Physics · Newton's Laws of Motion

JEE Main 2025 — 4 April, Morning Shift — Question 59

A body of mass $m$ is suspended by two strings making angles $\theta_{1}$ and $\theta_{2}$ with

the horizontal ceiling with tensions $T_{1}$ and $T_{2}$ simultaneously. $T_{1}$ and $T_{2}$ are

related by $T_{1}=\sqrt{3} T_{2}$ , the angles $\theta_{1}$ and $\theta_{2}$ are

Option A:
$\theta_{1}=30^{\circ} \quad \theta_{2}=60^{\circ}$ with $T_{2}=\frac{3 m g}{4}$
Option B:
$\theta_{1}=45^{\circ} \theta_{2}=45^{\circ}$ with $T_{2}=\frac{3 m g}{4}$
Option C:
$\theta_{1}=60^{\circ} \quad \theta_{2}=30^{\circ}$ with $T_{2}=\frac{m g}{2}$
Correct
Option D:
$\theta_{1}=30^{\circ} \quad \theta_{2}=60^{\circ}$ with $T_{2}=\frac{4 m g}{5}$

Answer: C

\begin{aligned} & T_{1} \sin \theta_{1}+T_{2} \sin \theta_{2}=m g \\ & T_{1} \cos \theta_{1}=T_{2} \cos \theta_{2} \\ & \sqrt{3} T_{2} \cos \theta_{1}=T_{2} \cos \theta_{2} \\ & \sqrt{3} \cos \theta_{1}=\cos \theta_{2} \\ & \theta_{1}=60^{\circ}, \theta_{2}=30^{\circ} \\ & \sqrt{3} T_{2} \cdot \frac{\sqrt{3}}{2}+T_{2} \cdot \frac{1}{2}=m g \\ & T_{2}=\frac{m g}{2} \end{aligned}

Answer key and solution verified before publishing.

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