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AITS-FT-I-(Paper-2)-PCM(S)-JEE(Advanced)/13

JEE(Advanced)-2013

FIITJEE

ANSWERS, HINTS & SOLUTIONS FULL TEST –I (Paper-2)

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. 1.

PHYSICS A

CHEMISTRY B

MATHEMATICS C

2.

B

B

B

3.

D

B

A

4.

D

B

C

A, D

A, C

A, C

A, B, C, D

A, B, C

A, C

7.

B, C

A, B, D

A, C

8.

A, C

A, B, D

C, D

9.

A, B, C

A, C, D

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

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

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

1.

8

5

4

2.

4

2

8

3.

9

2

3

4.

5

5

5

5.

3

8

3

6.

1

8

5

7.

5

6

3

8.

8

7

3

5. 6.

1.

2.

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AITS-FT-I-(Paper-2)-PCM(S)-JEE(Advanced)/13

2

Physics

PART – I SECTION – A

1.

By conservation of momentum J = mV + 2 mV V= J/3m.

2.

0 + 2 × 1 + 3 × 1 = VB ∴ VB = 5 volt

4.

6.

7.

1 2 gt 2 ⇒ S < 2g. (due to Lenz’s Law).

S<

The molar heat capacity has the general definition 1 ∆Q C= ⋅ n ∆T where n = number of moles, ∆Q = heat absorbed by the gas and ∆T = rise in temperature of gas. It is possible to obtain almost any set of values for ∆Q and ∆T by proper selection of a process.

2usin θ T= = g

2 × 10 × 10

1 2 = 1 sec.

30º 30º

30º 8.

M' g − T = M'a T = Ma M' g = a(M + M')

30º

R

…(i) …(ii)

T

M' g (M + M') ma sin θ = mgcos θ a = gcot θ

M

a=

Mg

M' g (M + M') cot θ M + cot θM' = M'

gcot θ =

Mcot θ (1 − cot θ) T=Ma = M. g cot θ Mg . T= tan θ

masinθ

M' =

ma θ mgcosθ

mgsinθ + macosθ

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3

FR = 2 2 iBa .

9.

AITS-FT-I-(Paper-2)-PCM(S)-JEE(Advanced)/13

2iBa

2iBa

SECTION –C 1.

P = β v2 mdv .v = β v 2 dt 2v 0

∫

v0

dv β t = ∫ dt v m0

β .t m mln2 t= β 4×2 t= × 0.693 0.693 t = 8 seconds. ln2 =

2.

3.

After 5 s, speed of detector = 50 – 10 × 5 = 0 and that of source = 0 + 10 × 5 = 50 m/s 1 and the source has fallen a distance = × 10 × (5)2 = 125 m 2 and the detector has rises a distance 1 = 50 × 5 − × 10(5)2 = 125 m 2  300 − 0  So, f ' = 130   = 156 Hz.  300 − 50  g= g' =

S 50 m/s

O

v=0

Ground

GM R2 G.2m

4R 2 g g' = 2 F = −mg' sin ∝

2R

θ

2R

θ

mgsinα α mg’

mgcosα

mg .sin 2θ 2 As θ is much small mg ma = − .2θ 2 a = −g.θ F=−

a = −g.

x 2R

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AITS-FT-I-(Paper-2)-PCM(S)-JEE(Advanced)/13

∴ ω=

4

g f 2R

T = 2π

2R g

2 × 6400 × 103 10 = 2 × 3 × 800 × 1.41 T = 6768 s. T = 2 × 3.14

4.

AB = v dt sin θ (along the circular arc) AN OA = sin θ ds = r dθ given vdt sin θ = OA dθ

…(i)

A dθ O

θ

B N 2a 3

AN .dθ sin θ a vdt sin θ = dθ sin θ dθ v = sin2 θ dt a dθ = 5 rad/s. dt vdt sin θ =

5.

