Iit Sol.paper2

IIT-JEE 2008 EXAMINATION PAPER (SOLUTIONS) PAPER – II Part – I (MATHEMATICS)

SECTION – I Straight Objective Type

13/04/08

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.

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, …

1  1 g′′  N +  – g′′   = 2  2  1 1  1 + ..... + (A) – 4 1 + +  2  9 25 (2 N – 1) 

 1 1  1 + .... + (B) 4 1 + +  2  9 25 (2 N – 1) 

 1 1  1 (C) – 4 1 + + + .... +  2  9 25 (2 N + 1) 

 1 1  1 (D) 4 1 + + + ..... +  2  9 25 (2 N + 1) 

[Ans. A] Sol.

g(x) = log f(x) & f(x + 1) = x f(x) Replece x by x –

1 2

So, f(x +

1 1 1 ) = (x – ). f(x – ) 2 2 2

∴ g(x +

1 1 1 ) = log (x – ) + g(x – ) 2 2 2

⇒ g (x +

1 1 1 ) – g (x – ) = log (x – ) 2 2 2

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⇒ g' (x +

1 1 1 ) – g' (x – ) = 1 2 2  x −  2 

⇒ g'' (x +

−1 1 1 ) – g'' (x – ) = 2 2 2 1  x −  2 

3  1  −1 ⇒ g''   – g''   2   2 1   4 −1 5 3 ⇒ g''   – g''   = 2 2 9   4 1  g''  N +  – g'' 2 

−4 1  N −  = 2   (2n − 1)2

Adding above n equations. 1  1 ⇒ So, g′′  N +  – g''   = – 4 2  2

2.

 1 1  1 + ..... + 1 + +  2 (2 N − 1)   9 25

Let two non-collinear unit vectors aˆ and bˆ 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 aˆ cos t + bˆ sin t. When P is farthest from origin →

→

O, let M be the length of OP and uˆ be the unit vector along OP . Then -

(A) uˆ =

(C) uˆ =

1 aˆ + bˆ and M = (1 + aˆ.bˆ) 2 | aˆ + bˆ | 1 aˆ + bˆ and M = (1 + 2aˆ.bˆ) 2 | aˆ + bˆ |

(B) uˆ =

1 aˆ – bˆ and M = (1 + aˆ.bˆ) 2 | aˆ – bˆ |

(D) uˆ =

1 aˆ – bˆ and M = (1 + 2aˆ.bˆ) 2 | aˆ – bˆ |

[Ans. A] Sol.

→

Given that, OP = M uˆ →

Where | OP | max = | aˆ cos t + bˆ sin t | = 1 + cos θ Where θ is angle between aˆ & bˆ and uˆ is the vector along the bisector of angle between aˆ and bˆ . 3.

Let

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2

I=

ex

∫ e 4x + e 2x + 1

dx, J =

e–x

∫ e – 4x + e – 2x + 1

dx

Then, for an arbitrary constant C, the value of J – I equals (A)

 e 4x – e 2x + 1  1  +C log   e 4x + e 2x + 1  2  

(B)

 e 2x + e x + 1  1  +C log   e 2x – e x + 1  2  

(C)

 e 2x – e x + 1  1  +C log   e 2x + e x + 1  2  

(D)

 e 4x + e 2x + 1  1  C log   e 4x – e 2x + 1  2  

[Ans. C] Sol.

J=

e 3x

∫ 1 + e 2x + e 4x

J–I=

dx

e 3x – e x

∫ e 4x + e 2x + 1 dx

Let ex = t

=

t2 –1

∫ t4 + t2 +1

dt =

∫

1–

1 t2

t 2 +1+

1

dt

t2

1

 1   t + –1 1 t  +C = ∫ dt = ln  2 1 2    1 t + + 1    t +  –1 t    t

1–

=

4.

t2

 e 2x – e x + 1  1  +C ln   e 2x + e x + 1  2  

Consider three points. P = (– sin (β – α), – cos β),

Q = (cos (β – α), sin β) and

R = (cos (β – α + θ), sin (β – θ)), where 0 < α, β, θ <

π . Then 4

(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 [Ans. D]

Sol.

On solving the determinant for some particular θ, we get answer.

5.

An experiment has 10 equally likely outcomes. Let A and B be two non-empty events of the experiment. 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

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(C) 4 or 8

(D) 5 or 10 [Ans. D]

Sol.

Let n(B) = x > 0 Now given that A and B are independent events. P(A).P(B) = P(A ∩ B) 4 x z . = where z = n(A ∩ B) ≤ minm (4, x) 10 10 10

⇒x=

5z where z may be equal to 0, 1, 2, 4. 2

⇒ x = 5 or 10.

6.

The area of the region between the curves y = and x =

∫ 0

∫ 0

2 –1

t 2

(1 + t ) 1 – t

2 +1

(C)

1 – sin x bounded by the lines x = 0 cos x

π is 4

2 –1

(A)

1 + sin x and y = cos x

2

(B)

dt

∫ 0

2 +1

4t (1 + t 2 ) 1 – t 2

dt

(D)

∫ 0

4t 2

(1 + t ) 1 – t 2

t (1 + t 2 ) 1 – t 2

dt

dt [Ans. B]

Sol.

Area bounded = π4 

 1 + sin − 1 − sin  cos x cos x 

∫0

  cos x + sin 2 0  x  cos − sin 2 

∫

π4 

π4

∫0

π4

∫0

 dx  

x x cos − sin 2 − 2 x x cos + sin 2 2

x 2 x 2

  dx   

x x x  x   cos + sin  −  cos − sin  2 2 2  2  dx cos x

x dx 2 x 2 cos 2 − 1 2 2 sin

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∫0

cos

π4

∫0

π4

∫0

x 2

2 sin

π4

x x 2 − sec 2 2 2

x dx 2 x 2 − sec 2 2

2 tan

x dx 2 x 1 − tan 2 2

2 tan

x 2

Put t = tan

1 x sec2 dx = dt 2 2

7.

