Concept Recapitulation Test I\\Advanced\\PAPER-1\\Answer\\Answer

JEE(Advanced)-2013 ANSWERS, HINTS & SOLUTIONS

CRT-I(Set-I) PAPER -1

ALL INDIA TEST SERIES

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FIITJEE

Q. No.

PHYSICS

CHEMISTRY

MATHEMATICS

ANSWER

ANSWER

ANSWER

1.

C

D

A

2.

C

B

B

3.

B

B

B

4.

C

C

D

5.

A

C

D

6.

B

B

A

7.

D

A

C

8.

C

D

B

9.

A

B

D

10.

D

D

A

11.

D

D

B

12.

C

C

B

13.

D

A

A

14.

B

D

B

15.

C

B

C

16.

A

B

B

17.

C

B

B

18.

A

A

B

19.

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

C

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

1.

2.

3.

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

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2

Physics

AITS-CRT(Set-I)-Paper-1-PCM(S)-JEE(Adv)/13

PART – I SECTION – A

1.

ρ + ρg (L − x ) − ρgL =

2.

v px = 5 , v p y = 5

d ( ρxu ) dt

5 2 = 7.1 m/s approx.

3.

Solve the problem using µ as the coefficient of kinetic friction. Linear velocity radius to zero before angular velocity becomes zero.

4.

II and IV quadrant field will be added.

5.

Conceptual.

6.

Leftward Force =

7.

Conceptual, torque of pseudo force.

8.

Effective length = (r2 + r1)

14.

I = I0 1 − e − t τ 

2I ∆P = C ∆t

2L E and τ = 3R R 3Rt   − E ⇒ I =  1 − e 2L   R   I 2 I2L = and IL = I 3 3 I0 =

15.

Magnetic energy stored at t = t0 is given by − 2 2E LI2  1 − e 2L , where I is value of IL at t = t0; IL = I =  3 3R 2 

3Rt

U=

2E2L  ⇒U= 1 − e 9R2 

−3Rt 0 2L

  

   

2

16.

Zero current since no path for current is available.

17.

dQ = 0, dW = 0

18.

For constant temperature T, dW = −nRT ln 2

19.

dQ = dU + dW = dW, ∴ dU = 0 as T = constant

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AITS-CRT(Set-I)-Paper-1-PCM(S)-JEE(Adv)/13

3

SECTION – B 2.

4V 2 , 9 8V 2 , Vf for C1 = 25

2V 2 9 2V 2 Vf for C2 = 25

Vi for C1 =

Vi for C2 =

Chemistry

PART – II SECTION – A

1.

Energy of photon corresponding to second line of Balmer series of Li2+ ion. 1 27  1 = 13.6 × 32  2 − 2  = 13.6 × 16 4  2

1  13.6  1 Energy needed to eject electron from n = 2 level in H-atom = = 13.6 × 12 ×  2 − 2  ⇒ 4 ∞  2 9 × 3 13.6 KE of ejected electron = 13.6 × − = 19.55 eV 16 4 2.

Cr +3 + 6H 2 O → [Cr (H 2 O) 6 ] +3 (violet) Cr +2 + 6H 2 O → [Cr(H 2 O) 6 ] +2 (blue )

Cr +3 + 3NH3 + 3H2O → Cr(OH)3 ↓ +3NH+4 grey −blue

Cr(OH)3 + 6NH3 → Cr (NH3 )6  ( OH)3   grey −blue

2Cr +3 + 3S2O82− + 8H2O → 2CrO24− + 16H+ + 6SO24−

CrO24− + 2H+ + H2O2 → CrO5 + 3H2O

3.

ρA =

4 Mx when V → volume of fcc unit cell. NA V

ρB = 8

My

M ρA 30 = x = = 0.3 NA V ρB 2My 2 × 50 ,

Total density ρA + ρB = 4.33 g/cc 7. Friedal − Craft  → Acylation

+

OH Ph

H →

Ph

Ph

Ph

Ph

Ph

O

8. CN CN CN CN

COOH +

H3 O  →

COOH → ∆ COOH COOH

O O

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4

AITS-CRT(Set-I)-Paper-1-PCM(S)-JEE(Adv)/13

9.

6 × 10−7 6 × 10−7 × 100 × 1000 6 × 10−2 H+  = × 1000 = = = 1.2 × 10−2 0.05 5 5

10.

O=C=C=C=O

11.

Group reagent for V group is (NH4)2CO3 in presence of NH4Cl and because of NH4OH, pH of solution is maintained.

12.

Lighter nuclei don’t have so high binding energy.

14.

MnO2 form solution when performing titration of KMnO4 in alkaline medium.

17.

In expansion of CO2 gas molecule, entropy of system ( ∆Ssys ) increases and ∆Ssurr decreases.

