m1 = 30 gm
m2 = 20 gm

Moment of inertia about the axis of rotation is
I = m1r12 + m2r22
Clearly r1 = 4 cm
And r2 = 6 cm
\(\therefore\) I = (30 × 10–3 × 16 × 10–4) + (20 × 10–3 × 36 × 10–4)
⇒ I = 1200 × 10–7 kg m2
If the system is rotated by small angle ‘\(\theta\)’, the restoring torque is \(\tau\)(R) = –k\(\theta\)
