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Vidyamandir Classes

VMC/2013/Solutions 1 Mock IIT Advanced/Test - 3/Paper-1

Solutions to Mock IIT Advanced/Test - 3[Paper-1]/2013

[CHEMISTRY]

Vidyamandir Classes


Vidyamandir Classes



[PHYSICS]

24.(A)

1 1 1

20 10V− =−

20 1V m= = −

Co-ordinate is (20 cm, 0.2 cm).

25.(C) For inductor di

V Ldt

= −

Or LV

di dtL

= −∫ ∫

V = (10 t − 20) where 0 4t ms≤ < ( )2

0

1 1 1010 20 20

2

t

Lt

i t dt tL L

⇒ = − + = − −

∫

25 20

Lt t

iL L

= − +

26.(B) Force of buoyancy

( )0 0BF A x L x gρ ρ = + − Weight sW LA gρ=

For equilibrium BF W=

( )0 0SLA g A x L x gρ ρ ρ ⇒ = + − ( )0s wL x L xρ ρ ρ⇒ = + −

( )00

1000 800 2

1000 300 7

w ss w w

w

x x

L L

ρ ρρ ρ ρ ρ

ρ ρ− −

⇒ = − + ⇒ = = =− −

27.(C) x a cos tω=

v a sin tω ω= −

1 11

2 2

aa cos t cos tω ω= ⇒ =

1 13 3 2 6

T Tt t

π πω

π⇒ = ⇒ = =

×

( )6

0

6

0

3

T /

T /

a sin t dt

av

Tdt

ω ω−

< >= =∫

∫

28.(A) Given 2 4y x= −

When 0 2x y= , = ±

The separation between open ends = 4.

Magnetic force on a current carrying element placed in a uniform field B is ( ) ɵ ɵ( )3 4 5 60i L B j k i× = × =��

ɵ

29.(B) Here at the instant of sliding limiting friction f Nµ= will act on the two blocks for limiting equilibrium, we use:

F = T Nµ+ …(i)

and T = Nµ …(ii)

and 2T + N = 50 …(iii)

On solving (ii) and (iii), we get

50

252 1

N Nµ

= =+

Now from (i) and (ii)

F = 2 Nµ = 2 0 5 25 25. N× × =

Vidyamandir Classes


30.(C)

31.(B) After 2 sec speed of boy will be 2 2 4v m / s= × =

At this moment centripetal force on boy is 2 30 16

806

rmv

F NR

×= = = .

Tangential force on boy is 30 2 60tF ma N= = × = .

Total friction acting on boy is 2 2 100r tF F F N= + = .

At the time of slipping F mgµ=

or 100 30 10µ= × × ⇒ 1

3µ = .

32.(B)

33.(C) By conserving angular momentum and linear momentum and also using concept of coefficient restitution, we get

1

21cmv v

ev cos

ω

θ

+ += =

ℓ

. . . . . .(i)

1cmmv cos mv mvθ = − . . . . . .(ii)

2

12 12 2

mmv cos mvθ ω= −

ℓ ℓ ℓ . . . . . .(iii)

Now solving, we get : 2

5cmv v cosθ= ,

12ω

5v cosθ=ℓ

, 13

5v v cosθ= −

34.(B) Change in angular momentum of the rod is 2 12 1

Δ12 5 5

cmm

L I v cos m v cosω θ θ= = × =ℓ

ℓℓ

35.(B) 13 2

Δ5 5

v v cos v v cos v cos v cosθ θ θ θ= + = − = 36.(AB)

37.(AB) ( )

( )11 1

2020 1

R

F R

µ

µ

−= = ⇒

− . . . . .(i)

If equiconvex lens of f = 20

Then ( )1 21

20 Rµ= −

From eq. (1) and (2) we conclude.

