Question 170 of 183

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 :



(Take acceleration due to gravity as $10 \mathrm{~m} / \mathrm{s}^2$ )

A$\frac{2+\sqrt{2}}{3}$B$\frac{2+\sqrt{3}}{3}$C$\frac{3+\sqrt{3}}{2}$D$\frac{3-\sqrt{2}}{2}$

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Answer: (C) $\frac{3+\sqrt{3}}{2}$

$\begin{aligned} & \frac{1}{2} \mathrm{~m} \times 100+0=\frac{1}{2} \mathrm{mV}_{\mathrm{B}}^2+\mathrm{mg}\left(\mathrm{R}-\frac{\mathrm{R} \sqrt{3}}{2}\right) \\ & 100=\mathrm{V}_{\mathrm{B}}^2+2 \mathrm{gR}\left(1-\frac{\sqrt{3}}{2}\right] \\ & \mathrm{V}_{\mathrm{B}}^2=100-20(2-\sqrt{3}) \\ & \left.\mathrm{V}_{\mathrm{B}}^2=60+20 \sqrt{3}\right) \\ & \mathrm{K} . \mathrm{E}_{\mathrm{B}}=\frac{1}{2} \mathrm{mV}_{\mathrm{B}}^2=\frac{\mathrm{m}}{2}(60+20 \sqrt{3}) \\ & \frac{1}{2} \mathrm{~m}(100)=\frac{1}{2} \mathrm{mV} \\ & 2 \\ & 100=\mathrm{mg}\left(\frac{3 \mathrm{R}}{2}\right) \\ & 100 \mathrm{~V}_{\mathrm{C}}^2=60 \\ & \mathrm{~V}_{\mathrm{C}}^2=40 \\ & \mathrm{~K} . \mathrm{E}_{\mathrm{C}}=\frac{1}{2} \mathrm{mV} \mathrm{V}_{\mathrm{C}}^2=\frac{1}{2} \mathrm{~m}(40) \\ & \mathrm{K} . \mathrm{E}_{\mathrm{B}}=\frac{60+20 \sqrt{3}}{40}=\frac{3}{2}+\frac{\sqrt{3}}{2}=\frac{3+\sqrt{3}}{2}\end{aligned}$