MHT CET · Physics · Oscillations
A small spherical ball of radius ' \(r\) ' is rolling on a curved surface which is frictionless and has a radius of curvature ' R '. Its motion is simple harmonic. Then its time period of oscillation is proportional to ( \(\mathrm{g}=\) acceleration due to gravity)
- A \(\sqrt{\frac{R}{g}}\)
- B \(\sqrt{\frac{\mathrm{r}}{\mathrm{g}}}\)
- C \(\sqrt{\frac{\mathrm{R}-\mathrm{r}}{\mathrm{g}}}\)
- D \(\sqrt{\frac{\mathrm{R}+\mathrm{r}}{\mathrm{g}}}\)
Answer & Solution
Correct Answer
(C) \(\sqrt{\frac{\mathrm{R}-\mathrm{r}}{\mathrm{g}}}\)
Step-by-step Solution
Detailed explanation
The effective length of the pendulum for the center of mass is \(L_{eff} = R-r\). The time period of a simple pendulum is \(T = 2\pi\sqrt{\frac{L_{eff}}{g}}\).
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