∴ R = 40 ( )1µ − . . . . .(ii)

That option (A) is correct

After Silvering :

10Feq cm= − (Combination will behave like a concave mirror)

(B) 15u = −

1 1 1 1 1 1

15 10eqv u F v+ = ⇒ − = −

1 1 1

30 So Bis correct15 10

v cmv= − ⇒ = −

(C) 20u = − 20v = − So C is wrong

Alternative :

For planoconvex ( )1 2

1 1 11

f R Rµ

= − −

1 2R R R= ∞ = r = 20 ( )1µ −

For equiconvex ( )1 1 11

20 R Rµ

= − +

( )40 1R µ= −

Vidyamandir Classes


38.(ABD)

Speed will be maximum when the block passes through the equilibrium position.

springF mg=

(B) From conservation of energy,

( ) 2 21 1

2 2maxmg l x Kx mv+ = + . . . .(i)

where 4 4

mg mg lx l

K mg= = =

Putting 4

lx = in equation (1), we get

3

2maxv gL=

(C) is incorrect as in equilibrium compression is 4l / which is more than 8l / .

(D) Time taken is 1

24 4 4

T m L

K g

ππ= =

39.(BD) Equivalent circuit :

Induced emf in a rotating rod with constant angular

Velocity 2ω

ω2

B ae; = , (here a = Radius)

Now use Kirchoff’s Law.

40.(3) 1v

mu n

= = −

( )0

1L

L

Xm

– X X n= = −

−

Lens formula ⇒( )0

1 1 1

L LX X X f− =− −

( )0

2 20

1 10L L

L L

dXdX dX.

dt dt dtX X X

− − − =

−

( ) ( )0

2 2 20 0

1 1 1L

L L L

dXdX

dt dtX X X X X

− + = − −

0

0; LdX dXV V

dt dt= =

⇒

2 2

01 1

1V Vn n

− + + = +

⇒ [ ]0 04

1 4 43

VV V V− + = ⇒ = ⇒ 03

4

VV =

41.(1) In first case 10

30 10 11

KR

× = − ×+

In second case 5

10 5 22

KR

× = − ×+

Solving above two equations we get : 1R =

42.(1) Work done ΔQ= (heat dissipated) Let e : induced emf

=2 2( )e N B v

t tR R

×= ×

ℓ=

2 2 32100 (0 4) 2 5 10

110 000

. .v

,

−× × ×× ×

2 32 5 10. −= ×ℓ

4 2

25 10 5 10 m− −= × = ×ℓ

25 10

1v

−×=

⇒ 86 25 10 16 1W . J Jµ−= × × =

Vidyamandir Classes


43.(9) 1

22 3

2 03

2

P

RkQ

kQE

RR= − =

⇒ 12

4

9 2

QQ =

2

1

9=

8

Q

Q

29 8

98

Q Cµ×

= =

44.(8) Initially 0100 2x g=

0 20x cm=

Finally, equilibrium position

0100 1x g′ =

0 10′ =x cm

1

2100 10

Tπ

π2

= =

2 10

Tt

π= =

21

10 502 100

S cmπ

= × × =

Total distance = 50 + 10 + 20 = 80cm.

So, x = 8

45.(4) Paschen series 7 6

2

1 1 11 097 10 10

9.

nλ

= × − ≤

2

1 1 1

9 10 97.n− ≤

2

1 1 97

9 10 97

.

.x≥

×

29 1097

197x

×≥

31097

7197

n≤ ≈ ⇒ 7 3 6 3 5 3 4 3 4lines− , → , − , − ⇒

46.(1) 2 2

3

24(ω )

10 10z L R

−= = +

×

22 2 2 1

(ω ) ωω

R L R LC

+ = + −

( ) 1ω ω

ωL L

C= − +

2 6

1 15

2ω 2 100 100 10L H

C π π −= = =

× × ×

( ) ( )2 2 22400 500 Rπ= +

R = 2 2(2400) (5 100)π− × = 2 210 (24) 25π−

= 10 326 180× ≈

I = 1A

Vidyamandir Classes



[MATHEMATICS]

