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JEE Mains · Physics · STD 11 - 13. oscillations

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 \(g=10\,m / s ^{2}\) )

  1. A \(5\)
  2. B \(1\)
  3. C \(2\)
  4. D \(7\)
Verified Solution

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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