Sol Aits 2013 Ft i Jeea Paper 1

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

JEE(Advanced)-2013

FIITJEE

ANSWERS, HINTS & SOLUTIONS FULL TEST –I (Paper-1)

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 C

CHEMISTRY A

MATHEMATICS D

2.

C

A

A

3.

A

B

B

4.

C

C

A

5.

D

B

D

6.

C

C

B

7.

A

D

B

8.

A

C

B

9.

C, D

A, C, D

A, C

10.

A, C

A, C

A, B

11.

A, C

A, B, C

B, C, D

12.

A, B, C, D

A, C

13.

A

A, B B

14.

A

A

C

15.

C

B

D

16.

B

D

B

17.

D

A

B

18.

A

D

A

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

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

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

1.

2.

B

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2

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

Physics

PART – I SECTION – A

1.

E2 = VR2 + (VL − VC )2 = VR2 + 0 As, E = VR ∴ VR = V = 220 V E 220 I= = = 2.2 amp R 100

3.

Equivalent circuit can be drawn as 79 Equivalent capacitance = C. 30

C

C 2C

C

C

C

C

2/3C

C

C C

⇒

C

C

2C

C

8/3C C

2C

⇒

C

C

2C

⇒ 19C/11 C

This is simply discharging of a capacitor.

q = Q 0e − t / CR where C = 4πε0r

∴ 5. 6.

When the circuit will reach to steady state the inductor will given zero resistance. Hence the entire current will only pass through it. H + 2H 3H = X= 2 2

8.

Here V ∝ 1/r r → distance between Sun and Planet.

9.

(y − h) + x 2 + h2 = A dy + dt

x

dx =0 x + h dt dy x dx =− 2 2 dt dt x +h 2

dt

2

= v 2A

aB = v 2A

A

y

2

dy 3 = − vA dt 5 3 | uB |= v A 5

d2 y

hh=4cm =4m

4.

B

hx=3cm =3m

…(i) h2

(x 2 + h2 ) 3

2

16 (5)3

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3

aB =

10.

16 2 vA 125

Mg – T = Ma T = ma Solving (i) and (ii) Mg a= (M + m) FBD of man Mg – N = Ma Mmg N= (M + m)

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

…(ii) …(i) …(ii)

N a Mg

11.

2πm T= = 3.14 s. qB

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4

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

Chemistry

PART – II SECTION – A O

1.

H

(1) O3 Et  →

OH Et

( ) 2 4 ( 2 ) NaBH4 → Et  2 Ag O/NH OH

O

2.

C PCl5 → O 

C

C

Cl

← 

−

CHO

COO

3.

HO H H

1 C 2 H C 3 HO C H 4C 5 H C HO

OH

H

CH2OH PCC

OH ← 

O

LiAlH 4

Cl

CHO

CH2OH

OH

Et

O

O

H

O

O

C

H

+

H →

OH

O

1 C 2 C 3 C 4 C 5 C

O

O

and

OH

6CH OH 2

CH2OH

H OH H

O

OH

Cyclic forms of diastereomers that differ only in the configuration of C1, are known as anomers.

6CH OH 2

The configuration differ only at carbon no. 1. 4.

For equimolar solutions, n1 = n2 ∴ x1 = 0.5 and 0.5 PB = xB × PBo = 0.5 × 160 = 80 mm PT = x T × PTo = 0.5 × 60 = 30 mm

PTotal = 80 + 30 = 110 mm

Mole fraction of toluene in vapour phase = 5.

PT 30 = = 0.27 PTotal 110

Entropy is a state function, i.e. the change entropy depends upon the initial and final states of the system and not on how that change is brought about.

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5

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

7. 0.6∆ o

d − orbitals in symmetrical

−0.4∆ o

field of ligands

high spin d4 configuration in an octahedral field

CFSE = 3(- 0.4) ∆o + 0.6 ∆o = - 1.2 ∆o + 0.6 ∆o = -0.6 ∆o Given (K sp )

8.

