JEE Advanced 2013 · Paper 1 · Physics Q6

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PhysicsSingle correct+2Question 6 of 60

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 $3\times10^8\ \mathrm{m\,s^{-1}}$ . The final momentum of the object is

0.3\times10^{-17}\ \mathrm{kg\,m\,s^{-1}}

1.0\times10^{-17}\ \mathrm{kg\,m\,s^{-1}}

3.0\times10^{-17}\ \mathrm{kg\,m\,s^{-1}}

9.0\times10^{-17}\ \mathrm{kg\,m\,s^{-1}}

Answer and worked solution

Official answer: B

Momentum of absorbed light. $p=\dfrac Ec=\dfrac{Pt}c=\dfrac{30\times10^{-3}\times100\times10^{-9}}{3\times10^8}=1.0\times10^{-17}\ \mathrm{kg\,m\,s^{-1}}$ .

(B)

Topics:Photon momentum and radiation pressure

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JEE Advanced 2013 · Paper 1, Physics question 6