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AITS-FT-II-(Paper-1)PCM(S)-JEE(Advanced)/13

JEE(Advanced)-2013

FIITJEE

ANSWERS, HINTS & SOLUTIONS FULL TEST – II (PAPER-1) ANSWERS KEY

ALL INDIA TEST SERIES

From Long Term Classroom Programs and Medium / Short Classroom Program 4 in Top 10, 10 in Top 20, 43 in Top 100, 75 in Top 200, 159 in Top 500 Ranks & 3542 t o t a l s e l e c t i o n s i n I I T - J E E 2 0 1 2

1

Q. No.

PHYSICS

CHEMISTRY

MATHEMATICS

ANSWER

ANSWER

ANSWER

1.

B

C

C

2.

B

B

A

3.

B

D

B

4.

B

C

A

5.

A

B

B

6.

A

A

D

7.

D

C

D

8.

C

C

A

9.

A, B

A, B

A, C

10.

B, D

A, C

A, B, C

11.

B, C, D

A, B, C, D

B, C

12.

A, D

A, D

B, C

13.

A, B, C, D

A, D

A, B, C

14.

A

B

D

15.

B

A

A

16.

B

B

A

17.

C

C

B

18.

C

B

C

1.

5

2

7

2.

4

4

4

3.

4

7

4

4.

3

0

5.

2 2

6

7

6.

8

8

4

7.

4

1

6

8.

3

2

2

9.

4

4

9

10.

1

4

1

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2

AITS-FT-II-(Paper-1)PCM(S)-JEE(Advanced)/13

Physics

PART – I SECTION – A

1.

di ε=L dt

2 dt ⇒ dt = 5 × 10−3 sec. Q = idt = 2 × 5 × 10-3 ∴ Q = 10-2 But Q = CV 10-2 = C × 400 ∴ C = 25µF.

400 = 1 ×

2.

Total flux through this area is = E.4π (1.5R)2  Q (1.5R)2 − R3 qnet 1 = Q+ ε0 ε0  (2R)3 − R3  Q  19  75Q 1+ = = ε0  56  56ε0

{ {

}

} 

R

R 1.5

2R

 

Gauss law E .4 π (1.5R)2 = Q (75/56ε0) 75Q E= 504πε0R2 3.

w0 =

1

=

1

4.

LC 2 × 10−2 op, 2π f = 50 25 op, f = Hz . π Use the definition of an elastic collision.

5.

Buoyant force = Weight of liquid displaced =

4 2 πr ρg 3

Viscous force = Stoke’s drag force = 6 πηrv 4 ∴ 6πηrv = πr 2ρg 3 6.

2(k2x0) = mg + iBL mg = 2kx0 ε i= R mgR . ∴ B= εL

8.

For a internal point gr g' = R

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3

AITS-FT-II-(Paper-1)PCM(S)-JEE(Advanced)/13

⇒ g′ ∝ r F as a external point GM 1 g' = 2 = g' ∝ 2 r r dx = t2 dt

9.

y=

…(i)

1 t3 23

dy t 2 = dt 2

t = 1, vx = 1, vy =

…(ii) 1 2

G 1 v = ˆi + ˆj 2 d2 x = 2t dt 2

…(iii)

d2 y

=t dt 2 at t = 1 s ax = 2 and ay = 1 G ∴ a = 2iˆ + ˆj

10.

11.

13.

…(iv)

F Acceleration of M, a =   M 1F 2 A= .t 2M 2MA t= . F G G G G G τ = M × B and U = −M.B G G here M and B are anti–parallel G G ∴ τ = O and U = +MB (maximum)

v+u=D and v – u = X D+ X D−X ⇒ v= ,u = 2 2 D2 − X 2 and f = 4D D+ X D−X m1 = , m2 = D−X D+ X

14. - 15.

The impedance is z = R2 + ω2L2 and the rms V current is irms = rms . z

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AITS-FT-II-(Paper-1)PCM(S)-JEE(Advanced)/13

16.

Since the center of mass will remain at rest. m(5A − x) = M.x

x= 17.

