The bob of a pendulum was released from a horizontal position. The length of the pendulum is $10 \mathrm{~m}$. If it dissipates $10 \%$ of its initial energy against air resistance, the speed with which the bob arrives at the lowest point is : [Use, $\mathrm{g}: 10 \mathrm{~ms}^{-2}$ ]
JEE MAIN 2024, Diffcult
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$ \ell=10 \mathrm{~m} $
$ \text { Initial energy }=\mathrm{mg} \ell $
$ \text { So, } \frac{9}{10} \mathrm{mg} \ell=\frac{1}{2} \mathrm{mv}^2 $
$ \Rightarrow \frac{9}{10} \times 10 \times 10=\frac{1}{2} \mathrm{v}^2 $
$ \mathrm{v}^2=180 $
$ \mathrm{v}=\sqrt{180}=6 \sqrt{5} \mathrm{~m} / \mathrm{s}$
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