A pendulum is suspended by a string of length 250 cm. The mass of the bob of the pendulum is 200 g. The bob is pulled aside until the string is at 60^o with vertical as shown in the figure. After releasing the bob, the maximum velocity attained by the bob will be __________ $ms^{- 1}$ . (if $g = 10 m / s^{2}$ )

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answer is 5.

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Detailed Solution

$V_{max} = \sqrt{2 g h}$

The speed will be highest at the lowest position
$\begin{array}{l} h = (ℓ - ℓ \cos 60^{\circ}) = \frac{ℓ}{2} \\ V_{max} = \sqrt{2 \times g \times \frac{ℓ}{2}} = \sqrt{10 \times 2.5} = 5 m / s \end{array}$

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