JEE Advanced 2019 — Paper 2 — Question 12

If $v$ is speed of mirror just after absorption of photons

$\mathrm{Mv}-\mathrm{N} \frac{2 \mathrm{~h}}{\lambda}=0 \Rightarrow \mathrm{v}=\frac{2 \mathrm{Nh}}{\mathrm{M} \lambda}$

If $\ell$ is the maximum compression in the spring

$\frac{1}{2} \mathrm{k} \ell^{2}=\frac{1}{2} \mathrm{mv}^{2}$

$\Rightarrow v=\sqrt{\frac{\mathrm{k}}{\mathrm{m}}} \ell$

$\Rightarrow \frac{2 \mathrm{Nh}}{\mathrm{M} \lambda}=\Omega \times 10^{-6}$

$\Rightarrow \mathrm{N}=\Omega \times 10^{-6} \times \frac{\mathrm{M} \lambda}{2 \mathrm{~h}}=8 \mathrm{~T} \times 10^{-6} \times 10^{-6} \times \frac{\mathrm{M} \lambda}{2 \mathrm{~h}}$

$\frac{4 \pi \mathrm{M} \lambda}{\mathrm{h}} \times 10^{-12}=10^{24} \times 10^{-12}=10^{12}$

$\Rightarrow \mathrm{x}=1$