Question

A uniform disc of radius $R$ lies in the $x-y$ plane, with its centre at origin. Its moment of inertia about $z$ -axis is equal to its moment of inertia about line $y=x+c$. The value of $c$ will be

• A. $-\frac{R}{2}$
• B. $\pm \frac{R}{\sqrt{2}}$
• C. $\frac{+R}{4}$
• D. -R

Answer 1

Answer: (B)

$I_{P Q R}=I_{A O B}+M \cdot(O N)^{2} I_{P Q R}$
$=\frac{1}{4} M R^{2}+M \cdot\left(\frac{c}{\sqrt{2}}\right)^{2}$

But $I_{P Q R}=\frac{1}{2} M R^{2} $
$\therefore c=\pm \frac{R}{\sqrt{2}}$