A body of mass m is rotated in a vertical circle with help of light string such that velocity of body at a point is equal to critical velocity at that point. If $T_{1}, T_{2}$ be the tensions in the string when the body is crossing the highest and the lowest points of vertical circle respectively, then which of the following relation is correct?

T_{2} - T_{1} = 6 m g

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T_{2} - T_{1} = 4 m g

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T_{2} - T_{1} = 3 m g

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T_{2} - T_{1} = 2 m g

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Solution

The correct option is A $T_{2} - T_{1} = 6 m g$
At lowest point $V_{2} = \sqrt{5 g R}$ and at highest point $V_{1} = \sqrt{g R}$
$∴$ At lowest point $T_{2} - m g = \frac{m ν^{2}}{R}$
$\to T_{2} = m g + 5 m g = 6 m g$
At highest point $T_{1} + m g = \frac{m ν^{2}}{R}$
$T_{1} = 0$
$∴ T_{2} - T_{1} = 6 m g$

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