Four spheres , each of mass M and radius r, are situated at the four corners of a square of side R, as shown in figure. The moment of inertia of the system about an axis perpendicular to the plane of square and passing through its centre will be

Options

A. `(5M(4r^2+5R^2))/2`
B. `(2M(4r^2+5R^2))/5`
C. `(5M(4R^2+5r^2))/2`
D. `(2M(4R^2+5r^2))/5`

Answer

B

Explanation

The situation of the four spheres is symmetrical about the given axis of rotation. Obviously each has same moment of inertia.
Now, for any one sphere, say at A,
`I=I_G+M(OA)^2" or " I=2/5Mr^2+M(R/sqrt2)^2" or " I=(2Mr^2)/5+(MR^2)/2=(M(4r^2+5R^2))/10`
For all the four spheres, taken together.
Moment of inertia `4I=(4M(4r^2+5R^2))/10=(2M(4r^2+5R^2))/5`