∫0

2 −1

∫0

2 −1

2.t 1− t 2

×

2dt sec 2

4t dt

(

1− t2 1+ t2

x 2

)

Consider a branch of the hyperbola x2 – 2y2 – 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 the point A, then the area of the triangle AB C is (A) 1 –

(C) 1+

2 3 2 3

(B)

3 –1 2

(D)

3 +1 2

[Ans. B] Sol. B C A

Hight =

b2 , a

Base = a e – a

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⇒∆=

b2 1 × × (ae – a) a 2

Hyperbola is (x –

2

2 ) – 2 (y +

(x – 2 ) 2 4

–

2 )2 = 4

(y + 2 )2 =1 2

2 = 4 (e2 – 1) 1 3 = e2 – 1 ⇒ e2 = ⇒e= 2 2

∆=

1 2 × ×2 2 2

3 2

 3   – 1  2   

 3  = – 1  2   

8.

A particle P starts from the point z0 = 1 + 2i, 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 z1 the particles moves π 2 units in the direction of the vector ˆi + ˆj and then it moves through an angle in anticlockwise 2

direction on a circle with centre at origin, to reach a point z2. The point z2 is given by (A) 6 + 7i

(B) – 7 + 6i

(C) 7 + 6i

(D) – 6 + 7i [Ans. D]

Sol. Z (6, 5)

2

P (7, 6) 45º

3 Z0 (1, 2)

(6, 2)

In diagram you can see P point is given by complex No. 7 + 6i and now it is rotated by

π angle in 2

anticlockwise, since so it will be i (7 + 6i) ⇒ z2= – 6 + 7i. 9.

π  π π Let the function g : (– ∞, ∞) →  – ,  be given by g(u) = 2 tan–1 (eu) – . Then g is 2 2 2  

(A) even and is strictly increasing in (0, ∞) CAREER POINT, 112, Shakti Nagar, Kota (Raj.) Ph.: 0744-2500092, 2500492 Website : www.careerpointgroup.com, Email: [email protected]

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(B) odd and is strictly decreasing in (–∞, ∞) (C) odd and is strictly increasing in (–∞, ∞) (D) neither even nor odd, but is strictly increasing in (–∞, ∞) [Ans. C] Sol.

g′ (u) =

2e u 1 + e 2u

>0

So g(u) increases Now g(–u) = 2 tan–1 (e–u) –  1 = 2 tan–1   eu

π 2

 π  – 2 

π  π = 2  – tan –1 (e u )  –   2   2

=

π – 2tan–1 eu = – g(u) 2

⇒ g(u) is odd increasing.

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

10.

Let a, b, c, p, q be real numbers. Suppose α, β are the roots of the equation x2 + 2px + q = 0 and α,

1 are β

the roots of the equation ax2 + 2bx + c = 0, where β2 ∉ {–1, 0, 1}. STATEMENT - 1

(p2 – q) (b2 – ac) ≥ 0. and STATEMENT – 2

b ≠ pa or c ≠ qa. (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 NOT a correct explanation for Statement-1

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(C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True. [Ans. B] Sol.

Given two equations has only one root in common, therefore roots can not be complex, now by using sum and product of roots in both equations are not equal, you can solve it.

11.

Consider L1 : 2x + 3y + p – 3 = 0 L2 : 2x + 3y + p + 3 = 0 where p is a real number, and C : x2 + y2 + 6x – 10y + 30 = 0. STATEMENT - 1

If line L1 is a chord of circle C, then line L2 is not always a diameter of circle C. and STATEMENT – 2

If line L1 is a diameter of circle C, then line L2 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 NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True. [Ans. C] Sol.

12.

Given that, L1 and L2 are two parallel lines, and distance between these two liens is less than radius.

Let a solution y = y(x) of the differential equation y 2 – 1 dx = 0

x x 2 – 1 dy – y 2

satisfy y (2) =

.

3

STATEMENT - 1

π  y (x ) = sec  sec –1 x –  6  and STATEMENT - 2

y(x) is given by 1 2 3 = – y x

1–

1 x2

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(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 NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True. [Ans. C] Sol.

x

x 2 – 1 dy – y

dy dx

=

∫

y 2 – 1 dx = 0

y y2 – 1 x x2 –1

dy

∫

=

y y2 – 1

dx x x2 –1

sec–1 y = sec–1 x + C 2

at x = 2, y = sec–1

2 3

3

= sec–1 2 + C

π π = +C 6 3 C=–

π 6

sec–1 y = sec-1 x –

π 6

y = sec (sec–1 x –

π ) 6

Again sec–1 y = sec–1 x – cos–1

π 6

π 1 1 – cos–1 = y 6 x

 1 1 1  π 1– = cos–1  + 1– 2 2   xy 6 x y   1 1 + 1– xy x2

13.

1–

1 y

2

=

3 2

Suppose for distinct positive numbers a1, a2, a3, a4 are in G.P. Let b1= a1, b2 = b1 + a2, b3 = b2 + a3 and b4 = b3 + a4. CAREER POINT, 112, Shakti Nagar, Kota (Raj.) Ph.: 0744-2500092, 2500492 Website : www.careerpointgroup.com, Email: [email protected]

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STATEMENT – 1

The number b1, b2, b3, b4 are neither in A.P. nor in G.P. and STATEMENT - 2

The numbers b1, b2, b3, b4 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 NOT a correct explanation for Statement-1 (C) Statement-1 is True, Statement-2 is False (D) Statement-1 is False, Statement-2 is True. [Ans. C] Sol.

Given that,

b1 = a1 b2 = a1 + a2 b3 = a1 + a2 + a3 b4 = a1 + a2 + a3 + a4

⇒

b1, b2, b3 and b4 are neither in A.P., G.P. nor in H.P.

SECTION – III

Linked Comprehension Type This section contains 2 paragraphs P14–16 and P17–19. Based upon each paragraph, 3 multiple choice questions have to be answered. Each question has 4 choices (A ), (B ), (C) and (D ), out of which O N LY O N E is correct. P14-16 : Paragraph for Question Nos. 14 to 16

Consider the lines L1 :

L2 :

14.

x +1 y+2 z +1 = = 3 1 2 y+2 x–2 z–3 = = 1 2 3

The unit vector perpendicular to both L1 and L2 is (A)

– ˆi + 7ˆj + 7kˆ 99

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(B)

– ˆi – 7ˆj + 5kˆ 5 3 – ˆi + 7ˆj + 5kˆ

(C)

(D)

Sol.