5mA . (M + m)

MvM − mv m = 0 1 2 1 1 2 KA = MvM2 + mv m 2 2 2 Solving (i) and (ii) Km vM = A. M(M + m)

18.

vm =

…(ii)

M Km A m M(M + m) =

KE =

…(i)

KM A m(M + m)

1 2 mv m 2

SECTION – C 1.

Let, V0 → velocity of sound when source is O at the velocity component of source towards listener is VS cos 90 = 0   V0 ∴ F' = F0   = FS  V0 − VS cos90  A time taken by sound to reach listener = . V0 Distance moved by source in this time  A  =  VS  V0  ∴ Distance between source and listener = A 2 + n2 A 2 = A 1 + n2 = 105 m. 2. If the plank is displaced slightly by x towards left, then a2 Kx Kx m2 f1 2 Kx – f1 = m a2 …(i) …(ii) f1 − f2 = 8 m a1 τcm = Icm α (8m)R2 α or ( f1 + f2 ) R = 2 (8m)a1 (f1 + f2 ) = …(iii) 2 From (ii) and (iii), 3a 2f1 = (8m) × 1 2 3a1 or f1 = (8m) × 4

α

f1 a1

f2 (Assumed)

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5

AITS-FT-II-(Paper-1)PCM(S)-JEE(Advanced)/13

and f2 = 2 ma1. Hence, f2 will be in forward direction. Also, for no slipping, at the point of contact between plank and cylinder. 2a1 = a2 …(iv)  3a  …(v) f1 = (8m) ×  2  = (3m)a2  8  from (i) and (v), 2Kx – (3m) a2 = m a2 2K a2 = x 4m  −d2 x   2K  or  2  =  x  dt   4m  or

 2 π2  = −  ×  x dt 2 4 2 

d2 x

π2 4 π or ω = 2 2π π = or T 2 or T = 4 seconds.

or

3.

ω2 =

When the two disc come in contact, they exert equal and opposite forces. …(i) ∫ Fadt = I1(ω0 − ω1 )

∫ Fbdt = I2 ω2

…(ii)

finally ω1a = ω2b I1(ω0 − ω1 ) = I2

ω1 =

I1ω0 I1 +

4

5.

µ=

sin 900 sin 300

a2 I2 b2

a2 b2

.ω1

= 4 rad / s .

=2 900

300

Maxn energy is liberated for transition En → E1 and minimum energy for En → En–1 Hence, E1 E1 − = 52.224 eV n2 12 E1 E1 and − = 1.224 eV 2 n (n − 1)2

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AITS-FT-II-(Paper-1)PCM(S)-JEE(Advanced)/13

6

Solving we get, E1 = – 54.4 eV and n=5 hence, 13.6Z2 E1 = − = −54.4 12 Z = 2. 6.

Magnetic field due to small ring at distance R from the centre B = 2

where M = i π a µ iπa2 µ ia2 ∴ B= 0 3 = 0 3 4π R 4R 2 µ ia dF = i0Rdθ 0 3 4R dFx = dFsinθ µ i.i a2 sin θdθ = 0 0 4R2 µ i i a2 π Fx = 0 02 ∫ sin θ dθ 4R 0 Fx = Fx =

µ0i i0 a2 4R2 µ0i i0 a2

µ0 M 4 π R3

θ

dθ

dF

×2

2R2

Fy = 0 ∴ Fnet = Fx =

µ0i i0 a2 2R2

= 8 Newton. 7.

F.B.D of m N = mg cos θ mg sin θ = ma a = g sin θ F.B.D. of M

N ….(i)

a mgsinθ

R = N cos θ + Mg µ R = N sin θ µ. (N cos θ + Mg) = N sin θ µ Mg = N [sin θ - µ cos θ] µ Mg = mg cos θ [sin θ - µ cos θ] µM m= cos θ[sin θ − µ cos θ] m = 4 kg.

mgcosθ

R Ncosθ

Nsinθ

µR Mg

8.

