From a circular disc of radius R and mass 9 M , a small disc of mass M and radius (R)/(3) is removed concentrically . The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is

From a circular disc of radius R and mass 9 M , a small disc of mass M

1.	From a circular disc of radius R and mass 9 M , a small disc of mass M and radius (R)/(3) is removed concentrically . The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its centre is
Answer» `(40)/(9) MR^(2)` `MR^(2)` ` 4 MR^(2)` `(4)/(9) MR^(2)` SOLUTION :Mass of the DISC = 9 M Mass of removed portion of disc = M The moment of inertia of the complete disc about an axis passing through its centre O and PERPENDICULAR to its plane is `I_(1) = (9)/(2) MR^(2)` Now, the moment of inertia of the disc with removed portion `I_(2) = (1)/(2) M ((R)/(3))^(2) = (1)/(18) MR^(2)` Therefore , moment of inertia of theremaining portion of disc about O is `I = I_(1) - I_(2) = (9 MR^(2))/(2) = (MR^(2))/(18) = (40 MR^(2))/(9)`