Fulltest II Advanced Paper 2 Answer Sol Aits 2013 Ft II Jeea Paper 2

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

JEE(Advanced)-2013

FIITJEE

ANSWERS, HINTS & SOLUTIONS FULL TEST – II (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

ANSWERS KEY Q. No.

PHYSICS

CHEMISTRY

MATHEMATICS

ANSWER

ANSWER

ANSWER

1.

C

A

C

2.

C

D

C

3.

B

A

C

4.

B

B

D

5.

B

B

B

6.

D

B

B

7.

A

C

A

8.

C

A

C

9.

B

A

C

10.

A

D

B

11.

B

C

D

12.

C

A

B

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

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

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

1.

5

3

3

2.

4

7

5

3.

2

3

9

4.

4

5

0

5.

4

4

3

1.

2.

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com

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

Physics

2

PART – I SECTION – A

3.

t  t    − − q = 2CV  1 − e 2CR  ; Wb = 2CV 2  1 − e 2CR         

4.

LC = 4/200 = 0.02

5.

Gravitational potential inside a shell = constant

6.

In nuclear, nuclear force play major role. They are charge independent forces. SECTION – C

1.

Current is 2Ω = 0.1 A current in circuit = 0.2 A equivalent resistance about AB = 5Ω e = I R = 1 volts and also e=B A V e 1 ∴ V= = = 5m / s . BA 2 × 0.1

2.

Force on the rod F – Fm = ma BA F− (BAV − E) = ma r BAE B2 A2 V mdV F+ − = r r dt solving B2 A 2 BAE F = −mV0 ω sin(ωt) + V0 cos ωt − r r Power expended by force is B2 A 2 2 BAE FV = −mV02 ω sin(ωt)cos(ωt) + V0 cos2 ωt − V0 cos ωt r r Its average over the cycle is B2 A 2 V02 . 2r

3.

Let A1 and A2 be the cross–sectional area of the pipe at points P and Q respectively. Let v1 and v2 be the velocities of oil at the points P and Q respectively. By conservation of mass, Q = A1v1 = A2v2 A  ⇒ v 2 =  1  (v1 ) = 4v1  A2  Applying Bernoullis’s equation between points P and Q, we have

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com

3

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

1 2 1 ρv1 = P2 + ρv 22 2 2 1 2 P2 = P1 + ρ(v1 − v 22 ) 2 1 = P1 + × 0.9[16 − 256] × 103 2 = 280 × 103 + ( −108) × 103 P2 = 172 × 103 Nm–2 P2 = 172 (KN m–2). P1 +

4.

πP0 v 0 4 πP0 v 0 Pv = P0 v 0 − = ( 4 − π ) 0 0 ; Put π = 3.14 4 4 0.86 = P0 v 0 = ( 0.22 ) (P0 v 0 ) 4

Wnet = ( 2P0 v 0 ) − (P0 v 0 ) − Wnet Wnet

Now, Pv  T1 = 0 0  R   4P0 v 0  T2 =  Thus, R    2P v T3 = 0 0  R 

∆U1→2 = 1×

3R [T2 − T1 ] 2

∆U2→3 = 1×

3R [T3 − T2 ] 2

∆U3→1 = 1×

3R [T1 − T3 ] 2

∆Q1→2 = ( 4.5 )(P0 v 0 ) + (1.22)(P0 v 0 ) = ( 5.72)(P0 v 0 ) ∆Q2→3 = −3P0 v 0 + 0 = −3 (P0 v 0 ) ∆Q3→1 = −1.5 (P0 v 0 ) − (P0 v 0 ) = −2.5 (P0 v 0 )

Thus efficiency η = η=

0.22 (P0 v 0 )

( 5.72 ) (P0 v 0 )

Wnet + veheat

= 0.04

Thus efficiency is 4% 5.

The total momentum of the system in the horizontal direction is conserved. We draw the F.B.D., assuming the displacement of the block to be x1 and x2 in opposite directions, and the total extension x is given by, x = x1 + x2 and m1x1 = m2x2 d2 x ∴ m1 21 = −k(x1 + x 2 ) dt

m1

kx x1

kx

m2

x2

d2 x 2

= −k(x1 + x 2 ) dt 2 after suitably manipulating the equations, we get,

and

m2

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com

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

d2 x 2 dt

2

=−

4

−k(m1 + m2 ) .x m1m2

i.e. the frequency =

1 k(m1 + m2 ) 2π m1m2

1 300(2 + 3) 2π 2×3  2.5Hz .

