Answer (6)

(1)

ANSWERS, HINTS & SOLUTIONS

CRT (Set-III)

(Paper–2)

PHYSICS CHEMISTRY MATHEMATICS

Q. No. ANSWER ANSWER ANSWER

1. D C B 2. B A C 3. C A C 4. C D C 5. C A B 6. D B B 7. A B D 8. A A B 9. B, C A, B, C, D A, C 10. A, C A, B, D A, B, C 11. A, B A, B, C A, C 12. A, C B, C B, C, D 1. (A) → (p, q, t) (B) → (p, s) (C) → (p, r, t) (D) → (p) (A) → (p, s) (B) → (p, r, s) (C) → (p, q r, s) (D) → (p, q, t) (A) → (s) (B) → (r) (C) → (q) (D) → (r) 2. (A) → (q) (B) → (p) (C) → (s) (D) → (r) (A) → (p, r) (B) → (q) (C) → (p, r) (D) → (q, s) (A) → (p) (B) → (p) (C) → (p) (D) → (q) 1. 3 4 5 2. 2 1 4 3. 1 3 3 4. 2 2 3 5. 3 6 1 6. 4 5 7

ALL INDIA TEST SERIES

FIITJEE

JEE(Advanced)-2013

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

ta

l

s

e

le

c

ti

o

n

s

i

n

I

IT

-J

E

2

0

1

2

(2)

P

h

y

s

i

c

s

PART – I

1. The z-component of the force and the x-component of displacement are ineffective here.

(

)

y dW=F dy 3xy.dy= ∵z 0=

(

)

4 2 6x dx y x = ∵ =

Integrating between x = 0 and x = 2 gives the result.

2.

6.77 10

2

−

λ

₌

_×

∴

v = 1000 × 2 × 6.77 × 10-2 m/s also 2

RT

MV

v

1.4 M

RT

γ

=

⇒

γ =

=

⇒

diatomic 3.

τ =

mg sin

θ× = α

L I

(

mgL

)

4mL

2

3 θ =

α

T 2

= π

θ

α

mgsinθ θ O

4. The speed is due to radial motion as well as due to angular motion.

5. Young’s modulus, Y F. A l l = × ∆

F is applied force, A is area of cross-section, l is length and ∆l is change in length. From graph, change in length for 20 N force is 1 × 10–4_m

∴ Y 20 1₆ ₄ A(10 )(1 10 )− − × = × = 2 × 1011 N m–2 6.

1 1

1 v u

− =

f

1

u

4

1

1 v

− = =

f

x

−

f

m

f u

=

+

7. Both a and b will be independent of material.

8.

M (I area)

2

1 10 5

4 π

π





=

_

× ×

_

×

=





1

1 W MB 1

5

4 10 J

2 π





 

=

_

−

_

=

×

_{ }

=





 

9. Here F _smg 1 m M   > µ _ + _  

(3)

For m k F− µ mg m.a= For M kmg MA µ = 2 A=0.4m / s

10. Impulse = change in momentum

2 1 2(v v ) = − ˆ ˆ 2(3i j) = +

As impulse is in the normal direction of colliding surface 1 tan 3 θ = 1 1 tan 3 −   θ = _{ }   2 6 θ α 1 1 90 tan 3 −   α = ° + _{ }  . 11. H KAd K(2 rL)d dr dr θ θ = = π 2 2 1 1 r r dr 2 KL _d r H θ θ π ∴

_∫

=

_∫

θ 2 1 2 1 2 KL( ) dQ H 80 dt r ln r π θ − θ = = = π       dm L 80 dt ⇒ = π dm 8 80 dt L 80 4200 4200 π π π ⇒ = = = × Kg/second.

12. Gravitational field inside the cavity is E 4 Gr 3

= πρ

where ρ is mass density and r is separation between centre of sphere and centre of cavity. Applying law of Conservation of energy :

2 esc 1 GMm GMm mV 0 2 − R + 45R = Solving V_esc 88GM 45R = .

