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A **uniform sphere** of radius $R$ and mass $M$ is placed inside a fixed hemispherical bowl of radius $5R$. For small oscillations (rolling without slipping), the **time period** is:
A$2\pi\sqrt{R/g}$
B$2\pi\sqrt{4R/g}$
C$2\pi\sqrt{28R/(5g)}$
D$2\pi\sqrt{5R/g}$
Answer & Solution
Correct answer: C. $2\pi\sqrt{28R/(5g)}$
For a sphere rolling inside a bowl of radius $a$, the effective pendulum length is $(a-R)$, and rolling factor multiplies by $(1 + I/MR^2) = 7/5$. Result: $T = 2\pi\sqrt{7(a-R)/(5g)}$. With $a = 5R$, $a - R = 4R$: $T = 2\pi\sqrt{28R/(5g)}$.
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