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Q.

A ball of mass ‘m’ is rotated in a vertical circle with constant speed. The difference in tension at the top and bottom would be

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a

6 mg

b

mg

c

2 mg

d

5 mg

answer is C.

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Detailed Solution

v = \sqrt{5 Rg}

$T_{max} = mg + \frac{{mv}^{2}}{R}$

$= mg + \frac{m5gr}{r}$

$= 6 mg$

$T_{\min} = \frac{{mv}^{2}}{R} - mg$

$= \frac{m5rg}{R} - mg$

$= 4mg$

$T_{1} - T_{2} = 6 mg - 4 mg$

$= 2mg$

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