Q142 / 160 · TS EAMCET 2014 · Physics
A thin wire of length $l$ having density $\rho$ is bent into a circular loop with $C$ as its centre, as shown in figure. The moment of inertia of the loop about the line $A B$ is
- A$\frac{5 \rho l^3}{16 \pi^2}$
- B$\frac{\rho l^3}{16 \pi^2}$
- C$\frac{\rho l^3}{8 \pi^2}$
- Correct answerD$\frac{3 \rho l^3}{8 \pi^2}$
Solution
The moment of inertia
$$
I=\frac{3}{2} m R^2
$$
$\left[\because\right.$ Linear density $\rho=\frac{m}{l}$
$$
\begin{aligned}
m & =I \rho \\
2 \pi R & =I \\
R & =\frac{I}{2 \pi} \\
I & =\frac{3}{2} \times I \rho \times\left(\frac{I}{2 \pi}\right)^2 \\
& =\frac{3}{2 \times 4} \cdot \frac{I^3 \rho}{\pi^2} \\
& =\frac{3 /{ }^3 \rho}{8 \pi^2}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.