Answer (8)

JEE(Advanced)-2013 ANSWERS, HINTS & SOLUTIONS CRT(Set-VI) (Paper-1) ANSWERS KEY

ALL INDIA TEST SERIES

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FIITJEE

Q. No.

PHYSICS

CHEMISTRY

MATHEMATICS

ANSWER

ANSWER

ANSWER

1.

A

D

B

2.

B

B

B

3.

A

C

B

4.

A

A

C

5.

C

A

D

6.

D

A

D

7.

B

C

D

8.

A, D

B, D

A, D

9.

B, D

A, B, D

B, C, D

10.

A, D

A, B

B, C

11.

A, B

A, C, D

C, D

12.

B

B

A

13.

B

A

D

14.

B

C

B

15.

C

B

C

16.

B

C

D

1.

4

1

2

2.

3

2

2

3.

1

9

3

4.

4

4

0

5.

2

5

2

6.

1

5

3

7.

4

3

1

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2

Physics

AITS-CRT(Set-VI)-Paper-1-PCM(S)-JEE(Adv)/13

PART – I SECTION – A

1.

For (1) particle, ut − t1 =

1 2 gt = −H 2

u ± u2 + 2gH u + u2 + 2gH = g g

For (2) particle, t2 =

−u ± u2 + 2gH −u + u2 + 2gH = g g

∴ ∆t = 2u/g 2.

∆W = −∆u m 2 2 ω =q E 4 qE ⇒ω= 2 m

6.

fav =

n1f1 + n2 f2 + n3 f3 n1 + n2 + n3

f  Cp =  + 1 R 2   7.

Potential drop = 2 volt ∴ q = CV

11.

If r1 and r2 are the resistances of the first and the second coils and v = mains voltage. V2 t2 r t V 2 t1 1 Q= = ⇒ 1 = 1 = r1 r2 t 2 2 r2 In series req = r1 + r2 v 2 t3 v 2 t1 ⇒ Q= ⇒ t3 = 3t = 45 min = r1 + r2 r1 In parallel v 2 t 4 ( r1 + r2 ) v 2 t1 = Q= r1r2 r1 ⇒ t4 =

12.

r2 t = 10 minutes. r1 + r2

Rsh(I – Ig) = Rg . Ig Rg 100 100 = 4 Ω = Rsh = Ω I 10 − 1 9999 −1 Ig

I

Ig

~

Rg

|I – Ig | Rsh

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AITS-CRT(Set-VI)-Paper-1-PCM(S)-JEE(Adv)/13

13.

3

-3

Ish = 10 – 10 = 9.999 A Power dissipated 100 = 0.9999 Watt. 9999 which is less than 1 watt. Hence it can be used.

P = I2shR sh = (9.999)2 ×

14.

15.

From the data given if orbital speed v = 2 × 105 m/s mv 2 GMm = , where M is mass of the massive object and m is mass of particle r r2 Rv 2 6 × 9.5 × 1015 × 4 × 1010 = ∼ 1037 kg . ⇒ M= G 6.67 × 10−11 Mass of the object Mass of sun

1037 = 107 > 50 1030

∴ It is a black hole. 16.

Rs =

2GM 2 × 6.67 × 10−11 × 6 × 1024 = ≈ 10−3 8 2 C2 ( 3 × 10 )

∴ Radius compression factor =

10−3 ≈ 10−9 6400 × 103

SECTION – C 2.

IR2 + IL2 = (1.5)2 + ( 2 ) = 2.5mA 2

Line current, I =

Circuit impedance, Z = 3.

V 15V = = 6kΩ I 2.5mA

Mirror formula 1 1 1 + = u u f 1 du 1 dv − 2 − =0 u dt v 2 dt 2

dv v 2 du  10  = 2 =   × 4 = 1m / sec dt u dt  20  4.

R

cos φ =

2

 1  R + − ωL  ω C   Putting the values, C = 500 µC 2

7.

