A disc has mass 9 m. A hole of radius R/3 is cut from it as shown in the figure. The moment of inertia of remaining part about an axis passing through the centre 'O' of the disc and perpendicular to the plane of the disc is :

Options

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

Answer

C

Explanation

Moment of inertia of complete disc of mass 9m and radius R, about an axis passing through its centre and perpendicular to its plane can be written as follows:
`I=(1)/(2)(9m)R^(2)=(9)/(2)mR^(2)`
We can assume distribution of mass proportional to area. And area of the hole is 1/9 times to that of area of complete disc. Hence mass of removed portion can be assume to be m. Moment of inertia of hole can be written about the same axis using parallel axis theorem.
`I_("hole")=I_(cm)+md^(2)=(1)/(2)m((R)/(3))^(2)+m((2R)/(3))^(2)=(1)/(2)mR^(2)`
Hence moment of inertia of the remaining system can be calculated as follows:
`I_("remaining")=I-I_("hole")=(9)/(2)mR^(2)-(1)/(2)mR^(2)=4mR^(2)`
Option (c) is correct.