Four spherical particles, each of mass m and radius r are kept at the four corners of a square of side a. Moment of inertia of the system along a side of the square is

Options

A. `4m[2/5 r^2 + a^2]`
B. `4m[2/7 r^2 + a^2]`
C. `2m [4/5r^2 + a^2]`
D. `2m[4/7 r^2 + a^2]`

Answer

C

Explanation

Moment of inertia of the two spheres lying on the axis of rotation it self is `2 xx 2/5 mr^2 mr^2.` Moment of inertia of the two spheres lying on other two corners (using parallel axes theorem) is `= 2(2/5 mr^2 + ma^2)`

Total moment of inertia = `2[2/5 mr^2 + 2/5 mr^2 + ma^2] = 8/5 mr^2 +2ma^2 = 2m [4/5 r^2 + a^2]`