Question

Two circular discs A and B are of equal masses and thickness but made of metals with densities $d_A$ and $d_B(d_A>d_B).$ If their moments of inertia about an axis passing through centres and normal to the circular faces be $I_A$ and $I_B$, then

• A. $I_A=I_B$
• B. $I_A>I_B$
• C. $I_A<I_B$
• D. $I_A>=<I_B$

Answer 1

The correct option is C $I_A<I_B$

Moment of inertia of circular disc about an axis passing through centre and normal to the circular face
$I=\frac{1}{2}M{R}^2=\frac{1}{2}M\left(\frac{M}{\pi t\rho }\right)$ $[AsM=V\rho =\pi R^{2}t\rho \therefore R^2=\frac{M}{\pi t\rho }]$
$\Rightarrow I=\frac{M^2}{2\pi t\rho }$ or $I\propto \frac{1}{\rho }$ If mass and thickness are constant.
So, in the problem $\frac{I_A}{I_B}=\frac{d_B}{d_A}$ $\therefore I_A<I_B$ $[As{d}_A>d_B]$