JEE Advanced 2025 — Paper 2 — Question 9

$\mathrm{dM}=\mathrm{dIA}$

$\mathrm{A}=\pi \mathrm{r}^{2}=\pi(\mathrm{R} \sin \theta)^{2}$

$\mathrm{dI}=\frac{\mathrm{da}}{\mathrm{T}}=\frac{\sigma(2 \pi \mathrm{r})(\mathrm{Rd} \theta) \omega}{2 \pi}$

$\mathrm{dI}=\frac{\sigma 2 \pi \mathrm{R}^{2} \omega \sin \theta \mathrm{~d} \theta}{2 \pi}$

$\mathrm{dI}=\sigma \mathrm{R}^{2} \omega \sin \theta \mathrm{~d} \theta$ Magnetic dipole moment :

$M=\int d M=\int_{0}^{\pi} \sigma R^{2} \omega \pi R^{2} \sin ^{3} \theta d \theta$

$M=\sigma R^{4} \omega \pi \int_{0}^{\pi} \sin ^{3} \theta d \theta \quad\left(\because \int_{0}^{\pi} \sin ^{3} \theta d \theta=\frac{4}{3}\right)$

$M=\left(\frac{Q}{4 \pi R^{2}}\right) R^{4} \omega \pi\left(\frac{4}{3}\right)$

Magnetic dipole moment $\mathrm{M}=\frac{\mathrm{QR}^{2} \omega}{3}$

Angular momentum $\mathrm{L}=\left(\frac{2}{5} \mathrm{MR}^{2}\right) \omega$

$\frac{\mathrm{M}}{\mathrm{L}}=\frac{\mathrm{QR}^{2} \omega}{3 \times \frac{2}{5} \mathrm{MR}^{2} \omega}=\frac{\mathrm{Q}}{2 \mathrm{M}}\left(\frac{5}{3}\right)$ $\alpha=\frac{5}{3}=1.67$