Answer (5)

JEE(Advanced)-2013

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ANSWERS, HINTS & SOLUTIONS CRT(Set-V) (Paper–1)

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1 AITS-CRT(Set-V)-Paper-1-PCM(S)-JEE(Advance)/13

ANSWERS KEY Q. No.

PHYSICS

CHEMISTRY

MATHEMATICS

ANSWER

ANSWER

ANSWER

1.

B

B

C

2.

D

B

A

3.

C

C

B

4.

C

B

A

5.

D

A

D

6.

C

A

D

7.

C

A

C

8.

D

D

C

9.

C

C

A

10.

D

C

C

11.

B, D

A, B, C

B, C

12.

A, B, D

A, B

A, B

13.

A, C, D

B, C, D

A, B

14.

B, C

A, B, C

A, B, C

15.

B, C, D

A, B, C, D

A, B, C

1.

2

2

2

2.

9

5

2

3.

6

3

6

4.

7

7

5

5.

2

5

2

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AITS-CRT(Set-V)-Paper-1-PCM(S)-JEE(Advance)/13

Physics 1.

R=

2

PART – I

u22 ⇒ u22 = Rg (1 + sin α ) .......... ( ii ) g (1 + sin α )

2R cos α =

u32 ⇒ u32 = 2Rgcos α g

........ ( iii )

Now, u1u2 = Rg 1 − sin2 α = Rgcos α =

u32 2

∴ u32 = 2u1u2

2.

a

a y1 − y 2 = g 4

tan θ =

∴ y1 − y 2 = 0.5 ........ ( i )

y1

Volume of water is 4 ( y1 + y 2 ) × 3 = 18 2 ∴ y1 + y 2 = 3 ........ ( ii )

θ y2 4m

2.5 2

∴ y2 =

∴ speed of liquid coming from orifice is = at + 2gy 2 = 10m / s 3.

By first law of thermodynamics Q = W + ∆U ⇒ Q/2 = ∆U n C T2 − T1 3R ⇒ = n T2 − T1 2 2

(

)

(

)

∴ C = 3R 4.

R = equivalent resistance across AB R    ar. 2  + ar  ar  R  ar +   2  =R R Now, ar. 2 + ar + ar R ar + 2 ⇒ 3R 2 + 2arR − 2a 2r 2 = 0

∴R =

(

2ar

C ar/2

ar/2

A

ar/2

R/2 ar

ar/2

B

)

7 +1


3 AITS-CRT(Set-V)-Paper-1-PCM(S)-JEE(Advance)/13 5.

For equilibrium torque about y-axis 2a πa2 mg − I B 2 π π 2 aBI ∴m = 4g

6.

Phase difference due to glass slab is ( µ − 1) t 0.5 × 0.2 × 10−6 π 2π = 2π = φ= −9 3 λ 600 × 10  3 φ  Since, I = Io cos   ⇒ I = Io   2  2   4I ∴ Io = 3

2

2

7.

By conservation of energy at A & B 2

1 1  3A  3A 3 mv 2 = K   − mg = mgA ⇒ ν = 2 2  2 2 4

3gA 2

ν natural length L

B A /2

mean position

∴ The period of oscillation is 2ν 2 6A 4π A = + T= + g 3 g 3 2g

A /2 A /2 A

8.

Since,

∴N =

9.

dN = K − λN .......... ( i ) dt N t dN ⇒ ∫ = ∫ dt K − λN o No

K

λ

(1 − e ) + N e −λt

−λt

o

Here

λ

= 42 + 0.3 × 10 ⇒ λ = 36cm 4 ∴pressure amplitude at middle of pipe is  2π  ∆P = ∆Po sinKx = ∆Po sin  × 24   36  5

=

3∆Po 2



10.

11.

4

ε = B νA = 11 V

11 11 − 11 Eeq = 1 1.5 = V 1 1 5 + 1 1.5 1× 1.5 req = = 0.6 1 + 1.5 11 Eeq 5 ∴I = = = 2A r + req 0.5 + 0.6

1.5Ω M

AI r = 0.5 Ω

N

1.0Ω

ε = 11 V

T cosθ = mgeff ..... ( i ) T sinθ =

mν 2 ........... ( ii) A sinθ

By (i) mgeff cos θ T= = cos2 θ

mg g2  2  qE 2  g +     m   

By (i) & (ii) mgeff mν 2 sinθ = ⇒ν = cosθ A sinθ

  qE 2  = mg 1 +      mg  

( geff sinθ ) A tanθ

=

qE A m g

2

∴ K.E =

12.

