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 ____________ ms ⁻¹. (if g = 10 m/s2) Numerical answer type. Work it out, then open the answer.

Answer: 5. Maximum speed is at the lowest point: v = √(2gl(1 - cos θ)). l = 2.5 m, θ = 60°, 1 - cos 60° = 1/2 v = √(2 × 10 × 2.5 × 0.5) = √(25) v = 5 m/s

A pendulum is suspended by a string of length 250 cm. The mass of the… | JEE Main 2022 (28 Jun, Shift 1)

Q52JEE Main 2022 · 28 JUN · SHIFT 1Easy

Simple Pendulum

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 $^\circ$ 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/s2)

Numerical answer type. Work it out, then open the answer.

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Answer: 5

Maximum speed is at the lowest point: $v = \sqrt{2gl(1 - \cos\theta)}$ .
$l = 2.5\text{ m}$ , $\theta = 60^\circ$ , $1 - \cos 60^\circ = \frac{1}{2}$
$v = \sqrt{2 \times 10 \times 2.5 \times 0.5} = \sqrt{25}$
$v = 5\text{ m/s}$

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