Given,              M = 3 kg

                        R = 40 cm = 0.40 m

                        F = 30 N

Moment of inertia of the hollow cylinder about its axis of symmetry

                         I = MR²

                           = 3 × (0.40)² = 0.48 kg-m²

Torque acting on the cylinder by the applied force,

                          $\mathrm{\tau }$ = F × R

                           = 30 × 0.40 = 12 N-m.

Angular acceleration produced in cylinder,

$\text{α}=\frac{\text{τ}}{\text{I}}=\frac{\text{12}}{\text{0.48}}={\text{25 rad/s}}^2$

Linear acceleration, a = R$\mathrm{\alpha }$

                           = 0.40 × 25 = 10.0 m/s²