Thermal resistance of the rod, A R= kA

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AITS-FT-II-(Paper-1)PCM(S)-JEE(Advanced)/13

When heat is transferred from first vessel to second, temperature of first vessel decreases while that of second vessel increases. Due to both there reasons, difference between temperature of vessels decreases. Let at an instant t, the temperature difference between two vessels be θ. θ KAθ H= = R A KAθ dQ = Hdt = dt …(i) A Since gases are contained in two vessels, therefore, processes on gases in two vessels are isochoric. Hence, decrease in temperature of gas in first vessel, dQ dQ dQ ∆θ1 = = = 5R 5R nC V 2× 2 Increase in temperature of gas in second vessel is dQ dQ ∆θ2 = = 3R 6R 4× 2 ∴ Decrease in temperature difference (–dθ) = ∆θ1 + ∆θ2 dQ 11 − dθ = × R 30 KAθ × 11 or, −dθ = dt 30RA 25

or

dθ KA × 11 t = dt θ 30AR 0∫ 50

−∫

or, An2 = or

9. 10.

t=

KA × 11 t 30AR

.693 × 30 × 242 × 10 −2 × 8.3 × 7 693 × 7 × 22 × 8.3 × 10 −4 × 11

= 3 seconds.

 di  d2 V 2dC dV  d2 q  d VL =  R − 2  = L iR −  C 2 +  = 4 µV.  dt dt   dt dt dt dt     From C.L.M. v M 11 v0 = Mv ⇒ v = 0 10 10 11 For block B to leave ground

…(i)

Mg K(x0 + x) = 2 Mg (where x0 = ) K Mg ∴ x= …(ii) K From COE 11 1 1 11 2 1  Mgx −  K(x0 + x)2 − Kx02  = 0 − Mv 10 2 2 2 10  

M 10

A K

M

2M

88M K putting the value v0 = 1 ms–1.

solving v 0 = g

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B

8

AITS-FT-II-(Paper-1)PCM(S)-JEE(Advanced)/13

Chemistry

PART – II SECTION – A

1.

0.0591 1 log Ecell = 0 − 1 2

After electrolysis Ecell = 0 −

(1 − 0.5 ) 0.0591 log 1 ( 2 − 1)

No change. 2.

10.

8RT = πMA

3RT 8 3 ⇒ = MB πMA MB

1 H+ + e −  Eo = 0.0V → H2 2 0.059 1 E = EH+ /H = − log + 2 1 H 

= - 0.059 × 7 = - 0.413 So, EH+ /H < ENi2+ /Ni 2

Ni will deposit at cathode At anode cell reaction 4OH−  → 2H2O + O2 + 4e 11.

Both complexes have Cr+3(d3), and coordination number is 6. Hence d2sp3 Pd+2(d8) complexes with coordination number → 4 are square plannar and magnetic moment is zero.

12.

∆ FeCl3  → FeCl2 + Cl2

2+

Dilute CuCl2 is blue due to Cu (H2 O )4  . PH3 is obtained be heating white P with NaOH. All are reducing because of weak P – H bond. 15.

Bottles 1,2,3,4 have Pb(NO3)2, HCl, Na2CO3 and CuSO4 When mixing Bottle 1 + Bottle 2 → PbSO4 Bottle 1 + Bottle 3 → PbCO3 Bottle 1 + Bottle 4 → PbSO4 Bottle 2 + Bottle 3 → CO2 Bottle 2 + Bottle 4 → No visible reaction Bottle 3 + Bottle 4 → CuCO3

18.

∆  → Li + N2 ↑

N2 + O2 Lithium  → Li2 O + Li3N ∆

H2 O LiOH← 

H2 O  → LiOH + NH3 ↑

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AITS-FT-II-(Paper-1)PCM(S)-JEE(Advanced)/13

SECTION – C CH3

CH3

1.

A=

B=

H2C

Br

Br

Br Inactive

Br

CH3

D=

C= CH3 H2CBr

Br

Inactive CH3

E= Br

H

4.

∆Tb = K b × m × i 2.08 = 0.52 × 1 × i i=4 K 3 FeIII ( CN)6 

6.

Cl

HO

Cl  →

Cl

Cl

O

OH ∆  →

OH

HO

(x)

O (z)

(y)

7.

logK = 0.1

10.

Glycinate is asymmetric bidentate ligand. AB type, so for M(AB)C2D2: C D

B

A

A

D M

D C

C

B

M c

D

C

B

M D

C

c

B

M A

C

D

A

C D

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AITS-FT-II-(Paper-1)PCM(S)-JEE(Advanced)/13

Mathematics

PART – III SECTION – A π/2

1.

