Data : Msph = 50 g, Rsph = 10 cm, Mrod = 60 g, Lrod = 20 cm
The MI of a solid sphere about its diameter is Isph,CM = 2/5 MsphRsph
The distance of the rotation axis (transverse symmetry axis of the dumbbell) from the centre of sphere, h = 30 cm.
The MI of a solid sphere about the rotation axis, Isph = Isph, CM + Msphh2
For the rod, the rotation axis is its transverse symmetry axis through CM.
The MI of a rod about this axis,
Irod = 1/12 MrodL2rod
Since there are two solid spheres, the MI of the dumbbell about the rotation axis is
![](https://www.sarthaks.com/?qa=blob&qa_blobid=12871265011797217819)