MHT CET · Physics · Oscillations
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, \(\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^2\) )
- A \(0 \cdot 8 \pi\)
- B \(0 \cdot 6 \pi\)
- C \(0 \cdot 4 \pi\)
- D \(0 \cdot 2 \pi\)
Answer & Solution
Correct Answer
(A) \(0 \cdot 8 \pi\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \mathrm{T} & =2 \pi \sqrt{\frac{l}{\mathrm{~g}}} \\ & =2 \pi \sqrt{\frac{1.6}{10}} \quad \ldots(\because \mathrm{R}=l) \\ & =0.8 \pi\end{aligned}\)
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