A uniform rod of length $2.0\mathrm{ }\mathrm{m}$ is suspended through its endpoint about which it performs small angular oscillations in the vertical plane. Its time period is nearly

• A

  $1.6\mathrm{ }\mathrm{s}$
• B

  $1.8\mathrm{ }\mathrm{s}$
• C

  $2.0\mathrm{ }\mathrm{s}$
• Correct answer
  D

  $2.3\mathrm{ }\mathrm{s}$

Solution

It is a physical pendulum, and its time period $T=2\pi \sqrt{\frac{I}{mgx}}$

here I = moment of inertia of the pendulum about point of suspension

$I=\frac{M{L}^2}{12}+\frac{M{L}^2}{4}=\frac{M{L}^2}{3}$

and x = distance of point of suspension from its COM.

x = L/2

$\Rightarrow T=2\pi \sqrt{\frac{M{L}^2/3}{mgL/2}}=2\pi \sqrt{\frac{2L}{3g}}$

$\Rightarrow T=2\pi \sqrt{\frac{2\times 2}{3\times 10}}=2.3\text{ sec}\text{.}$

Hence, the correct answer is option: 'D'.