Laws of Motion

Question 569
CBSEENPH11026238

A bullet of mass 20 g and moving with 600 m/s collides with a block of mass 4 kg hanging with the string. What is velocity of bullet when it comes out of block, if block rises to height 0.2 m after collision?

  • 200 m/s

  • 150 m/s

  • 400 m/s

  • 300 m/s

Solution

A.

200 m/s

Linear momentum is a product of the mass (m) of an object and the velocity (v) of the object.

According to conservation of linear momentum

          m1 v1 = m1 v + m2 v2

where v1 is the velocity of the bullet before the collision, v  is the velocity of bullet after the collision and v2 is the velocity of the block.

∴     0.02 × 600 = 0.02 v + 4 v2

We have

       v2 = 2gh

           = 2 × 10 × 0.2

       v2 = 2 m/s

∴      0.02 × 600 = 0.02 v + 4 × 2

⇒             0.02 v = 12 - 8

⇒                    v = 40.02

⇒                      v = 200 m/s

Question 570
CBSEENPH11026239

Voltage in the secondary coil of a transformer does not depend upon

  • frequency of the source

  • voltage in the primary coil

  • ratio of number of turns in the two coils

  • Both (b) and (c)

Solution

A.

frequency of the source

NPNS = VPVS = n = Turns ratio

Ratio of number of  turns in the two coils results induced emf. 

Voltage in the primary coil also affects the induced emf.

Question 571
CBSEENPH11026251

A shell of mass 20 kg at rest explodes into two fragments whose masses are in the ratio 2 : 3. The smaller fragment moves with a velocity of 6 m/s. The kinetic energy of the larger fragment is

  • 96 J

  • 216 J

  • 144 J

  • 360 J

Solution

A.

96 J

Total mass of the shell = 20 kg

Ratio of the masses of the fragments are 8 kg and 12 kg 

Now according to the conservation of momentum 

             m1 v1 = m2 v2

∴            8 × 6 = 12  × v

v (velocity of the larger fragment) = 4 m/s

  Kinetic energy = 12mv2

                        = 12 × 12 × 42

   Kinetic energy = 96 J

Question 572
CBSEENPH11026252

An electric bulb has a rated power of 50 W at 100 V. If it is used on an AC source 200 V, 50 Hz, a choke has to be used in series with it. This choke should have an inductance of

  • 0.1 mH

  • 1 mH

  • 0.1 mH

  • 1.1 H

Solution

D.

1.1 H

Power P = I × V

             = V R × V

           P = V2 R

Resistance of bulb

          R = V2P

           R = 100250

               = 200 Ω 

Current through bulb

            (I) = VR

                 = 100200

              I  =  0.5 A

In a circuit containing inductive reactance (XL ) and resistance (R ), impedance (Z) of the circuit is

             Z = R2 + ω2 L2              ...... (i)

             Z = 2000.5

             Z = 400 Ω

Now,      XL2 = Z2 - R2

             XL2  = ( 400 )2  - ( 200 )2

           2 πf L2 = 12 × 104

             L = 23 × 1002π × 50

                  = 23 π

              L = 1.1 H

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