Answer (6)

JEE(Advanced)-2013

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ANSWERS, HINTS & SOLUTIONS CRT (Set-III) (Paper–2)

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1 AITS-CRT(Set-III)-Paper-2-PCM(S)-JEE(Advance)/13

Q. No.

PHYSICS

CHEMISTRY

MATHEMATICS

ANSWER

ANSWER

ANSWER

1.

D

C

B

2.

B

A

C

3.

C

A

C

4.

C

D

C

5.

C

A

B

6.

D

B

B

7.

A

B

D

8.

A

A

B

9.

B, C

A, B, C, D

A, C

10.

A, C

A, B, D

A, B, C

11.

A, B

A, B, C

A, C

12.

A, C

B, C

B, C, D

1.

(A) → (p, q, t) (B) → (p, s) (C) → (p, r, t) (D) → (p) (A) → (q) (B) → (p) (C) → (s) (D) → (r) 3

(A) → (p, s) (B) → (p, r, s) (C) → (p, q r, s) (D) → (p, q, t) (A) → (p, r) (B) → (q) (C) → (p, r) (D) → (q, s) 4

(A) → (s) (B) → (r) (C) → (q) (D) → (r) (A) → (p) (B) → (p) (C) → (p) (D) → (q) 5

2.

2

1

4

3.

1

3

3

4.

2

2

3

5.

3

6

1

6.

4

5

7

1.

2.

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AITS-CRT(Set-III)-Paper-2-PCM(S)-JEE(Advance)/13

Physics 1.

2

PART – I

The z-component of the force and the x-component of displacement are ineffective here. dW = Fy dy = 3xy.dy (∵ z = 0 )

(

= 6x 4 dx ∵ y = x 2

)

Integrating between x = 0 and x = 2 gives the result. 2.

3.

λ = 6.77 × 10−2 2 ∴ v = 1000 × 2 × 6.77 × 10-2 m/s γRT MV 2 ⇒ γ= = 1.4 also v = M RT ⇒ diatomic O

τ = mg sin θ× L = Iα

( mgL ) θ =

4mL2 α 3

θ mgsinθ

θ T = 2π α 4.

The speed is due to radial motion as well as due to angular motion.

5.

Young’s modulus, Y =

F.l A × ∆l

F is applied force, A is area of cross-section, l is length and ∆l is change in length. From graph, change in length for 20 N force is 1 × 10–4 m ∴

Y=

20 × 1 A(10 )(1 × 10−4 ) −6

= 2 × 1011 N m–2

6.

7. 8.

9.

1 1 1 − = v u f f m= f +u

u 4 1 −1 = = −1 v f x1

Both a and b will be independent of material.

1  M = (I area) =  2 × π ×  × 10 = 5π 4   1 1 W = MB  1 −  = 5π × 4   = 10π J  2 2  m Here F > µsmg  1 +   M

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3 AITS-CRT(Set-III)-Paper-2-PCM(S)-JEE(Advance)/13 For m F − µk mg = m.a For M µk mg = MA A = 0.4m / s2

10.

Impulse = change in momentum = 2(v 2 − v1 ) = 2(3iˆ + ˆj)

2

As impulse is in the normal direction of colliding surface 1 tan θ = 3  1 θ = tan−1   3  1 α = 90° + tan−1   . 3 11.

H= ∴

θ 6 α

KAdθ dθ = K(2πrL) dr dr r2

θ

dr 2πKL 2 ∫ r = H ∫ dθ θ r 1

1

2πKL(θ2 − θ1 ) dQ =H= = 80π dt  r2  ln    r1  dm ⇒ L = 80π dt dm 8π 80π π Kg/second. ⇒ = = = dt L 80 × 4200 4200

12.

4 πρGr 3 where ρ is mass density and r is separation between centre of sphere and centre of cavity. Applying law of Conservation of energy : 1 GMm GMm mVesc 2 − + =0 2 R 45R

Gravitational field inside the cavity is E =

Solving Vesc =

88GM . 45R

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AITS-CRT(Set-III)-Paper-2-PCM(S)-JEE(Advance)/13

4

SECTION – B 2.

