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 is (acceleration due to gravity \(g=10 \mathrm{~m} / \mathrm{s}^2\) )
- A \(0 \cdot 2 \pi \mathrm{~s}\)
- B \(0.4 \pi \mathrm{~s}\)
- C \(0 \cdot 6 \pi \mathrm{~s}\)
- D \(0 \cdot 8 \pi \mathrm{~s}\)
Answer & Solution
Correct Answer
(D) \(0 \cdot 8 \pi \mathrm{~s}\)
Step-by-step Solution
Detailed explanation
\(T = 2\pi\sqrt{\frac{R}{g}}\) \(T = 2\pi\sqrt{\frac{1.6}{10}} = 2\pi\sqrt{0.16} = 2\pi(0.4) = 0.8\pi \mathrm{~s}\)
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