47.(B) Let ( )22P t , t and

2

1 2Q ,

tt

−

Circle with OP as diameter is 2 2 21 2 0S x y t x ty≡ + − − =

Circle with OQ as diameter is 2 22 2

1 20S x y x y

tt≡ + − + =

R lies on 1 2 0S S− = i.e. 1

2

= −

y t xt

∴ 1

1tt

− = −

Point of intersection of tangents is 1

1, tt

− −

i.e. ( )1 1,− −

48.(D) ( ) nA Adj A A I=

Clearly, 4 3A , n= =

( ) ( )21 44 256n

Adj Adj A A−

= = =

1 24 16

nAdj A A

−= = =

∴ ( ) 256

1616

Adj Adj A

Adj A= =

49.(C) 2 1

1 2 3x . . .x

+ + +

=

2 11

1

2

xx

x

+

Now using the property that 1 1 1

1x x x

− < ≤

(i.e. using sandwich theorem)

We get : ( ) ( )

2 11

1 1 11 1

2 2 2

+ − < ≤ +

xx

x xx

Now applying sandwich theorem the required limit is 1

2

50.(C) 2a c b+ = and 22ac b bc− = .

Eliminating ‘b’ from these two equations, we get ( )2 24 3 0 3a ac c a c a c− + = ⇒ = ≠∵

51.(B)

2

5 2 3 2 43 2

x xa b

x x x x

ln a ln bI x dx

a b a b

= + ∫

2 3

2 36

x x

x x

ln a bdx

a b= ∫

But 2 3x xa b t=

2 3t ln a b dx dt=

2 3 2 2 3 2

1 1 1

6 6

lnt lntI dt dt

tln a b t ln a b t

− − = = −

∫ ∫

2 3

1

6

lnetk

tln a b

= − +

2 3

2 3 2 3

1

6

= − +

x x

x x

lnea bk

ln a b a b

Vidyamandir Classes


52.(C) P(unequal no. of heads and tails) = 1 P− (equal no. of heads and tails) = ( )

2

2

1 1 2 !1 1

2 2 ! 4

n nn

n n

nC

n

− = −

53.(A) ( ) ( )

93

257

53 64

625 3

3 37 5

409

loglog

loglog

N+

= −

33 325 6

3 325 6 6

409

log log

N+ = −

( ) ( )25 6 6 25 6 6

409N

+ −=

N = 1

2 2 1 0log N log= =

54.(B) The auxiliary equation is 3 27 16 12 0D D D− + − =

or ( ) ( )23 4 4 0D D D− − + = or D = 3, 2, 2

∴ by rule (ii), ( ) 2 31 2 3

x xy c x c e c e= + +

55.(C) The general solution tells that the auxiliary equation has roots 1 1 1i, i,− + − − . So, the auxiliary equation is

( ) ( ) ( )1 1 1 0D i D i D+ − + + − = or ( ){ } ( )21 1 1 0D D+ + − =

or ( )( )2 2 2 1 0D D D+ + − = or 3 22 0D D+ − =

∴ the differential equation is 3 2

3 22 0

d y d yy

dx dx+ − =

56.(D) Statement 1 : Locus of point of intersection of only perpendicular lines is a circle and other vertices B and C do

not form a circle

Statement 2 : Obvious (standard Definition)

