the gravitational

field of particle. Its speed V as a function of distance r (0 < r < ) from the centre of the system isrepresented by

V

Rr

V

Rr

V

Rr

V

Rr

(A) (B) (C) (D)

12

Sol. (C)

Gravitational field , E = 3

rG4 0 r R

= 2

30

r3

RG4 r > R

For revolving test mass mE = r

mV2

V = ErOr, V r r R

V r

1 r > R


27. Which one of the following statements is WRONG in the context of X-rays generated from a X-raytube?(A) Wavelength of characteristic X-rays decreases when the atomic number of the target increases(B) Cut-off wavelength of the continuous X-rays depends on the atomic number of the target(C) Intensity of the characteristic X-rays depends on the electrical power given to the X-rays

tube(D) Cut-off wavelength of the continuous X-rays depends on the energy of the electrons in the X-

rays tubeSol. (B)

28. Students I, II and III perform an experiment for measuring the acceleration due to gravity (g) using asimple pendulum. They use different lengths of the pendulum and/or record time for different numberof oscillations. The observations are shown in the table.Least count for length =0.1 cmLeast count for time = 0.1 s

Student Length of the Number of Total time for Time pendulum (cm) oscillations ( ) ( ) oscillations (s) period (s)

64.0 8 128.0 16.0

n n

I

II

III

64.0 4 64.0 16.0 20.0 4 36.0 9.0

If EI, E

II and E

III are the percentage errors in g, ie.

g100

g

for students I, II and III respectively..

(A) EI = 0 (B) E

I is minimum (C) E

I = E

II(D) E

II is maximum

Sol. (B)

For simple pendulum T = 2 g

or g = 2

2

T

4

gg

=

+

TT2

E1 = 100

161.0

264

1.0

= 1.40 (approx)

E2 = 100

161.0

264

1.0

= 1.40 (approx)

13

E3 = 100

9

1.02

20

1.0

= 2.72 (approx)

Numerically, E1 = E

2 and E

3 > E

2 = E

1

But more observations give less error to student I Correct option is (B)

29. Figure shows three resistor configurations R1, R

2 and R

3 connected to 3 V battery. If the power

dissipated by the configuration R1, R

2 and R

3 is P

1, P

2 and P

3, respectively, then

Figure

1

11

1

1

3V

3V

1

11

1

1

3V1

11

1

1

R1

R2

R3

(A) P1 > P

2 > P

3(B) P

1 > P

3 > P

2(C) P

2 > P

1 > P

3(D) P

3 > P

2 > P

1

Sol. (C)Resistance in three Cases are respectively

R1 = 1

R2 =

21

R3 = 2

Power dissipated in three cases are respectively

P1 =

132

= 2W

P2 = 18 W

P3 = 4.5 W

P2 > P

1 > P

3




30. In a Young�s double slit experiment, the separation between the two slits is d and the wavelength of

the light is . The intensity of light falling on slit 1 is four times the intensity of light falling on slit 2.Choose the correct choice (s) .(A) If d = , the screen will contain only one maximum(B) If < d < 2, at least one more maximum (besides the central maximum) will be observed

on the screen(c) If the intensity of light falling on slit 1 is reduced so that it becomes equal to that of slit 2, the

intensities of the observed dark and bright fringes will increase(D) If the intensity of light falling on slit 2 is increased so that it becomes equal to that of slit 1,

the intensities of the observed dark and bright fringes will increase

14

Sol. (A, B)As x = dsinwhen, d = , for dsin = n

sin = n n = 0 is the only possible value

(A) is correctWhen, < d < 2for maxima, dsin = nMany values of n are possible depending on the exact value of d.

(B) is correctInitially I

1 = 4I

0, I

2 = I

0, I

max = 9I

0, I

min = I

0

after modification I1' = I

0, I

2' = I

0Imax

' = 4I0, I

min' = 0

Hence intensity of maxima and minima decreases (C) is incorrectInitially I

1 = 4I

0, I

2 = I

0, I

max = 9I

0, I

min = I

0

after modification I1' = 4I

0, I

2' = 4I

0Imax

' = 16I0, I

min' = 0

(D) is incorrect

31. A particle of mass m and charge q, moving with velocity Venters Region II normal to the boundary as shown in thefigure. Region II has a uniform magnetic field B perpendicularto the plane of the paper. The length of the Region II is .Choose the correct choice(s)Figure.

