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A bob of mass 100 g tied at the end of a string of length 50 cm is revolved in a vertical circle with a constant speed of $1 {ms}^{- 1}$ . When the tension in the string is 0.7N, the angle made by the string with the vertical is $(g = 10 {ms}^{- 2})$

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60 °

180 °

90 °

0 °

answer is D.

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Detailed Solution

m = 0.1 kg

$l = \frac{50}{100} = 0.5 m$

$T = \frac{{mv}^{2}}{r} + mg \cos θ$

$0.7 = \frac{0.1 \times 1^{2}}{0.5} + 0.1 \times 10 \cos θ$

$0.7 = \frac{1}{5} + \cos θ$

$\cos θ = \frac{1}{2}$

$θ = 60 °$

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