brain storming comprehensions (physics)

8/10/2019 Brain Storming Comprehensions (Physics)

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Brain Storming

Comprehensions UNIT 4Linked Comprehension Type

C1. 1. Answer (3)

Fringe width = F =d

Dλ

d

v

dt

dF R

λ==⇒

And a = kt

kt dt

dv =⇒

2

2kt v =⇒

22

2t R

d

kt

dt

dF R α⇒

λ==⇒

2. Answer (1)

x component of velocity = velocity of screen =2

2kt

Also, y =d Dnλ

⇒ y component of velocity =dt

dy

=d

kt n

2

2λ

⎟ ⎠

⎞⎜⎝

⎛ λ+=⇒ j

d

ni

kt v ˆˆ

2

2

3. Answer (4)

Position vector of nth maxima relative to (n – 1)th maxima = j

d

D ˆλ

j d

kt j

dt

Dd

d a ˆˆ

2

2 λ=⋅

λ=⇒

C2. 1. Answer (1)

For minimum energy, j = 0 ⇒π

=2

hL . Use

I

Lk

2

2

=

2. Answer (1)

If KT < (KE)R, there will be no rotation.

3. Answer (2)For nitrogen, minimum KE of rotation is very low.



330 Brain Storming Comprehensions Success Magnet-Solutions (Part-I)


C3. 1. Answer (1)

N

A B

4I I

N

For the disc A

NR Δt = 4Iω1

NR Δt = I (ω + ω2)

⇒ 4ω1 = ω + ω

2...(i)

21

12

uu

v v e

−

−=

ω=ω+ω⇒−ωω+ω

= 2121 2202

1

R

R R ...(ii)

4ω 1 – ω 2

= ω

2ω 1 + 2 ω2

= ω

1031 ω=ω⇒

52

ω=ω

If the next collision take place of n1 revolutions of the first disc and n

2 revolution of the second disc then

n1 ω2

= n2 ω1

2

3

2

1

n

n⇒

2. Answer (2)

So the time after which next collison will take place

= 32

1

×ωπ or 22

2×ωπ

= 3103

2××

ωπ

=ωπ

20

and T =ωπ

So = 20 T

3. Answer (2)

If I 1 = I

2 and e = 1 then

I ω 1 = I (ω + ω 2

)

10

21 =−ωω+ω

ω 1 – ω 2

= ω

ω1 + ω 2

= ω

⇒ ω1 = ω and ω2

= 0

So next collision will take place after

T 22 =ωπ



331Success Magnet-Solutions (Part-I) Brain Storming Comprehensions


C4. 1. Answer (1)

(V – V ′)5 = V × 1 ⇒5

4=

′

V

V

2. Answer (2)

Acceleration = drag force / mass = 6πηr

m

V V ⎟ ⎠ ⎞⎜

⎝ ⎛ − 2

00

3. Answer (4)

Velocity increases exponentially from zero to V 0.

C5. 1 Answer (3)

For a horizontal axis through C , ∫ ⎟ ⎠

⎞⎜⎝

⎛ θ

===2

22 l mmR dmR I

2. Answer (1)

For the position of centre of mass, we have

θ

θ−=

θ

θθ

=

∫

∫

π

θ−π

π

θ−π

)cos1(

sin

2

2

2

20 R

d

d R

x

Also,θ

= l

R

3. Answer (3)

Mgd

I T π= 2

Here d = R – x 0, I = I cm + md 2 = (MR 2 – mx 02) + md 2

C6. The situation is shown in figure.

3 0 °

θr a

B′

C ′

A′

A

B C

v

v

v

O

3a1. Answer (3)

dt

dr v −=30cos ⇒ vt ar

2

3−=

2. Answer (1)

dt

d

r

v θ=

30sin⇒

⎟⎟ ⎠

⎞⎜⎜⎝

⎛

−=θ⇒

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ −

=θt v a

a

vt a

vdt d

32

2ln

3

1

2

32

3. Answer (1)

All the sides of Δ ABC turn through same angle θ.

The particles converge, when r = 0

⇒v

at 3

2=



332 Brain Storming Comprehensions Success Magnet-Solutions (Part-I)


C7. 1. Answer (3)

Relative to cart, velocity of small body will be at an angle 45° above horizontal.

Let V C be the speed of cart

V c

V 0

45°

Cart

By conservation of linear momentum along horizontal,

v mmv v v m c c ×=+−2

)( 0

22 0v

v v c −=

22

2 0v v v c

−=

2. Answer (2)

The vertical component of the small body is2

0v

So, height above the point of leaving is ( ) g v

g v

422

2

0

2

0 =

Above the cart surface, height =g

v R R

42

20+−

3. Answer (2)

Momentum along horizontal remains conserved,

So, mv = (m + m)V f

⇒ v f =

2

v

Loss = K i – K

f =

42)2(

2

1

2

1 222 mv v

mmv =⎟ ⎠

⎞⎜⎝

⎛ −

C8. 1. Answer (3)

The phase difference, if the wave travelled along x-axis is k Δ x . Now the phase difference will be )(·)( r k Δ

= ⎥⎦

⎤⎢⎣

⎡+

⎥⎥⎦

⎤

⎢⎢⎣

⎡++

λπ

k j i k j i ˆ2ˆ

2

1 –ˆ32·ˆ

2

1ˆ

2

1ˆ2

32

=λπ7

2. Answer (4)

For a point source, wave has spherical symmetry and the wave representation would be like

Displacement Y = Asin(ωt – kr )

So, Δφ = k Δr = ) –(2

12 r r λπ

= 002

=×λπ

∴ r 2 = r

1 =

222 132 ++

3. Answer (3)

]ˆcosˆcosˆcos[ j j i v

k γ+β+αω

=

v v

v v

v v z y x =γ=β=α cos,cos,cos



333Success Magnet-Solutions (Part-I) Brain Storming Comprehensions


C9. 1. Answer (1)

For the object, not to hit the planet, its minimum distance ( i .e., when it is at perigee of its path) must

be slightly greater than the radius of planet.

2. Answer (3)

In the above case, semi-major axis of the orbit would be a =2

3R

Total mechanical energy =R

GMm

a

GMm

3 –

2 – =

So,R

GMm

R

GMmmv

3 –

)2( –

2

1 20 =

R

GM

v 30 =

3. Answer (1)

When v < v 0, object falls on planet, so it will not orbit. When v > v

escape object will not orbit.

R

GM

R R

GM

hR

GM v e =

+=

+=

22

brain storming comprehensions (physics)

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