N  N  2  N′ =  n + 2n 2  2 2  Where n is no of half lives elapsed for species A N′ 5  1 1  = =  n +1 + n  Now, +1 N 32 2  2 2 

⇒ n=4

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4

Chemistry

PART – II SECTION – A

1.

Heat capacity and internal energy are exclusive property as their value depends upon mass of sample.

2.

Top to bottom stability increases among sulphates of alkaline earth metals.

3.

N2 has small size, no polarity and smaller surface area.

4.

Both have still a replacable hydrogen thus behaves as an acid.

5.

Due to maximum number of unpaired electron,

6.

Due to angle strain cyclopropane will also give Br2-water test but cyclohexane is free from angle strain

7.

CH3

(A)

8.

CH3

CH3

OH

Cl

(B)

(C)

(A) B – F bond length is smaller in BF3 than in BF4− due to back bonding in BF3. S

S

(B) is more stable due to back bonding. (C) Triplet carbene is more stable than singlet carbene. (D) 9.

( CH 3 )3 Si − OH

is more acidic because

( CH 3 )3 SiO−

is more stable due to pack bonding.

A  → B + 3C

(1− 0.1)

( 0.1)

( 0.3 )

PT = (1 + 3 × 0.1) = 1.3 atm ∆P = 0.2 atm or 76×0.3 cm of Hg or 760×0.3 mm Hg = 228 mm of Hg. 10.

(A) Due to strong salvation of smaller Li+, much energy is released which makes LiClO4 more soluble (B) Na reacts with ethanol but not with diethyl ether

C2 H 5OH + Na  → C2 H 5ONa +

1 H2 2

(C) Top to bottom basicity increases in alkali metal hydrides. 11.

+ + Since HCOOH is weak its  H  is always less than  H  given by HCl solution having same

concentration (ii) HCOONa and NaOH cannot form buffer a mixing 12.

E has abnormally higher IE1 value

13.

Due to very high IE values.

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AITS-CRT(Set-VI)-Paper-1-PCM(S)-JEE(Adv)/13

16.

5

H

Solution for the Q. No. 14 to 15 (P)

(Q) (R)

SECTION – C 1.

Let n atoms be ionized

6 × 1023 × 10 −22 × EA = n × IP 6 ×1023 × 10−22 × 3.6 n= 13 = 16.67 ⇒ 1.66 × 101 2.

trans & cis

3.

Degree of unsaturation in product is 7. Hence, reactant must be having 7 + 2 = 9. C 2 H5

4.

H3C

H

H3C

Br

4-product

C 2 H5

5.

λ2 = 30.4 × 10 −7 cm, λ1 = 108.5 × 10−7 cm + Let excited state of He be n2. It comes from n2 to n1 and then n2 to 1 to emit two successive Given

photon. 1 1  1 = RH × Z 2  2 − 2  λ2  1 n1 

1 1  1 = 109678 × 4 ×  2 − 2  ⇒ n1 = 2 −7 30.4 ×10 1 n1  Now for λ1 : n1 = 2 and n2 = ?

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AITS-CRT(Set-VI)-Paper-1-PCM(S)-JEE(Adv)/13

1 1 = RH × Z 2  2 − 2  λ1  2 n1  1

1 1 1 = 109678 × 4 ×  2 − 2  ⇒ n2 = 5 −7 108.5 ×10  2 n1 

6.

All oxygen atoms are peroxide oxygen.

7.

M.wt. of

Fe ( SCN )3 is 230 g.

19 g water and 81 g  Fe ( SCN )3  is present in 100 g of compound.

∵ 81 g  Fe ( SCN )3  combines with 19 g H2O 19 ∴230 g  Fe ( SCN )3  = × 230 = 54 g 81 54 No. of water molecules = =3 18

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AITS-CRT(Set-VI)-Paper-1-PCM(S)-JEE(Adv)/13

Mathematics

7

PART – III SECTION – A

1.

