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 m/s. The ratio of its kinetic energies at point B and C is :

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

Kinetic energy at any point on the circle can be calculated using conservation of energy.

$$KE_A\ =\ \frac{1}{2}mv_A^2\ =\ \frac{1}{2}\ \times\ 0.1\ \times\ 10^2\ =\ 5J$$

$$PE_A\ =\ 0\ $$

At point B:-

Height $$H_B\ =\ R\left(1-\cos30^{\circ\ }\right)\ =\ 2-\sqrt{\ 3}\ m$$

$$PE_B\ =\ mgH_B\ =\ 0.1\times10\times\left(2-\sqrt{\ 3}\ \right)$$

$$KE_A\ +\ PE_A\ =\ KE_{B\ }+\ PE_B$$ 

⇒ $$5+0=KE_B+2-\sqrt{\ 3}$$

$$\therefore\ KE_B\ =\ 3+\sqrt{\ 3}\ J$$

At point C:-

Height $$H_C\ =\ R\left(1+\cos\ 60^{\circ\ }\right)\ =\ 2\left(1+\frac{1}{2}\right)=3\ m$$

$$PE_C\ =\ mgH_C=0.1\times10\times3\ =3\ J\ $$

$$KE_A\ +\ PE_A\ =\ KE_C\ +\ PE_C$$

⇒ $$5+0=KE_C+3$$

$$\therefore\ KE_C\ =\ 5-3\ =2\ J$$

Ratio of Kinetic energy at point B and C is therefore, $$Ratio=\frac{\left(3+\sqrt{\ 3}\right)}{2}$$