For central maxima, path diff (∆x) = 0 for any point P on the screen. ∆x = µ m (S 2P ) − [µ m (S1P − x ) + µ x ] masses x = thickness of glass slab. = µ m [S 2P − S1P] − (µ m − µ ) x

µ = 20 – 4t S1

y

µ

 y = µ m  d.  − (µ m − µ ) x = 0  D

Here, Dx  20 − 4t − 5  Dµ −µ  x = y =  m d  20 − 4t  d  µm 

=

Dx d

S2

 15 − 4t   20 − 4t  …..(i)  

D

At time, t = 0 15 1 15 3 15 Dx 15 y= = × 0 .2 × = = m= cm = 7.5cm . × d 20 2 20 200 40 2 R.I. of medium cannot be less than 1 which become 19 = 4.755. Here after this time R.I. of medium will not change. So position of central At time t = 4 maxima at time t = 5 s will be same as at time t = 4.75 s. ∴y=

1 Dx  − 4  = × 0.2 × −4 = - 0.4 m . 2 d  1 

|y| =40 cm. For speed of central maxima, differentiating equation (i), w.r.t time we get   dy Dx  −20  = dt d  ( 20 − 4t )2   

Central maxima will be at the centre of geometrical centre of screen when R.I. of medium is 5. 15 . Hence at this t = 4 ∴

dy  Dx  20  1 20 2 = = m / s = 8 cm/s.  −  = x 0 .2 x − 15 dt  t = d  25  2 25 25 4

Fringe width β =

Dλ 1 100 x10 −10 x = =10-6 m = 1µm. −3 dµ 5 2x10

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5 AITS-CRT(Set-III)-Paper-2-PCM(S)-JEE(Advance)/13 SECTION – C

1.

y=

x3 x 2 − 30 10

N

dy x2 x dy = − and at x = 2; =0 dx 10 5 dx d2 y dx2

=

d2y dx 2

Mg

x 1 d2 y − andat x = 2; 2 = +ve 5 5 dx

Approximate graph

Hence the particle is at it’s lowest (minimum) position, in it’s path at x = 2 m d2 y dx dx = 2x −2 2 dt dt dt  d2 y  = 20m / s2  2  dt  at x =2 ⇒ N − mg = m

d2 y dt 2

P = f.v = ( µN) V =

⇒ N = 1(10 + 20 ) = 30N 1 × 30 × 10 = 3 watt 100

2.

Using conservation of charge principle and induction, do charge distribution on each plate.

3.

N = mg µ N = m r ω2 µ mg = m. 2 sin θ ω2 µg ω= 2 sin θ

N mrω

µN

2

mg

0.1× 10 2 × sin30 ω = 1 rad / s. ω=

4.

z

m = I2S S = BA × AD BA = 2diˆ − 2ajˆ

D

AD = 2bkˆ ˆ × 2bkˆ S = 2(diˆ − aj)

x

A

C

B

ˆ M = −4bI(djˆ + ai) | M |= 4bI d2 + a2 = 2 Tesla. 5.

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y

AITS-CRT(Set-III)-Paper-2-PCM(S)-JEE(Advance)/13

6

We get, K2 = 1 .44 MeV. K1 = 3 MeV. 6.

x = A cos ωt dx v= = − Aω sin ωt dt dx v= = − Aω sin ωt dt a=

d2 x dt 2

= − Aω2 cos ωt

amax = Aω2 τmax = I αmax a = I max R 1 Aω2 = MR2 2 R 1 = MR Aω2 2 τmax = 4N.m

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7 AITS-CRT(Set-III)-Paper-2-PCM(S)-JEE(Advance)/13

Chemistry

PART – II SECTION – A

3.

H+  HCO−  3    = 10−4 Ka =    1 [H2CO3 ] H+  CO2−     3  = 10−7 Ka =    2 HCO−  3  

CO2−  3  1 2−  f CO3 = = 2 − − H CO  + HCO  + CO  3   3  [H2CO3 ]  2 3   + 1+ CO2−  3  

(

=

)

1 10 −10 10 −11

5.

+

10 −5 10 −7

HCO−  3   CO2−  3  

=

1 2

H+  H +    +   + 1 Ka Ka Ka 1 2 2

= 0.01 +1

This diasaccharide is sucrose which is formed by 6

CH2OH H 4

5

OH

O H

6

H

CH2OH

1 & OH

OH 3

H

2

OH

5 HO

H 4

OH

O

OH 2

OH 3

CH2OH 1

H

α-D-(+)-Glucopyranose. 6.