5 3 7ˆi – 7ˆj – kˆ

[Ans. B]

99 →

→

Let b = 3ˆi + ˆj + 2kˆ and d = ˆi + 2ˆj + 3kˆ →

ˆi

→

ˆj kˆ

⇒ b × d = 3 1 2 = ˆi (3 – 4) – ˆj (9 – 2) + kˆ (6 – 1) 1 2 3 =

15.

– ˆi – 7ˆj + 5kˆ 1 + 49 + 25

The shortest distance between L1 and L2 is (A) 0 (C)

(B) 41

(D)

5 3

17 3 17 5 3

[Ans. D] Sol.

Let A (–1, –2, –1) and C(2, –2, 3) →

AC = (– ˆi + 2ˆi ) + (–2 ˆj + 2 ˆj ) + (– kˆ + 3 kˆ ) = – 3 ˆi – 4 kˆ →

Shortest distance =

→ →

AC .( b × d ) → →

| b× d | →

=

17 5 3

→

Where b and d are vectors along the lines. 16.

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 L2 is (A) (B) (C)

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(D)

Sol.

23

[Ans. C]

75

Plane passing through given point and ⊥ to L1 and L2 is given by, – (x + 1) – 7(y + 2) + 5(z + 1) = 0 – x – 7y + 5z – 1 – 14 + 5 = 0 – x – 7y + 5z – 10 = 0 x + 7y – 5z + 10 = 0 1 + 7 – 5 + 10

Now ⊥ distance =

49 + 1 + 25

=

13 75

=

13 5 3

P17-19 : Paragraph for Question Nos. 17 to 19 Consider the function f : (–∞, ∞) → (–∞, ∞) defined by f(x) =

17.

x 2 – ax + 1 x 2 + ax + 1

, 0 < a < 2.

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

18.

[Ans. A]

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 [Ans. A]

19.

Let ex

g(x) =

f ' (t )

∫ 1+ t2

dt.

0

Which of the following is true ?

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(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 (–∞, ∞)

[Ans. B]

17 to 19

f : (–∞, ∞) → (–∞, +∞) Given that f(x) =

x 2 – ax + 1 x 2 + ax + 1

0
Now differentiating f(x); f′(x) =

2a ( x 2 – 1) ( x 2 + ax + 1) 2 ( x 2 + ax + 1) 2 4ax – 4a ( x 2 – 1)(2 x + a )( x 2 + ax + 1)

and f′′ (x) =

( x 2 + ax + 1) 4

Sol.17

Now find f′′(1) and f′′ (–1) and put in given choices.

Sol.18

According to f’(x) we can check the increasing and decreasing intervals.

Sol.19

g’(x) =

f ' (e x ) 1+ e

2x

. ex

Now use f′(x) and find the intervals where g′(x) is +ve and –ve.

SECTION – IV Matrix - Match Type This section contains 3 questions. Each question contains statements given in two columns which have to be matched. Statements (A ,B ,C ,D ) in Column Ι have to be matched with statements (p,q,r,s) in Column ΙΙ. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A -p , A -s , B -q , B -r , C -p , C -q and D -s , then the correctly bubbled 4 x 4 matrix should be as follows:

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13

20.

q q

r

s

A

p p

r

s

B

p

q

r

s

C

p

q

r

s

D p

q

r

s

Match the Statements/Expression in Column Ι with the Statements/Expression in Column ΙΙ and indicate your answer by darkening appropriate bubbles in the 4 × 4 matrix given in the ORS.

Column Ι

Column ΙΙ

(A ) The minimum value of

x 2 + 2x + 4 x+2

is

(p) 0

(B ) Let A and B be 3 × 3 matrices of real numbers,

(q) 1

Where A 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 the possible values of k are (C ) Let a = log3 log3 2. An integer k satisfying 1 < 2 (– k +3

–a

)

(r) 2

< 2, must be less than

(D ) If sin θ = cos ϕ, then the possible values of

(s) 3

1 π (θ ± ϕ – ) are π 2 [Ans. A : (r); B : (q,s); C: (q,s); D: (p,r) ] Sol.(A) On solving, we get

y ( x + 2) = x2 + 2x + 4 x2 + (2 –y) x + 4 – 2y = 0 4 + y2 – 4y – 4 (4 – 2y) ≥ 0 y2 + 4y – 12 ≥ 0 y2 + 6y – 12 ≥ 0 y (y + 6y) –2 ( y + 6y) ≥ 0 (y – 2) (y + 6) ≥ 0 (B)

given that, (A + B) (A – B) = (A – B) (A +B) AT = A (Q A is symm.) BT = – B (Q B is skew symm.) CAREER POINT, 112, Shakti Nagar, Kota (Raj.) Ph.: 0744-2500092, 2500492 Website : www.careerpointgroup.com, Email: [email protected]

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(AB)T = BT.AT = A (– B) = – AB (C)

( )

a = log3log32 | 3–a = log 32

⇒ 20 <

   −k + 1  2  log 3  2

−1

< 21

0 < – k + log23 < 1 – log23 < – k < 1 – log23 log23 rel="nofollow"> k > log23 – 1 (D)

sinθ = sin(π/2 – φ) cos (π/2 – θ) = cosφ

π/2 – θ = ± φ + 2nπ θ ± φ = π/2 + 2nπ 1 (θ ± φ – π/2) = 0 or 2 π

21.

Consider all possible permutations of the letters of the work ENDEANOEL. Match the Statements/Expressions in Column Ι with the Statements/Expressions in Column ΙΙ and indicate your answer by darkening appropriate bubbles in the 4 × 4 matrix given in the ORS.

Column Ι

Column ΙΙ

(A ) The number of permutations containing the

(p) 5!

Word ENDEA is (B ) The number of permutations in which the

(q) 2 × 5!

Letter E occurs in the first and the last positions is (C ) The number of permutations in which none of

(r) 7 × 5!

The letters D, L, N occurs in the last five positions is (D) The number of permutations in which the letters

(s) 21 × 5!

A, E, O occur only in odd positions is [Ans. A : (p); B : (s); C: (q); D: (q) ] Sol.