We have, 10.2 = W + Kmax, 1 and 10.2 Z2 = W + Kmax, 2 h h Also λ de −Broglie = = ρ 2mK ∴

λ1 K2 = = 2.3 λ2 K1

⇒

…(i) …(ii)

K 2 = 5.25 K1

…(iii)

Also 10.2 Z2 = energy corresponding to longest wavelength of the Lyman series = 3 × 13.6 ⇒ Z = 2. ∴ From equations (i), (ii) and (iii) W = 3 eV. 6.

Velocity of efflux v = 2gy Range x = 2gy ×

A

2h g

y h

 dx  The velocity of the block must be   .  dt  ∴ Vb = Vb =

dx = dt

a x

2h 1 dy × 2g × g 2 y dt

h dy . y dt

…(i)

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V

5 Using equation of continuity Ady = a 2gy dt equation (i) and (ii) h a Vb = × 2gy y A Vb = 2gh ×

AITS-FT-I-(Paper-2)-PCM(S)-JEE(Advanced)/13

…(ii)

a A

1 –1 = 1 ms . 20 If there were no hole GM g0 = 2 R decrease in g due to hole (absent mass) GM / 8 GM g1 = = 2 2R2 R 2 effective g = g0 – g1 GM = 2R2 = 5.5 m/s2. = 20 ×

7.

( )

8.

Rdθ

q q .Rdθ = .dθ 2πR 2π dq qdθω di = = T 2π 2π dq =

di =

qω 4π2

dB =

dθ θ

.dθ

µ0 di(R sin θ)2

2R3 π µ0 sin2 θ  qω  dB = ∫ ∫ 2R  4π2  dθ 0 µ qω B= 0 16πR φ = Bπa2 µ qω φ = πa2 . 0 16πR µ0 qωa2 16R dφ | ε |= dt φ=

µ0 qa2 α. 16R = 8 volt 8 i = = 8 A. 1 | ε |=

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6

AITS-FT-I-(Paper-2)-PCM(S)-JEE(Advanced)/13

Chemistry 1.

PART – II

SECTION – A Wsolute = 3 gms, Wsolvent = 50 gm = 75 – 25

WOH− =

3 x + . m=5= 40 56

( 3 40 + x 56 ) × 1000 47

X=9 2.

Major product is formed via E2. H anti to halogen will be eliminated.

3.

nHx = nNaOH = 0.25 × VNaOH, pH = pKa = 5 (acid buffer)

1 1 pK a + logC = 9 2 2 0.25V C= ⇒ VNaOH = 40 ml 60 + VNaOH

⇒ logC = −1 ⇒ C = 10−1

pH = 7 +

nHx = 10 mm = 0.82 gm

%= 4.

0.82 × 100 = 80% 1.025

Time required for equifractional change in a first order reaction is constant. SECTION – C

O

1.

N

C

H

C H

H P 2.

Q Cl

Cl Cl

H3 N

Cl

–

NH3

H3 N

NH3

Na2B 4 O7

+

H2O

1 mol NaOH

+ H5BO3

2 mol

2 mol

Cl

Cl

H3 N

+

Co

NH3

3.

Cl

H3 N

Co

Cl

Co H3 N

NH3

NH3

Cl

(Fac)

(Mer)

→ 2Na B ( OH)4  + 2H3BO3 2 mol →

NaB ( OH)4 

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7 4.

A = 2y1/3 nm. Density = m/a3

= 5.

6.

7.

AITS-FT-I-(Paper-2)-PCM(S)-JEE(Advanced)/13

M = 6.023 × y × 4 amu

6.023 × 4 × 1.66 × 10−27 kg / m³ = 8y × 10 −27

40 = 5 kg / m³ 8

Two double bonds capable exhibiting geometrical and one chiral carbon molecule asymmetric. 23 = 8

Pvm = 58.33 gm RT 58.33 %= × 100 = 77% 75 w=

O

O B–O–B

O

O

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8

AITS-FT-I-(Paper-2)-PCM(S)-JEE(Advanced)/13

Mathematics

PART – III SECTION – A

1.

