Answer (8)

(1)

ANSWERS, HINTS & SOLUTIONS

CRT(Set-VI)

(Paper-1)

ANSWERS KEY

PHYSICS CHEMISTRY MATHEMATICS Q. No. 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

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FIITJEE

JEE(Advanced)-2013

/ 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)

2

P

h

y

s

i

c

s

PART – I

SECTION – A

1. For (1) particle, ut − 1_gt2 2 = −H t1 = 2 2 u u 2gH u u 2gH g g ± + + + = For (2) particle, t2 = 2 2 u u 2gH u u 2gH g g − ± + − + + = ∴ ∆t = 2u/g 2. ∆W = −∆u 2 2 m _{q E} 4 ω = ⇒ ω = 2 qE m 6. fav = 1 1 2 2 3 3 1 2 3 n f n f n f n n n + + + + Cp = f 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.

Q = 2 1 1 V t r = 2 2 2 V t r ⇒ 1 1 2 2 r t 1 r = t = 2 In series req = r1 + r2 ⇒ Q = 2 3 2 1 1 2 1 v t v t r +r = r ⇒ t3 = 3t = 45 min In parallel Q =

(

)

2 2 4 1 2 1 1 2 1 v t r r v t r r r + = ⇒ t4 = 2 1 2 r _t r +r = 10 minutes. 12. Rsh(I – Ig) = Rg . Ig Rsh = g 4 g R ₁₀₀ I ₁ 10 1 I = Ω − − = 100 9999Ω ~ I Ig Rg |I – Ig | Rsh

(3)

13. Ish = 10 – 10-3 = 9.999 A Power dissipated P = 2 2 sh sh 100 I R (9.999) 9999 = × = 0.9999 Watt.

which is less than 1 watt. Hence it can be used. 14. From the data given if orbital speed v = 2 × 105_m/s

2 2

mv GMm

r = r , where M is mass of the massive object and m is mass of particle ⇒ M Rv2 G = 15 10 37 11 6 9.5 10 4 10 _{10 kg} 6.67 10− × × × × = × ∼ . 15. 37 7 30

Mass of the object 10 ₁₀ ₅₀

Mass of sun 10 = > ∴ It is a black hole. 16.

(

)

11 24 3 s 2 ₈ 2 2GM 2 6.67 10 6 10 R 10 C _{3 10} − − × × × × = = ≈ ×

∴ Radius compression factor

3 9 3 10 ₁₀ 6400 10 − − = ≈ ×

SECTION – C

2. Line current, I = 2 2 R L I +I = _(1.5)2 ₊

( )

₂ 2 ₌_2.5mA Circuit impedance, Z V 15V 6k I 2.5mA = = = Ω 3. Mirror formula 1 1 1 u u+ =f 2 2 1 du 1 dv ₀ dt dt u v − − = 2 2 2 dv v du 10 _{4 1m / sec} dt u dt 20   = =_ _ × =   4. ₂ 2 R cos 1 R L C φ =   +_ − ω _ ω  

Putting the values, C = 500 Cµ

7. N′ = _n _{n 2} N N 2 2 2 2     +    

Where n is no of half lives elapsed for species A

Now, N 5 _{n 1}1 _n1

N 32 2 +

′₌ ₌ ₊ 

(4)

4

C

h

e

m

i

s

t

r

y

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 CH₃ OH CH₃ Cl (A) (B) (C)

8. (A) B – F bond length is smaller in BF3 than in BF4

−_{due to back bonding in BF}

3.

(B) S S

is more stable due to back bonding.

(C) Triplet carbene is more stable than singlet carbene.

