Q102 / 150 · MHT CET 2024 (03 May Shift 1) · Physics

A small sphere oscillates simple harmonically in a watch glass whose radius of curvature is 1.6 m . The period of oscillation of the sphere in second is (acceleration due to gravity, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2$ )

  • Correct answer
    A
    $0 \cdot 8 \pi$
  • B
    $0 \cdot 6 \pi$
  • C
    $0 \cdot 4 \pi$
  • D
    $0 \cdot 2 \pi$

Solution

$\begin{aligned} \mathrm{T} & =2 \pi \sqrt{\frac{l}{\mathrm{~g}}} \\ & =2 \pi \sqrt{\frac{1.6}{10}} \quad \ldots(\because \mathrm{R}=l) \\ & =0.8 \pi\end{aligned}$

Hence, the correct answer is option: 'A'.

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