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MHT CET 2026 (15 Apr, Shift 1) · Other SHM Systems

Other SHM Systems question from MHT CET 2026 (15 Apr, Shift 1)

Q2MHT CET 2026 · 15 APR · SHIFT 1Easy

Other SHM Systems

A small sphere oscillates simple harmonically in a watch glass whose radius of curvature is 1.61.6 m. The period of Oscillation of the sphere is (Acceleration due to gravity g=10g = 10 m/s2^{2})

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Answer: (c)

  1. A small sphere in a bowl moves like a pendulum of length RR: T=2πRgT = 2\pi\sqrt{\frac{R}{g}}.
  2. T=2π1.610=2π×0.4=0.8π sT = 2\pi\sqrt{\frac{1.6}{10}} = 2\pi \times 0.4 = 0.8\pi\text{ s}
  3. So the answer is (c).
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