(D)

(

CH

_{3 3}

)

Si OH

−

is more acidic because

(

CH

_{3 3}

)

SiO

− is more stable due to pack bonding. 9. (1 0.1−A )→( ) ( )0.1B 3C+ 0.3

(

1 3 0.1

)

1.3 = + ×

=

T

P

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

2 5 2 5 2

1

2 C H OH Na

+



→

C H ONa

+

H

(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

IE

₁ value

(5)

16. _H

Solution for the Q. No. 14 to 15

(P) (Q) (R)

SECTION – C

1. Let n atoms be ionized

23 22 6 10_× _×10− _×EA n IP_{= ×} 23 22

6 10

10

3.6

13 n

−

×

=

1

16.67 1.66 10

=

⇒

×

2. trans & cis

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

4. C2H5 C₂H₅ C H₃ H C H₃ Br 4-product 5. Given

λ

₂

₌

30.4 10

_×

−7

cm

,

λ

₁

₌

108.5 10

_×

−7

cm

Let excited state of

He

+_{be n}

2. It comes from

n

2 to

n

1 and then

n

2 to 1 to emit two successive photon. 2 2 2 2 1 1 1 1 1 H R Z n λ   = × _ − _   1 7 2 2 1 1 1 1 109678 4 2 30.4 10− 1 n n   = × ×_ − _⇒ = × _ _

(6)

6 2 2 2 1 1 1 1 1 2 H R Z n λ   = × _ − _   2 7 2 2 1 1 1 1 109678 4 5 108.5 10− 2 n n   = × ×_ − _⇒ = × _ _

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

(

)

3

230 g Fe SCN





∴

_

_

=

19 230 54

81 ×

=

g

No. of water molecules =

54

3

18 =

(7)

M

a

t

h

e

m

a

t

i

c

s

PART – III

SECTION – A

1. sin x1 2 2 − π π − ≤ ≤ 1 sin y 2 2 − π π − ≤ ≤ 1 1 sin x sin y− − −π ≤ + ≤ π

so for sin x sin y−1 + −1 = −π ⇒ _{sin x}1 _{and sin y}1

2 2 − _{= −}π − _{= −}π x = −1 and y = −1 (–1,0) (0, –1) (0,0) y = – (x + 1)2 area required 0 2 1 1 (x 1) dx − − − +

∫

1 1 3 = − 2 3 = sq. units. 2. ∴ b 3 4 a I=

∫

(x a) (b x) dx− − / 2 0 2(b a) π

= −

∫

(a cos2 θ + b sin2_{θ − a)}3_{(b – a cos}2_{θ − b sin}2_θ)4_{sin θ cos θ dθ}

/ 2 8 7 9 0 2(b a) sin cos d π = −

∫

θ θ θ / 2 8 7 2 4 0

2(b a) sin (1 sin ) cos d

π

= −

∫

θ − θ θ θ

Let sin θ = t ⇒ cos θ dθ = dt ⇒ 1 8 7 2 4 0 I 2(b a)= −

∫

t (1 t ) dt− 1 ₈ 8 7 2 4 6 8 0 (b a) 2(b a) t (1 4t 6t 4t t )dt 280 − = −

∫

− + − + = 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.

(

)

cos ec 2 1 dt B t 1 t θ = +

∫

sin 2 1 1 udu let u B t 1 u θ₋ = ⇒ = +

∫

A B 0 A B ⇒ + = ⇒ = − 2 0 2 2 A A A e A 1 0 1 2A 1 − ∴ − = −

(8)

8

5. As g x is a curve which is obtained by the reflection of

( )

f x

( )

ex e x on y x 2 − − = =

( )

g x ⇒ is inverse of f x .

