Question

Two circular discs $A$ and $B$ are of equal masses and thickness but made of metal 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 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 $I_A<I_B$

$\text{Step 1: Moment of inertia of circular discs [Ref. Fig.]}$

Moment of inertia of circular discs about an axis passing through their centres and perpendicular to circular faces:

$I_A=\frac{m_A{R}_A^2}{2}$ $....\left(1\right)$

$I_B=\frac{m_B{R}_B^2}{2}$ $....\left(2\right)$

$\text{Step 2: Solving equations}$

Divide equation $\left(1\right)$ and $\left(2\right)$

$\frac{I_A}{I_B}=\frac{m_A}{m_B}.{\left(\frac{R_A}{R_B}\right)}^2={\left(\frac{R_A}{R_B}\right)}^2$ $[\because m_A=m_B]$

Since, $m_A=m_B\Rightarrow d_A\times 4\pi R_A^2=d_B.4\pi R_B^2$

$\Rightarrow {\left(\frac{R_A}{R_B}\right)}^2=\left(\frac{d_B}{d_A}\right)<1$ $[\because d_A>d_B]$

$\Rightarrow \frac{I_A}{I_B}=\left(\frac{d_B}{d_A}\right)<1\Rightarrow I_A<I_B$

Hence, Option (C) is correct.