JEE Main 2026 — 21 January, Morning Shift — Question 27

Area of the loop $=1 \mathrm{~m}^{2}$ $B=\sin (100 t)$ $\therefore \quad \phi=\mathrm{BA}=\sin (100 \mathrm{t})$ $\therefore \quad \frac{\mathrm{d} \phi}{\mathrm{dt}}=100 \cos (100 \mathrm{t})$ $\therefore \quad \mathrm{P}=\frac{\mathrm{V}^{2}}{\mathrm{R}}=\frac{10^{4} \cos ^{2}(100 \mathrm{t})}{100}$ $\therefore \quad$ Thermal energy dissipated in 1 time period

$$\begin{aligned} & =\int_{0}^{\mathrm{T}} \mathrm{Pdt}=\int_{0}^{\mathrm{T}} 100 \cos ^{2}(100 \mathrm{t}) \mathrm{dt} & \mathrm{~T}=\frac{2 \pi}{100}=\frac{\pi}{50} \mathrm{sec} & \therefore \quad \mathrm{Q}=100 \int_{0}^{\pi / 50} \cos ^{2}(100 \mathrm{t}) \mathrm{dt} & \quad=100 \int_{0}^{\pi / 50} \frac{1+\cos 200 \mathrm{t}}{2} \mathrm{dt} & \quad=100\left[\frac{\pi}{100}\right]=\pi \end{aligned}$$