For SHE, E0=0 at every temperature.

7.

No of molecules of K2S in 10 L water = 30.11×4×1022 No. of moles of K2S = 2 mol Molality = 2×10−1 m ∆Tf = iKfm = 1.116oC Tf = 272.034 K

8.

MO2 + 4H+ + PO43− + e− → MPO4 + 2H2O M + PO43− → MPO4 + 3e− The net cell reaction will be 3MO2 + 12H+ +4PO43− + M → 4MPO4 + 6H2O

11.

T → constant PV = nRT (where R and T → constant) n P= V n PI = V

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AITS-CRT(Set-III)-Paper-2-PCM(S)-JEE(Advance)/13

8

n 2V n PIII = V 2n PIV = 2V ∴ PII < PIV = P1 = PII ⇒ Average K.E. depends only on absolute T. ∴ (Av. KE)I = (Av. KE)II =(Av. KE)III=(Av. KE)IV Mass ⇒ Density = Volume ∴ dII < dI < dIV = dIII PII =

12.

E = −13.6

Z2 n2

For Li Z=3 9 1 9 Energy of 2s for Li E2 = −13.6 × 4 ∴ E1 < E2 So statement (A) is correct. Shape of all the d-orbital are not same. So statement (B) is wrong. Energy of 3d > energy of 4s So (C) is wrong statement. SECTION – B

Energy of 1s for Li E1 = −13.6 ×

1.

O

(A)

Gives positive test with 2, 4-dinitrophenyl hydrazine and form highly stable hydrate.

O O O

(B) CH OH

O

(C)

CI3 (D)

C

C

Gives positive test with 2, 4-dinitrophenyl hydrazine (>C=O group test) and Fehling solution ( C OH group test). It forms highly stable hydrate also. Gives positive test with 2, 4-dinitrophenylhydrazine, O

C

H

H

Tollen’s reagent and Fehling’s solution (due to group). It forms stable hydrate also.

O

Gives positive test with 2,4-dinitrophenylhydrazine and Tollen’s reagent (due to –CHO group). It gives Perkin’s reaction (being an aromatic aldehyde)

H

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9 AITS-CRT(Set-III)-Paper-2-PCM(S)-JEE(Advance)/13 2.

Use relations: 1 pH = ( pKw + pKa + logC ) 2  Salt  pOH = pKb + log    Base  1 pH = ( pKw + pKa − pKb ) 2 pH + pOH = 14 SECTION – C

1.

 → CO(g) + 2H2(g) CH3OH(g) ←  Initially At eq.

rH

2

rmix

=

0.2 mole (0.2−x)

0 x

0 2x

Mmix =2 2 2

Mmix = 16 Mmix =

32(0.2 − X) + X(28) + 2X(2) 0.2 − X + X + 2X

X = 0.1 x(2x)2 0.1× 0.04 = = 0.04 0.2 − x 0.1 100 Kc = 4

Kc =

2.

It is a pyrosilicate. Pyrosilicate ion is Si2 O7−6 O O

O

Si

O

Si

O O O In this two units of Si2 O7−6 joined along a corner containing oxygen atom. 3.

CH3

2 H3C

O ∆ COOH  →

HO

CH3

O CH3

O

O H

and OH

H

H

O CH

3

O

O cis

trans

Cis form is optically active. It gives two stereoisomers → d−cis, l−cis. Trans form is optically inactive because it contains center of symmetry. 4. +

H /H2O CH2=C=CH2  → CH

3

O ||

− C− CH3

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AITS-CRT(Set-III)-Paper-2-PCM(S)-JEE(Advance)/13

5. CO

6.

CO

CO

CO

Fe

CO

Fe

CO

CO

CO

10

CO

Hg2Cl2 → Hg + HgCl2 ∆H = ? 2Hg + Cl2 → Hg2Cl2 ∆Hf1 = −125 KJ / mole Hg + Cl2 → HgCl2 ∆Hf2= −640 KJ / mole ∆H = ∆Hf2 − ∆Hf1 = −640 + 125 = −515 KJ/ mol −103 x = − 515 ⇒x=5

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11 AITS-CRT(Set-III)-Paper-2-PCM(S)-JEE(Advance)/13

Mathematics

PART – III SECTION – A

e

1.

e

I = ∫ f ′′ ( x ) ln x dx = ln x.f ( x ) − ∫ e

1

1

I

II

f′(x)

1

x

dx

I = I − I1 e 1 1 1 I1 = ∫ f ′ ( x ) dx = − e 2 x 1 ∴ I = 1−

1 1 3 1 + = − e 2 2 e

2.

