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Q.

A pendulum bob has a mass $m = \frac{1}{2} k g$ and length of string $l = 1 m .$ The bob is released from position A. Then maximum tension in the string is ____

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a

$10 \sqrt{3} N$

b

20 N

c

$20 \sqrt{3} N$

d

10 N

answer is A.

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Detailed Solution

$\begin{array}{l} V_{\max} = \sqrt{2 g l (1 - \cos θ)} = \sqrt{2 \times 10 \times 1 \times (1 - \cos 60^{0})} m / s \\ = \sqrt{10} m / s \\ ∴ T_{\max} = m g + \frac{m V_{\max}^{2}}{l} = (\frac{1}{2} \times 10 + \frac{\frac{1}{2} \times 10}{1}) N = 10 N \end{array}$

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