
The moment of inertia of a sphere is given by the formula I=52MR2.
Since there are two masses, the total moment of inertia about the axis perpendicular to the rod passing through the centre is
I=2×[52MR2+M(2l+r)2]...(i)
The given data is
M=2kgl=(0.4−0.2)m=0.2mR=0.1m
Substituting the values in equation (i)
I=(54(2×0.12)+2×2(20.2+0.1)2)kgm2⇒I=((54×0.02)+(4×0.04))kgm2⇒I=(0.016+0.16)kgm2=0.176kgm2