JEE MAIN - Physics (2024 - 29th January Evening Shift - No. 15)
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, $$\mathrm{g}: 10 \mathrm{~ms}^{-2}$$]
$$5 \sqrt{6} \mathrm{~ms}^{-1}$$
$$5 \sqrt{5} \mathrm{~ms}^{-1}$$
$$2 \sqrt{5} \mathrm{~ms}^{-1}$$
$$6 \sqrt{5} \mathrm{~ms}^{-1}$$
Explanation
_29th_January_Evening_Shift_en_15_1.png)
$$\ell=10 \mathrm{~m}$$,
Initial energy $$=\mathrm{mg} \ell$$
$$\begin{aligned} & \text { So, } \frac{9}{10} \mathrm{mg} \ell=\frac{1}{2} \mathrm{mv}^2 \\ & \Rightarrow \frac{9}{10} \times 10 \times 10=\frac{1}{2} \mathrm{v}^2 \\ & \mathrm{v}^2=180 \\ & \mathrm{v}=\sqrt{180}=6 \sqrt{5} \mathrm{~m} / \mathrm{s} \end{aligned}$$
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