Question

A uniform circular disc of radius 'R' lies in the X-Y plane with the centre coinciding with the origin, The moment of inertia about an axis passing through a point on the X-axis at a distance x = 2R and perpendicular to the X-Y plane is equal to its moment of inertia about an axis pasising through a point on the Y-axis at a distance y =d and parallel to the X-axis in the X-Y plane.,The value of 'd' is

• A. $\frac{4R}{3}$
• B. $\sqrt{17}\left(\frac{R}{2}\right)$
• C. $\sqrt{15}\left(\frac{R}{2}\right)$
• D. $\sqrt{13}\left(\frac{R}{2}\right)$

Answer 1

Answer: (B)

$\sqrt{17}\left(\frac{R}{2}\right)$

Answer 2

On x-axis at a distance 2R, $I = \frac{mR^2}{2} + m (4R^2) = \frac{9}{2}mR^2$ On y-axis at a distance d ' $I = \frac{mR^2}{4} + md^2$ equating both, $d = \sqrt{17} \frac{R}{2}$