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A uniform rod of length $L$ is suspended from one end. Its **time period** for small oscillations is:
A$2\pi\sqrt{L/g}$
B$2\pi\sqrt{2L/(3g)}$
C$2\pi\sqrt{L/(2g)}$
D$2\pi\sqrt{3L/(2g)}$
Answer & Solution
Correct answer: B. $2\pi\sqrt{2L/(3g)}$
Compound (physical) pendulum: $T = 2\pi\sqrt{I/(Mgd)}$. Here $I = ML^2/3$ (about end), $d = L/2$. $T = 2\pi\sqrt{(ML^2/3)/(Mg \cdot L/2)} = 2\pi\sqrt{2L/(3g)}$.
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