(4)

SECTION – B

2. For central maxima, path diff (∆x) = 0 for any point P on the screen.

(

S P

)

[

(

SP x

)

x

]

x=µ_m ₂ − µ_m ₁ − +µ

∆

masses x = thickness of glass slab. = µ_m

[

S₂P−S₁P

]

−

(

µ_m−µ

)

x =

(

)

x 0 D y . d m m − µ −µ =      µ Here,     − − − =       µ µ − µ = t 4 20 5 t 4 20 d Dx x d D y m m =     − − t 4 20 t 4 15 d Dx _…..(i) At time, t = 0 y = 20 15 d Dx × = 20 15 2 . 0 2 1_× _× ₌ _cm ₇_.₅_cm 2 15 m 40 3 200 15 ₌ ₌ ₌ _.

R.I. of medium cannot be less than 1 which become At time

4 19

t= = 4.755. Here after this time R.I. of medium will not change. So position of central maxima at time t = 5 s will be same as at time t = 4.75 s.

∴ y =     − 1 4 . d Dx = 0.2 4 2 1_× _×₋ = - 0.4 m |y| =40 cm.

For speed of central maxima, differentiating equation (i), w.r.t time we get

(

)

2 dy Dx 20 dt d _{20 4t}  ₋    =  ₋   

Central maxima will be at the centre of geometrical centre of screen when R.I. of medium is 5. Hence at this 4 15 t= . ∴      − =    = 25 20 d Dx dt dy 4 15 t = 25 20 x 2 . 0 x 2 1 ₋ ₌ _m_/_s 25 2 _{= 8 cm/s.} Fringe width µ λ = β d D = 5 10 x 100 x 10 x 2 1 10 3 − − =10 -6_{m = 1µm.} y D S1 S2 µ = 20 – 4t µ

(5)

SECTION – C 1. 3 2 2 2 2 2 2 x x y 30 10 dy x x dy and at x 2; 0 dx 10 5 dx d y x 1_{andat x 2;}d y _ve 5 5 dx dx = − = − = = = − = = +

Hence the particle is at it’s lowest (minimum) position, in it’s path at x = 2 m

(

)

( )

2 2 2 2 2 at x 2 2 2 d y dx dx 2x 2 dt dt dt d y 20m / s dt d y N mg m N 1 10 20 30N dt 1 P f.v N V 30 10 3 watt 100 = = −   =       ⇒ − = ⇒ = + = = = µ = × × =

N

Mg

2 2 d y dx

Approximate graph

2. Using conservation of charge principle and induction, do charge distribution on each plate.

3. N = mg µ N = m r ω2 µ mg = m. 2 sin θ ω2 g 2sin µ ω = θ µN N mg mrω2 0.1 10 2 sin30 × ω = × ω = 1 rad / s. 4. m I S= ₂ S BA AD= × ˆ ˆ BA=2di 2aj− ˆ AD 2bk= ˆ ˆ ˆ S 2(di aj) 2bk= − × x y z A B C D ˆ ˆ M= −4bI(dj ai)+ 2 2 | M | 4bI d= +a = 2 Tesla. 5. Nuclear reaction is 1 3 3 1 1H +1H →2He +0n + Q Q = – 1.2745 MeV

(6)

We get, K2 = 1 .44 MeV. K1 = 3 MeV. 6. x = A cos ωt dx v A sin t dt = = − ω ω dx v A sin t dt = = − ω ω 2 2 2 d x a A cos t dt = = − ω ω 2 max a = ω A τmax = I αmax Iamax R = 2 2 1 A MR 2 R ω = 1_{MR A} 2 2 = ω τ_max =4N.m

(7)

C

h

e

m

i

s

t

r

y

PART – II

SECTION – A 3.

[

]

1 2 3 ₄ a 2 3 2 3 ₇ a 3 H HCO K 10 H CO H CO K 10 HCO + − − + − − −             = =             = =      

(

)

[

]

1 2 2 2 3 2 3 2 2 2 ₃ 3 ₃ ₂ ₃ 3 2 2 a a a 3 3 CO ₁ ₁ f CO H CO HCO CO _{H CO} HCO _H _H 1 ₁ CO CO K K K − − − − − ₊ ₊ − −       = = =   ₊  ₊    _ _ _ _               _{+ +}  __ __ __ __ + +             10 5 11 7 1 0.01 10 10 ₁ 10 10 − − − − = = + +

5. This diasaccharide is sucrose which is formed by

O OH OH OH CH₂OH H H H H OH O OH OH CH₂OH H H OH CH₂OH O H 6 4 3 1 5 2 6 2 1 5 4 3 & α-D-(+)-Glucopyranose.