(A) No. of permutations containing word ENDEA = 5! As consider ENDEA as one letter & remaining four

∴ total letters to be arranged = 5. (B) ∴ Remaining 7 letters can be arranged in = 7!/2! = 21 × 5! (C) D, I, N can occupy only four positions & remaining 5 alphabets can occupy remaining places, in

4! 5! × = 2 × 5! 2! 3!

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15

(D) odd positions are 5 so A, E,O can occupy these places in

22.

5! 4! × = 2 × 5! . 3! 2!

Consider the lines given by L1 : x + 3y – 5 = 0 L2: 3x – ky – 1= 0 L3 : 5x + 2y – 12 = 0 Match the Statements/Expressions in Column Ι with the Statements/Expressions in Column ΙΙ and indicate your answer by darkening appropriate bubbles in the 4 × 4 matrix given in the ORS.

Column Ι

Column ΙΙ

(A ) L1, L2, L3 are concurrent, if

(p) k = – 9

(B ) One of L1, L2, L3 is parallel to at least one of

(q) k = –

6 5

the other two, if 5 6

(C ) L1, L2, L3 form a triangle, if

(r) k =

(D) L1, L2, L3 do not form a triangle, if

(s) k = 5 [Ans. A : (s); B : (p,q); C: (r); D: (p,q,s) ]

1

Sol.

3

(A) On solving 3 – k 5 2

–5 –1 = 0 – 12

We get k = 5 (B) For any two lines to be parallel K will be equal to – 9 and

–6 . 5

(C&D) Given three lines form ∆ when lines are not parallel as well as concurrent.

Part – II (PHYSICS) (Section – I)

Straight Objective Type This section contains 9 multiple choice questions numbered 1 to 9. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

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23.

A light beam is traveling from Region I to Region IV (Refer Figure). The refractive index in Regions I, II, III and IV are n0,

n0 n0 n ¸ and 0 , respectively. The angle of incidence θ for which the beam just 2 6 8

misses entering Region IV is Figure : Region II Region III Region IV

Region I

n0 2

θ n0 0

3 (A) sin–1   4

Sol.

n0 sin θ =

n0 sin θc 6

0.2m

n0 8

0.6m

1 (C) sin–1   4

1 (D) sin–1   3

[Ans.B]

(Snell's law)

6 8

Sin θc =

n0 sin θ = sin θ =

1 (B) sin–1   8

n0 6

n0 6 × 6 8

1 8

1 θ = sin–1   8

24.

A vibrating string of certain length l under a tension T resonates with a mode corresponding to the first overtone (third harmonic) of an air column of length 75 cm inside a tube closed at one end. The string also generates 4 beats per second when excited along with a tuning fork of frequency n. Now when 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

Sol.

(B) 336

n–4=ν

(C) 117.3

(D) 109.3

[Ans.A]

ν frequency of string. ν=

3v 4L

L = length of tube v = velocity of sound in air

∴ n=

3v 4L

+ 4 = 344

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25.

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

d initially. Suppose the liquid level decreases at a constant speed 3

V, the time constant as a function of time t is Figure : C d

(A)

6ε 0 R 5d + 3Vt

(B)

d 3

(15d + 9Vt )ε 0 R 2

2 2

2d − 3dVt − 9V t

(C)

R

6ε 0 R 5d − 3Vt

(D)

(15d − 9Vt )ε 0 R

2d 2 + 3dVt − 9V 2 t 2

[Ans.A] Sol. C1

K =1

2 d + vt 3

C2

K =2

d − vt 3

at any time t level of liquid is

d – vt 3

these can be seen as series combination of two capacitor of C1 & C2 1 Ceq

=

Ceq =

2d d + vt − vt 3 + 3 ε0 2ε 0 6 ∈0 3vt + 5d

τ = R × Ceq τ=

26.

6∈0 R 3vt + 5d

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 it reach the point B. The angle θ at which the speed of the bob is half of that at A, satisfies Figure :

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B

θ

L A

(A) θ = Sol.

π 4

(B)

π π <θ< 4 2

(C)

V

π 3π < θ< 2 4

(D)

3π <θ<π 4

[Ans.D]

using energy conservation 1 mv 2 2

=

1 m(v / 2 )2 2

1 mv 2 2

=

11 2  mv  + mg L(1 − cos θ) 42 

+ mgh

3 mv 2 = mg L(1 – cosθ) 8 3 m. 5gL = (1 – cosθ)mg 8 15 8

= 1 – cosθ

Cosθ = –7/8 3π/4 < θ < π 27.

A glass tube of uniform internal radius (r) has a valve separating the two identical ends. Initially, the valve is in a tightly closed position. End 1 has hemispherical soap bubble of radius r. End 2 has subhemispherical soap-bubble as shown in figure. Just after opening the valve, Figure :

2

1

(A) air from end 1 flows towards end 2. No change in the volume of the soap bubbles (B) air from end 1 flows towards end 2. Volume of the soap bubble at end 1 decreases (C) no change occurs (D) air from end 2 flows towards end 1. Volume of the soap bubble at end 1 increases CAREER POINT, 112, Shakti Nagar, Kota (Raj.) Ph.: 0744-2500092, 2500492 Website : www.careerpointgroup.com, Email: [email protected]

19

[Ans.B] Sol.

excess pressure in side soap bubble Pexcess =

4γ r

γ → surface tension r → radius of bubble (Pexcess)1 > (Pexcess)2 ; since r1 < r2 28.

A block (B) is attached to two unstretched spring S1 and S2 with spring constants k and 4k, respectively (see figure I). The other ends are attached to identical supports M1 and M2 not attached to the walls. The springs and supports have negligible mass. There is no friction anywhere. The block B is displaced towards wall 1 by a small distance x (figure II) and released. The block returns and moves 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

y isx

Figure : 2 M2

S2

S1

M1 1 I

S1

M1 1

B 2

M2

x

S2

B

II

x

(A) 4 Sol.

(B) 2

(C)

1 2

(D)

1 4

[Ans.C]

using energy conservation 1 1 k(x2) = 4k(y2) 2 2

(y/x)2 =

1 4

y/x = ½ 29.

