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 $$60°$$ with vertical as shown in the figure. After releasing the bob, the maximum velocity attained by the bob will be ______ m s$$^{-1}$$. (if $$g = 10$$ m s$$^{-2}$$)

From conservation of energy,

Loss in Potential Energy (PE) = Gain in Kinetic Energy (KE)

$$PE_{initial}\ =\ mgh_i$$

From figure, $$h_i\ =\ l\left(1-\sin\left(30^{\circ\ }\right)\right)$$ = $$250\left(1-\frac{1}{2}\right)\ =\ \frac{250}{2}\ =\ 125\ cm\ =\ 1.25\ m$$

Therefore, $$PE_i\ =\ 0.2\ \times\ 10\ \times\ 1.25\ =\ 2.5\ J$$

$$PE_{final}\ =\ 0\ $$

$$KE_{initial}\ =\ 0\ $$ (as the bob is at rest)

Therefore, $$PE_{initial}\ -\ PE_{final}\ =\ KE_{final}\ -\ KE_{initial}$$

$$\therefore\ \ KE_{final}\ =\ PE_{initial}\ =\ 2.5\ J$$

$$\therefore\ \ \frac{1}{2}mv_{\max}^2\ =\ 2.5\ J\ $$

$$\therefore\ \ v_{\max}\ =\ \sqrt{\frac{\left(\ 2\times\ 2.5\right)}{0.2}\ }=\ 5$$m/s