6. For SHE, E0=0 at every temperature.

7. No of molecules of K2S in 10 L water = 30.11×4×1022

No. of moles of K2S = 2 mol

Molality = 2×10−1_m

∆Tf = iKfm = 1.116oC

Tf = 272.034 K

8. MO2 + 4H+ + PO43− + e− → MPO4 + 2H2O

M + PO43− → MPO4 + 3e−

The net cell reaction will be

3MO2 + 12H+ +4PO43− + M → 4MPO4 + 6H2O

11. T → constant

PV = nRT (where R and T → constant) n P V = I n P V =

(8)

II n P 2V = III n P V = IV 2n P 2V = II IV 1 II P P P P ∴ < = =

⇒ Average K.E. depends only on absolute T. ∴(Av. KE)I = (Av. KE)II =(Av. KE)III=(Av. KE)IV

Mass Density Volume ⇒ = ∴dII < dI < dIV = dIII 12. E 13.6Z₂2 n = − For Li Z = 3 Energy of 1s for Li E₁ 13.6 9 1 = − × Energy of 2s for Li E₂ 13.6 9 4 = − × 1 2 E E ∴ <

So statement (A) is correct.

Shape of all the d-orbital are not same. So statement (B) is wrong. Energy of 3d > energy of 4s So (C) is wrong statement. SECTION – B 1. (A) O O O

Gives positive test with 2, 4-dinitrophenyl hydrazine and form highly stable hydrate.

(B)

CH C

OH

O Gives positive test with 2, 4-dinitrophenyl hydrazine

(>C=O group test) and Fehling solution ( C OH

group test). It forms highly stable hydrate also.

(C) O

C H CI3

Gives positive test with 2, 4-dinitrophenylhydrazine, Tollen’s reagent and Fehling’s solution (due to C

O H group). It forms stable hydrate also.

(D)

H

O Gives positive test with 2,4-dinitrophenylhydrazine and

Tollen’s reagent (due to –CHO group). It gives Perkin’s reaction (being an aromatic aldehyde)

(9)

2. Use relations:

(

)

1 pH pKw pKa logC 2 = + + Salt pOH pKb log Base   = + _ _  

(

)

1 pH pKw pKa pKb 2 = + − pH pOH 14+ = SECTION – C 1. CH3OH(g) ←→ CO(g) + 2H2(g) Initially 0.2 mole 0 0 At eq. (0.2−x) x 2x 2 H _mix mix mix mix r _M 2 2 r 2 M 16 32(0.2 X) X(28) 2X(2) M 0.2 X X 2X X 0.1 = = = − + + = − + + = Kc = = × = − 2 x(2x) 0.1 0.04 0.04 0.2 x 0.1 100 Kc = 4 2. It is a pyrosilicate. Pyrosilicate ion is 6 2 7 Si O− Si O O O O Si O O O In this two units of 6

2 7

Si O− _{joined along a corner containing oxygen atom.}

3. 2 O O O CH₃ H CH₃ H O ∆ → COOH C H₃ O H and O O O CH₃ H H CH₃ O cis trans

Cis form is optically active. It gives two stereoisomers → d−cis, l−cis. Trans form is optically inactive because it contains center of symmetry.

4. CH2=C=CH2 H /H O2 ₃ ₃ O ||

CH

C CH

+

→

− −

(10)

5. CO CO CO Fe Fe CO CO CO CO CO CO 6. Hg2Cl2 → Hg + HgCl2 ∆H = ? 2Hg + Cl2 → Hg2Cl2 ∆Hf1 = −125 KJ / mole Hg + Cl2 → HgCl2 ∆Hf2= −640 KJ / mole ∆H = ∆Hf2 − ∆Hf1 = −640 + 125 = −515 KJ/ mol −103 x = − 515 ⇒ x = 5

(11)

M

a

t

h

e

m

a

t

i

c

s

PART – III

SECTION – A 1.

