Question bank/JEE Main/Physics

A uniform disc of radius R lies in the x-y plane, with its
centre at origin. its moment of inertia about zaxis 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$- R$

Show answer and solution▾

$I_{PQR} = \frac{1}{4}{MR}^{2} + M{(\frac{C}{\sqrt{2}})}^{2}$
But $I_{PQR} = \frac{1}{2}{MR}^{2}$
$\therefore C = \pm \frac{R}{\sqrt{2}}$
Hence (2) is correct.

Source: Grass-G / P-18 Physics paper 27 Nov Third.docx