AgBr

= 5 × 10−13

KBr + AgNO3  → AgBr + KNO3

Given, [AgNO3] = 0.05 M ⇒  Ag+  = NO3−  = 0.05M

 Ag+  Br −  = (K sp ) AgBr ⇒ 0.05 × Br −  = 5 × 10 −13 5 × 10 −13 ⇒ Br −  = = 1× 10 −11M 0.05 ∴ Br −  = [KBr ] = 1× 10 −11M wt. of KBr = 1× 10−11 × 120 = 1.20 × 10−9 gm / litre

10. H H Br Br Plane of symmetry

11.

(A)

anh.ZnCl2 Ph − CH2 − OH  → Ph − CH2 − Cl + Zn ( OH) Cl ( immediate ) conc.HCl

( white turbidity ) CH3 − CH2 − OH → CH3 − CH2 − Cl + Zn ( OH) Cl

( slow )

Lucas reagent

(B)

Lucas Ph − CH2 − OH  → PhCH2 Cl + Zn ( OH) Cl reagent

(immediate )

( White turbidity ) Ph − OH  → No reaction under normal condition Lucas reagent

(C)

Lucas → ( CH3 )2 CH − Cl + Zn ( OH) Cl ( CH3 )2 CH − OH  reagent ( White turbidity )

(Moderate rate )

Lucas CH3 − CH2 − CH2 − CH2 − OH  → CH3 − CH2 − CH2 − CH2 − Cl + Zn ( OH) Cl reagent

( Slow )

( White turbidity )

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

12.

6

H2S + O3  →H2O + S + O2 PbS + 4O3  →PbSO4 + 4O2 3SO2 + O3 → 3SO3

3SnCl2 + 6HCl + O3  → 3SnCl4 + 3H2O

13.

Zn + Fe+2  → Fe + Zn+2 0.0591 logK eq n Given, E = 0.2905 0.0591 0.01 ⇒ 0.2905 = Eo − log 2 0.001 ⇒ 0.02905 + 0.0295 log10 = 0.2905 + 0.0295 = 0.32 0.0591 ∴ Eo = logK eq 2 0.0591 ⇒ 0.32 = logK eq 2 E = Eo −

0.32

⇒ K eq = 10 0.0295

14.

For 1 mole of O2 2   O 2 → 2O2 −  i.e. × 3O2 −  3   4 Or mol of Al to change into Al+3 ions. 3 n=4 Thus, ∆G = −nEF ⇒E= −

∆G +827000 = = 2.14 volt nF 4 × 96500

0.059 1.1× 2 logK eq ⇒ = 37.2881 = logK eq 2 0.059 ⇒ K eq = 1.9 × 10 37

15.

Eocell =

16.

Copper sulphate when reacts with hypophosphorus acid, gets reduced to cuprous hydride. 4CuSO 4 + 3H3PO 2 + 6H2 O → 2Cu2H2 + 3H3PO 4 + 4H2 SO 4

17.

3Ba ( OH)2 + 2P4 + 6H2O  → 3Ba (H2PO2 )2 + 2PH3

( J) (L ) (K ) → BaSO4 + 2H3PO2 Ba (H2PO2 )2 + H2 SO4  (L ) ( N) ( M) gives apple green colour in the flame

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7 18.

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

H4P2O7 (pyrophosphoric acid) is a tetrabasic acid, i.e. 4-hydroxyl groups are present. It is prepared by removing one water molecule from two molecules of orthophosphoric acid. Each phosphorus atom lies in same tetrahedral environment, so called isopolyacid. O O O O HO

P

P

HO

+

OH

−H2 O OH  → HO

OH

OH

P

P

O

OH

OH

OH

SECTION-B

1.

(A)

H3C

CH

(1) OH COOH  → H3C ( 2 ) H+ −

CH

Br (B)

H3C

COOH

OH

CH

CH2

COOH

(1) OH−  → ( 2 ) H+

H3C

CH

CH

COOH

Br O

(C)

C H 2C

CH2

CH2

OH−

→ H2 C OH 

C

Br

CH2

CH2

Br

O

C

O

O

O

O

(D)

C H2C

CH2

CH2

CH2

Br

C

−

OH → OH 

O

O

Reactions A, C, and D show SN2 reactions. 2.

