JEE MAIN - Physics (2025 - 22nd January Evening Shift - No. 7)
A body of mass 100 g is moving in circular path of radius 2 m on vertical plane as shown in figure. The velocity of the body at point $A$ is $10 \mathrm{~m} / \mathrm{s}$. The ratio of its kinetic energies at point B and C is :
_22nd_January_Evening_Shift_en_7_1.png)
(Take acceleration due to gravity as $10 \mathrm{~m} / \mathrm{s}^2$)
Explanation
_22nd_January_Evening_Shift_en_7_2.png)
$$\begin{aligned} & \frac{1}{2} \mathrm{~m} \times 100+0=\frac{1}{2} \mathrm{mV}_{\mathrm{B}}^2+\mathrm{mg}\left(\mathrm{R}-\frac{\mathrm{R} \sqrt{3}}{2}\right) \\ & 100=\mathrm{V}_{\mathrm{B}}^2+2 \mathrm{gR}\left(1-\frac{\sqrt{3}}{2}\right] \\ & \mathrm{V}_{\mathrm{B}}^2=100-20(2-\sqrt{3}) \\ & \left.\mathrm{V}_{\mathrm{B}}^2=60+20 \sqrt{3}\right) \\ & \mathrm{K}_{\mathrm{E}} \mathrm{E}_{\mathrm{B}}=\frac{1}{2} \mathrm{mV}_{\mathrm{B}}^2=\frac{\mathrm{m}}{2}(60+20 \sqrt{3}) \\ & \frac{1}{2} \mathrm{~m}(100)=\frac{1}{2} \mathrm{mV}_{\mathrm{C}}^2+\mathrm{mg}\left(\frac{3 \mathrm{R}}{2}\right) \end{aligned}$$
$$\begin{aligned} & 100=V_{\mathrm{C}}^2=60 \\ & \mathrm{~V}_{\mathrm{C}}^2=40 \\ & \mathrm{~K} \cdot \mathrm{E}_{\mathrm{C}}=\frac{1}{2} m V_{\mathrm{C}}^2=\frac{1}{2} \mathrm{~m}(40) \\ & \mathrm{K} \cdot \mathrm{E}_{\mathrm{B}}=\frac{60+20 \sqrt{3}}{40}=\frac{3}{2}+\frac{\sqrt{3}}{2}=\frac{3+\sqrt{3}}{2} \end{aligned}$$
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