Solutions to Mock IIT Advanced/Test -...

Vidyamandir Classes

VMC/2013/Solutions 16 Mock IIT Advanced/Test - 3/Paper-2

Solutions to Mock IIT Advanced/Test - 3[Paper-2]/2013

[CHEMISTRY]

Vidyamandir Classes


Vidyamandir Classes



[PHYSICS]

21.(B) K = Wg + Wf

fW mg cos S mg xµ θ µ= − = −

x is smaller for 60°

22.(C) 34100

3r gπ× – 34

500 63

r g rvπ πη× − = 34500

3r aπ×

30 26 1 10

3

.v

−= × ×

200

18v=

11 m/s Ans

23.(C) 3

378 4

rtan= ° =

r = 6m

( ) 20F P h g rρ π= +

( )510 10 800 10 36π= + × × × ×

51 8 36 10. π× × × 72 10= ×

24.(D) Imagine 3 loops.

2 2 20 0 0 03B B B B B= + + = towards F

25.(A) kx = 0.06 g

k × 0.6 = 0.6

k = 1 N/m

Wext

+ W

g + W

string

Δ 0K= =

( )2 210 06 0 6 1 0 0 6 0

2extW . g . .− × × + × × − =

Wext = 0.54 J

26.(D) ( )2 2 21 10 06 1 1 0 6 0 0 06

2 2. g . . v× × + × × − = ×

⇒ v2 = 26 ⇒ 26v = m/s

27.(A) Reading of C = V {i in that branch = 0}

Reading of V

AR

= ⇒ R

28.(C)

At t = 0 inductor is open circuit 2

Vi

R=

3

2 22 2

a bV V

P.d . V V iR V V−

= − =− + = + =

Vidyamandir Classes


29.(ABC) Moving plane is like a moving observer. Therefore, number of waves encountered by moving plane,

01

v v c vf f f

v c

+ + = =

Frequency of reflected wave, 2 1s

vf f

v v

=

−

c vf

c v

+=

−

Wavelength of reflected wave 22 2

v c c c v

f f f c vλ

−= = =

+

30.(BD) C is decreased = ⇒Q CV Q is decreased

and 21

2=U CV is also decreased

and charge flows back in battery.

31.(B)

32.(BCD) 3ω 90 10 2 1000 180L π π−= × × × −

6

1 1 10002 1000

ω 0 5 10C .π

π−= × × =

×

⇒ circuit is inductive L RV V> ⇒ voltage leads the current

1ω

80πωL

CtanR R

φ−

= = ; 80

3R

π=

At resonance, 3 6

1ω

90 10 0 5 10LC .− −

1= =

× × ×

54

9

1 10 210

315 245 19−

= = = × ××

33.(1) If rod is in middle, i = 0 ⇒ F = 0

Eq. emf =

2 2

2 2

2

2 ( ) 2 ( )

1 1 2

2 ( ) 2 ( )

x

L x xL x L x

L

L x L x L x

ε ε εερ ρ

ρ ρ

−

− − + = ⇒+

− + −

ℓ

1 1 1

2 ( ) 2 ( )eq.R L x L xρ ρ= +

− +=

1

2ρ×

2 2

2L

L x−

⇒ 2 2( )

eq.L x

RL

ρ −=

2 2( )

x

LiL x

RL

ε

ρ=

−+

= 2 2

( )

x

L x RL

ε

ρ − +

2 2 2( )

x B Bma F i B x

L x RL L RL

ε ε

ρ ρ

− −= = − = ≈

− + +

ℓ ℓℓ

2

Ba x

m( L RL )

ε

ρ

−=

+

ℓ ⇒

2( )2

m L RLT

B

ρπ

ε+

=ℓ

⇒ T = 1 sec.

34.(6) By linear momentum conservation impulse (J) = mV.

By angular momentum conservation, angular impulse = J ω2

I=ℓ

So mv 2

ℓ = I ω or ω =

2

mv

I

ℓ =

2

212

mv

m

ℓ

ℓ

= 6v

ℓ = 6 rad/s

Vidyamandir Classes


35.(2) kq2

1 1

2r r

−

=

1

2m(V

2)

2

2

kq

r=

1

2mV

2

2kq

r=

1

2m ( )22V =

1

2m 2( )maxV

Vmax

= 2V = ( )10 2 2 = 20 m/s.

So, V0 = 2 m/s

36.(1) ( )464 2T rσ π= ℓ

310 1r m mm−= =

37.(3) By Gaus's law of magnetism.

0 1 2 3 4 5Tφ φ= = − + − + − +

3Wbφ =

38.(9) x y

t = 0 16% = N0 0

t = t 2% = N 14%

N = 0

8

N ⇒

1

8 =

1

2n ⇒ n = 3

t = 3T1/2

= 135 yrs.

