Four spheres of diameter 2a and mass M are placed with their centres on the four corners of a square of side b. Then the moment of inertia of the system about an axis along one of the sides of the square is

Options

A. `(4)/(5) Ma^(2) + 2 Mb^(2)`
B. `(8)/(5) Ma^(2) + 2 Mb^(2)`
C. `(8)/(5) Ma^2 + Mb^2`
D. `(4)/(5) Ma^(2) + 4 Mb^(2)`

Answer

B

Explanation

ABCD is a square of side b . Four spheres , each of mass M and radius a are placed at the four corners of the square as shown in the figure .
We calculate moment of inertia of the system about AD .
Moment of inertia of each of the spheres A and D about AD = `(2)/(5) Ma^(2)`
Moment of inertia of each of the spheres B and C about AD = `((2)/(5) Ma^(2) + Mb^(2))` , using theorem of parallel axes.
`therefore` Total moment of inertia of the system about AD
` = 2[(2)/(5) Ma^(2) + ((2)/(5) Ma^(2) + Mb^(2))] = (8)/(5) Ma^(2) + 2 Mb^(2)`