From a disc of mass ' M ' and radius ' R ', a circular hole of diameter ' R ' is cut whose rim passes through the centre. The moment of inertia of the remaining part of the disc about perpendicular axis passing through the centre is

Step-by-step solution

Moment of inertia of disc is given by I _ disc = I _ f + I _ hole .... I _ f = . M.I. of remaining part aligned & I _ r = I _ dise - I _ hole & I _ disc = MR ^2 2 aligned By parallel axes theorem we get, I _ hole = [ M 4 ( R 2 )^2 2 + M 4 ( R 2 )^2 ] array l M _ bole = M _ dive 4 the surface density is same array I _ hole = [ MR ^2 32 + MR ^2 16 ] Substituting eq (iii) and eq (ii) in eq (i) we get, aligned I _ r & = MR ^2 2 - MR ^2 32 - MR ^2 16 & = MR ^2 [ 1 2 - 1 32 - 1 16 ] & = 13 32 MR ^2 aligned