39. [A-p, s] [B-p, s] [C-r, t] [D-q, r, t]

P = K = constant F = k = constant

Fv = k ma = k

mav = k

m dv

vds

v = k mv dv

ds= k

3

3

mv = kS 3v S∝

3v S∝ m dv

ds= k

m dv

vds

v = k v t∝

2

2

mv= kt P = Fv

2v t∝ P v∝

1

Fv

∝ P S∝

1/3

1 1and∝ ∝F F

E S P t∝

40. [A-r, s] [B-q, s] [C-p, t] [D-p, s, t]

Vidyamandir Classes



[MATHEMATICS]

41-42. 41.(D) 42.(B)

2 21 1 1 1

2 3 6 4 9

x y xy x ycos cos cos .θ θ− −+ = ⇒ = − − −

⇒

22 2

1 14 9 6

x y xycosθ

− − = −

⇒ 2 2

214 9 3

x y xy coscos

θθ− = + −

⇒ 2 2 29 4 12 36 6x y xy cos sin nθ θ+ − = ⇒ = and ( ) ( )2 21 1 0cos x sin x− −− >

( ) [ )1 1 10 1

2 2

− − − > ⇒ ∈ − ⇔

cos x sin x x , p, q

π

∴ 1

12

p , q= − =

Now 1

6 0 1 02

n , α

− = ∈ − ⇒ =

and ( ) ( ) [ ]2 22

1 2 2− −− = = ∈ −

q p . x xsin sin x e e x ,π π

Number of solution 4 10 nα= = + −

43.(C) 2

2 2 2

2

22 0

c gcc t gct a

tt+ + + − = ⇒ 2 4 3 2 2 22 2 0c t gct a t gct c+ − + + =

(i) 11 1 2 2

xx c t g

g

∑∑ = ∑ = − ⇒ = (ii) 2 2 1 2

2 1 2 211

x xx x c t t a

a

∑∑ = ∑ = − ⇒ =

(iii) 1 2 3 1 2 3 4 1 2 3

1 2 3 1

1x x x t t t t t t t

y y y t

∑ ∑= =

∑ ∑

(iv) G is the point 1 1

4 4 2 2

∑ ∑ − − =

x y g g, ,

C is the point ( )g, g− −

∴ M is the point2 2

g g,

− −

. Thus M and G coincide ∴ 2

2OG

OM=

∴ 1 2 31 1 22

1 2 3

26

x x xx x x OG

g y y y OMa

∑∑ ∑+ + + =

∑

44.(B) Let ( )21

xf x

x=

+ and ( )

( )2

1

1g x

x x=

+

∴

( )1

1 2 2 21 1

1

1 1

e

e

xA A . dx . dx

x x x

π

π

− = −+ +∫ ∫

1Put x in IInd int egral

t

=

( )2

2 2 2

1 1 1

11

21 1 1

ee e

e e e

x t t. dx . dt . dt log t

x t t

π

π

= + = = ++ + +∫ ∫ ∫ 21

12

log e= =

Vidyamandir Classes


45.(B)

1 2 3

2 3 1

3 1 2

0

z z z

z z z

z z z

= ⇒ ( ) ( ) ( )2 22

1 2 2 3 3 1

10

2z z z z z z − + − + − =

⇒ 1 2 3z z z= =

∴ z1, z2, z3 lies on the circle with centre at the origin (0, 0)

Let 3 1 3

2 1 2

2z z z

arg argz z z

θ θ θ θ −

= ⇒ = = + −

2

3 1 3 1 3 1

2 1 2 1 2 1

z z z z z zarg arg arg

z z z z z z

− − −= + =

− − −

46.(D) The unit vector along the angle bisector of vectors a��

and b��

is 1

2

a b

a b

±

��

��

Consider positive sign and the vector ( )a b a b ˆˆ ˆl m n la mb nca b a b

× + + = + + ×

��

��

ˆˆ ˆa, b, c are unit vectors along a , b , c��

∴ The required projection ( ) ( )1

2 2

+= + + + =

l mˆ ˆˆ ˆ ˆla mb nc . a b

47.(C) Case I : 0 < x < 1 ⇒ 0n

x → as ( ) 0n f x→∞ ⇒ =

Case II : 1 nx x> ⇒ → +∞ as ( ) 1n f x→∞ ⇒ =

∴ f has a jump (finite) discontinuity at x = 1

48.(B) Equation of tangent is 12 2 1

x sec ytanθ θ− =

If it cuts the coordinate axes at A and B, then

( )2 2 0A cos ,θ≡ ; ( )0B , cotθ≡ −

2 2S cos cotθ θ= −

21

2 2 02 4

dSsin cos ec sin

d

πθ θ θ θ

θ= − + = ⇒ = ⇒ =

2

20

d S

dθ< for

4

πθ =

∴ S is maximum

49.(ABC) Let y mx c= + be a chord of the given curve. Equation of the pair of lines through the origin and the points of

intersection of the chord and the curve is

( )2 23 2 4 0

y mxx y x y

c

− − − − =

If these are at right angles then coefficient of 2x + coefficient of 2 0y =

2 4

3 1 0m

c c

+ + − + =

⇒ ( )2m c= − +

So the equation of the chord is ( ) ( ) ( )2 0 2 1 0y c x c y x c x+ + − = ⇒ + + − =

