Q89 / 90 · JEE Main 2021 (25 Jul Shift 1) · Physics

A pendulum bob has a speed of $3 m s^{- 1}$ at its lowest position. The pendulum is $50 cm$ long. The speed of bob, when the length makes an angle of $60 °$ to the vertical will be $(g = 10 m s^{- 2})$ ______ $m s^{- 1}$ .

Correct answer: 2

Solution

Applying work energy theorem:

$w_{g} + w_{T} = Δ K$

$- m g l (1 - \cos 60 °) = \frac{1}{2} m v^{2} - \frac{1}{2} m u^{2}$

$v^{2} = u^{2} - 2 g l (1 - \cos 60 °)$

$v^{2} = 9 - 2 \times 10 \times 0.5 (\frac{1}{2})$

$v^{2} = 4$

$v = 2 m s^{- 1}$

Hence, the correct answer is 2.