Four holes of radius R are cut from a thin square plate of side 4 R and mass M. The moment of inertia of the remaining portion about Z-axis is

Options

A. `(pi)/(12)MR^(2)`
B. `((4)/(3)-(pi)/(4))MR^(2)`
C. `((4)/(3)-(pi)/(6))MR^(2)`
D. `((8)/(3)-(10pi)/(16))MR^(2)`

Answer

D

Explanation

If M mass of the square plate before cutting the holes, them mass of portion of each hole.
`m=(M)/(16R^(2))xxpiR^(2)=(pi)/(16)M`
`therefore` Moment of inertia of remaining portion,
`l=l_("square")-4l_("hole")`
`=(M)/(12)(16R^(2)+16R^(2))-4[(mR^(2))/(2)+m(sqrt2R)^(2)]`
`=(M)/(12)xx32R^(2)-10mR^(2)`
`=(8)/(3)MR^(2)-(10pi)/(16)MR^(2)=((8)/(3)-(10pi)/(16))MR^(2)`'