In case of (A), compound given in (A) exchanges – H with – D and becomes racemic. In case (B), compound in (B) reacts through a carbocation and gives racemic mixture. In case of (C), no bond attached to the stereocentre breaks in reaction of (C), so retention. In case of (D), compound in (D) gives Hoffmann elimination and it is an example of antielimination. Product does not have any stereogenic centre. D Reaction (B ) gives H3C

OH

( d, A − form )

C2 H 5 D Reaction ( C ) gives H3C

COOH C2H5

H3C Re action (D ) gives

CH3

(No stereocentre )

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8

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

Mathematics

PART – III SECTION – A

1.

We must have f(a2 + 2a + 5) > f(a + 11) ⇒ a2 + 2a + 5 < a + 11 ⇒ a ∈ (–3, 2) ∴ a can take values –2, –1, 0, 1

2.

∆ABC is equilateral ⇒a=b=c

3.

AC2 = AB2 + BC2 ….. (1) BD2 = BC2 + CD2 ….. (2) CA2 = CD2 + DA2 ….. (3) ….. (4) DB2 = DA2 + AB2 Equation (1) + (3) and (2) + (4) gives AC = BD So, equation (1) and (2) give AB = CD and equation (2) and (3) give BC = AD So ∆′s CBD and ADB are congruent ⇒ ∠CBD = ∠ ADB = 90º – ∠ABD ⇒ ∠CBD + ∠ABD = ∠CBA ⇒ DB lies in the plane of ∆ABC, so the points are coplanar

4.

Let a – d = 3t; then bc = (d + zt)(d + t) = d2 + 3td + 2t2 a 3t  3t  Also = 1 + , so hk = d2 1 +  = d2 t + 3td < bc d d d  ⇒ bc > hk

5.

Clearly P ≡ (1, 2) Equation of tangent at P is 4x(1) + y(2) = 8 ⇒ A ≡ (2, 0) and B ≡ (0, 4) Similarly normal at P is 2x – 4y = 6  3 ⇒ A '( −3, 0) and B '  0,   2

B

P A A′

B′

0 4 1 1 5 ∴ Area BPB' = 1 2 1 = and Area APA ' = 5 2 4 3 0 1 2 ∴ Ratio = 4 6.

As sin x + sin y ≥ cos α cos x ∀ x ∈ R π Let x = − ⇒ sin y ≥ 1 ⇒ sin y = 1 2 ⇒ 1 + sin x ≥ cos α cos x ⇒ cos α cos x – sin x ≤ 1 ⇒

cos2 α + 1 = 1 ⇒ cos α = 0 ⇒ sin y + cos α = 1 + 0 = 1

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9

1

7.

I=

∫

sin x x

0

⇒ I<

1

dx <

∫ 0

2 and J = 3

x x 1

∫ 0

dx =

cos x x

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

2 3 1

∫

< x −1/ 2 dx = 2 0

⇒J a ⇒ b + c > 6 ⇒ b + 2b > 6 ⇒ b > 2 b2 + 4b2 − 6 Again < 1 ⇒ b < 6 ⇒ b ∈ (2, 6) and consequently c ∈ (4, 12) 4b2

Let AD be angle bisector ⇒

y

BE 3 = , take BE = 3K and EC = 4K EC 4 ⇒ A must be (3K, 6) 6 Now equation of line BA is y = x 3K ⇒ Ky – 2x = 0 Since perpendicular from c on AB = 7 0 − 14K 4 = 7 ⇒ K2 = ⇒ 2 3 K +4 2

A(3K, 6) E

x′

B

E

C(7K, 0)

x

y′

2

Now c 2 = ( 6 − 0 ) + ( 3K − 0 ) = 36 + 9K2 = 36 + 12 ⇒ c = 48 = 4 3 1

1

∫

16–18. Let 12 yf(y)dy = a and 20 0

∫ ( y f(y)) dy + 4 = b 2

0

2

⇒ f(x) = ax + bx ⇒ f(y) = ay2 + by 1

1

a b a b  ⇒ f(x) = 12x 2 y ( ay 2 + by ) dy + 20x y 2 ( ay 2 + by ) dy + 4x = 12x 2  +  + 20x  +  + 4x 4 3 5 4 

∫ 0

⇒ a = 3a + 4b ⇒ a = –2b and 4a + 5b + 4 = b ⇒ a + b = –1 ⇒ a = –2, b = 1 ∴ f(x) = −2x 2 + x

∫ 0

….. (1) ….. (2)

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x

11

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

SECTION – B

1.