Region II Region IIIRegion I

V

(A) The particle enters Region III only if its velocity V > q Bm

(B) The particle enters Region III only if its velocity V<q Bm

(C) Path length of the particle in Region II is maximum when velocity V = q Bm

(D) Time spent in Region II is same for any velocity V as long as the particle returns toRegion I

Sol. (A, C, D)Particle undergoes circular motion of radius

R = qBmV

For the particle to go in region III, < R

< qBmV

or v > m

qB

If v < m

qB

path of particle is semicircular. Also when, R < time spent is m/qB, which is

independent of initial velocity.

V = qBM

path of particle is semicircular, Also R =

V > qBM

path is circular arc

correct options are A, C, D

15

32. Assume that the nuclear binding energy per nucleon (B/A)versus mass number (A) is as shown in figure. Use this plotto choose the correct choice(s) given below.Figure.

02

4

68

100 200 A

B/A

(A) Fusion of two nuclei with mass numbers lying in therange of 1< A < 50 will release energy

(B) Fusion of two nuclei with mass numbers lying in therange of 51< A < 100 will release energy

(C) Fission of a nucleus lying in the mass range of100 < A < 200 will release energy when broken into two equal fragments

(D) Fission of a nucleus lying in the mass range of200 < A < 260 will release energy when broken into two equal fragments

Sol. (B, D)In fusion, X + Y ZEnergy released = Binding energy of product � BE of reactant

In case I, 1 < A < 50, for X and Yfor Z, A < 100Energy released = 0Option A is incorrect.In case II, 51 < A < 100, for X and Yfor Z, 100 < A < 200Energy is releasedOption B is correctIn Fission reaction, X Y + ZIn case III, 100 < A < 200 for Xfor Y and Z A < 100Energy is absorbedOption C is incorrectIn Fission reaction X Y + ZIn case IV, 200 < A < 260 for Xfor Y and Z, 100 < A 130Energy is released.Option D is correct

33. Two balls, having linear momenta 1�p pi

and 2�p pi

, undergo a collision in free space. There is

no external force acting on the balls. Let 1p '

and 2p'

be their final momenta. The following option(s)

is (are) NOT ALLOWED for any non-zero value of p, a1, a

2, b

1, b

2, c

1 and c

2.

(A) 1 1 1 1� � �p' a i b j c k

, 2 2 2� �p' a i b j

(B) 1 1�p' c k

, 2 2�p' c k

(C) 1 1 1 1� � �p' a i b j c k

, 2 2 2 1� � �p' a i b j c k

(D) 1 1 1� �p' a i b j

, 2 2 1� �p' a i b j

Sol. (A, D)From Conservation of Linear momentum

1 2 1 2p' p ' p p

1 2p ' p ' 0

or, 1 2p ' p '

(A) is not allowed as c1 = 0 (B) is allowed

(C) is allowed (D) is not allowed as b1 = �b

1 or b

1 = 0

16

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.

34. Statement - 1 : The stream of water flowing at high speed from a garden hose pipe tends to spreadlike a fountain when held vertically up, but tends to narrow down when held verticallydown.

Statement - 2 : In any steady flow of an incompressible fluid, the volume flow rate of the fluid remainsconstant.







Sol. (A)

35. Statement - 1 : Two cylinders, one hollow (metal) and the other solid (wood) with the same massand identical dimensions are simultaneously allowed to roll without slipping downan inclined plane from the same height. The hollow cylinder will reach the bottom ofthe inclined plane first.

Statement - 2 : By the principle of conservation of energy, the total kinetic energies of both thecylinders are identical when they reach the bottom of the incline.







Sol. (D)

36. Statement - 1 : An astronaut in an orbiting space station above the Earth experiencesweightlessness.

Statement - 2 : An object moving around the Earth under the influence of Earth�s gravitational force

is in a state of �free-fall�.







Sol. (A)

37. Statement - 1 : In a Meter Bridge experiment, null point for an unknown resistance is measured.Now, the unknown resistance is put inside an enclosure maintained at a highertemperature. The null point can be obtained at the same point as before by decreasingthe value of the standard resistance.

Statement - 2 : Resistance of a metal increases with increase in temperature.(A) Statement - 1 is True







Ans. (D)

17


This section contains 3 paragraphs. Based upon each paragraph, 3 multiple choice questionshave to be answered. Each question has 4 choices (A) , (B) , (C) and (D) , out of which ONLYONE is correct.

