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)
(A)mg/2
(B)mg/4✓ Correct
(C)mg/3
(D)mg
Answer: (B)
Solution
Immediately after one string is cut, the end still held by the remaining vertical string has zero acceleration along that string, so the centre falls with a=2lα. Taking torque about the centre, T⋅2l=12ml2α, which gives α=ml6T and hence a=m3T. Newton's law for the centre, mg−T=ma, then reads mg−T=3T, so the tension is T=4mg.
JEE Main · 2026 · 21 Jan 2026 · Shift 1 · Single correct option