−2 ; x ∈ [ −2, 1) π  Clearly we have f(x) =  −1 ; x ∈ [ − 1, 0) and g(x) = sec x; x ∈ R – (2x + 1) 2  x ; x ∈ [0, 2] 

 −2 ; sec x ∈ [ −2, − 1)  We have fog ( x ) =  −1 ; sec x ∈ [ −1, 0) secx ; secx ∈ [0, 2]   2π   2π 4π   4π  −2 ; x ∈  − 3 , − 3  ∪  3 , 3  − {−π, π}      x = −π, π ∴ fog ( x ) =  −1 ;   π π secx ; x ∈ − ,   3 3  Limit of fog exist at x = –π, π, –1, points of discontinuity of fog are –π, π and points of 5π differentiability of fog are −1, 6

{ }

 π sec( −2) ; x ∈ [ −2, − 1) − − 2  x ∈ [ −1, 0) Again g(f(x)) =  sec( −1) ;  π  secx ; x ∈ [0, 2] −  2 Limit of g(f(x)) does not exist at x = –1

{}

2.

(A) Let Ei ; i = 1, 2, 3, 4 represents the events that the bag contains I white balls: Clearly 1 P (Ei ) = . Let W be the event that the ball drawn is white then, 4  W  1 1 2 3 4 5 P(W ) = P (Ei ) ⋅ P   =  + + +  =  Ei  4  4 4 4 4  8

∑

 E  P (E 4 ) P ( W / E 4 ) 1/ 4 2 p Now P  4  = ⇒p=6 = = = W P(W ) 5 / 8 5 15 (B) Required number of ways = 12 C1 + 12 C2 ( 2 C1 + 2 ⋅ 2 C2 ) + 12 C3 ( 3 C1 + 2 ⋅ 3 C2 ) + ..... +

(

= 12

C1 + 2 ⋅ 12

=

12

∑r ⋅ r =1

12

C2 + 3 ⋅

12

C3 + ..... + 12 ⋅

12

∑

Cr + 12

10

12

C2 ) + 2 ( 12 C2 ⋅ 2 C2 +

12

12

C12 ( 12 C1 + 2 ⋅

C3 ⋅ 3 C2 + ..... +

12

12

C2 )

C2 ⋅

12

C2 )

Cr − 2 = 12 × 211 + 12 × 11× 210

r =2

= 13 × 210 × 12 = 13 × 210 × m ⇒ m = 12 5x 5y 5z y z   x (C) + + = 5 + + 2−x 2−y 2−z 2 − x 2 − y 2 − z  

 1 1   x − 2 + 2 y − 2 + 2 z − 2 + 2  1 + + = 5  −3 + 2  + + = 5   2−y 2−z   2−x  2 − x 2 − y 2 − z   Now apply A.M ≥ H.M on 2 – x, 2– y and 2 – z

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

12

2−x+2−y+2−z ≥ 3

3 1 1 1 9 ⇒ + + ≥ 1 1 1 2−x 2−y 2−z 5 + + 2−x 2−y 2−z 5x 5y 5z 9  + + ≥ 5  −3 + 2 ⋅  = 3 Hence 2−x 2−y 2−z 5 

⇒

∴ Least value = 3 n

(D)

∑12 ⋅ K K =1

12

CK ⋅

11

CK −1 = 122

12

∑(

11

K =1

= 12 ⋅

CK −1 ) = 122 ⋅ 2

22! 11! 11!

21⋅ 19 ⋅ 17.....3 − 212 ⋅ 6 ⇒ p = 6 11!

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