( )

e e e 1 I 1 _II 1 f x I f x ln x dx ln x.f x dx x ′ ′′ =

∫

= −

∫

I = I − I1 I1 =

( )

e 1 1_{f x dx} x ′

∫

= 1 1 e−2 ∴ I 1 1 1 3 1 e 2 2 e = − + = − 2.

( )

( )( )( )

4x 2a p x a p x y z a b c det B 4t 2b q 4 2 1 y b q 8 a b c 8 p q r 4z 2c r z c r p q r x y z − = − = − = − = − − = −8 × 2 = −16. 3. Replace x → 1

x and solve to get

( )

x 1 f x 2 + =

(

)

10099 1 10100 f 10099 5050 2 2 + = = = 4. cubic is x3 − 6x2_{+ 11x − 30 = 0} (x − 5) (x2_{− x + 6) = 0} Hence a = 5 ; bc = 6 ⇒ bc 6 a =5 5. r = 7 sin30° = R r 1 R r 2 − = + 2R − 2r = R + r R = 3r = 21. 6. 2 0 0 1 1 1 c A 0 c / b 1 0 2 2 ab c / a 0 1 = = −

If A is independent of a, b and c then c2 = ab ⇒ a, c, b are G.P.

7.

( )

1 1 n 1 n n 1 tan t 1 C dt n sin t − − + =

_∫

(12)

(

)

1 1 1 n 1 1 2 n 1 n 1 a a n n 1 tan t_dt sin t tan t L lim n C lim n dt 0 ; L 1 sin t n − − − + − →∞ →∞ + = ⋅ = ⋅ ∞ × =

∫

applying Leibnitz rule.

8. 2f(x) and f x

2    

  have the same domain as f(x) and f(2x), f(x + 2),

f f

2    

  have the range a f(x). ⇒ m = 2, n = 3

Verify by considering f(x) = sin−1x ; d : |x| ≤ 1 ; R , 2 2 π π ₋      10. (A)

[

]

x 0 x x f(x) lim ; x → + = [ ]2x _{if x 0} x f(x) 0 if x 0 x  >  =   _<  ⇒ x 0lim f(x) 0→ = (B) 1/ x_{1/ x} x 0 xe lim 1 e → ₊ ; x 0 1/ x x lim f(x) 0 1 e + − → = ₊ = − ; ( ) x 0 0 h lim f(x) 0 1 − → − = = (C) ( )1/ 5

x 3lim x 3→ − Sgn (x – 3) = 0 where sgn is the signum function (bvious)

(D) 1 x 0 tan x lim x −

→ does not exist as RHL = 1; LHL = –1

11. Area (T) = 2 3 c.c c 2 = 2 Area (R) = e 3 2 0 c _{x dx} 2 −

∫

= 3 3 3 c c c 2 − 3 = 6 ∴ 3 ₃ c 0 c 0 area (T) c 6 lim lim 3 area (R) 2 c + + → = → ⋅ = 12. y x 3 0 x 1 + = > + ⇒ x < –3 or x > –1 As x → –3, y → 0; x → ∞, y → 1 range (0, 1) ∪ (1, ∞) SECTION – B 1. We have

(

)

( ) i2 i2 n 1 n _n _n z 1 z 1 z e ... z e π π ₋         − = − − −        

(

)

i2 (n 1) n 1 _n 1 z ... z z 1 .... z e π ₋ − _ _ + + + = − −     ……(1)

Put z = 1 and take modulus _{n 2}n 1_sin _sin2 _{... sin n 1}

(

)

n n n

− π π π

= −

(A) if n = 21

20 2 9 10 20

21 2 sin sub ...sin sin ...sin

21 21 21 21 21

π π π π π

(13)

21 = 2 20 10 sin ...sin 2 21 21 π π  _{ ⋅}     ∴ sin ... sin10 ₁₀21 21 21 2 π π = (B) If n = 22 2 21 2 10

22 2 sin sin .... sin

22 22 22 π π π   = _ _   ∴ sin ... sin10 22₁₁ 11₁₀ 22 22 2 2 π π₌ ₌

(C) Again put z = –1 is ……..(1) and take modulus

(

)

n 1

2

1 cos cos ... cos n 1 2

n n n − π π π = − . n = 21 20 2 20

1 cos cos .... cos 2

21 21 21 π π π = ∴ cos ... cos10 1₁₀ 21 21 2 π π₌

(D) sin , sin2 ... sin10 11₁₀

22 22 22 2

π π π

=

10

10 9 11

cos cos .... cos

22 22 22 2

π π π ₌

∵sinθ =cos 90

(

− θ

)

