A pendulum bob has a speed of 3 m/s at its lowest position. The pendulum is 0.5 m long. The speed of the bob, when the length makes an angle of ${60}^{\mathrm{o}}$ to the vertical, will be (If  $\mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2$ )

Let bob velocity be v at point B where it makes an angle of 60^o with the vertical, then using conservation of mechanical energy

${\mathrm{KE}}_{\mathrm{A}}+{\mathrm{PE}}_{\mathrm{A}}={\mathrm{KE}}_{\mathrm{B}}+{\mathrm{PE}}_{\mathrm{B}}$

$\Rightarrow \frac{1}{2}\mathrm{m}\times 3^2=\frac{1}{2}{\mathrm{mv}}^2+\mathrm{mgl}(1-\mathrm{cos\theta })$

$\Rightarrow 9={\mathrm{v}}^2+2\times 10\times 0.5\times \frac{1}{2}\Rightarrow \mathrm{v}=2\mathrm{m}/\mathrm{s}$