JEE Advance - Physics (2013 - Paper 1 Offline - No. 15)
A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30 mW and the speed of light is 3 $$\times$$ 108 ms$$-$$1. The final momentum of the object is
0.3 $$\times$$ 10$$-$$17 kg-ms$$-$$1
1.0 $$\times$$ 10$$-$$17 kg-ms$$-$$1
3.0 $$\times$$ 10$$-$$17 kg-ms$$-$$1
9.0 $$\times$$ 10$$-$$17 kg-ms$$-$$1
Explanation
Duration of pulse, t = 100 ns = 100 $$\times$$ 10$$-$$9 s.
Power of the pulse, P = 30 mW = 30 $$\times$$ 10$$-$$3 W.
Speed of light, c = 3 $$\times$$ 108 ms$$-$$1
The final momentum of the object is
$$p = {E \over c} = {{P \times t} \over c} = {{30 \times {{10}^{ - 3}} \times 100 \times {{10}^{ - 9}}} \over {3 \times {{10}^8}}} = {10^{ - 17}}$$ kg-ms$$-$$1
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