A circular disc of mass 9m, has a hole of radius R/3, cut from it as shown in given figure. The moment of inertia of the remaining part of the disc about the axis passing through the centre of the disc and perpendicular to its plane will be :

Options

A. `(37)/(9)mR^(2)`
B. `(40)/(9)mR^(2)`
C. `4mR^(2)`
D. `8mR^(2)`

Answer

C

Explanation

Mass per unit area of the disc `sigma=(9m)/(piR^(2))`
`therefore` Mass of removed disc `M=sigma={pi((R)/(3))^(2)}=(9m)/(piR^(2)).(piR^(2))/(9)=m`
Moment of inertia of second disc about the given axis (From theorem of parallel axis)
`I.=(1)/(2)m((R)/(3))^(2)+m((2R)/(3))^(2)`
`=(1)/(2)m((R^(2))/(9))+m((4R^(2))/(9))=(mR^(2))/(2)`
Moment of inertia of remaining part = Moment of inertia of complete disc - Moment of inertia of removed disc
`=(1)/(2)(9mR^(2))-(mR^(2))/(2)=4mR^(2)`