1  qE  A q2E2 A m = 2  m  g 2mg

by first low of thermodynamics Q = W + ∆U ⇒ 2Q = ∆U ……………. (i) 5R TB − TA ⇒ 2 n C TB − TA = n 2 5R ∴C = 4 By (i) 1 1 5  5 Q = ∆U =  ( 6Po Vo − 4Po Vo )  = Po Vo 2 2 2  2 since∆U = -2W, therefore temperature increases from A to B

(

)

(

)



C

using Kirchhoff’s loop law di −3i − 1 − 3i + 36 = 0 dt di ∴ = 36 − 6i .............. ( i ) dt −3i − 6 ( i − i1 ) + 36 = 0

R1 = 3 Ω

A

i - i1

i1

R2 = 6 Ω

i B

1H

i - i1

⇒ 3i − 2i1 = 12 ........ ( ii )

i

R3 = 6 Ω

di1 = 12 − 4i1 by (i) and (ii) dt on solving i1 = 3 (1 − e −4t )

R4 = 3 Ω

D

36 V

i = 2 (1 − e ) + 4 from (ii) ∴ power supply by battery = 36(2i – i1) = 36(9 – e-4t) (i − i1 )t =∞ = 3A −4t

14. 15.

IY =

1 3 mR2 & IX = IZ = mR2 2 4

Given x = 3t 3 − 18t 2 + 36t ........ ( i ) ∴ν = 9 ( t − 2 ) ................ ( ii ) 2

a = 18(t – 2) ……………..(iii)

SECTION –C 1.

u

speed of ball B after elastic collision u = ucos 60o = 2

60

A 2.

3.

o

B

If angle of incident at A is less than 90o, the ray will also reflected at B in air. For T.I.R, i = 90o at A. 1 2 ∝ ( Z − 1) for Kα line since, λ 1 2 (11 − 1) 10 2 λ ⇒ = ⇒ 4 = 2 2 1 ( Z − 1) ( Z − 1) 4λ ∴Z=6



4.

Given that m1 – m2 = 6 unit …….(i) equation of motion Gm1m2 m2r = m1ω 2r1 = m1ω 2 m1 + m2 r2

∴ m1 + m2 =

ω 2r 3

G by (i) & (ii) m1 =7 m2

5.

6

r r1 m1

ω c

r2

m2

= 8 unit ...... ( ii)

GM r [ for r ≤ R ] R3 Now A 1 T = 2π ⇒T∝ g r g=

∴

T2 = T1

R =2 R 4



Chemistry

PART – II SECTION – A

1.

V4+ has 1 unpaired e–, Cu2+ has 1unpaired e–.

4.

LiHCO3 exist only in solution, rest all alkali metal bicarbonates exist in solid state. Bicarbonates of alkaline earth metal exist only in solution. Reactivity of alkaline earth metal for N2 decreases down the group.

5.

C=C of enol is more reactive towards bromine than aromatic ring. _

_

+

O

O

AlCl 3

R

R

R H

7.

O

AlCl 3

Br

Br Br

4ZnCl2 + 4Na 2 CO3 + 3H2 O → ZnCO 3 .3Zn ( OH )2 + 8NaCl + 3CO 2 Basic zinc carbonate

ZnCl2 + H2 S  → ZnS + 2HCl

9.

In PH3 the lone pair of phosphorous remains almost in an orbital having more s-character.

10.

In case of (C) the benzyl cation formed which is stable in comparison to the phenyl, vinyl and bridged carbocation.

12.

Conductance of H+ > Na+

13. A=

OH

B=

I

C=

O

14. R

C

CH2R +

R2 C

CH2 + N2 O

SECTION – C

1.

Be2C and Al4C3 are methanides.

2. Fe

(

 η5 − C5H5 

)

2

Fe  



3.

8

O H (A)

(B)

O

O

Br

(C)

O

(D)

4.

3HgS + 2HNO3 + 6HCl  → 3HgCl2 + 3S + 2NO + 4H2O

5.

RCN RNHCHO N2

O OH O (E)

O O (F)

O

O (G) O



Mathematics

PART – III SECTION – A

1.

Circle has (2, −1) as its centre and radius of this circle is 2. Thus if P(x, y) be any point on it, then x ∈ [0, 4]. x3 x Let g(x) = + 100 50 3x 2 1 + > 0 ∀x ∈ (0,4] 100 50 The g(x) is increasing in [0, 4] 18 g(0) = 0, g(4) = 25  18  g(x) = 0,  ∀x ∈ [0,4]  25  ⇒ [g(x)] = 0 ∀ x ∈ [0, 4]  x3 x  y= +  , represents the x-axis. 100 50  CA = CB = 2, CD = 1 CD 1 π cos θ = = θ= CA 2 3 2π 1 2π ∠ACB = , ∆ACB = × 22 × sin = 3 3 2 3 1 2π 4π = sq. units Area of sector ACB = × 22 × 2 3 3 g'(x) =

2.

A, AR, AR2 ARE 3 SIDES OF ∆ABC. Sum of any 2 sides of a +le ABC> third side of the

+le ABC

5 +1 2 π sec x = 2

⇒r < 3.

 π  The equation has only one solution in  − ,0  and no solution in  2  π Because for 0 < x < 4 1 < sec x < 2

 π  0, 4  .  

2

4.