18.

19.

V  30 Work done = q = nRT n  2  = 3 × 8.314 × 288.15 n V 10  1 q q ∴ ∆S = = = 27.4 JK −1 T 288.15 ∆Suni will be zero for all reversible process.

SECTION-B

2.

4Zn + 10HNO3  → 4Zn (NO3 )2 + NO2 ↑ +5H2 O

Mg + 2HNO3  → Mg (NO3 )2 + H2 ↑

4Sn + 10HNO3  → 4Sn (NO3 )2 + NH4NO3 + 3H2 O

3Pb + 8HNO3  → 3b (NO3 )2 + 2NO ↑ +4H2 O

Mathematics

PART – III SECTION – A

101

1.

I1 =

∫

dx

−100

( 5 + 2x − 2x )(1 + e 2

101

= 2I1 =

∫

−100

2.

101

dx 5 + 2x − 2x

2

2 − 4x

= I2 ⇒

)

=

dx

∫ (5 + 2 (1 − x ) − 2 (1 − x ) ) (1 + e

−100

2

2 − 4(1− x )

)

I1 1 = . I2 2

Since incident ray is perpendicular to the line mirror y – x = 1 so after refection reflected ray goes back and intersect x axis at A A ≡ (1, 0) Whose equation is x – y = 1.

B

C

(0, 1) 45°

45° A (1, 0)

(0, –1)

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AITS-CRT(Set-I)-Paper-1-PCM(S)-JEE(Adv)/13

3.

5

Let OA = a , OB = b and OC = c , then AB = b − a and OP =

1 1 1 a , OQ = b , OR = c 3 3 2

Since P, Q, R and S are coplanar, then PS = αPQ + βPR ( PS can be written as a linear combination of PQ and PR )

(

) (

= α OQ − OP + β OR − OP

)

a α β i.e., OS − OP = − ( α + β ) + b + c 3 2 3 a α β ⇒ OS = (1 − α + β ) + b + c 3 2 3

(

Given OS = λ AB = λ b − a From (1) and (2), β = 0, 4.

….. (1)

)

….. (2)

1− α α = −λ and = λ ⇒ 2λ = 1 + 3 λ ⇒ λ = –1 3 2

Let the numbers be x and y | x – y | ≤ 10 If x = 1 x=2 x=3 : : : : x = 11 x = 12 x = 13 : : x = 25

No. of ways 1 ≤ y ≤ 11 1 ≤ y ≤ 12 1 ≤ y ≤ 13 : : : : 1 ≤ y ≤ 21 2 ≤ y ≤ 22 3 ≤ y ≤ 23 : : 15 ≤ y ≤ 25

∴ Now of ways = (11 + 12 +….. + 20) + 21 × 15 =

20

10

k =1

k =1

11 12 13

21 21 21 21

∑ k −∑ k + 315 =

20 × 21 10 × 11 − + 315 2 2

= 210 – 55 + 315 = 470. 5.

f′(x) f (x)

= 1 ⇒ ln f ( x ) = x + c.ln f ( 0 ) = c ⇒ c = 0

f(x) = ex 1

∫ e (x x

1

2

−e

0

x

1

) dx = ∫ x e dx − ∫ ( e ) 2 x

0

( )

 ex  2 x x x =  x e − 2xe + 2e − 2 

6.

(3 (3

) 3 − 5) 3 +5

2n +1

2n +1

x

2

dx

0 2 1

e2 1 e2 3  = e − 2e + 2e − − 2 + = − − e  2 2 2 2  0

= α+β

(

 = β1 < 1 ⇒ α + β − β1 = 2  2n+1C1.5 3 3  = even integer ∴ β – β1 = 0 (∵ –1 < β – β1 < 0) ∴α = even integer

)

2n

(

+ 2n+1 C3 53 3 3

)

2n − 2

 + .......  

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AITS-CRT(Set-I)-Paper-1-PCM(S)-JEE(Adv)/13

6

7.

cos β = 1 4x2 – 10x – 6x + 15 < 0 ⇒ (2x – 3) (2x – 5) < 0 (2x – 3) (2x – 5)

–∞ +ve

3/2

–ve

+ve ∞

5/2

∴ tan α = 2 ∴ sin(α + β) sin(α – β) = cos2 β – cos2α = 1 – 8.

1 4 = 5 5

∠BOC = 2π – (π + A) =π–A ∵ a2 = R2 + R2 – 2R2 cos(π – A) a2 = 2R2 (1 + cos A) a a R= = sec A / 2 2 2.2cos A / 2 2

A

B

C π–(B+C)/2

O

9.

( a + b + c )n + ( a + b − c )n

= 2.