=

∴ number of complete oscillations in 1 minute =

60 = 24 . 2.5

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com

5

Chemistry

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

PART – II SECTION – A

1. 2.

Gem-dihalides on substitution by OH produces carbonyl compound, which will reacts with 2,4-DNP but 1,2-dichloroethane will give vic-diol.

MnO 4−

+ Fe+2

0.1 M 0.01M E0  MnO

−2 4

R.P. 

+8H⊕  →

Mn+2

+ Fe +3

0.09 M   Mn+2 

>E

0 +  R.P. H   H2 

So permanganate electrode behave as cathode. Ecell = 1.51 – 0 = 1.51. Using Nernst’s equation :- (n = 5) EMnO−

4

Mn+2

MnO −4  H+  0.059 log  = 1.51 + 5 Mn+2 

8

= 1.51 + 0.11 = 1.4 V 3.

Amine inversion will change the configuration of N only and not that of carbon.

5.

Compound (A) will first undergo Cannizzaro reaction and the product B then will undergo esterification.

6.

NH + HCl  → NH4 Cl ( s ) at room temperature.

3 500×0.3 760×RT

600×0.2 760×RT

0 Moles

150 120 760RT 760RT 30 120 0 760RT 760RT

Now total moles =

30 100 × 1 + 760RT 760RT

Total P = 130 × RT

760RT × 1.5

P in torr =

(Total volume = 1. 5 Litre)

130 = 86.66 torr 1.5

SECTION – C

5.

(S)-2,3-dimethylpentane, (R)-2,3-dimethylpentane, (S)-3-methylhexane and (R)-3-methylhexane.

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com

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

Mathematics

6

PART – III SECTION – A

3

1.

1 1 1 1  f(x) =  x −  +  x −  +  2 4 2 2 3/4

∫

I=

g(x) dx let x = t +

1/ 4

1 ⇒ I= 2

1/ 4

∫

−1/ 4

1  g  t +  dt  2 3

  1   1  1 1  1 1 1 1  f  t +  = t3 + t + ⇒ f  f  t +   =  t3 + t  +  t3 + t  +  4  4 4  2 2 4 2   2   1 1  ⇒ g  t +  = p(t) + where p(–t) + p(t) = 0  2 2 1/ 4

⇒ I=

∫

−1/ 4

2.

1 1 dt = 2 4

Required area is α α  2  ex ln x + 1 − 1 − (x − 1)2 dx   0  1

∫(

∫

)

1 −1

n  r −1 r  1 n ( r n t ⋅ ⋅ = ⋅ Cr 4 − 3r ) C C 3 lim   r t n n   n→∞ n →∞ 5 5 r =1 r =1  t =o  n n   1 1 n n Cr 4r − Cr 3r  = lim n ( 5n − 4n ) = 1 . = lim n   n→∞ 5 n→∞ 5  r =1  r =1  n

3.

lim

∑

1

∑

∑

4.

5.

−1

∑

∑

In triangle OA1A2 2 1  π  2R − 4 cos   = = 2 4 2R 2 2 ⇒ R = 4+2 2 Also 2 2 ⇒ ( r + 1) − ( r − 1) = 4 + 2 2 ⇒ 4r = 4 + 2 2 1   ⇒ r = 1 +  2 

A8

O

A3

R π/4 R A1

A2

Vertex of octagen O1 r (r–1) 1O

1 A2 R

1

 3π  f  = 2π  2 

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com

7

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

 3π  ⇒ f '   ⋅ g' ( 2π ) = 1  2 

⇒ f ' ( x ) = 25(2) ( 2x − 3π )

24

+

4 − sin x 3

3  3π  7 ⇒ f '   = ⇒ g' ( 2π ) =  2  3 7

6.

Let the age of friends be a, ar, ar2, r > 1 and money distributed is x, xr, xr2 respectively. ⇒ (1 + r + r2)x = (a(1 + r + r2) + a)t Also after 3 years ar2 + 3 = 2(a + 3) ⇒ a(r2 – 2) = 3 x + 105 = (a + 3)t and xr + 15 = (ar + 3)t (given) ⇒ (1 + r)x + 120 = (a(1 + r) + 6)t ⇒ 2(a + 3)t = r2x – 120 = 2x + 210 ⇒ (r2 – 2)x = 330 Using (r2 – 2)x = 330, a(r2 – 2) = 3 ⇒ x = 110a ⇒ 110a + 105 = (a + 3)t ⇒ 5(22a + 21) = (a + 3)t and 110ar + 15 = (ar + 3)t ⇒ 5(22ar + 3) = (ar + 3)t 22a + 21 a + 3 ⇒ 5ar = 7a + 6 ⇒ = 22ar + 3 ar + 3 Where 5ar = 7a + 2(a(r2 – 2)) {Using a(r2 – 2) = 3} 3 ⇒ 2r2 – 5r + 3 = 0 ⇒ r = . Hence a = 12 years 2

7.

