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/s²)

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(4) (3+√3)/2. K_A=1/2×0.1×100=5 J. B is 30° from the bottom: h_B=R(1-cos 30°)=2-√3 m, so K_B=5-mgh_B=5-(2-√3)=3+√3 J. C is 90° beyond B, i.e. 120° from the bottom: h_C=R(1-cos 120°)=3 m, so K_C=5-3=2 J. (K_B)/(K_C)=(3+√3)/2.

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/s²)

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