A body of mass $1 k g$ is moving in a vertical circular path of radius $1 m$ . The difference between the kinetic energies at its highest and lowest positions is :

4 \sqrt{5} J

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20 J

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10 J

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30 J

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Solution

The correct option is A $20 J$
Kinetic energy is the energy possessed by a body of mass $m$ due to its velocity $v$ .
$K E = \frac{1}{2} m v^{2}$
At the highest point velocity, $v_{A} = \sqrt{r g}$
$∴ K E = \frac{1}{2} m v_{A}^{2}$
At the lowest point velocity, $v_{B} = \sqrt{5 r g}$
$K E = \frac{1}{2} m v_{B}^{2}$
Difference in $K E = \frac{1}{2} m (v_{B}^{2} - v_{A}^{2}) = \frac{1}{2} m (5 r g - r g)$
$= 2 m r g = 2 \times 1 \times 1 \times 10 = 20 J$

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