A transverse sinusoidal wave moves along a string in the positive x –direction at a speed of 10 cm/s. The wavelength of the wave is 0.5 m and its amplitude is 10 cm. At a particular time t, the snap-shot of the wave is shown in figure. The velocity of point P when its displacement is 5 cm isFigure :

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20

y P x

(A)

3π ˆ j m/s 50

(B) −

3π ˆ j m/s 50

(C)

3π ˆ i m/s 50

(D) −

3π ˆ i m/s [Ans.A] 50

Wax velocity : v = 10 cm/s

Sol.

λ = 0.5 m/s = 50 cm/s A = 10 cm Y = 5 cm vy = ω A 2 − y 2 = 2 3π cm/sec =

30.

3π m/s (in +ve y. direction) 50

Consider a system of three charges

q q 2q , and – placed at points A, B and C, respectively, as shown 3 3 3

in the figure. Take O to be the centre of the circle of radius R and angle CAB = 60°. Figure : y B C O

60°

x

A

(A)The electric field at point O is

q 8πε0 R 2

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 (D) The potential at point O is

q2 54πε0 R 2

q 12πε0 R

[Ans.C]

Sol.

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21

B

−2q 90º C 3 60º

q/3

2R

O

q/3 A

E at o =

=

2q 1 × 3 4π ∈0 R 2

q 6π ∈0 R 2

option (A) is wrong Potential energy of system =

q q 1 q  − 2q 1   − 2q  q 1  1 + × × + × ×  × ×  4π ∈0  3 3 2R 3  3 BC   3  3 AC 

AC = 2R cos 60 = R BC = 2R sin 60 =

=

1 4π ∈0

3R

 q2 2q 2 2q 2  − −   ≠0 12R 9 3R 9R 

Option (B) is wrong Potential at point 0 =

1 q 2q   q + − 4π ∈0  3R 3R 3R 

Option (D) is wrong Option C force =

1 q 4π ∈0 3

=

31.

×

q2 54π∈0 R 2

2q 3 (2R sin 60) 2

Option (C) is correct

A radio active sample S1 having an activity of 5µCi has twice the number of nuclei as another sample S2 which has an activity of 10µCi. The half lives of S1 and S2 can be (A) 20 years and 5 years, respectively (B) 20 years and 10 years, respectively CAREER POINT, 112, Shakti Nagar, Kota (Raj.) Ph.: 0744-2500092, 2500492 Website : www.careerpointgroup.com, Email: [email protected]

22

(C) 10 years each (D) 5 years each Sol.

[Ans.A]

λ1 N1 = A 01 = 5µCi λ 2 N 2 = A 02 = 10µCi

N1 = 2N2 5 10 = 2. λ1 λ2

5[t1/2]1 = 20[t1/2]2 (t1/2)1 = 20 [t1/2]2 = 5

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. 32.

STATEMENT - 1 It is easier to pull a heavy object than to push it on a level ground. and

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 NOT a correct explanation for STATEMENT-1 (C) STATEMENT-1 is True, STATEMENT-2 is False (D) STATEMENT-1 is False, STATEMENT-2 is True.

[Ans.B]

Sol. 33.

STATEMENT-1 For practical purposes the earth is used as a reference at zero potential in electrical circuits. and

STATEMENT-2

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23

The electrical potential of a sphere of radius R with charge Q uniformly distributed on the surface is given by

Q . 4πε0 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 NOT a correct explanation for STATEMENT-1 (C) STATEMENT-1 is True, STATEMENT-2 is False (D) STATEMENT-1 is False, STATEMENT-2 is True. Sol.

As radius of earth is very large, therefore potential on the earth can be assumed zero

34.

STATEMENT-1

[Ans.A]

The sensitivity of a moving coil galvanometer is increased by placing a suitable magnetic material as a core inside the coil. and

STATEMENT-2 Soft iron has a high magnetic permeability and cannot be easily magnetized or demagnetized. (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 NOT a correct explanation for STATEMENT-1 (C) STATEMENT-1 is True, STATEMENT-2 is False (D) STATEMENT-1 is False, STATEMENT-2 is True. Sol.

Z = NiAB

....(1)

Cθ = NiAB

....(2)

[Ans.C]

θ = sensitivity i θ NAB = from (2) i C

Sensitivity proportional to B. As soft iron can be easily magnetised (µr is high) so B will increase when core is of soft iron ∴ using soft iron core sensitivity will increase statement 2 is wrong

35.

STATEMENT-1 For an observer looking out through the window of a fast moving train, the nearby objects appear to move in the opposite direction to the train, while the distant object appear to be stationary.

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and

STATEMENT-2 r r If the observer and the object are moving at velocities V1 and V2 respectively with reference to a r r laboratory frame, the velocity of the object with respect to the observer is V2 − V1 .

(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 NOT a correct explanation for STATEMENT-1 (C) STATEMENT-1 is True, STATEMENT-2 is False (D) STATEMENT-1 is False, STATEMENT-2 is True.

[Ans.B]

SECTION – III

Linked Comprehension Type This section contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be answered. Each question has 4 choices (A), (B), (C) and (D), out of which ONLY ONE is correct.

Paragraph for Question Nos. 36 to 38 A uniform thin cylindrical disk of mass M and radius R is attached to two identical massless springs of spring constant k which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance d from its centre. The axle is massless and both the springs and the axle are in a horizontal plane. The unstretched length of each spring is L. The disk is initially at its equilibrium position r with its centre of mass (CM) at a distance L from the wall. The disk rolls without slipping with velocity V0 = V0ˆi . The coefficient of friction is µ. Figure :

y 2d

d d

V0

R

x

36.

The net external force acting on the disk when its centre of mass is at displacement x with respect to its equilibrium position is-

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25

(A) – kx

Sol.

(C) −

(B) – 2kx

a ← 2kx

2kx 3

(D) −

4kx 3

[Ans.B]

V0 f

fnet = 2kx – f = mac

...(1)

fR = Icα

...(2)

ac = αR

...(3)

from (1) & (2) 2kx –

ICα = mac R

...(4)

from (4) & (3) 2kx – IC aC =

aC =

aC

= mac

R2

2kx IC +m R2 2kx m +m 2

=

4kx 3m

→

Fnet = Net force = ma(– ˆi ) →

Fnet = +

37.