Paragraph - I (Questions numbers 38 to 40)

A small spherical monatomic ideal gas bubble 5

3

is trapped

inside a liquid of density

, (see figure), Assume that the bubbledoes not exchange any heat with the liquid. The bubble containsn moles of gas. The temperature of the gas when the bubble isat the bottom is T

0, the height of the liquid is H and the

atmospheric pressure is P0 (Neglect surface tension)

Figure.

liquid

y

H

P0

38. As the bubble moves upwards, besides the buoyancy force the following forces are acting on it(A) Only the force of gravity(B) The force due to gravity and the force due to the pressure of the liquid(C) The force due to gravity, the force due to the pressure of the liquid and the force due to

viscosity of the liquid(D) The force due to gravity and the force due to viscosity of the liquid

Sol. (D)

39. When the gas bubble is at a height y from the bottom, its temperature is

(A) 2 / 5

00

0

P gHT

P gy

(B) 2 / 5

00

0

P g(H y)T

P gH

(C) 3 / 5

00

0

P gHT

P gy

(D) 3 / 5

00

0

P g(H y)T

P gH

Sol. (B)

P1 = P

0 +

gH P

2 = P

0 +

g (H � y)

T1 = T

0T

2 = ?

PV = K

nRTP K

P

11P K T

111

22

TPTP

5 51

3 30 0

0 2

P gH T

P g(H y) T

2 53 30 0

0 2

P gH T

P g(H y) T

T2 = T

0 ×

250

0

P g(H y)

P gH

correct option is (B)

18

40. The buoyancy force acting on the gas bubble is (Assume R is the universal gas constant)

(A) 2 / 5

00 7 / 5

0

(P gH)nRgT

(P gy)

(B) 02 / 5 3 / 5

0 0

nRgT

(P gH) [P g(H y)]

(C) 3 / 5

00 8 / 5

0

(P gH)nRgT

(P gy)

(D) 03 / 5 2 / 5

0 0

nRgT

(P gH) [P g(H y)]

Sol. (B)

BF V g

= nRT

gP

=

21

50

0 0

P g(H y)nR g T

[P g(H y)] P gH

=

35

00 2

50

P g(H y)nR g T

(P gH)


Paragraph - II (Questions numbers 41 to 43)

A small block of mass M moves on a frictionless surface of aninclined plane, as shown in figure. The angle of the inclinesuddenly changes from 60º to 30º at point B. The block is initially

at rest at A. Assume that collisions between the block and theincline are totally inelastic (g = 10m/s2)Figure.

3 3m 3m

60º

A

v

M

B

30º C

41. The speed of the block at point B immediately after it strikes the second incline is

(A) 60 m/s (B) 45 m/s (C) 30 m/s (D) 15 m/s

Sol. (B)

60º

30º

u sin30ºB

u cos30ºB

uB

B

A

C

Velocity just before reaching B = 2gh

= 2 10 3 3

= Bu 60

= uB cos 30º

Velocity just after reaching B = 60 cos 30º

= 3

602

VB = 45 ms�1


19

42. The speed of the block at point C, immediately before it leaves the second incline is

(A) 120 m/s (B) 105 m/s (C) 90 m/s (D) 75 m/s

Sol. (B)Velocity of block when it leaves at point C

= 2BV 2g 3 3 tan30º = 105 ms�1


43. If collision between the block and the incline is completely elastic, then the vertical (upward)component of the velocity of the block at point B, immediately after it strikes the second incline is

(A) 30 m/s (B) 15 m/s (C) 0 (D) 15 m/s

Sol. (C) u sin30ºB

u cos30ºB

30º

30ºB

C

Velocity in vertical direction= u

B sin30º cos30º � u

B cos30º sin 30º = 0

correct option is (C)

Paragraph - III (Questions numbers 44 to 46)In a mixture of H�He+ gas (He+ is singly ionized He atom), H atoms and He+ ions are excited to theirrespective first excited states. Subsequently, H atoms transfer their total excitation energy to He+

ions (by collisions). Assume that the Bohr model of atom is exactly valid.44. The quantum number n of the state finally populated in He+ ions is

(A) 2 (B) 3 (C) 4 (D) 5Sol. (C)

For H atom, E1 = �13.6 ×

2

2

1

2 = �3.4eV

For He+ ion, E1� = �13.6 ×

2

2

2

2 = �13.6eV

After collision, hydrogen atom , E2 = �13.6 eV

For He+ ion E2� = �13.6

2

2

2

neV

E1 + E

1� = E

2 + E

2�

n = 4 correct option is (C)

45. The wavelength of light emitted in the visible region be He+ ions after collisions with H atoms is(A) 6.5 × 10�7 m (B) 5.6 × 10�7 m (C) 4.8 × 10�7 m (D) 4.0 × 10�7 m

Sol. (C)Visible light is obtained for transition from n = 4 to n = 3.