2a   1  a PQ is a focal chord Q  −  =  2 , −  t   t  t −2   Normal drawn at P(t) meets the curve at Q  − t, t   1  1  ∴ Normal drawn at Q  −  meets the parabola again at R  2t +  t t    2   1 1  R a  2t +  , 2a  2t +   t t    

1 1   Equation of tangent at R is y  2t +  = x + a  2t +  t t   2 Equation of focal chord PQ is ( t − 1) y = 2t ( x − a )

2

….. (1) ….. (2)

T is the point of intersection of line (1) and (2) a  1  2a  1 ⇒ T   4t 2 − 2  ,  4t +   t  t  3  3  ∴ Area of ∆ TQR = 2.

8a2 3

 1  t + t 

3

Let a point (3λ + 1, λ + 2, 2λ + 3) of the first line also lies on the second line. Then 3λ + 1 − 3 λ + 2 − 1 2λ + 3 − 2 ⇒λ=1 = = 1 2 3 Hence the point of intersection P of the two lines is (4, 3, 5). Equation of plane perpendicular to OP where OP is (0, 0, 0) and passing through P is 4x + 3y + 5z = 50 a b c

3.

For non triangle solution ∆ = b c a = 0 c a b ⇒ a + b + c = 0 ⇒ ‘1’ is a root of ax2 + bx + c = 0

4.

Clearly we have xdy + ydx xdy − ydx = e xy x2 y y ⇒ d(xy)e − xy = d   ⇒ + e− xy = c x x

∫

5.

∫

Let the roots of az3 + bz2 + cz + d = 0, z1 = x1, z2, z3 = x2 ± iy2 b ⇒ z 1 + z 2 + z3 = − a ⇒ ab > 0 d Also z1 z2 z3 = x1 ( x 22 + y 22 ) = − a ⇒ ad > 0

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9

6.

7.

AITS-FT-I-(Paper-2)-PCM(S)-JEE(Advanced)/13

3 1 3 and β2 = ⇒ β = ± 2 2 4 ∴ New roots are ω and ω2 which are roots of x3 – 1 = 0 or x2 + x + 1 = 0 On solving we have α =

 x 2 4x 8 + ; x 0 , f ( α3 ) < 0 and f ( θ )θ→π / 2 → ∞

⇒ There will be 3 such values of ‘θ’

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11

4.

Let the equation be K(x – α)(x – β) = 0 ∴ Sum of coefficient = K(α – 1)(β – 1) is prime ⇒ K is 1 and α – 1 is 1 ⇒ One root is 2 and β – 1 is 2 Hence roots are 2 and 3.

5.

Let t = x3 in I2 1

We have I2 =

AITS-FT-I-(Paper-2)-PCM(S)-JEE(Advanced)/13

1

1 1 1 1 dt = dt 3 e t (2 − t) 3 e1− t (1 + t) 0 0

∫

∫

1

I 1 et 1 ⇒ I2 = dt = I1 ⇒ 1 = 3 3e (1 + t) 3e eI2

∫ 0

6.

Let 3, 4 and r be radii of the circles inscribed into the ∆’s ACD, BCD and ∆ABC respectively, we 3c r AB c get = = ⇒ b= 3 AC b r 4c r c Similarly ∆ABC and ∆BCD are similar, we get = ⇒ a = 4 a r Now as c 2 = a2 + b2 ⇒ c2 =

7.

9c 2 r2

+

Putting x = 2,

16c 2 r2

⇒r=5

1 and – 1 successively 2

 1 f(2) + f   = 3 2  1 ( ) 3 f   + f −1 = 2 2 and f(–1) + f(2) = 0

….. (1) ….. (2) ..... (3)

3 Solving, we get f(2) = 4 2

8.

1 7  7 On solving we have A = cos4 θ + 1 − cos2 θ + 1 =  cos2 θ −  + ⇒ A ∈  ,  2 4 4

 2 

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