4x 2a −p x a p x y z a b c det (B ) = 4t 2b −q = ( 4 )( 2 )( −1) y b q = −8 a b c = −8 p q r = −8 × 2 = −16. 4z 2c −r z c r p q r x y z

3.

Replace x → f (x) =

1 and solve to get x

x +1 2

f (10099 ) =

10099 + 1 10100 = = 5050 2 2

4.

cubic is x3 − 6x2 + 11x − 30 = 0 (x − 5) (x2 − x + 6) = 0 bc 6 Hence a = 5 ; bc = 6 ⇒ = a 5

5.

r=7 R−r 1 = R+r 2 2R − 2r = R + r R = 3r = 21.

sin30° =

6.

0 0 1 1 1 c2 A= 0 c/b 1 = 0− 2 2 ab c/a 0 1 If A is independent of a, b and c then c2 = ab ⇒ a, c, b are G.P.

7.

Cn =

1 n

1

∫n

tan−1 ( t ) sin−1 ( t )

dt

n +1

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AITS-CRT(Set-III)-Paper-2-PCM(S)-JEE(Advance)/13

12

1

L = lim n2 ⋅ Cn = lim n ⋅ a →∞

a →∞

1

∫n

tan−1 t dt ( ∞ × 0 ) ; L = sin−1 t

∫n

tan−1 t dt sin−1 t

n +1

n +1

1 n

applying Leibnitz rule. 8.

10.

x 2f(x) and f   have the same domain as f(x) and f(2x), f(x + 2), 2 ⇒ m = 2, n = 3  π π Verify by considering f(x) = sin−1x ; d : |x| ≤ 1 ; R  − ,   2 2

f f   have the range a f(x). 2

 [ 2x ] if x > 0  (A) f(x) = lim ⇒ lim f(x) = 0 ; f(x) =  x x →0 x →0 x  0 if x < 0  x (0 − h) x xe1/ x = −0 ; lim− f(x) = (B) lim ; lim+ f(x) = =0 − 1/ x 1/ x x →0 1 + e x →0 x →0 1 1+ e

[x + x ]

1/ 5

(C) lim ( x − 3 ) x →3

Sgn (x – 3) = 0 where sgn is the signum function (bvious)

tan−1 x does not exist as RHL = 1; LHL = –1 x →0 x

(D) lim

11.

Area (T) =

c.c 2 c 3 = 2 2

Area (R) =

c3 c3 c3 c3 − = − x 2 dx = 2 3 6 2

e

∫ 0

∴ lim+ c →0

12.

area (T) c3 6 = lim+ ⋅ =3 area (R) c →0 2 c 3

x+3 > 0 ⇒ x < –3 or x > –1 x +1 As x → –3, y → 0; x → ∞, y → 1 range (0, 1) ∪ (1, ∞) y=

SECTION – B

1.

i2 π  i2 π   ( n−1)   We have zn − 1 = ( z − 1)  z − e n  .....  z − e n         i2 π  ( n−1)   ……(1) 1 + z + ..... + zn−1 = ( z − 1) ....  z − e n     π 2π π ..... sin ( n − 1) Put z = 1 and take modulus n = 2n−1 sin sin n n n (A) if n = 21 π 2π 9π 10π 20π ⋅ sin 21 = 220 sin sub .....sin .....sin 21 21 21 21 21

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13 AITS-CRT(Set-III)-Paper-2-PCM(S)-JEE(Advance)/13 2

π 10π   ⋅ 220 21 =  sin ......sin  21 21   π 10π 21 ..... sin = 10 21 21 2 (B) If n = 22

∴ sin

2

π 2π 10π   22 = 221  sin sin .... sin 22 22 22   π 10π 22 11 ∴ sin ..... sin = 11 = 10 22 22 2 2 (C) Again put z = –1 is ……..(1) and take modulus π 2π π 1 = cos cos ..... cos ( n − 1) 2n−1 . n n n n = 21 π 2π 20π 20 1 = cos cos .... cos 2 21 21 21 π 10π 1 ∴ cos ...... cos = 21 21 210 π 2π 10π 11 (D) sin , sin ..... sin = 10 22 22 22 2 10π 9π π 11 cos cos .... cos = 22 22 22 210 ∵ sin θ = cos ( 90 − θ ) 2.

