JEE Main 2025 — 22 January, Morning Shift — Question 62

$\quad \mathrm{v}=\frac{\mathrm{uf}}{\mathrm{u}-\mathrm{f}}=\frac{-24 \cdot 1}{-24-1}=\frac{24}{25}$ $\mathrm{m}=\frac{-\mathrm{v}}{\mathrm{u}}=-\frac{24}{25(-24)}=\frac{1}{25}$ $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\mathrm{v}_{\mathrm{I}}=-\mathrm{m}^{2} \cdot \mathrm{v}_{0}=\frac{-1}{(25)^{2}} \cdot 25=\frac{-1}{25}$

Diff. $\frac{-1}{\mathrm{v}^{2}}\left(\frac{\mathrm{dv}}{\mathrm{dt}}\right)+\frac{1}{\mathrm{u}^{2}}\left(\frac{\mathrm{du}}{\mathrm{dt}}\right)=0 \quad\left[\frac{\mathrm{dv}}{\mathrm{dt}}=\mathrm{v}_{\mathrm{i}} ; \frac{\mathrm{du}}{\mathrm{dt}}=\mathrm{v}_{0}\right]$

$\frac{+2}{\mathrm{v}^{3}} \cdot\left(\mathrm{v}_{\mathrm{I}}\right)^{2}-\frac{1}{\mathrm{v}^{2}} \cdot \mathrm{a}_{\mathrm{I}}-\frac{2}{\mathrm{u}^{3}} \cdot\left(\mathrm{v}_{0}\right)^{2}+\frac{1}{\mathrm{u}^{2}} \cdot \mathrm{a}_{0}=0$

$\frac{\mathrm{a}_{\mathrm{I}}}{\mathrm{v}^{2}}=\frac{2}{\mathrm{v}^{3}} \cdot \mathrm{v}_{\mathrm{I}}^{2}-\frac{2}{\mathrm{u}^{3}} \cdot \mathrm{v}_{0}^{2}$ $\mathrm{a}_{\mathrm{I}}=\frac{2}{\mathrm{v}} \cdot \mathrm{v}_{\mathrm{I}}^{2}-\frac{2 \mathrm{v}^{2}}{\mathrm{u}^{3}} \cdot \mathrm{v}_{0}^{2}$

$=\frac{2 \cdot 25}{24} \cdot \frac{1}{25} \cdot \frac{1}{25}-\frac{2}{(24)^{3}} \cdot \frac{24}{25} \cdot \frac{24}{25} \cdot 25 \cdot 25$ $\mathrm{a}_{\mathrm{I}}=\frac{2}{24.25}-\frac{2}{24}$

$\mathrm{a}_{\mathrm{I}}=\frac{2}{24} \cdot \frac{-24}{25}=\frac{-2}{25}$ $100 \mathrm{a}_{\mathrm{I}}=\frac{2}{25} \times 100=8$