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 \(10 \mathrm{~m} / \mathrm{s}^2\) )

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