(A) 1st, 4th, 7th terms are a, a+3d, a+6d ax+by+c=0 ax+(a+3d)y+(a+6d)=0 a(x+y+1)+3d(y+2)=0 passes through (1, –2) (B) a, b, c are three consecutive terms of A.P. a = A+(m–1)d, b = A+md, c = A+(m+1)d. (A+(m–1)d)x+A+md)y+A+(m+1)d = 0 A ( x + y + 1) + d mx + my + m − x + 1 = 0 ∴ x + y + 1 = 0, –x + 1 = 0 x = 1, y = 2

(

)

(

)

(C) a = A + (r–1)d, b = A + r 2 − 1 d , c = A + 2r 2 − r − 1 d

( A + (r − 1) d) x +  A + (r 2 − 1) d y + A + ( 2r 2 − r − 1) d = 0 A ( x + y + 1) + d ( r − 1)  x + ( r + 1) y + ( 2r + 1)  = 0

X + y + 1 = 0, x + y + 1 + r(y + 2) = 0 Y = –2, x = 1

(

)

(

)

(D) a = A + (r–1)d, b = A + r 2 − 1 d , c = A + 3r 2 − 2r − 1 d

( A + (r − 1) d) x +  A + (r 2 − 1) d y + A + ( 3r 2 − 2r − 1) d = 0 A ( x + y + 1) + d ( r − 1)  x + ( r + 1) y + ( 3r + 1)  = 0

x + y + 1 = 0, x + y + 1 + r(y + 3) = 0 y = –3, x = 2 SECTION – C FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com

AITS-CRT(Set-III)-Paper-2-PCM(S)-JEE(Advance)/13

14

tan θ1 / 2 3 AB = implies PB – PA = . This 5 tan θ2 / 2 2 implies locus of P is branch of hyperbola of eccentricity = 5

1.

Let A (z1), B(z2) and P(z) forms a triangle ABC then

2.

tan7x = 0 ⇒

S1 − S3 + S5 − S7 =0 1 − S2 + S 4 − S 6

⇒ 7 tan x − 35 tan3 x + 21tan5 x − tan7 x = 0 3π 2π π π 2π 3π , − tan , − tan ,0,tan ,tan ,tan are the roots of equation 7 7 7 7 7 7 7x − 35x 3 + 21x 5 − x 7 = 0 π 2π 3π ⇒ tan2 ,tan2 ,tan2 are the roots of equation 7 − 35x + 21x 2 − x 3 = 0 7 7 7 π 2π 3π are roots of equation 7x 3 − 35x 2 + 21x − 1 = 0 ⇒ cot 2 ,cot 2 ,cot 2 7 7 7 π 2π 3π ⇒ cot 2 + cot 2 + cot 2 =5 7 7 7

⇒ − tan

3.

(A) 2x + 2yy′ = 0 ⇒ y ' = −

x y

∴ y ' ( 2 ) = −1 = A (B) cos y.y′ + cos x = sin x.cos y.y′ + sin y cos x when x = y = p –y′ – 1 = 0 + 0 (C) 2exy (xy′ + y) + exey y′ + eyex – ex – eyy′ = e.exy(xy′ + y) at x = 1, y = 1 2e(y′ + 1) + e2y′ + e2 – e – ey′ = e2(y′ + 1) ey′ + e = 0 ⇒ y′ = – 1 Hence |A + B + C| = 3 4.

5  11  ⇒ a−b = 3 Focus is  ,0  ⇒ Distance of focus from line = 2 6 

5.

Put an = tan θn

6.

f K −2 ( x ) = 0 has K distinct roots, f K −1 ( x ) = 0 has K + 1 distinct roots, f K ( x ) = 0 has K distinct

roots ⇒ K + 1 = 8 ⇒ K = 7

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