A uniform rod of mass m and length l suspended by means of two identical inextensible light strings as shown in figure. Tension in one string immediately after the other string is cut, is ____ . (g acceleration due to gravity)

Solution

Physics sir · 5 steps

The moment string B goes, the rod starts rotating about A, so it is no longer in equilibrium.

1. Rotation about A. Just after the cut, end A is held by an inextensible string, so A cannot move vertically and the rod starts to rotate about it. The rod has not picked up any speed yet, so ω = 0 and there is no centripetal term.
2. Torque about A. Gravity acts at the centre, a distance l/2 from A. The moment of inertia of the rod about its end is ml²/3.

   mg l/2 = ml²/3 α ⇒ α = 3g/2l

3. Centre's acceleration. The centre is l/2 from A, so it accelerates downward at a = α l/2 = 3g/4.
4. Newton's second law. For the whole rod, mg - T = m × 3g/4, which gives T = mg - 3mg/4 = mg/4.
5. So. The tension in the remaining string immediately after the cut is mg/4, option (b).

The idea to keep. Right after a support is cut, the body is no longer in equilibrium. Take torque about the remaining support to get α, find the centre's acceleration, then use F = ma to get the tension.

Answer from the NTA final answer key. Solution by PrepWiser.