( )

∴ _{g x}

( )

₌_{log x}

(

₊ _{1 x}₊ 2

)

₌_f−1

( )

_x

( )

2 2 1 2x g' x . 1 x 1 x 2 1 x   ⇒ = _ + _ + +  +  2 1 0, x R 1 x = ≠ ∀ ∈ +

⇒ g x has no tangent parallel to x-axis also

( )

g' x is always defined, x R

( )

∀ ∈

⇒ g x has no tangent parallel to y-axis since,

( )

g' x

( )

> ⇒0 g x

( )

has not any Extremum 6. _sin x−1 _=0, cos x_ −1 _=0

(

cos1,sin1

)

x

⇒ ∈

7. DA (1 x)i= − ˆ ˆ ˆ+ + j k ˆ ˆ ˆ DB= + −i (1 y)j k+ ˆ ˆ ˆ DC= + + −i j (1 z)k

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 0 y z − − = − –x (–z + zy – y) – y (–z) = 0 zx – zxy + xy + yz = 0 ⇒ xy + yz + zx = xyz ⇒ 1 1 1 1 x+ + = y z 8. Let the focus be

( )

0,α

SM SP= SM 1 5 α + =

∴ P lies on the line 2x y− + α = at a distance 0 1 5 α + _{units from S.}

(

1

)

x 0 y 1 2 ₅ 5 5 α + − − α = = ± • 2x y 1− = P • S M (0,α)

( )

x,y 1, 2

(

1

)

5 5 α + α +  =_ α + _  or

( )

(

1

)

2

(

1

)

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

(9)

9. Slope of normal tan dx dy θ = −

∴The given equation becomes at a general point

( )

x,y is

(

)

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

(

)

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

(

)

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

(

) (

)

' ' ' yy y −xy −x y −xy = 0 ∴dy xy dx = or dy x dx= y 2 x logy c 2 = + (or) _x2₋_y2 ₌_c 2 x 2

y ke= (or) logy2 =x2−logk 2 2

logky =x

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

n 3 4       .

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

For K = 3, Probability that intersection is singleton = n 8 n 1 7 8 −       11. f(x) = 2 2 sin x, x 0 0, x 0 tan x _, _{1 x 0} x >   ₌    _{− < <}  Hence, ( )

x 0lim f x→ + = and 0 x 0lim f x→ − ( )= . 1 12–13. Equation of tangent at (x,y) is Y y p(X x)− = −

Where p dy dx = . Then OA x y p = − and OB y px= − p OA OB 1 y px p 1 + = ⇒ = + − OA.OB 4= ⇒ =y px 2+ − p 2 p AB 1 y px 1 p = ⇒ = − + 14. 2 2 2 2 2 2 2a b x y a b + = +

(10)

10

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

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

SECTION – C

1. 2 2 h ab 2 1 3 tan 2 a b 1 3 − + θ = = = + − 2. 2

(

)

2 2 tan x 2 6 2 tan x 2 1 tan x

I 4 sin x cos x e dx 4 tan x sec x cos x 1 tan x e 1 tan x  −  = _ _ = − +  

∫

2 t=tan x

(

)

(

)

(

)

t t 3 1 t e 2e I 2 dt c 1 t 1 t 2 − − ⇒ = = + + +

∫

2 4 tan x 2cos xe c = − + 3. y 1 1

(

2

)

1 1

(

1

)

y 1 1

ysec x tan x 1 dx ysec x sec x dx 1 sec x sec x dx

− − − − − − − − − − = − π − + −

∫

1 1 1 1 y y 2 − sec x dx− − sec x− − − =

∫

−

∫

y

(

1

)

(

)

(

)

1 2 sec x dx− y 1 y 1 2 =

∫

π − − π − = π − − λ a b 3 ∴ + =

4. taking dot product with a→=_abc_= +p qcosθ +r cosθ − − −(1) taking dot product with c→=_abc_=pcosθ +qcosθ + − − −r (2) from (1) and (2) p r= . 5. sin–1_{x = cos}–1_y 2 x= 1 y− ... (1) or sin–1 y = cos–1 x 2 y= 1 x− ... (2)

then from equation (1) and (2) & graph

y = x 1 O Required area is = 8 π_{= λ} 16 16 ₂ 8 λ π = ⋅ = π π 7. –π ≤ θ1 + θ2 – 2π ≤ 0 if θ1 + θ2 ≥ π