Moment of inertia about the axis of rotation is I=m1r12+m2r22 Clearly r1=4‌cm And r2=6‌cm ∴‌‌I=(30×10−3×16×10−4)+(20×10−3×36×10−4) ⇒‌‌I=1200×10−7‌kg‌m2 If the system is rotated by small angle ' θ ', the restoring torque is τ(R)=−kθ And ‌