A solid sphere of radius r has moment of inertia I about its axis.....?

Question

A solid sphere of radius r has moment of inertia I about its geometrical axis. If it is melted into a disc of radius r and thickness t. If its moment of inertia about the tangential axis( which is perpendicular to plane of the disc), is also equal to I, then the value of r is equal to : a) (2/(15^1/2))*R
b) (2/(5^1/2))*R
c) (3/(15^1/2))*R
d) ( (3/15)^1/2)*R
Please explain.

Answer 1

The way you stated the problem is a bit confusing. Here's my interpretation:

The sphere has radius R, and the disc has radius r. The tangential anis is through the center of the disc and perpendicular to the plane of the disc.

Under these conditions the I of the sphere
I=2/5 *R^2
is equal to the I of the disc
I=1/2*r^2

solving for r
4/5 *R^2=r^2
r=(2/(5^(1/2)))*R

j