A uniform rod of mass m and length l suspended by means of two id… — Solution | JEE Physics

MCQMediumJEE 2026Torque & Angular Momentum

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 horizontal uniform rod AB is suspended by two identical vertical light strings from a fixed support above, one at each end of the rod, with the rod length marked as l.

A
$\frac{mg}{3}$
B
$\frac{mg}{2}$
C
$\frac{mg}{4}$
D
$mg$

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: A uniform rod of mass $m$ and length $l$ is supported by two identical vertical strings. One string is cut, and we need the tension $T$ in the remaining string immediately after the cut.

Find: The value of $T$ just after one string is cut.

Immediately after the cut, the rod starts rotating about the end where the remaining string is attached. Use rotational motion about that end and then relate it to the acceleration of the center of mass.

The torque about the pivot is due only to weight:

\tau = mg \cdot \frac{l}{2}

The moment of inertia of a uniform rod about one end is:

I = \frac{1}{3}ml^2

Using:

\tau = I\alpha

we get

mg\frac{l}{2} = \left(\frac{1}{3}ml^2\right)\alpha

So,

\alpha = \frac{3g}{2l}

Force and Rotation Relation

Now relate angular acceleration to the linear acceleration of the center of mass. Since the center of mass is at distance $\frac{l}{2}$ from the pivot,

a_{cm} = \frac{l}{2}\alpha

Substituting $\alpha = \frac{3g}{2l}$ ,

a_{cm} = \frac{l}{2}\left(\frac{3g}{2l}\right) = \frac{3g}{4}

Net Force on Center of Mass

For vertical motion of the center of mass,

mg - T = ma_{cm}

Substitute $a_{cm} = \frac{3g}{4}$ :

mg - T = m\left(\frac{3g}{4}\right)

Hence,

T = mg - \frac{3mg}{4} = \frac{mg}{4}

Therefore, the tension in the remaining string immediately after the other is cut is $\frac{mg}{4}$ . The correct option is C.

Common mistakes

Assuming the center of mass falls with acceleration $g$ is incorrect. The rod is a rigid body constrained to rotate about the remaining end, so the center of mass has acceleration less than $g$ . First find $\alpha$ , then use $a_{cm} = \frac{l}{2}\alpha$ .
Including the tension as a torque about the pivot is wrong. The remaining string acts at the pivot point, so its moment arm is zero. Only weight $mg$ produces torque about that point.
Using the moment of inertia of the rod about its center, $\frac{1}{12}ml^2$ , is incorrect here. Since the rod rotates about one end immediately after the cut, the correct moment of inertia is $\frac{1}{3}ml^2$ .

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Step-by-step solution

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