15 8

22 2

hc 4.1 10 3 10E 1 1

13.6 24 2

74.8 10 m correct option is (C)

46. The ratio of the kinetic energy of the n = 2 electron for the H atom of that of He+ ion is

(A) 14

(B) 12

(C) 1 (D) 2

Sol. (A)

2

2

ZK

n

2H

2He

K 1 1K 42

correct option is (A)

20

JEE - 2008 PAPER CHEMISTRY

PAPER � I (CODE 3)


This section contains 6 multiple choice questions. Each question has 4 choices (A) , (B) , (C) and (D)out of which ONLY ONE is correct.

47. Under the same reaction conditions, initial concentration of 1.386 mol dm�3 of a substance becomeshalf in 40 seconds and 20 seconds through first order and zero order kinetics, respectively. Ratio

0

1

k

k of the rate constants for first order (k1) and zero order (k0) of the reactions is

(A) 0.5 mol�1 dm3 (B) 1.0 mol dm�3

(C) 1.5 mol dm�3 (D) 2.0 mol�1 dm3

Sol. (A)

k1 = 2/1t

693.0 =

40693.0

k0 = 2/1

0

t2

A =

202386.1

= 20693.0

0

1

kk

= 20/693.040/693.0

= 4020

= 0.5

48. 2.5 mL of 52

M weak monoacidic base (Kb = 1 × 10�12 at 25ºC) is titrated with 152

M HCl in water at

25ºC. The concentration of H+ at equivalence point is (Kw = 1 × 10�14 at 25ºC)

(A) 3.7 × 10�13 M (B) 3.2 × 10�7 M(C) 3.2 × 10�2 M (D) 2.7 × 10�2 M

Sol. (D)meq. of base = meq. of HCl

2.5 × 52

= 152

× V VHCl = 7.5 ml

Vsolution = 2.5 × 7.5 = 10 ml

BOH + HCl BCl + H2O

2.5 × 52

= 1

At equivalence point, only BCl is present , [BCl] = 101

= 0.1

Which undergoes hydrolysis:

B+ + H2O BOH + H+ Kh = b

w

KK

= 12

14

10

10

= 10�2

Kh = )h1(

Ch2

10�2 Þ

)h1(h1.0 2

10h2 + h � 1 = 0

h = 0.27 [H+] = 0.27 × 0.1 = 2.7 × 10�2

21

49. Native silver metal forms a water soluble complex with a dilute aqueous solution of NaCN in thepresence of(A) nitrogen (B) oxygen (C) carbon dioxide (D) argon

Sol. (B)4Ag + 8NaCN = 2H2O + O2 4Na[Ag(CN)2] + 4NaOH

50. Aqueous solution of Na2S2O3 on reaction with Cl2 gives(A) Na2S4O6 (B) NaHSO4 (C) NaCl (D) NaOH

Sol. (B)Na2S2O3 + 4Cl2 + 4H2O 2NaHSO4 + 8HCl

51. Hyperconjugation involves overlap of the following orbitals(A) � (B) � p (C) p � p (D) �

Sol. (B)Hygperconjugation involves overlappiong of C�H bond with p-orbital of double bond, carbocation,

and free radical.

52. The major product of the following reaction is

Me Br

F

NO2

PhS Na

dimethylformamide

(A)

Me SPh

F

NO2

(B)

Me SPh

F

NO2

(C)

Me Br

SPh

NO2

(D)

Me SPh

SPh

NO2

Sol. (A)

Me Br

F

NO2

PhS Na

dimethylformamide

Me

NO2

SPh

F

DMF is aprotic solvent which favours SN2 and SN2 involves inversion.

22



53. A gas described by van der Waals equation(A) behaves similar to an ideal gas in the limit of large molar volumes(B) behaves similar to an ideal gas in the limit of large pressures(C) is characterised by van der Waals coefficients that are dependent on the identity of the gas

but are independent of the temperature.(D) has the pressure that is lower than the pressure exerted by the same gas behaving ideally.

Sol. (A, C, D)At low pressure, when the sample occupies a large volume, the molecules are so far apart for mostof the time that the intermolecular forces play no significant role, and the gas behaves virtuallyperfectly. a and b are characteristic of a gas and are independent of temperature.

