Q 12-08-074JEE MainJEE Main 2023 (30 Jan, Shift 1)Easy
A small object at rest absorbs a light pulse of power $20\ \text{mW}$ and duration $300\ \text{ns}$. Assuming speed of light as $3\times10^8\ \text{m s}^{-1}$, the momentum of the object becomes equal to
Answer: (B) $2\times10^{-17}\ \text{kg m s}^{-1}$
Energy absorbed $E=Pt=20\times10^{-3}\times300\times10^{-9}=6\times10^{-9}\ \text{J}$.
For complete absorption $p=\dfrac Ec=\dfrac{6\times10^{-9}}{3\times10^8}=2\times10^{-17}\ \text{kg m s}^{-1}$.
Solution by Sreeraj P, M.Sc Physics