4 4 kx(– ˆi ) = – kx 3 3

The centre of mass of the disk undergoes simple harmonic motion with angular frequency ω equal to(Α)

k M

(B)

2k M

(C)

2k 3M

(D)

4k 3M

[Ans.D] Sol.

Energy of system E=

1 1 1 1 mv2 + ICω2 + K (x)2 + kx2 2 2 2 2

E=

1 1 Ic v 2 mv2 + 2 2 R2

E=

 Ic  1  1 mv2 1 + + 2  Kx 2  2 2 2    mR 

+ 2  Kx 2  1 2



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dE =0 dt

dv  Ic  m dx × 2v + 2 Kx  = 0 1 + 2 2 dt dt    mR  dv dt

2(K / m)

+

1 + I c / mR 2

.x=0

dv = a = – ω2x dt

ω=

2K / m 1 + I c / mR 2

ω=

38.

Ic =

mR 2 2

4K 3m

The maximum value of V0 for which the disk will roll without slipping is(A) µg

M k

(B) µg

M 2k

(C) µg

3M k

(D) µg

5M 2k

[Ans.C]

Sol. ω

V0

f ≤ µN Icα ≤ µN R

µNR Ic

α≤

α≤ ac R

µmgR mR 2 / 2

=α≤

2µg R

4 KA = ac ≤ 2µg 3 m A≤

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27

v ≤ ωA 4K 3m

v≤A

=

4 K 3 µgm . 3m 2 K

Paragraph for Question Nos. 39 to 41 The nuclear charge (Ze) is non-uniformly distributed within a nucleus of radius R. The charge density ρ(r) [charge per unit volume] is dependent only on the radial distance r from the centre of the nucleus as shown in figure. The electric field is only along the radial direction. Figure : ρ(r) d

a

39.

Sol.

R

r

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

[Ans.A]

Total charge inside is Ze E.4πR2 =

Ze ∈0

E is independent of distribution rather is depends on total charge. 40.

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

3Ze 4πR

(B)

3

3Ze

πR

3

(C)

4 Ze 3πR

3

(D)

Ze 3πR 3

[Ans.B] d

Sol.

θ R

r

ρr = (R – r)

d R

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28

Ze = Q =

⇒d=

41.

∫ dQ = ∫ ρ

R

r

.dV = ∫ (R − r ) 0

d 4πr2 dr R

3Ze

πR 3

The electric field within the nucleus is generally observed to be linearly dependent on r. This implies (B) a =

(A) a = 0

R 2

(C) a = R

(D) a =

2R 3

[Ans.C] Sol.

For uniform volumetric charge density E=

ρr i.e. E ∝ r 3 ∈0

SECTION – IV Matrix - Match Type This section contains 3 questions. Each question contains statements given in two columns, which have to be matched. Statements in Column I are labelled as A, B, C and D whereas statements in Column II are labelled as p, q, r and s. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A -q , A -r , B -p , B -s , C -r , C -s and D -q , then the correctly bubbled matrix will look like the following :

42.

q q

r

s

A

p p

r

s

B

p

q

r

s

C

p

q

r

s

D p

q

r

s

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

Column II

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29

S

(A)

(p) Real image

S

(B)

(q) Virtual image

(C) S

(r) Magnified image

S

(D)

(s) Image at infinity

[Ans. A: (p,q,r,s); B:(q); C:(p,q,r,s); D(p,q,r,s)] Sol.

for option (D) R1 R2 n2 n1

n1

R2 > R1

n2 > n1

1 f 1 f

= =

n 2 − n1  1 1  −   n1  R 1 R 2  n 2 − n1  1 1  −   n1  R 1 R 2 

f = +ve ∴ less is converging ∴ Its option is P, Q, R, S

43.

Column I contains a list of processess involving expansion of an ideal gas. Match this with Column II

describing the thermodynamic change during this process. Indicate your answer by darkening the appropriate 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 gas decreases

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

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30

I

II

ideal gas

vacuum

(B) An ideal monoatomic gas expands to twice its original volume such that its pressure P ∝

1 V2

(q) The temperature of the gas increases ,

or remains constant

where V is the volume of the gas (C) An ideal monoatomic gas expands to

(r) The gas loses heat

twice its original volume such that its pressure P∝

1 V4 / 3

, where V is its volume

(D) An ideal monoatomic gas expands such that

(s) The gas gains heat

its pressure P and volume V follows the behaviour shown in the graph P

2V1 V

V1

Sol.

[Ans. A: (q); B:(p,r); C:(p,s); D(q,s)]

[A] work done by gas W = 0

∆Q = 0 ⇒ ∆u = 0 ⇒ ∆T = 0 [B]

2

PV = const. ⇒

T.V = const

⇒

∆T = –ve

C=

R R – γ −1 n −1

∴ ∆V = + ve for PVn = const.

⇒ For PV2 = const. C = +ve ⇒ ∆Q = nC∆T = –ve [C]

PV 4 / 3 = const. ⇒ TV1 / 3 = const. ⇒ ∆T = –ve ∴ ∆v = +ve

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31

C=

R R – γ − 1 n −1

⇒ C = –ve for PV 4 / 3 = const. ⇒ ∆θ = nC∆T = +ve. [D]

P2V2 – P1V1 = +ve ⇒ ∆T = +ve ⇒ ∆U = +ve volume is increasing ⇒ W = +ve

44.

Column I gives a list of possible set of parameters measured in some experiments. The variations of the

parameters in the form of graphs are shown in Column II. Match the set of parameters given in Column I with the graphs given in Column II. Indicate your answer by darkening the appropriate bubbles of the 4 × 4 matrix given in the ORS. Column I

Column II y

(A) Potential energy of a simple pendulum (y-axis)

(p)

O

x

as a function of displacement (x axis) y

(B) Displacement (y axis) as a function of time

(q)

O

x

(x axis) for a one dimensional motion at zero or constant acceleration when the body is moving along the positive x –direction y

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

(r)

O

x

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32

y

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

(s)

O

x

pendulum as a function of its length (x axis) [Ans. A: (p); B:(q,s); C:(s); D(q)] Sol.