54. A solution of colourless salt H on boiling with excess NaOH produces a non-flammable gas. The gasevolution ceases after sometime. Upon addition of Zn dust to the same solution, the gas evolutionrestarts. The colourless salts(s) H is (are)(A) NH4NO3 (B) NH4NO2 (C) NH4Cl (D) (NH4)2SO4

Sol. (A, B)All ammonium compounds on reacting with any base releases NH3 gas which is non-infalmmble. Onaddition of Zn dust, Zn becomes oxidises and releases NH3 gas which is possible by nitrate andnitrite.Hence NH4NO3/ NH4NO2NH4NO3 + NaOH NH3 + NaNO3 + H2O7NaOH + NaNO3 + 4Zn 4Na2ZnO2 + NH3 + 2H2ONH4NO2 + NaOH NaNO2 + NH3 + H2O3Zn + 5NaOH + NaNO2 3Na2ZnO2 + NH3 + H2O

55. The correct statement(s) concerning the structure E, F and G is (are)

(E)

OH3C

H C3 CH3

(F)

H3C

H C3 CH3

OH(G)

H3C

H C3

CH3

OH

(A) E, F and G are resonance structures (B) E, F and E, G are tautomers(C) F and G are geometrical isomers (D) F and G are diastereomers

Sol. (B, C, D)E, F are tautomers so those are not resonating structures.E, F and E, G are tautomers. F, G are geometrical isomers and all geometrical isomers arediastereomers.

56. The correct statement(s) about the compound given below is (are)

CH3

Cl H

Cl H

H C3

(A) The compound is optically active(B) The compound possess centre of symmetry(C) The compound possesses plane of symmetry(D) The compound possesses axis of symmetry

23

Sol. (A)

Sol. Compound is H Cl

CH3

CH3

Cl H

so it is optically active.

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.

57. Statement - 1 : For every chemical reaction at equilibrium, standard Gibbs energy of reaction iszero.

Statement - 2 : At constant temperature and pressure, chemical reactions are spontaneous in thedirection of decreasing Gibbs energy.







Sol/ (D)G = 0 at eq. (not G0)G < 0 reaction spontaneous

58. Statement - 1 : The plot of atomic number (y-axis) versus number of neutrons (x-axis) for stablenuclei shows a curvature towards x-axis from the line of 45º slope as the atomic

number is increased.Statement - 2 : Proton-proton electrostatic repulsions begin to overcome attractive forces involving

protons and neutrons in heavier nuclides.(A) Statement - 1 is True







Sol. (A)

Z

no. of neutron

When Z > 20, the number of neutron must increase above the number of protons so as to overcome

protonproton repulsion i.e., 1pn

24

59. Statement - 1 : Pb4+ compounds are stronger oxidizing agents than Sn4+ compounds.Statement - 2 : The higher oxidation states for the group 14 elements are more stable for the heavier

members of the group due to �inert pair effect�.







Sol (C)Pb+4 is stronger oxidising agent then Sn+4 and for heavier elements, lowr oxidation state is morestable.

60. Statement - 1 : Bromobenzene upon reaction with Br2/Fe gives 1,4-dibromobenzene as the majorproduct.

Statement - 2 : In bromobenzene, the inductive effect of the bromo group is more dominant than themesomeric effect in direction the incoming electrophile.







Sol. (C)Directive nature of electrophile is determine by mesomeric effect.


This section contains 3 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. 61 to 63

In the following reaction sequence, products I, J and L are formed. K represents a reagent.

Hex-3-ynal 3

4

PBr.2

NaBH.1 I

OH.3

CO.2ether/Mg.1

3

2 J K Cl

O

Me

quinolineBaSO/Pd

H

4

2 L

61. The structure of the product I is

(A) Me Br (B)

Br

Me

(C) Me Br (D) BrMe

Sol. (D)

H

O

Me

4NaBH.1 OH

Me

Br

Me

(I)

PBr3

25

62. The structure of compounds J and K, respectively, are

(A) COOHMe and SOCl2

(B) MeOH

O

and SO2Cl2

(C) Me

COOH

and SOCl2

(d)COOHMe

and CH3SO2Cl

Sol. (A)

Br

Me

COOH

Me

(J)

1. Mg/ether

2. CO2

3. H O3+

(Formation of GR)

(Formation of acid)

SOCl2 (K)

COCl

Me

63. The structure of product L is

(A) COOHMe

(B) CHOMe

(C) Me

CHO(D) CHOMe

Sol. (C)