Showing that the chord passes through the point of intersection of the lines 2 0y x+ = and 1 0x − = ⇒ through

the point ( )1 2, − and the equation of the parabola in (C) is ( ) ( )21 4 2x y− = + whose vertex is also ( )1 2, − .

Note : The centre of the circle in (D) is ( )1 2,− .

Vidyamandir Classes


50.(BC) Let = +��

ˆ ˆb xi yj . Since ��a is perpendicular to

��b so 4 3 0x y+ = . Thus

4

3

= −

��ˆ ˆb x i j . Let = +

��ˆ ˆc ui vj be the

required vector. According to the given condition

1 4 3 5= ⇒ + =

��

��c . a

u va

. Also

2

4

32 2 3 4 10

161

9

− = ⇒ = ⇒ − = ±

+

��

��

ux vxc . b

u vb

x

Solving these equations we have u = 2 and 1v = − or 2 11

5 5u , v= − = .

51.(BCD)

(A) False .

( )1 1 1

or8 8 4

P TTT HHH = + =

(B) ( )( )

( )( )

( ) ( )( )1 1

cP A BP A B P A P A B

P B P B P B

∩∩ − ∩= =

− −

( ) ( ) ( ) ( ) ( ) ( )1P A B P B P B P A P B P A B ∩ − = − ∩

( ) ( ) ( )P A B P A P B∩ =

Hence, the given statement is true.

(C) Let E1 be the event that white ball is drawn in first draw ; E2 be the event that black ball is drawn in first

draw ; E be the event that white ball is drawn in second draw.

∴ ( ) ( ) ( )1 2

1 2

E EP E P P E P P E

E E

= +

d w w w b

w b d w b w b d w b

+ = + + + + + + +

w d w b

w b w b d w b d

+ = + + + + + +

w

w b

= +

Which is independent of d.

(D) To prove that A, B, C are pairwise independent only.

Now,

( ) ( ) ( )P A B P I P II∩ = +

( ) ( )P A B C P A B C= ∩ ∩ + ∩ ∩

( ) ( ) ( ) ( ) ( ) ( )P A P B P C P A P B P C= + (given)

( ) ( ) ( ) ( ) = × + P A P B P C P C ( ) ( )P A P B= ×

Similarly, for the other two. Hence, this statement is correct.

52.(AC) 1 1

142 522 2

tan cot° °= −

1

22 302

cot°

= − + °

122 30 1

2

122 30

2

cot . cot

cot cot

° ° − = −

°+ °

( ) ( )( )2 1 3 1

2 3 1

+ −=

− + +

A B

C

I

II

Vidyamandir Classes


( )( ) ( )( )6 3 1 3 2 1 6 3 1 2 1 3

2 2 2 2

+ − − − + − + −= =

+ −

2 3 6 3 2 6 3 3 2 1 3

2 2

+ − + + − − − +=

−

4 3 2 6 4 2 4

2 2

+ − −=

−

( )18 6 8 3 16 8 2

8= − + − − 2 2 3 6= + − −

Hence, 3 6 3, ,µ λ µ λ= = + =

53.(4) ( )

2

2

2

1if 1

1

1if 1 1

1

if 1

xx

x

xf x x

x

x x

− ≥ +−

= − < <+

≤ −

( )

( )

( )

2

22

2

22

2 11

1

2 11 1

1

2 1

− +>

+ − −′ = − < < + < −

x x, x

x

x xf x , x

x

x , x

( ) 0f x′ = gives 2 1x = + or 1 2−

The function has a continuity at 1x = −

As ( ) 0x , f x→∞ →

The graph is as shown below :

Where, 0dy

dx= i.e. it exists or

dy

dx is non-existent.

The points 1 1 2 1x , ,= − − and 1 2+ are the four critical points on the graph of this function.