θ

x

h

(A) U = mgh + U0 U = mgl (1– cosθ) + U0 U = mgl 2sin2θ/2 + U0 ⇒ U = 2mgl

θ2 + U0; for small θ 4

(Β) S = S0 + ut +

1 2 at 2

(C) R ∝ v2 (D) T2 ∝ l

Part – III (Chemistry) 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.

Solubility product constants (Ksp) of salts of types MX, MX2 and M3X at temeperature 'T' are 4.0 × 10–8, 3.2 × 10–14 and 2.7 × 10–15, respectively. solubilities (mol dm–3) of the salts at tremperature 'T' are in the order (A) MX > MX2 > M3X CAREER POINT, 112, Shakti Nagar, Kota (Raj.) Ph.: 0744-2500092, 2500492 Website : www.careerpointgroup.com, Email: [email protected]

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(B) M3X > MX2 > MX (C) MX2 > M3X > MX (D) MX > M3X > MX2 [Ans. D] Sol.

S1 = solubility of MX S2 = solubility of MX2 S3 = solubility of M3X ∴ S12 = 4 × 10–8 or S1 = 2 × 10–4 (M) 4 S23 = 32 × 10–15 Or S23 = 23 × 10–15 or S2 = 2 × 10–5 (M) 3M+ + X3–

M3X (S)

3S3 27S34

S3

–16

= 27 × 10

S3 = 10–4 (M)

46.

Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milli ampere current. The time required 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 [Ans. B]

0.01 mol of H2 gas ≡ 0.01 × 2 equiv H+

Sol.

∴

Current required = 0.01 × 2 × 96500 C

∴

0.01 × 2 × 96500 = 10–3 × 10 × t t=

47.

0.01× 2 × 96500 0.01

= 19.3 ×104 s

Among the following, the surfactant that will form micelles in aqueous solution at the lowest molar concentration at ambient conditions is (A) CH3(CH2)15N+(CH3)3Br–

(B) CH3(CH2)11OSO 3− Na+

(C) CH3(CH2)6COO–Na+

(D) CH3(CH2)11N+(CH3)3Br– [Ans. A]

48.

Both [Ni(CO)4] and [Ni(CN)4]2– are diamagnetic. The hybridisations of nickel in these complexes, respectively, are (A) sp3, sp3

(B) sp3, dsp2

(C) dsp2, sp3

(D) dsp2, dsp2 [Ans. B]

49.

The IUPAC name of [Ni(NH3)4][NiCl4] is CAREER POINT, 112, Shakti Nagar, Kota (Raj.) Ph.: 0744-2500092, 2500492 Website : www.careerpointgroup.com, Email: [email protected]

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(A) Tetrachloronickel (II) – tetraamminenickel (II) (B) Tetraamminenickel (II) – tetrachloronickel (II) (C) Tetraamminenickel (II) – tetrachloronickelate (II) (D) Tetrachloronickel (II) – tetrachloronickelate (0) [Ans. C] 50.

Among the following, the coloured compound is (B) K3[Cu(CN)4]

(A) CuCl

(C) CuF2

(D) [Cu(CH3CN)4]BF4 [Ans. C]

2+

Sol.

Because Cu has only unpaired electron

51.

In the following reaction sequence, the correct structures of E, F and G are

O O Ph

*

(* implies

Heat

I2 NaOH

[E]

OH 13

C labelled carbon) O

(A) E =

[F] + [G]

O

Ph * CH3

F=

–+ Ph * O Na

O (B) E =

O * CH3

Ph

F= Ph

O (C) E =

F= Ph

O (D) E =

Ph

–+ O Na

G = CHI3 Ph

–+ O Na

* G = CHI3 Ph

–+ O Na

* G = CH3I Ph

O * CH3

Ph

G = CHI3

O * CH3

F= Ph

[Ans. C] Θ

Sol.51

∗ I2 ∆ PhCO O + *CHI Ph C CH 2 COOH → Ph C C H 3 → 3 NaOH || || [F] [G ] O O

[E] 52.

The correct stability order for the following species is –

⊕

⊕ O (I)

(II)

⊕ O (III)

⊕ (IV)

(A) (II) > (IV) > (I) > (III) CAREER POINT, 112, Shakti Nagar, Kota (Raj.) Ph.: 0744-2500092, 2500492 Website : www.careerpointgroup.com, Email: [email protected]

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(B) (I) > (II) > (III) > (IV) (C) (II) > (I) > (IV) > (III) (D) (I) > (III) > (II) > (IV) [Ans.D] 53.

Cellulose upon acetylation with excess acetic anhydride / H2SO4 (catalytic) gives cellulose triacetate whose structure is –

AcO AcO AcO H (A)

H O H OAc H

O H

O

OAc

O O H OAc H H H OAc

H O H OAc H H

O

OAc

H

H

AcO AcO AcO

H

H H OH

(B)

O H

O

O

H OH

H H OH

O

H

OH

H OH

O

H

H

O O H H OH

H

H AcO

AcO

AcO H (C)

H H O H O O O H H H H OAc H OAc H OAc H O O H O H H H OAc OAc OAc AcO

AcO H (D)

H H

H

O H H

O OAc OAc

O

AcO H H

H

O H H

OAc OAc

O

H H

O O H

H OAc OAc [Ans. A ]

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36

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 [Fe(H2O)5NO]SO4 is paramagnetic. and STATEMENT - 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 NOT a correct explanation for STATEMENT -1 (C) STATEMENT -1 is True, STATEMENT -2 is False (D) STATEMENT -1 is False, STATEMENT -2 is True. [Ans. A] Sol.

6

2

Fe = 3d 4s

Fe+2 = 3d6 : NO donates one electron to Fe2+, thereby number of unpaired electrons remained in Fe is three.

55.

STATEMENT-1 The geometrical isomers of the complex [M(NH3)4Cl2] are optically inactive. and

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 NOT a correct explanation for STATEMENT -1 (C) STATEMENT -1 is True, STATEMENT -2 is False (D) STATEMENT -1 is False, STATEMENT -2 is True. [Ans. B] Sol. CAREER POINT, 112, Shakti Nagar, Kota (Raj.) Ph.: 0744-2500092, 2500492 Website : www.careerpointgroup.com, Email: [email protected]

37

Cl NH3

NH3

Cl

NH3 Cl

NH3 M

M NH3

Cl

NH3

NH3 NH3

Trans isomer has the axis of Cis isomer have axis of symmetry Inactive symmetry Inactive

56.