57.(C) 1+

−

−→∞ →

+=

+

n n

n nn

| sin x | | cos x |a lim lim

α

α α

α α

2

21 1

n

nn

| sin x | | cos x |lim lim | sin x|

α

α

α+

−

−→∞ →

+= =

+

1−

−

−→∞ →

+=

+

n n

n nn

| sin x | | cos x |b lim lim

α

α α

α α

2

21 1−→∞ →

+= =

+

n

nn

| sin x | | cos x |lim lim | cos x |

α

α

α

( )2 1

14 2 2 2n

nc lim cos cos . . . . . . cos

n n n n

π π π π

→∞

−= + + + +

( )114 4

4 4 4 2

4

n

nsin cos

nlim sin cosn

sinn

π ππ π π

π→∞

− = = =

OR alternatively

11

0 0

1 1

4 2 4 2 2

−

→∞=

= = =∑ ∫n

nr

r xc lim cos cos dx

n n

π π π π

( )1

2f x max | sin x |,| cos x |,

=

∴ range of f (x) is

11

2,

58.(D) If circum centre is ( )S α

1 8| z | SA Rα− = = = And 1 2 3 4r r r r R+ + − =

Vidyamandir Classes


59.(ABD) 5 21 1 10p C C= × =

. . . .i.e. select one match out of 5 which is wrongly forcast in 5C1 ways. Suppose its true results is win,

then it can be wrongly forcast as loss or draw in 2C1 ways.

Similarly ( )35 23 1 80q C C= =

Similarly ( )55 25 1 32r C C= = ⇒ 2 5 8q r, p q= = and ( )2 p r q+ >

60.(AC)

61.(ACD)

(A) ( )2 2 2x iy i x y i xy− = − + ⇒ 2x xy= − and 2 2x y y− = −

0x = or 1

2y

−= ;

For x = 0, y = 0 or 1

For1

2y

−=

3

2x = ±

(B)

71

1 0+

+ =

z

z ⇒

( ) ( )1 2 1 2 11

7 7

k kcos i sin

z

π π+ ++ = +

⇒ 1

1 cos i sinz

θ θ− = − − where ( )2 1

7

k πθ

+=

⇒ 1

1− =

− −z

co s i sinθ θ

1

1z

cos i sinθ θ

−⇒ =

− −

1

22 2 2

sin sin icosθ θ θ

−=

−

( )12 2

22

2

sin i cos

Re z

sin

θ θ

θ

− + −

= ⇒ = ⇒ ( )6

0

7

2k

k

Re z

=

−=∑

(C) 21 1

−= i nz z e π / ⇒ ( ) ( )2 2

x iy x iy cos i sinn n

π π − = + −

2 2

xcos y sin xn n

π π+ = and

2 2x sin y cos y

n n

π π− + = −

2 2

1x cos y sinn n

π π − =

and 2 2

1x sin y cosn n

π π = +

and also

2 1y

x= − (given) ⇒ ( )2 1

8tan tan

n

π π= − = ⇒ n = 8

(D) 3 2

1 2

i sin

i sin

θ

θ

+

− is purely real

⇒ 3 2

01 2

i sinIm

i sin

θ

θ

+ = −

( ) ( )1 2 3 2 0sin sinθ θ+ =

0sin ; nθ θ π= =

( ) 0, ;θ π π θ∈ − ∴ =

Vidyamandir Classes


62.(AC) As 0 1b a a c c b , x− + − + − = = is a root. So, the other root is 2 or 1

2

∴ 1 2c b

b a

−⋅ =

− or

11

2

c b

b a

−⋅ =

−

∴ 2 2b a c b− = − or 2 2b a c b− = −

∴ 3 2b a c= + or 3 2b a c= +

∴ 2

2 1

a cb

+=

+ or

2

1 2

a cb

+=

+

∴ B divides AC in the ratio 2 : 1 or 1 : 2 internally.

63.(0) ∵ | z | = real and positive, imaginary part is zero

∴ arg | z | = 0 ⇒ [ ] 0arg | z | =

∴ [ ]100 100

0 0

0 0

x x

arg | z | dx .dx

= =

= =∫ ∫

64.(6) Since, 2 3 0a b c+ + = . . . .(i)

∴ 2 3 0a c b c c c× + × + × = taking cross product with c

⇒ ( )2 0 0c a b c− × + × + =

∴ ( )2c a b c× = × . . . .(ii)

Again from eq. (i)

2 3 0× + × + × =a b b b c b taking cross product with b

∴ ( )3a b b c× = × . . . .(iii)

Given ( )a b b c c a b cλ× + × + × = ×

⇒ ( ) ( ) ( ) ( )3 2b c b c b c b cλ× + × + × = × [from equations (ii), (iii)]

