JEE Main 2024 Physics question
A uniform rod AB of mass 2 kg and length 30 cm at rest on a smooth horizontal surface. An impulse of force 0.2 Ns is applied to end B. The time taken by the rod to turn through at right angles will be s, where x = .
Answer: 4
Solution
ω = J(L/2)/(mL²/12) = 6J/(mL) = 6 × 0.2/(2 × 0.3) = 2 rad/s; t = (π/2)/ω = π/4 s.