JEE Advanced 2024 — Paper 2 — Question 41

$y=\frac{8 \times 6000 \times 10^{-10} \times 1}{(0.8+0.04 \sin \omega t) \times 10^{-3}}$ $y=\frac{48 \times 10^{-4}}{(0.8+0.04 \sin \omega t)}$

$$\begin{aligned} & \left|\frac{d y}{d t}\right|=+\frac{48 \times 10^{-4}}{(0.8+0.04 \sin \omega t)^{2}}(0.04 \omega \cos \omega t) \\ & t=0, v=v_{\max } \\ & v_{\max }=\frac{48 \times 10^{-4} \times 0.04 \times 0.08}{(0.8)^{2}}=24 \times 10^{-6} \mathrm{~m} / \mathrm{s}=24 \mu \mathrm{~m} / \mathrm{s} \end{aligned}$$