Laws of Motion

• Question 569

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 = 2 m/s

∴      0.02 × 600 = 0.02 v + 4 × 2

⇒             0.02 v = 12 - 8

⇒                    v = $\frac{4}{0.02}$

⇒                      v = 200 m/s

Question 570

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

= 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

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 is96 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 = $\frac{1}{2}{\mathrm{mv}}^{2}$

=

Kinetic energy = 96 J

Question 572

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 of0.1 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 = $\frac{{\mathrm{V}}^{2}}{\mathrm{P}}$

R = $\frac{{\left(100\right)}^{2}}{50}$

= 200 Ω

Current through bulb

(I) = $\frac{\mathrm{V}}{\mathrm{R}}$

= $\frac{100}{200}$

I  =  0.5 A

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

Z =               ...... (i)

Z = $\frac{200}{0.5}$

Z = 400 Ω

Now,

${\mathrm{X}}_{\mathrm{L}}^{2}$  = ( 400 )2  - ( 200 )2

=

L = 1.1 H