2. (A) 1st, 4th, 7th terms are a, a+3d, a+6d

ax+by+c=0

ax+(a+3d)y+(a+6d)=0

a(x+y+1)+3d(y+2)=0 passes through (1, –2) (B) a, b, c are three consecutive terms of A.P.

a = A+(m–1)d, b = A+md, c = A+(m+1)d.

(A+(m–1)d)x+A+md)y+A+(m+1)d = 0

(

)

x A x y 1+ + +d m_ +my m x 1+ − + _=0 ∴ x + y + 1 = 0, –x + 1 = 0 x = 1, y = 2 (C) a = A + (r–1)d, b= +A

(

r2−1 d

)

, c= +A

(

2r2− −r 1 d

)

(

)

(

_A_{+ −}_{r 1 d x}

)

₊__A₊

(

_r2₋_{1 d y A}

)

_ _{+ +}

(

_2r2_{− −}_{r 1 d 0}

)

₌  

(

) (

)

(

)

(

)

A x y 1+ + +d r 1 x− _ + +r 1 y+ 2r 1+ _=0 X + y + 1 = 0, x + y + 1 + r(y + 2) = 0 Y = –2, x = 1 (D) a = A + (r–1)d, b= +A

(

r2−1 d

)

, c= +A

(

3r2−2r 1 d−

)

(

)

(

_A_{+ −}_{r 1 d x}

)

₊__A₊

(

_r2₋_{1 d y A}

)

_ _{+ +}

(

_3r2₋_{2r 1 d 0}₋

)

₌  

(

) (

)

(

)

(

)

A x y 1+ + +d r 1 x− _ + +r 1 y+ 3r 1+ _=0 x + y + 1 = 0, x + y + 1 + r(y + 3) = 0 y = –3, x = 2 SECTION – C

(14)

1. Let A (z1), B(z2) and P(z) forms a triangle ABC then 1 2 tan / 2 3 tan / 2 2 θ = θ implies PB – PA = AB 5 . This

implies locus of P is branch of hyperbola of eccentricity = 5

2. 1 3 5 7 2 4 6 S S S S tan7x 0 0 1 S S S − + − = ⇒ = − + − 3 5 7

7 tan x 35 tan x 21tan x tan x 0

⇒ − + − =

3 2 2 3

tan , tan , tan ,0,tan ,tan ,tan

7 7 7 7 7 7

π π π π π π

⇒ − − − are the roots of equation

3 5 7

7x 35x− +21x −x = 0

2 22 23

tan ,tan ,tan

7 7 7

π π π

⇒ are the roots of equation _{7 35x 21x}₋ ₊ 2₋_x3 ₌₀

2 22 23

cot ,cot ,cot

7 7 7

π π π

⇒ are roots of equation 7x3−35x2+21x 1 0− =

2 22 23

cot cot cot 5

7 7 7 π π π ⇒ + + = 3. (A) 2x + 2yy′ = 0 ⇒ y ' x y = − ∴ y '

( )

2 = − = 1 A

(B) cos y.y′ + cos x = sin x.cos y.y′ + sin y cos x when x = y = p

–y′ – 1 = 0 + 0

(C) 2exy (xy′ + y) + ex_ey_{y′ + e}y_ex_{– e}x_{– e}y_{y′ = e.e}xy_{(xy′ + y)}

at x = 1, y = 1

2e(y′ + 1) + e2_{y′ + e}2_{– e – ey′ = e}2_{(y′ + 1)}

ey′ + e = 0 ⇒ y′ = – 1 Hence |A + B + C| = 3 4. Focus is 11,0 6  _⇒  

  Distance of focus from line =

5

2 ⇒ − = a b 3

5. Put a_n=tanθ _n

6. fK 2−

( )

x =0 has K distinct roots, fK 1−

( )

x =0 has K+ 1 distinct roots, fK

( )

x = has K distinct 0