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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