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

STATEMENT-2 No process is possible in which the sole result is the absorption of heat from a reservoir and 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 NOT a correct explanation for STATEMENT -1 (C) STATEMENT -1 is True, STATEMENT -2 is False (D) STATEMENT -1 is False, STATEMENT -2 is True. [Ans. A]

57.

STATEMENT-1 Aniline on reaction with NaNO2/HCl at 0ºC followed by coupling with β-naphthol gives a dark blue coloured precipitate.

and STATEMENT-2 The colour of the compound formed in the reaction of aniline with NaNO2 / HCl at 0ºC followed by coupling with β-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 NOT a correct explanation for STATEMENT -1 (C) STATEMENT -1 is True, STATEMENT -2 is False (D) STATEMENT -1 is False, STATEMENT -2 is True. CAREER POINT, 112, Shakti Nagar, Kota (Raj.) Ph.: 0744-2500092, 2500492 Website : www.careerpointgroup.com, Email: [email protected]

38

[Ans. D] OH

Sol.

N = N – Ph OH

NaNO HCl

2→ Ph NH2  

SECTION – III Linked Comprehension Type This section contains 2 paragraphs. Based upon each paragraph, 3 multiple choice questions have to be answered. Each question has 4 choices (A ), (B ), (C) and (D ), out of which ONLY ONE is correct. Paragraph for Question Nos. 58 to 60

In hexagonal systems of crystals, a frequently encountred arrangement of atoms is described as a hexagonal prism. Here, the top and bottom of the cell are regular hexagons and three atoms are sandwiched in between them. A space-filling model of this structure, called hexagonal close-packed (HCP), is constituted of a sphere on a flat surface surrounded in the same plane by six identical spheres as closly as possible. Three spheres are then placed over the first layer so that they touch each other and represent the second layer. Each one of these three spheres touches three spheres of the bottom layer. Finally, the second layer is covered with a third layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be 'r'.. 58.

The number of atoms in this HCP unit cell is (A) 4

(B) 6

(C) 12

(D) 17 [Ans. B]

59.

The volume of this HCP unit cell is (A) 24 2 r

3

(B) 16 2 r

3

(C ) 12 2 r

3

(D)

64 3 3

r3 [Ans. A]

Sol.

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39

h/2

3 × (2r)2 4 3 × 4r2 = 6 3 r 2 =6× 4

Area of the base = 6 ×

h = 4r

2/3

vol. of the HCP unit cell = 6 3 r 2 × 4r

60.

2 3

= 24 2 r3

The empty space in this HCP unit cell is (A) 74%

(B) 47.6%

(C)32%

(D) 26% [Ans. D]

Paragraph for Question Nos. 61 to 63 A tertiary alcohol H upon acid catalysed dehydration gives a product I. Ozonolysis of I leads to compounds J and K. Compound J upon reaction with KOH gives benzyl alcohol and a compound L, whereas K on reaction with KOH gives only M.

M=

H3C

O Ph H

Ph

61.

Compound H is formed by the reaction of O

O

(A) Ph

CH3

+

PhMgBr

(B)

O

(C) Ph

CH3

Ph

+

PhCH2MgBr

O H

+

PhCH2MgBr

(D) Ph

Me H

+ Ph

MgBr

[Ans. B]

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40

62.

The structure of compound I is CH3

Ph

(A) H

(B)

Ph

Ph

H

Ph

CH3

H

CH2Ph

(C )

Ph

H3C

CH3

H3C

(D) H

Ph

[Ans. A] 63.

The structures of compounds 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+ [Ans. D] PhCH2OH + PhCOO [L]

Ph

Sol.61 to 63

H+

Ph CH2 C Me OH

KOH

Ph

 → Ph CH=C – H 2O

Me

O 3 → PhCHO

[I]

+

PhCOCH3

[K]

[J]

KOH

[H]

CH3

COPh C=C

Ph

H [M]

SECTION – IV Matrix Match Type This section contains 3 questions. Each question contains statements given in two columns which have to be matched. Statements in Column-I are labelled as A, B, C and D whereas statements in Column-II are labelled as p, q, r and s. The answers to these questions have to be appropriately bubbled as illustrated in the following example. If the correct matches are A-p , A-s , B-q, B-r , C-p , C-q and D-s, then the correctly bubbled matrix will look like the following: CAREER POINT, 112, Shakti Nagar, Kota (Raj.) Ph.: 0744-2500092, 2500492 Website : www.careerpointgroup.com, Email: [email protected]

41

64.

q q

r

s

A

p p

r

s

B

p

q

r

s

C

p

q

r

s

D p

q

r

s

Match the conversion in Column-I with the type(s) of rection(s) given in Column-II. Indicate your answer 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 [Ans. A : (p); B : (p,q); C: (p,r); D: (p,r,s)]

65.

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

Column-I

Column-II

(A ) Orbital angular momentum of the electron

(p) Principal quantum number

in a hydrogen-like atomic orbital (B ) A hydrogen-like one-electron wave function

(q) Azimuthal quantum number

obeying Pauli principle (C ) Shape, size and orientation of hydrogen like

(r) Magentic quantum number

atomic orbitals (D ) Probability density of electron at the nucleus

(s) Electron spin quantun number

in hydrogen-like atom [Ans. A : (q); B : (s); C: (p,q,r); D: (p,q)] 66.

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

Column II

⊕ Θ (A) H2N – NH3Cl

(p) Sodium fusion extract of the compound gives Prussian blue colour with FeSO4

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42

(B) HO

(C) HO

(D) O2N

⊕ Θ NH3I COOH

⊕ Θ NH3Cl

⊕ Θ NH-NH3Br

(q) Gives positive FeCl3 test

(r ) Gives white precipitate with AgNO3

(s) Reacts with aldehydes to form the

NO2

corresponding hydrazone derivative [Ans. A : (r,s); B : (p,q); C: (p,q,r); D: (p,s) ]

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