Question

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

$I_{X{X}^{\prime }}=\frac{M{R}^2}{2}+M{R}^2=\frac{3}{2}M{R}^2$...(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
$I_{x{x}^{\prime }}=\frac{3}{2}(\rho L)\frac{L^2}{4{\pi }^2}$
$\Rightarrow \frac{3\rho L^3}{8{\pi }^2}$
Hence, option a is correct.