( C .c .(a + b) n

0

n

0

n−2

+n C2 .c 2 ( a + b )

+ ........ +n Cn .c n ( a + b )

0

)

Number of term = ( n + 1) + ( n − 1) + ( n − 3 ) + ...... + 3 + 1

(n + 2 )(n + 2 ) (n + 2 ) n+2 =  . (1 + n + 1) = =  2 2  2  If n ∈ even n4 + 4n ∈ even and n4 + 4n > 2 ⇒ n4 + 4n is not a prime if n ∈ odd, let n = 2k + 1 ⇒ n4 + 4n = (2k + 1)4 + 42k + 4 = (2k + 1)4 + 4(2k)4 + 4(2k)2(2k + 1)2 – 4(2k)2(2k + 1)2 = ((2k + 1)2 + 2(2k)2)2 – (2(2k)(2k + 1))2 = ((2k + 1)2 + 2(2k)2 + 2(2k)(2k + 1))((2k + 1)2 + 2(2k)2 – 2(2k)(2k + 1)). 2

10.

11.

px2 + 4µxy + qy2 + 4a(x + y + 1) = 0 represents pair of straight lines iff 4apq + 16a2µ – 4a2p – 4a2q – 16 aµ2 = 0 ⇒ 4µ2 – 4aµ + ap + aq – pq = 0. For real µ, 16a2 – 4.4(ap + aq – pq) ≥ 0 ⇒ (a – p)(a – q) ≥ 0 ⇒ a ≤ p or a ≥ q.

12.

2f(x) + f(−x) = 2f(−x) + f(x) ⇒ f(x) = f(−x) 1 1  f(x) = sin  x −  3x  x e

I=

∫

1/ e

1 1  sin  x −  dx = − 3x  x

e

1



1

∫ 3t sin  t − t  dt = − I

1/ e

⇒ I = 0.

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AITS-CRT(Set-I)-Paper-1-PCM(S)-JEE(Adv)/13

13.

7

2

x + 8x – 16 = 0 ⇒ (x + 4)2 = 32

P

x = –4 ± 4 2 x = –4 + 4 2 x = 1, y = +2 2 , (1, 0) (1, ±1) (1, ±2) → 5 integral points x = 2, y = 2 3 (2, 0) (2, ±1, (2, ±2), (2, ±3) → 7 integral points

Q

x = 3, y = ± 7 , (3, 0), (3, ±1), (3, ±2) → 5 integral points

πr 2 =π 4

14.

Area =

15.

Area = π –

16.

Area =

1 ×2×2=π–2 2

3

1

∫ ( x − 2) dx = 2 . 2

17.

According to graph f (a + h) f″ (a − h) > 0.

x=a

18.

According to above graph if f(a) = 0 ⇒ f″(a) = 0 (it is point of inflection) ⇒ f″′(γ) = 0 α ≤ γ < β (using Rolle ’s Theorem).

19.

f′(x) ≠ 0 ⇒ f (x) = 0 has atmost one real. ⇒ f″(x) = 0 has atmost one solution.

SECTION-B 1.

sin1 sin5 (A) − sin 2 sin 6 sin x sin ( x + 1) sin ( x + 1) cos x − cos ( x + 1) sin x

take f(x) = f′(x) =

sin ( x + 1) ⇒ f(x) is increasing. ⇒ f(1) < f(5). (B) take f(x) = tanx – x2 f′(x) = sec2x – 2x f″(x) = 2 sec2x tanx – 2 f′(1) > 0 and f″(x) > 0 for x > 1 2

=

sin1 2(

sin

x + 1)

>0

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AITS-CRT(Set-I)-Paper-1-PCM(S)-JEE(Adv)/13

 π ⇒ f(x) is increasing in  1,   2 3 9 ⇒ tan > . 2 4 x  e − 1 (C) lim   does not exist as left hand and right hand limits are not equal. x →0  x  (D) f′(α) = 0 and f′(x) > 0 ∀ x ∈ R – {α} ⇒ f′(x) = (x – α)2n g(x) ⇒ f″(α) = 0.

2.

(A) f2(x) = f2(– x) ⇒ f(x) = f(– x) or f(x) = – f(– x) if f(x) is not continuous then f(x) may be neither even nor odd. 1 (B) f(x) = x[x2] + domain of f(x) is (– 1, 1) 1 − x2 1 which is an even function ⇒ f(x) = 1 − x2 x3 − x 2 e x −1

2x 2 − x3 e x −2

2

(C) f(x) = + = e x + e− x ⇒ f(x) is neither even nor odd. (D) 2x f(x) = f(2x2 – 1) put x = – x ⇒ –2x f(– x) = f(2x2 – 1) ⇒ f(x) is odd.

2

but (x ≠ 1, 2)

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