Question Detail

Moment of inertia for solid cylinder I = ½MR² = ½ × 50 × (0.5)² = 25 × 0.25 = 6.25 kg·m². Initial angular velocity ω₀ = 0, final ω = 300 rev/min = 300 × 2π/60 = 300 × 0.10472 = 31.416 rad/s. Angular acceleration α = (ω - ω₀)/t = 31.416/20 = 1.5708 rad/s². Torque τ = Iα = 6.25 × 1.5708 = 9.8175 Nm. This is about 9.8 Nm, not matching options. If I used ω = 300 rev/min = 300/60 = 5 rev/s × 2π = 31.4 rad/s, correct. 9.8 Nm is not in options. If I = MR² = 50 × 0.25 = 12.5, then τ = 12.5 × 1.5708 = 19.635 Nm ≈ 19.6 Nm. So they likely used I = MR² (thin hoop) instead of solid cylinder.