3  f ( x ) = 0 if  tan−1 2x  > 1 π  π ⇒ tan−1 2x > or tan−1 2x < −π / 3 3

⇒ 2x > 3 or 2x < − 3 ⇒ 2x > 3



5.

f is continuous at x = 0 ∴ 0 = lim f(x) = lim x a An x forces " a > 0 " is necessary. x →0 +

6.

10

x →0 +

2

Let t = sin x; t ∈ [0,1]

f(x) = g(t) = te−2t 1  > 0 if t ∈ [0, )   2 g'(t) = (1 − 2t) e−2t  1 < 0 if t ∈ ( ,1]  2  1 1 max f = max g = g   =  2  2e

min f = ming = min {g(0),g(1)} = 0

max f − min f =

8.

1 . 2e

f ' ( x ) = 2 ( log2x ) − ( log x ) 2

2

f ′ ( x ) foes not exist for all x in (–1,2) ∴ f ′ ( x ) does not exist for all x in (–1, 2) Let g ( x ) =

2x

∫ x

sin x dx , x

 sin2x  sin x ⇒ g′ ( x ) = 2  − x  2x  ∴ g′ ( x ) does not exist at x = 0 and so is not differentiable in (–1, 2). x

 1 − t + t2 let h ( x ) =   1 + t + t2 0

∫

then, h′ ( x ) =

 dt 

1 − x + x2

9.

, which is defined for all x in (–1, 2) as 1 + x + x 2 ≠ 0 . 1 + x + x2 G G G G G G G G G G G G G2 G2 G2 [a b c] = [b × c c × a a × b] = [a b c]2 = a = b = c

10.

Re ( w ) =

11.

1  z − 1 z − 1 +  =0 2  z + 1 z + 1

5 −1 4 3 +1 sin750 = 2 2 sin180 =

sin540 = sin720 =

So, a =

5 +1 4 10 + 2 5 4 5 −1 5 +1 or 4 4



sin 470 + sin610 − sin110 − sin250 = 2sin540 cos70 − 2 sin180 cos70 = 2cos70 sin540 − sin180   5 + 1   5 − 1   = 2cos70   −      4   4   = cos70 ∴ λ = cos70 Clearly, λ = cos70 > cos150

Also, λ > cos360 and λ > cos1 (as 1 is in radian) [ x ]+{x}

13.

I=

∫ 0

1   {x} − 2 dx =  

1

1  = [ x ]  {x} − dx + 2  0

∫

[ x]

{x}

1

∫ 0

∫ 0

[ x ]+{x} 1 1   − + x dx { }    {x} − 2 dx 2     [ x]

∫

1

1 1    {x} − 2 dx = [ x ]  x − 2 dx +     0

∫

{x}

 x2 x   x2 x  = [ x]  − + −  2 2   2 2   0  0 14.

=0+

{x} ({x} − 1) {x} 2

=

2

{x}



1

∫  x − 2 dx 0

({x} − 1)

 1  1 Replace x by 2, 2f ( 2 ) + 2f   − 2f (1) = 4 ⇒ f ( 2 ) + f   = 2 + f (1) − − − − − (1) 2   2 Replace x by 1, f (1) = −1 − − − − − (2) 5 1  1 1 , 2f   + f(2) + 2 = − − − − − (3) 2 2 2 2  1 Solve (1) and (3) ⇒ f   = 0;f(2) = 1 2

Replace x by

15.

(A) according to the given condition a1 = a2 = a3 cos α = ±

1 3

a1 = ±

1 3

⇒ tan α = ± 2

(B) | a − b |2 = 1 or 49 (C) | b − c |2 = b2 + c 2 − 2.b.c = 1 or 81 (D) | a + b + c |2 = 50 + 0 ⇒| a + b + c |= 5 2



12

SECTION – C

2.

There are

10

C6 ways to pick = 210. There are

( C )( 2

1

7

)

C4 = 70 favorable ways. So the

70 1 = 210 3 The slope of line is −a b . Assume that a > 0 and b < 0 so that a ≠ b . Also a ≠ c ≠ b . If c = 0 there are 3 ways to choose a and 3 ways to choose b. Note that line x − y = 0,2x − 2y = 0 and3x − 3y = 0 are the same. Then the number of lines = 32 − 2 = 7 . If c ≠ 0 , there are 3 ways to choose a, 3 ways to choose b and 4 ways to choose c. Then the

probability is 3.

number of lines in this case = 32 x 4 = 36. The total number of lines = 7 + 36 = 43 5.

P = infinite G.P Where a = 1, r = –tan2x a 1 1 ∴ p= = = cos2 x, q = = sin2 y 2 1 − r 1 + tan x 1 + cot 2 y ∴ S=

S=

1 2

2

1 − tan x cot y

=

1  1 − cos x  1 − sin2 y  1−     cos2 x   sin2 y  2

pq 1 = p + q −1 1 1 1   + −   p q pq 


Answer (5)

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