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MHT CET 2025 (22 Apr, Shift 2) · Other SHM Systems

Other SHM Systems question from MHT CET 2025 (22 Apr, Shift 2)

Q9MHT CET 2025 · 22 APR · SHIFT 2Medium

Other SHM Systems

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 ( $g =$ acceleration due to gravity)

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Answer: (c)

The centre of the ball moves on a circle of radius $R - r$ , so it acts like a pendulum of length $R - r$ .
$T = 2\pi\sqrt{\frac{R-r}{g}} \Rightarrow T \propto \sqrt{\frac{R-r}{g}}$ , answer (c).

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