JEE Advanced 2025 — Paper 2 — Question 16

$\cos \theta_{0}=1-\frac{\theta_{0}^{2}}{2}=0.9$

$\frac{\theta_{0}^{2}}{2}=0.1 \Rightarrow \theta_{0}=10.2=\frac{1}{\sqrt{5}}$

$\overrightarrow{\underset{\mathrm{v}^{\prime}}{\rightleftarrows}} \underset{\underset{\mathrm{v}^{\prime}}{\rightarrow}}{\stackrel{-}{\leftrightarrows}}$

$f_{\text {max }}=\frac{v+v^{\prime}}{v-v^{\prime}} f$

$f_{\text {min }}=\frac{v-v^{\prime}}{v+v^{\prime}} f$

$\Delta \mathrm{f}_{\text {max }}=\mathrm{f}_{\text {max }}-\mathrm{f}_{\text {min }}=\frac{\mathrm{v}+\mathrm{v}^{\prime}}{\mathrm{v}-\mathrm{v}^{\prime}} \mathrm{f}-\frac{\mathrm{v}-\mathrm{v}^{\prime}}{\mathrm{v}+\mathrm{v}^{\prime}} \mathrm{f}$

$=\frac{\left(v+v^{\prime}\right)^{2}-\left(v-v^{\prime}\right)^{2}}{v^{2}-v^{\prime 2}} \mathrm{f}$

$\Delta \mathrm{f}_{\text {max }}=\frac{4 \mathrm{vv}^{\prime}}{\mathrm{v}^{2}-\mathrm{v}^{\prime 2}} \mathrm{f}$

Here, $\mathrm{v}^{\prime}=\ell \Omega_{\max }$

$=\ell \cdot \theta_{0} \cdot \omega \quad(\omega=$ angular frequency $)$

$=\ell \theta_{0} \sqrt{\frac{g}{\ell}}$

$v^{\prime}=\theta_{0} \sqrt{g \ell}$

$\mathrm{v}^{\prime}=\frac{1}{\sqrt{5}} \sqrt{10 \times 8}$

$\mathrm{v}^{\prime}=4$

Put in equation (i) $\Delta \mathrm{f}_{\max }=\frac{4 \times 330 \times 4 \times 660}{330^{2}-4^{2}}$

$\approx \frac{16 \times 330 \times 660}{330} \approx 32$