54.(6) d xa yb zc= + +��

Take dot product with b c , c a , a b× × ×��

one by one

∴ d b c a d c a b d a b c

da b c

+ + =

��

��

�� ⇒ d b c a d c a b d a b c

da b c

+ + =

��

��

��

∴ 36 + + = �� d b c a d c a b d a b c

Vidyamandir Classes


55.(0) (i) 0x x⇒ ≥ (ii) [ ]216 0 4 4x x ,− ≥ ⇒ ∈ − (iii)

( ) ( ) ( )2 0 0 2x x x , ,− > ⇒ ∈ −∞ ∪ ∞

(iv) ( )0 2 2 12 2

x xsin n , n

π ππ π ≥ ⇒ ∈ +

∴ domain is {4} and Range is {5}

∴ q = 5 and p = 4 ∴ 1q

p

=

and [ ]1 1 3

1 04 4 4

q q qk k

p p p

= ⇒ − = − = = ⇒ =

56.(2) ( )

1

1

01x

dxI

e x=

+∫ and

24

2 2 3

02

tane sinI . d

tan cos

π θ θθ

θ θ=

−∫

( ) ( )

1 1 11

2

0 0 0

1 1

2 2 2 2 1 2 1

x x

x

e e e dxI dx dx

x x e x

−

= = =− − − +∫ ∫ ∫

∴ 1 1

2 2

22

I eI

I e I= ⇒ =

57.(9) 1 1 3 3z , PA PB≤ − ≤ = =

Greatest value of w OA OP PA= = +

∴ 5 3G = +

And least value of 0w L= =

⇒ ( )25 9G L+ − =

58.(6) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )F x f x g x h x f x g x h x f x g x h x= + +′ ′ ′ ′

So ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )0 0 0 0 0 0 0 0 0 0 021F x F x f x g x h x f x g x h x f x g x h x= = + +′ ′ ′ ′

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )0 0 0 0 0 0 0 0 04 7f x g x h x f x g x h x kf x g x h x= − +

( ) ( ) ( ) ( )0 0 03 k f x g x h x= − + ( ) ( )03 k F x= − +

Hence 21 3 24k k= − + ⇒ =

59. [A-q] [B-s] [C-p] [D-q]

(A) Equation of common chord: 0+ − λ =x y

Replace ( )= λ −y x in equation of circle I :

( ) ( )22 2 2 0+ λ − − − λ − =x x x x ⇒ ( )2 22 2 2 0− λ + λ − λ =x x …(i)

As exactly two common tangent possible ⇒ roots of equation (i) must be real and unequal.

⇒ ( ) ( )2 22 4 2 2 0λ − × × λ − λ > ⇒ 0 4< λ <

(B) ( )2

2

π

− π

+∫ | sin x| | cos x | dx =

2

0 0

4 8 16

π π

+ =∫ ∫/

sin x dx cos x dx

(C) +

=−dy x y

dx x ⇒ 1+ = −

dy y

dx x [Linear DE]

= =∫dx

xI .F . e x ⇒ 2

2= − +

xxy C

As curve passes through ( )3 1 6 4− ⇒ =, C/ ⇒ 2

42

= − +x

xy

As it also passes through ( )1λ, ⇒ 2

4 4 22

λλ = − + ⇒ λ =− ,

Vidyamandir Classes


(D) DRs of line of intersection of planes ( )5 6 3 5 3 4 are 11 9 4− − = + + = −x y z x y z , ,and

DRs of normal to required plane: ( )5 3 7−, ,

A point in this plane: ( )1 1 3− −, ,

⇒ equation of required plane: ( ) ( ) ( )5 1 3 1 7 3 0+ + − − + =x y z

5 3 7 19 0+ − − =x y z …(i)

As, ( )6 2λ, , lies on (i) : ⇒ 5 18 14 19 0 3λ + − − = ⇒ λ =

60. [A-r] [B-s] [C-p] [D-q]

(A) 12

1 2

iz

z

−=

+ is a circle

(B) ( ) ( )1 1 11

2 2 2

i z i zz i

+ + − +− =

SP = e.PM, where S is the focus and ( ) ( )1 1 1i z i z+ + − + is equation of directrix

Note : Distance of point z0 from line a 0az az b+ + = is 0 0

2

az az b

a

+ +

1< ⇒e ellipse

(C) According to the question

( ) ( )2 6 6 4 3 0+ − − + − <k k

⇒ ( ) ( )2 7 0− − <k k

⇒ ( )2 7∈k ,

⇒ { }3 4 5 6∈k , , ,

(D) According to the question

1 1 11 2 30

2 2 2

+ + + + + =

x y z

⇒ 1 1 1 6 0+ + + =x y z . . . .(i)

and 1 1 11 2 3

1 1 1

− − −= =

x y z

⇒ 1 1 1− = −x y . . . .(ii)

and 1 1 1− = −y z . . . .(iii)

Solving (i), (ii) and (iii) we get :

1 1 13 2 1= − = − = −x , y , z

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