JEE Main 2026 — Jan 21, Shift 1 — Physics solution

JEE Main2026Jan 21, Shift 1Physics

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
C
mg/3
D
mg

Solution

Immediately after one string is cut, the rod begins rotating about the point where the remaining string meets the rod (call it $O$ ). Take torques about $O$ :

$m g \cdot \frac{l}{2} = I_{O} α = \frac{m l ^{2}}{3} α \Rightarrow α = \frac{3 g}{2 l}$ .

The centre of mass is at $l /2$ from $O$ , so its tangential (downward) acceleration is $a_{c} = (l /2) α = 3 g /4$ .

Newton's second law for the COM in the vertical direction: $m g - T = m a_{c} \Rightarrow T = m (g - 3 g /4) = \frac{m g}{4}$ .

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