π π ≤ sin−1 x ≤ 2 2 π π − ≤ sin−1 y ≤ 2 2 −π ≤ sin−1 x + sin−1 y ≤ π −

(–1,0) (0,0)

so for sin−1 x + sin−1 y = −π

(0, –1)

π π ⇒ sin x = − and sin−1 y = − 2 2 x = −1 and y = −1 −1

0

∫ −(x + 1) dx 2

area required 1 −

−1

= 1−

2

y = – (x + 1) 1 2 = sq. units. 3 3

b

2.

∫

∴ I = (x − a)3 (b − x)4 dx a

π/2

= 2(b − a)

∫

(a cos2 θ + b sin2 θ − a)3(b – a cos2 θ − b sin2 θ)4 sin θ cos θ dθ

0 π/2

= 2(b − a)8

∫

π/2

sin7 θ cos9 θdθ = 2(b − a)8

0

∫ sin

7

θ(1 − sin2 θ)4 cos θdθ

0

Let sin θ = t ⇒ cos θ dθ = dt 1

1

∫

∫

⇒ I = 2(b − a)8 t 7 (1 − t 2 )4 dt = 2(b − a)8 t 7 (1 − 4t 2 + 6t 4 − 4t 6 + t 8 )dt = 0

0

3.

x1 + x2 + x3 + x4 = 8 ∴ A.M. = 2 x1x2x3x4 = 16 ∴ G.M. = 2 Since all are positive and A.M. = G.M. ∴ x1 = x2 = x3 = x4 = 2

4.

B=

∫

cos ecθ

(

dt

t 1 + t2

1

(b − a)8 280

)

1 −udu =u⇒B= 1 t 1 + u2 ⇒ A + B = 0 ⇒ A = −B

∫

let

A

A2

0

A

2

1

2A 2

∴e

sin θ

−A −1 = 0 −1

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AITS-CRT(Set-VI)-Paper-1-PCM(S)-JEE(Adv)/13

8

5.

As g ( x ) is a curve which is obtained by the reflection of f ( x ) =

⇒ g ( x ) is inverse of f ( x ) .

(

e x − e− x on y = x 2

)

∴ g ( x ) = log x + 1 + x 2 = f −1 ( x )

 2x  1 .1 + ≠ 0, ∀x ∈ R  =   x + 1 + x2  2 1 + x2  1 + x2 ⇒ g ( x ) has no tangent parallel to x-axis also g' ( x ) is always defined, ∀x ∈ R 1

⇒ g' ( x ) =

⇒ g ( x ) has no tangent parallel to y-axis since, g' ( x ) > 0 ⇒ g ( x ) has not any Extremum 6.

 sin−1 x  = 0, cos−1 x  = 0     ⇒ x ∈ ( cos1,sin1)

7.

DA = (1 − x)iˆ + ˆj + kˆ DB = ˆi + (1 − y)jˆ + kˆ DC = ˆi + ˆj + (1 − z)kˆ For all the three vectors DA , DB and DC vectors to be co–planer 1− x 1 1 1 1− y 1 =0 1 1 1− z R1 → R1 – R2 R3 → R3 – R2 −x y 0 1 1− y 1 = 0 −z 0 y –x (–z + zy – y) – y (–z) = 0 zx – zxy + xy + yz = 0 1 1 1 ⇒ xy + yz + zx = xyz ⇒ + + = 1 x y z

8.

Let the focus be ( 0,α )

SM =

SM = SP

2x − y = 1

α +1

M

5

∴ P lies on the line 2x − y + α = 0 at a distance

x −0 y−α = =± 1 2 5

( α + 1)

α +1 5

units from S.