⇒ ( ) ( )6 b c b cλ× = ×

∴ 6λ =

65.(1) { }

1

1 1

1

− −

→∞=

+∑

a a a aan

ank

k n . n k . klim

n n

/ /

1

1

1

→∞=

= +

∑

a an

nk

k klim .

n n n

/

( )1

1

0

= +∫ a ax x dx

/

( )1

1 1 1

0

1 11

+ +

= + + +

a ax x

a

a

/1

11 1

a

a a= + =

+ +

66.(1) ( ) ( ) ( )2 2 0t f x tf x f '' x− + =′ , has equal roots

Discriminant = ( )( ) ( ) ( )24 4 0f x f x f x− =′ ′′

( )( )

( )( )

f x f x

f x f x

′′ ′=

′ . . . . integrating both side.

( )( ) ( )ln f x ln f x ln c= −′

⇒ ( ) ( )f x cf x= ′

Vidyamandir Classes


( ) ( )1

0 02

f cf c= ⇒ =′

( )( )

2f x

f x

′=

ln f (x) = 2x + k

⇒ (∵ f (0) = 1) ⇒ k = 0

⇒ ln f (x) = 2x

∴ ( ) 2xf x e=

2 2 2 24 4 0x x xt e te e− + =

⇒ 2 4 4 0t t− + =

t = 2

( ) 2

0 0

1 1 22 2 1 1

2 2 2

x

x x

f x t elim lim

x x→ →

− −∴ − = × − = − =

67.(4) ( ) ( ) ( )2 22 1f x kx x x= − +′ ∵ f (x) is divisible by x3, so it does not have the terms of x

2, x

1, x

0

( )5 4 3 22 2k x x x x= − + −

⇒ ( ) ( )3

3 210 24 15 4060

kxf x x x x= − + −

( )( )

( )3 2 3 8 264

4 1 4 39

f

f

−= × =

−

68.(1) x > 0, y > 0 & 8 2| x iy | r+ + = & ( )84 8 4

yarg x iy tan

x

π π− + = ⇒ =

−

⇒ (x + 8)2 + y

2 = 2r

2, 8y x= − ⇒ x > 8 ∵ y is also positive

⇒ ( ) ( )2 2 28 8 2x x r+ + − = ⇒ 2 22 2 64 2x r+ × =

⇒ 2 264x r+ = ⇒

2 2 64r x− =

⇒ ( ) ( ) 64r x r x− + =

∵ r N∈ and 8 0x , y> > , the only possibility is

⇒ 2 32r x , r x− = + = ⇒ 15 17x , r= =

69.(5) ( ) ( ) ( ) ( ) ( )5 5 51 1 1 1 1

++ = + + = + +

n n nx x x x x

( )5 5 5 4 5 3 5 2 5 1 5 00 1 2 3 4 5C x C x C x C x C x C x= + + + + +

2 1 5

0 1 2 1 5n n n n r n r n r n n

r r r nC C x C x . . . C x C x . . . C x . . . C x+ ++ +

× + + + + + +

Comparing coefficient of xr + 5

⇒ 5 5 5 55 0 1 1 5 5

n n n nr r r rC C C C C . . . C C++ + += + + +

1 2 3 4 55 10 10 5n n n n n nr r r r r rC C C C C C+ + + + += + + + + +

⇒ LHS =( )( )

55 55

4 444

coeff. of in 1 5

5coeff. of in 1

++ ++

+ +++

+ + += = =

+ ++

nr nr

n nrr

x x C n n k

r r kCx x (Given)

⇒ k = 5

Download - Solutions to Mock IIT Advanced/Test - 3[Paper-1]/2013100p.s3.amazonaws.com/.../SOLUTIONS-MOCKIIT-TEST-3-PAPER-1.… · Solutions to Mock IIT Advanced/Test - 3[Paper-1]/2013 [MATHEMATICS]

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