JEE Mains · Physics · STD 11 - Oscillations
- JEE Mains 2022
- Medium
- Numerical
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 \(g=10\,m / s ^{2}\) )

Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(5\)
Step-by-step Solution
Detailed explanation
\(V _{\max }=\sqrt{2 gh }\)
The speed will be highest at the lowest position.
\(h =\left(\ell-\ell \cos 60^{\circ}\right)=\frac{\ell}{2}\)
\(V _{\max }=\sqrt{2 \times g \times \frac{\ell}{2}}=\sqrt{10 \times 2.5}=5\,m / s\)
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