COCl

MeH2

Pd/BaSO quinoline4

(Rosenmund's reactionand Lindlar's catalyst)

CHO

Me

26

Paragraph for Question Nos. 64 to 66

Properties such as boiling point, freezing point and vapour pressure of a pure solvent change whensolute molecules are added to get homogeneous solution. These are called colligative properties.Applications of colligative properties are very useful in day-today life. One of tis examples is the useof ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobilesA solution M is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is0.9.Given Freezing point depression constant for water (Kf

water) = 1.86 K kg mol�1

Freezing point depression constant of ethanol (Kfethanol) = 2.0 K kg mol�1

Boiling point elevation constant of water (Kbwater) = 0.52 K kg mol�1

Boiling point elevation constant of ethanol (Kbethanol) = 1.2 K kg mol�1

Standard freezing point of water = 273 KStandard freezing point of ethanol = 155.7 KStandard boiling point of water = 373 KStandard boiling point of ethanol = 351.5 KVapour pressure of pure water = 32.8 mm HgVapour pressure of pure ethanol = 40 mm HgMolecular weight of water = 18 g mol�1

Molecular weight of ethanol = 46 g mol�1

In answering the following question, consider the solutions to be ideal dilute solutions and solutes tobe non-volatile and non-dissociative..

64. The freezing point of the solution M is(A) 268.7 K (B) 268.5 K (C) 234.2 K (D) 150.9 K

Sol. (D)For solution M, water is solute and ethanol is solvent Tf = (kf)ethanol × m.

= 2 × 1000/469.0

1.0

= 469

2000

= 4.8

Tf = Tf0 � Tf = 155.7 � 4.8 = 150.9 K

65. The vapour pressure of the solution M is(A) 39.3 mm Hg (B) 36.0 mm Hg (C) 29.5 mm Hg (D) 28.8 mmHg

Sol. (B)P = PA

0xA

= 40 × 0.9 = 36.0 mm

66. Water is added to the solution M such that the mole fraction of water in the solution becomes 0.9.The boiling point of this solution is:(A) 380.4 K (B) 376.2 K (C) 375.5 K (D) 354.7 K

Sol. (B)Now water is treated as solvent so

Tb = (Kb)water × m

= 0.52 × 1000/189.0

1.0

= 189

520

= 3.2

Tb = Tb0 + Tb = 373 + 3.2 = 376.2

27

Paragraph for Question Nos. 67 to 69.

There are some deposits of nitrates and phosphates in earth�s crust. Nitrates are more soluble in

water. Nitrates are difficult to reduce under the laboratory conditions but microbes do it easily.Ammonia fors large number of complexes with transition metal ions. Hybridization easily explainsthe ease of sigma donation capability of NH3 and PH3. Phosphine is a flammable gas and is preparedfrom white phosphorous.

67. Among the following, the correct statement is(A) Phosphates have no biological significance in humans(B) Between nitrates and phosphates, phosphates are less abundant in earth�s crust

(C) Between nitrates and phosphates, nitrates are less abundant in earth�s crust

(D) Oxidation of nitrates is possible in soil.Sol. (C)

Nitrates are soluble

68. Among the following, the correct statement is(A) Between NH3 and PH3, NH3 is a better electron donor because the lone pair of electrons

occupies spherical �s� orbital and is less directional

(B) Between NH3 and PH3, PH3 is a better electron donor because the lone pair of electronsoccupies sp3 orbital and is more directional

(C) Between NH3 and PH3, NH3 is a better electron donor because the lone pair of electronsoccupies sp3 orbital and is more directional

(D) Between NH3 and PH3, PH3 is a better electron donor because the lone pair of electronsoccupies spherical �s� orbital and is less directional.

Sol. (C)Between NH3 and PH3, NH3 is a better electron donor because it is more basic and the lone pair ofelectrons occupies sp3 orbital and is more directional.

69. White phosphorus on reaction with NaOH gives PH3 as one of the products. This is a(A) dimerization reaction (B) disproportionation reaction(C) condensation reaction (D) precipitation reaction

Sol. (B)3H2O + P4 + 3NaOH 3NaH2PO2 + PH3So disproportionation reaction

SOLUTIONS TO JEE - 2008 PAPERS PAPER Œ I (CODE 3) MATHS - Rao IIT · 2011-12-02 · SOLUTIONS TO...

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Transcript of SOLUTIONS TO JEE - 2008 PAPERS PAPER Œ I (CODE 3) MATHS - Rao IIT · 2011-12-02 · SOLUTIONS TO...