 2  sin3x 1   k  3 + 2 sin x  − cos x − 2k sin x     0

≤1

 2  −1 1   k  +  − 0 − 2k  − {−1} ≤ 1 3 2    

k2 − 2k + 1 ≤ 1 6 k (k − 12) ≤ 0 0 ≤ k ≤ 12 k = 1, 2, 3, 4, …, 12. 101

2.

I1 =

∫

( 5 + 2x − 2x 2 )(1 + e2− 4x ) −100 101

= 2I1 =

3.

dx

∫

5 + 2x − 2x 2 −100

⇒

I1 1 = . 2 I2

I=

∫ ( 6f ( x ) − x ) ( x

I=−

4.

101

dx

∫ (x

2x 2

− f ( x ))

2

− f ( x ))

dx +

2

−100

2 − 4(1− x )

)

= I2

2x ( x − 6f ( x ) ) + f(x) 2

dx

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

=

2

dx

f ( x)

∫ ( 6f(x) − x ) ( x

2

− f ( x ))

2

dx =

 f ' ( x ) − 2x  1 +c  dx = 2 2 2 x − f(x) − f ( x ) )  

∫  ( x

 ( sin ( x + h ) )ln(x +h) − (sin x)ln x  ln x ( )  Let g(x) = sin x ⇒ g'(x) = lim  h→0  h   ln ( sin x )  π π + cot x ln x  ⇒ f   = g'   = 0 ⇒ f(x) = g'(x) = g(x)   x  2 2

5.

n(x1) = 3, x1 = 2, 3, 4. 4 ≤ x2 < x3 ≤ 8 ⇒ from 4, 5, 6, 7, 8 Selecting any 2 ⇒ n(x2, x3) = 5C2 = 10 and n(x4) = 2 x4 = 0, 5 ⇒ Number of such 4 digit numbers 3 × 10 × 2 = 60

6.

Let X denotes the number of toss required. 3

2 1   ⋅ 2 1 2 1 2 1 3 3 = 8 P(X ≥ 4) =   ⋅ +   ⋅ +   ⋅ + ...... ∞ = 2 27 3 3 3 3 3 3 1− 3 3

4

5

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AITS-FT-II-(Paper-1)PCM(S)-JEE(Advanced)/13

  2 6 1    ⋅  6 2 1 2 1 3 3  = 2 P(X ≥ 7) =   ⋅ +   ⋅ + ..... ∞ =       2  3 3 3 3 3  1−  3   6

7

6

2   8 Required probability =  3 3 = 27 2   3

8.

R = P'Q8P = A 8  3 −2  3 A2 =   1  0 0

3 − 2  1 

 3 A3 =  0

3) 0

−2   3 − 2 = 1  0 −2   3 − 2 3 − 2   (  = 1  0 1  

3

 −2 ( 3 ) − 2 3 − 2   1

= Ak then λ11 = ( 3 ) ⇒ r11 = ( 3 ) = 81 ⇒ If λij  2×2 k

9.

8

 ∆   ∆   abc  r1 + r2 = 3R ⇒   +  = 3  4∆  s−a s−b

∆2 3ab = ⇒ 4s(s – c) = 3ab ( s − a )( s − b ) 4 1 ⇒ ∠c = 60º ⇒ cos c = 2 Also r2 + r3 = 2R  ∆   ∆  2abc ⇒  + = 4∆ s−b s−c π π and ∠B = ⇒ 2s(s – a) = bc ⇒ ∠A = 6 2

⇒

10.

⇒ ∆1 = 2 2 ( 4 − 10 )

and ∆ 2 = 2 2 ( 4 + 10 )

x2 2 + ( y − 4) = 1 9

P2 P1 ∆1

∆2 A B 11.

(3a + 4b)(3a – 2b) = 0 and (3a – 2b)(a + b) = 0 ⇒ 3a = 2b

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AITS-FT-II-(Paper-1)PCM(S)-JEE(Advanced)/13

12.

12

f(x) is non differential at x = 1 (Point A in figure) But f(x) is cont. ∀ x ∈ [0, 2] ⇒ g′(t) = f(t) – f(t – 1), t ∈ [1, 2]

y = f(x)

= ( 2t 2 − 6t + 6 ) − ( ( t − 1) + 1) = 2t 2 − 7t + 6 = ( 2t − 3 )( t − 2 )

3 2

1

y=x+1

2

⇒ Maximum when t =

14.

