Q 12-08-201JEE MainJEE Main 2019 (10 Apr, Shift 2)Easy
Light is incident normally on a completely absorbing surface with an energy flux of $25\ \text{W cm}^{-2}$. If the surface has an area of $25\ \text{cm}^2$, the momentum transferred to the surface in $40\ \text{min}$ time duration will be:
Answer: (B) $5.0\times10^{-3}\ \text{N s}$
Power absorbed $= 25\times25 = 625\ \text{W}$. Energy in $40\ \text{min}$: $U = 625\times2400 = 1.5\times10^6\ \text{J}$.
For complete absorption $p = \dfrac Uc = \dfrac{1.5\times10^6}{3\times10^8} = 5.0\times10^{-3}\ \text{N s}$.
Solution by Sreeraj P, M.Sc Physics