Given cn = a1 + a2 + a3 + ….. + an where a1, a2, ….., an are in A.P. with d = 2 and dn = b1 + b2 + b3 + …… + bn where b1, b2, b3, ….., bn are in A.P. with d = 2 Also (an, cn) lies on y = px2 + qx + r Now cn = pan2 + qan + r

pan2−1

….. (1)

+ qan−1 + r cn – 1 = ∴ From (1) and (2), we get cn – cn – 1 = p ( an2 − an2−1 ) + q ( an − an−1 )

….. (2)

an = ( an − an−1 ) p ( an + an−1 ) + q

….. (3)

⇒ an = p ( an + an−1 )( an − an−1 ) + q ( an − an−1 )

( an − an−1 = d) 10.

y ⇒ at A, t = – 1 x 1 − 9t 2 4 Slope of tangent at A, m = =− . −6t 3 ⇒ Equation of tangent at A ⇒ 3y + 4x + 2 = 0 Since tangent passes through the point B(t1) hence 3(t1 – 3t13) + 4(1 – 3t12) + 2 = 0 2 ⇒ (t1 + 1)2(3t1 – 2) = 0 ⇒ t1 = 3 2  1 ⇒ B ≡ − , −  . 9  3

y = t(1 – 3t2) ⇒ t =

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com

8

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

11.

Slope of tangents at x = 0 and m1 =

1 3

and m2 = –

1 3

π ⇒ tan φ = . 3

SECTION – B

2.

(A) We have f(x) =

1 lncos−1 x

For domain of f(x), ln ( cos −1 x ) > 0 ⇒ cos−1 x > 1 and cos−1 x ≤ π ⇒ cos π ≤ x ≤ cos 1 ⇒ –1 ≤ x < cos 1 Hence number of integers in the domain of f(x) are 2 i.e., –1 and 0 G G (B) Vector normal to the plane is n = ˆi − 3ˆj + 2kˆ and vector along the line is V = 2iˆ + ˆj − 3kˆ G G x⋅v 2−3−6 7 Now sin θ = G G = = x v 14 14 14 Hence cosec θ = 2 −4

(C) We have J =

∫ (3 − x

2

) tan ( 3 − x 2 ) dx . Put (x + 5) = t, we get

−5 1

2 2 J = ∫ ( 3 − ( t − 5 ) ) tan ( 3 − ( t − 5 ) ) dt

1

=

0

−1

Now K =

∫ ( 6 − 6x + x

2

∫ ( −22 + 10t − t

2

) tan ( −22 + 10t − t 2 ) dt

0

) tan ( 6x − x 2 − 6 ) dx . Put (x + 2) = z, we get

−2 1

K=

∫ ( 6 − 6 ( z − 2) + ( z − 2) ) tan ( 6 ( z − 2) − ( z − 2) 2

2

− 6 ) dz

0

1

∫ ( 22 − 10z + z

2

) tan ( −22 + 10z − z2 ) dz

0

Hence (J + K) = 0 SECTION – C

1.

7   1   7  A  − , − 4, 0  , B  0, − , 1 and C  , 0, 8  .  2   3  3  56 Hence V = 18 10

2.

1+

∑ (3 ·

r 10

Cr + r · 10 Cr

r =1

) =4

10

+ 10 × 29

⇒ α = 1 and β = 5 ⇒k>4 3.

n−6

x=

n

C6 ( 31/ 3 )

y=

n

Cn− 6 ( 31/ 3 )

6

( 4−1/ 3 )6 ( 4−1/ 3 )x −6

and

x 1 n −12 / 3 −1 = (12 ) ⇒ n = 9 ⇒ (12 ) = y 12

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com

9 4.

z2 + z 3 = 0

5.

Let y =

n

∑ r =0

2n+1

2n+1

Cr =

∑

2n+1

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

Cr

r =n +1

⇒ 2y = 22n+1 ⇒ y = 22n ⇒n=3

FIITJEE Ltd., FIITJEE House, 29-A, Kalu Sarai, Sarvapriya Vihar, New Delhi -110016, Ph 46106000, 26569493, Fax 26513942 website: www.fiitjee.com