A uniform rod of length $L$ is suspended from one end. Its **time period** for small oscillations is:
- $2\pi\sqrt{L/g}$
- $2\pi\sqrt{2L/(3g)}$
- $2\pi\sqrt{L/(2g)}$
- $2\pi\sqrt{3L/(2g)}$
Show answer and 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), $d = L/2$.
$T = 2\pi\sqrt{(ML^2/3)/(Mg \cdot L/2)} = 2\pi\sqrt{2L/(3g)}$.