Q102 / 150 · MHT CET 2016 · Physics

In order to just complete the vertical circular motion, the ratio of kinetic energy of a particle at highest point to that at lowest point is

A
5
B
2
C
0.5
Correct answer
D
0.2

Solution

K.E

= \frac{1}{2} m v^{2}

At lowest point in vertical circular motion

V_{l} = \sqrt{5 r g}

and at highest point

V_{h} = \sqrt{r g}

∴ \frac{{(K . E)}_{h}}{{(K . E)}_{l}} = \frac{1}{5} = 0.2

Hence, the correct answer is option: 'D'.

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