JEE Main 2025 — Dual Nature of Matter Question with Solution

Force due to beam assuming complete reflection $\mathrm{F}=\frac{2\mathrm{P}}{\mathrm{C}}=\frac{2}{\mathrm{C}}\frac{\mathrm{dE}}{\mathrm{dt}};\mathrm{P}$ is power
So change in momentum of mirror.
$\mathrm{m}(\mathrm{V}-0)=\int \mathrm{Fdt}=\frac{2}{\mathrm{C}}\int \mathrm{dE}=\frac{2\mathrm{E}}{\mathrm{C}}$
Now using work energy theorem ...(1)
$\begin{aligned}
& \mathrm{W}_{\mathrm{g}}=\Delta \mathrm{k} \\ & -\mathrm{mg} \ell(1-\cos \theta)=0-\frac{1}{2} \mathrm{mv}^2 \\ & \mathrm{~g} \ell\left(2 \sin ^2 \frac{\theta}{2}\right)=\frac{\mathrm{v}^2}{2}
\end{aligned}$<br/>as$\theta$issmall<br/>$\begin{aligned} & \mathrm{g} \ell 2\left(\frac{\theta}{2}\right)^2=\frac{1}{2} \frac{4 \mathrm{E}^2}{\mathrm{~m}^2 \mathrm{c}^2} \\ & \mathrm{~g} \ell \theta^2=\frac{4 \mathrm{E}^2}{\mathrm{~m}^2 \mathrm{c}^2} \\ & \theta=\frac{2 \mathrm{E}}{\mathrm{mc} \sqrt{\mathrm{g} \ell}}\end{aligned}$