QJEEAdv06Ph49b7: JEE Advanced 2006 Physics - Class 11 JEEAdv06Ph49b7 Solution: A solid sphere of radius has moment of inertia about its…

To solve this problem, we need to compare the moments of inertia of the sphere and the disc. For a solid sphere of radius $R$ , the moment of inertia about its geometrical axis is given by:

$I_{\text{sphere}} = \frac{2}{5} M R^2$

where $M$ is the mass of the sphere.

When the sphere is melted into a disc of radius $r$ and thickness $t$ , the volume (and hence mass) remains the same. So, the volume of the sphere is:

$\frac{4}{3} \pi R^3$

And the volume of the disc is:

$\pi r^2 t$

Equating the volumes, we get:

$\frac{4}{3} \pi R^3 = \pi r^2 t$

This simplifies to:

$t = \frac{4 R^3}{3 r^2}$

Next, we need to find the moment of inertia of the disc about the tangential axis (which is perpendicular to the plane of the disc). The moment of inertia of a disc of mass $M$ and radius $r$ about an axis through its center and perpendicular to the plane of the disc is:

$I_{\text{disc, center}} = \frac{1}{2} M r^2$

Using the parallel axis theorem, the moment of inertia about a tangential axis is:

$I_{\text{disc, tangential}} = I_{\text{disc, center}} + M r^2 = \frac{1}{2} M r^2 + M r^2 = \frac{3}{2} M r^2$

Since we are given that the moment of inertia of the sphere and the disc about their respective axes are equal, we set:

$\frac{2}{5} M R^2 = \frac{3}{2} M r^2$

Canceling the mass $M$ from both sides and solving for $r$ , we get:

$\frac{2}{5} R^2 = \frac{3}{2} r^2$

Solving for $r$ , we have:

$r^2 = \frac{4}{15} R^2$

Taking the square root of both sides:

$r = \frac{2}{\sqrt{15}} R$

So, the correct answer is:

Option A: $\frac{2}{\sqrt{15}} R$