# solutions to mock iit advanced/test - 3[paper-1]/2013100p.s3. solutions to mock iit advanced/test -

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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

VMC/2013/Solutions 2 Mock IIT Advanced/Test - 3/Paper-1

• Vidyamandir Classes

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

• Vidyamandir Classes

VMC/2013/Solutions 4 Mock IIT Advanced/Test - 3/Paper-1

• Vidyamandir Classes

VMC/2013/Solutions 5 Mock IIT Advanced/Test - 3/Paper-1

• Vidyamandir Classes

VMC/2013/Solutions 6 Mock IIT Advanced/Test - 3/Paper-1

• Vidyamandir Classes

VMC/2013/Solutions 7 Mock IIT Advanced/Test - 3/Paper-1

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

[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 L dt

= −

Or L V

di dt L

= −∫ ∫

V = (10 t − 20) where 0 4t ms≤ < ( ) 2

0

1 1 10 10 20 20

2

t

L t

i t dt t L L

  ⇒ = − + = − − 

   ∫

25 20

L t t

i L 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⇒ = + −

( )0 0

1000 800 2

1000 300 7

w s s w w

w

x x

L L

ρ ρ ρ ρ ρ ρ

ρ ρ − −

⇒ = − + ⇒ = = = − −

27.(C) x a cos tω= v a sin tω ω= −

1 1 1

2 2

a a cos t cos tω ω= ⇒ =

1 1 3 3 2 6

T T t t

π π ω

π ⇒ = ⇒ = =

×

( ) 6

0

6

0

3

T /

T /

a sin t dt

a v

T dt

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

25 2 1

N N µ

= = +

Now from (i) and (ii)

F = 2 Nµ = 2 0 5 25 25. N× × =

• Vidyamandir Classes

VMC/2013/Solutions 8 Mock IIT Advanced/Test - 3/Paper-1

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

80 6

r mv

F N R

× = = = .

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

3 µ = .

32.(B)

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

1

21 cmv v

e v cos

ω

θ

+ + = =

. . . . . .(i)

1cmmv cos mv mvθ = − . . . . . .(ii)

2

1 2 12 2

m mv cos mvθ ω= −

ℓ ℓ ℓ . . . . . .(iii)

Now solving, we get : 2

5 cmv v cosθ= ,

12 ω

5 v cosθ= ℓ

, 1 3

5 v v cosθ= −

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

Δ 12 5 5

cm m

L I v cos m v cosω θ θ= = × = ℓ

ℓ ℓ

35.(B) 1 3 2

Δ 5 5

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

37.(AB) ( )

( ) 11 1

20 20 1

R

F R

µ

µ

− = = ⇒

− . . . . .(i)

If equiconvex lens of f = 20

Then ( )1 21 20 R

µ= −

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 correct 15 10

v cm v = − ⇒ = −

(C) 20u = − 20v = − So C is wrong

Alternative :

For planoconvex ( ) 1 2

1 1 1 1

f R R µ

  = − − 

 

1 2R R R= ∞ = r = 20 ( )1µ −

For equiconvex ( )1 1 11 20 R R

µ  = − +   

( )40 1R µ= −

• Vidyamandir Classes

VMC/2013/Solutions 9 Mock IIT Advanced/Test - 3/Paper-1

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 2

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

where 4 4

mg mg l x l

K mg = = =

Putting 4

l x = in equation (1), we get

3

2 maxv gL=

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

(D) Time taken is 1

2 4 4 4

T m L

K g

π π= =

39.(BD) Equivalent circuit :

Induced emf in a rotating rod with constant angular

Velocity 2ω

ω 2

B a e; = , (here a = Radius)

Now use Kirchoff’s Law.

40.(3) 1v

m u n

= = −

( )0

1L

L

X m

– X X n = = −

Lens formula ⇒ ( )0

1 1 1

L LX X X f − = − −

( ) 0

2 2 0

1 1 0L L

L L

dXdX dX .

dt dt dtX X X

  − − − = 

 −

( ) ( ) 0

2 2 2 0 0

1 1 1L

L L L

dXdX

dt dtX X X X X

   − + =  − − 

0

0; LdX dXV V

dt dt = =

⇒ 2 2

0 1 1

1V V n n

     − + + = +         

⇒ [ ]0 0 4

1 4 4 3

V V V V− + = ⇒ = ⇒ 0

3

4

V V =

41.(1) In first case 10

30 10 1 1

K R

× = − × +

In second case 5

10 5 2 2

K R

× = − × +

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 t R R

× = ×

ℓ =

2 2 3 2100 (0 4) 2 5 10 1

10 000

. . v

,

−× × × × ×

2 32 5 10. −= ×ℓ

4 2

25 10 5 10 m − −= × = ×ℓ

25 10

1 v

−× =

⇒ 86 25 10 16 1W . J Jµ−= × × =

• Vidyamandir Classes

VMC/2013/Solutions 10 Mock IIT Advanced/Test - 3/Paper-1

43.(9) 1

2 2 3

2 0 3

2

P

R kQ

kQ E

RR = − =      

⇒ 12 4

9 2

Q Q =

2

1

9 =

8

Q

Q

2 9 8

9 8

Q Cµ ×

= =

44.(8) Initially 0100 2x g=

0 20x cm=

Finally, equilibrium position

0100 1x g′ =

0 10′ =x cm

1

2 100 10

T π

π 2

= =

2 10

T t

π = =

21

10 50 2 100

S cm π

= × × =

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

So, x = 8

45.(4) Paschen series 7 6 2

1 1 1 1 097 10 10

9 .

nλ  

= × − ≤   

2

1 1 1

9 10 97.n − ≤

2

1 1 97

9 10 97

.

.x ≥

×

29 1097

197 x

× ≥

3 1097

7 197

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

46.(1) 2 2 3

24 (ω )

10 10 z L R

− = = +

×

2 2 2 2 1(ω ) ω

ω R L R L

C

  + = + − 

 

( ) 1ω ω ω

L L C

= − +

2 6

1 1 5

2ω 2 100 100 10 L 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

VMC/2013/Solutions 11 Mock IIT Advanced/Test - 3/Paper-1

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

[MATHEMATICS]

47.(B) Let ( )2 2P t , t and 21 2Q , tt  

−   

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