P

• S •

(0,α)

5

5

 ( α + 1)  α +1 2 ( α + 1)  2 ( α + 1)  ,α − ,α +   or ( x, y ) =  − 5 5 5    5  α +1 7α + 2 −α − 1 3α − 2 or x = x= ,y = ,y = 5 5 5 5 5y − 2 5y + 2 α = 5x − 1 = α = −5x − 1 = 3 7 35x − 7 = 5y − 2 3x + y + 1 = 0 7x − y − 1 = 0

( x, y ) = 

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Slope of normal tan θ = −

9.

9

dx dy

∴ The given equation becomes at a general point ( x,y ) is   dx  dy  2 y  x2  −  +  = x y +1   dy  dx  2

(

)

(

)

dy  dy  − yx 2 + y   = .x y2 + 1 dx  dx  2

(

)

dy  dy  y   − x y2 + 1 − yx 2 = 0 dx  dx  yy'2 − xy 2 y' − xy' − yx 2 = 0

(

) (

)

yy ' y ' − xy − x y ' − xy = 0 ∴

dy x dy = = xy or dx dx y

log y = y=

x2 + c (or) x 2 − y 2 = c 2

x2 ke 2

(or) logy 2 = x 2 − logk

logky 2 = x 2 10.

For K = 2, each element has four possibility and three out of four are favorable for the event. n

3 Hence required probability =   . 4 n

7 For K = 3, Probability that intersection is empty is   8

n −1

For K = 3, Probability that intersection is singleton =

11.

n 7   8 8

x>0  sin x,  x=0 f(x) =  0,  2  tan x , −1 < x < 0  x 2 Hence, lim+ f ( x ) = 0 and lim− f ( x ) = 1 . x →0

x →0

12–13. Equation of tangent at (x, y) is Y − y = p(X − x)

y dy . Then OA = x − and OB = y − px p dx p OA + OB = 1 ⇒ y = px + p −1

Where p =

OA.OB = 4 ⇒ y = px + 2 −p AB = 1 ⇒ y = px −

14.

x2 + y2 =

p 1 + p2

2a2b2 a 2 + b2

∴ radius of the circle

2a2b2 2

a +b

2

=

2 ab a2 + b2

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AITS-CRT(Set-VI)-Paper-1-PCM(S)-JEE(Adv)/13

15.

As the lines joining common point of intersection must be equally inclined to the axis tan α = − tan β ⇒ α + β = π

16.

Both the curves intersects at three distinct points Hence number of distinct lines formed will be 3.

SECTION – C 1.

tan θ =

2.

I=4

∫

2

2 h − ab 2 1+ 3 = =2 a+b 1− 3

 1 − tan2 x  tan2 x sin x cos x  e dx = 4  1 + tan2 x   

∫

(

)

2

tan x sec 2 x cos6 x 1 − tan2 x etan

x

t = tan2 x ⇒I= 2

∫

(1 − t ) et −2e t dt = (1 + t )3 (1 + t )2 2

= −2cos4 xetan 3.

∫

y

−y

=2

x

+c

sec −1 x − tan−1

∫

−1

−y

+c

(

sec −1 x dx −

)

x 2 − 1 dx =

∫

−1

−y

∫

−1

−y

sec −1 x = 2

(

)

sec −1 x π − sec −1 x dx +

∫

y

1

sec −1 x − sec −1 x dx

∫ ( π − sec x ) dx − π ( y − 1) = π ( y − 1) − 2λ y

−1

1

∴a + b = 3

4.

→

taking dot product with a = abc  = p + qcos θ + r cos θ − − − (1) →

taking dot product with c = abc  = pcos θ + qcos θ + r − − − (2) from (1) and (2) p = r . 5.

sin–1 x = cos–1 y

1

x = 1 − y2 ..... (1) or sin–1 y = cos–1 x y = 1 − x2 ..... (2) then from equation (1) and (2) & graph π Required area is = = λ 8 16λ 16 π = ⋅ =2 π π 8 7.

y=x

O

–π ≤ θ1 + θ2 – 2π ≤ 0 if θ1 + θ2 ≥ π

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