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 their centres and perpendicular to the circular faces are $I_A$ and $I_B$ then

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

Answer 1

The correct option is C $I_A<I_B$

Given,
$m_A=m_B=m\text{kg}$
Let $R_A$ and $R_B$ be the radii of discs $A$ and $B$ respectively.
Then, $m=\left(\pi R_A^{2}t\right)d_A=\left(\pi R_B^{2}t\right)d_B$
where $t\to$ thickness
As $d_A>d_B$, so $R_A^2<R_B^2$ ... (1)

Now,
Moment of inertia of disc $A$
$\left(I_A\right)=\frac{1}{2}m{R}_A^2$
Moment of inertia of disc $B$
$\left(I_B\right)=\frac{1}{2}m{R}_B^2$
$\therefore \frac{I_A}{I_B}=\frac{R_A^2}{R_B^2}$
Since, $\frac{R_A^2}{R_B^2}<1$, we get $I_A<I_B$