Question

A thin metal wire of length $'L'$ and uniform linear mass density $'\rho'$ is bent into a circular coil with $'O'$ as centre. The moment of inertia of a coil about the axis $XX'$ is

• A. $\frac{3\rho L^{3}}{8\pi^{2}}$
• B. $\frac{\rho L^{3}}{4\pi^{2}}$
• C. $\frac{3\rho L^{2}}{4\pi^{2}}$
• D. $\frac{\rho L^{3}}{8\pi^{2}}$

Answer 1

Answer: (A)

Key Idea Moment of inertia of a thin circular coil about its diameter,
$I=\frac{M R^{2}}{2}$
Moment of inertia of a thin circular coil,
$I=\frac{M R^{2}}{2}$
Now, moment of inertia of a ring about axis $X X^{\prime}$ as in figure below,

$I_{X X'}=\frac{M R^{2}}{2}+M R^{2}=\frac{3}{2} M R^{2}$...(i)
(Using theorem of parallel axis)
Given, $L=$ length of wire of ring
and $\rho=$ linear mass density
Then, mass of the ring = linear density $\times$ length
$\Rightarrow M=\rho L$...(ii)
and $L = 2\pi R$
$\Rightarrow R=\frac{L}{2 \pi}$
Now, putting the value from Eqs. (ii) and (iii) in (i), we get
$I_{x x'}=\frac{3}{2}(\rho L) \frac{L^{2}}{4 \pi^{2}}$
$\Rightarrow \frac{3 \rho L^{3}}{8 \pi^{2}}$
Hence, option a is correct.