A

2

1

3 2

2

3 and minimum when t = 2. 2

 A  P ( A ∩ BE ) P , where A = sum is 4. = P B ( E)  BE  1 1 1 And P (BE ) = P (B2 ) + P (B4 ) + ..... = + + + .....∞ = 4 16 64

1 4

=

1 4 1 ⋅ = 4 3 3

1 4  A   A  P ( A ∩ BE ) = P ( A ∩ B2 ) + P ( A ∩ B 4 ) = P (B2 ) P  + P (B 4 ) P     B2   B4  1−

1  3  1  1  ( 4 ⋅ 3 ⋅ 62 + 1) 433 =  +  4  = 4 4  36  16  6  16.6 16.64 433 433 1  A  Hence P  = ⋅3 = ≅  4 32 ⋅ 216 16  BE  16.6

=

15.

n(S) = 6 ⋅ 6 ⋅ 6 = 63 = 216 Now greatest number is 4, so atleast one of the dice shows up 4. ∴ n(A) = 43 – 33 = 37 37 Hence P(A) = 216

16-18. Equation of tangent of slope m to y2 = 4x is 1 y = mx + m 16.

As (1) passes through P(6, 5), so 5 = 6m +

….. (1) 1 m

1 1 or m = 3 2 2  2   1  1 and  2 , Points of contact are  2 ,    m1 m1   m2 m2  Hence R(4, 4) and Q(9, 6) 6 5 1 1 1 Area of ∆PQR = 4 4 1 = 2 2 9 6 1

⇒ 6m2 − 5m + 1 = 0 ⇒ m =

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13

17.

AITS-FT-II-(Paper-1)PCM(S)-JEE(Advanced)/13

1 x + 2 ⇒ x – 2y + 4 = 0 ….. (2) 2 1 and y = x + 3 ⇒ x – 3y + 9 = 0 3 Now equation of circle C2 touching x – 3y + 9 = 0 at (9, 6), is ( x − 9 )2 + ( y − 6 )2 + λ ( x − 3y + 9 ) = 0 y=

As above circle passes through (1, 0), so 64 + 36 + 10λ = 0 ⇒ λ = –10 Circle C2 is x2 + y2 – 28x + 18y + 27 = 0 ….. (3) 2 Radius of C2 is r2 = 196 + 81 − 27 = 277 − 27 − 250 ⇒ r2 = 5 10 SECTION – C

1.

⇒ p – 3 = 0, 2, 3, 4 ⇒ Σp = 5 + 6 + 7 + 3 = 21

5

y

y = |x + 1/x – 3|

4 3 2 1

⇒n=7

0 3

y=3

1 2.

1  1   x + 2  x + 2 α β   2

Let I =

∫

x 2 dx +

−1

⇒

 α 2 + β2  1 2  + αβ = x +  2   ( αβ )

 t 4 + 4t   1 1 1 1 2 x x+ 2 − + + + x =   2  2  2t αβ 4t 4t   ( αβ ) 

3  t2 1 1  1  +  + 3 2 −  2 4 t  2t  4t 

dI 3  t 1 1 1  1/ 4 =  − 2 − 3 + 2  = 0 ⇒ t 4 = 2 ⇒ t = ± ( 2) dt 2  2 t t t 

⇒ Min I(t) = 3 +

3 2 9 3 2 1 1 1  =3+ + 1/ 4 + − 1/ 4  = 3 + ⋅  2 2 8 2 4 2 2 2 2 

9 ⇒   + 3 = 1+ 3 = 4 8

3.

 3 1   3 1  The population after 2n days = P   ⋅  ⋅  ⋅  ⋅ ..........n times  = 310  2 2   2 2   n

10

3 3   =  4 4

4.

⇒ n = 10 ⇒ Total number of days = 20 days

Let f(x) = 2x3 – 3x2 + p ⇒ f ' ( x ) = 6x 2 − 6x = 0 at x = 0, 1 ⇒ f(0) > 0 ⇒ p > 0 f(1) < 0 ⇒ 2 – 3 + p < 0 ⇒p