JEE MAIN - Physics (2024 - 29th January Evening Shift - No. 6)
A bob of mass '$$m$$' is suspended by a light string of length '$$L$$'. It is imparted a minimum horizontal velocity at the lowest point $$A$$ such that it just completes half circle reaching the top most position B. The ratio of kinetic energies $$\frac{(K . E)_A}{(K . E)_B}$$ is :
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Explanation
Apply energy conservation between A & B
$$\begin{aligned} & \frac{1}{2} \mathrm{mV}_{\mathrm{L}}^2=\frac{1}{2} \mathrm{mV}_{\mathrm{H}}^2+\mathrm{mg}(2 \mathrm{~L}) \\ & \because \mathrm{V}_{\mathrm{L}}=\sqrt{5 \mathrm{gL}} \end{aligned}$$
So, $$\mathrm{V}_{\mathrm{H}}=\sqrt{\mathrm{gL}}$$
$$\frac{(\mathrm{K} . \mathrm{E})_{\mathrm{A}}}{(\mathrm{K} . \mathrm{E})_{\mathrm{B}}}=\frac{\frac{1}{2} \mathrm{~m}(\sqrt{5 \mathrm{gL}})^2}{\frac{1}{2} \mathrm{~m}(\sqrt{\mathrm{gL}})^2}=\frac{5}{1}$$
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