JEE Physics 2025 Question with Solution

Given: A body of mass $m$ is in equilibrium under tensions $T_{1}$ and $T_{2}$, with $T_{1} = \sqrt{3}\, T_{2}$.

Find: The values of $\theta_{1}$, $\theta_{2}$, and $T_{2}$.

For equilibrium, resolve forces horizontally and vertically.

Horizontal equilibrium:

$$T_{1} \cos\theta_{1} = T_{2} \cos\theta_{2}$$

Using $T_{1} = \sqrt{3}\, T_{2}$,

$$\sqrt{3}\, T_{2} \cos\theta_{1} = T_{2} \cos\theta_{2}$$ $$\cos\theta_{2} = \sqrt{3}\, \cos\theta_{1}$$

Now try the standard angle pair $\theta_{1} = 60^{\circ}$ and $\theta_{2} = 30^{\circ}$:

$$\cos 30^{\circ} = \sqrt{3} \cos 60^{\circ}$$ $$\frac{\sqrt{3}}{2} = \sqrt{3} \times \frac{1}{2}$$

This is satisfied.

Vertical equilibrium:

$$T_{1} \sin\theta_{1} + T_{2} \sin\theta_{2} = mg$$

Substitute $T_{1} = \sqrt{3}\, T_{2}$, $\theta_{1} = 60^{\circ}$, and $\theta_{2} = 30^{\circ}$:

$$\sqrt{3}\, T_{2} \sin 60^{\circ} + T_{2} \sin 30^{\circ} = mg$$ $$\sqrt{3}\, T_{2} \left(\frac{\sqrt{3}}{2}\right) + T_{2} \left(\frac{1}{2}\right) = mg$$ $$\left(\frac{3}{2} + \frac{1}{2}\right) T_{2} = mg$$ $$2T_{2} = mg$$ $$T_{2} = \frac{mg}{2}$$

Therefore, the correct option is B.

Given: The body is suspended by two strings making angles $\theta_{1}$ and $\theta_{2}$ with the horizontal. The tensions satisfy $T_{1} = \sqrt{3}\, T_{2}$.

Find: The correct angle combination and the corresponding value of $T_{2}$.

Concept Used: For a body in equilibrium,

• Horizontally: $T_{1} \cos\theta_{1} = T_{2} \cos\theta_{2}$
• Vertically: $T_{1} \sin\theta_{1} + T_{2} \sin\theta_{2} = mg$

From horizontal equilibrium,

$$T_{1} \cos\theta_{1} = T_{2} \cos\theta_{2}$$ $$\sqrt{3}\, T_{2} \cos\theta_{1} = T_{2} \cos\theta_{2}$$ $$\cos\theta_{2} = \sqrt{3}\, \cos\theta_{1}$$

This implies $\theta_{2} < \theta_{1}$. Among the given options, $\theta_{1} = 60^{\circ}$ and $\theta_{2} = 30^{\circ}$ satisfy this relation.

Check:

$$\cos 30^{\circ} = \sqrt{3} \cos 60^{\circ}$$ $$\frac{\sqrt{3}}{2} = \sqrt{3} \times \frac{1}{2}$$

So the horizontal condition is correct.

Now use vertical equilibrium:

$$T_{1} \sin\theta_{1} + T_{2} \sin\theta_{2} = mg$$ $$\sqrt{3}\, T_{2} \sin 60^{\circ} + T_{2} \sin 30^{\circ} = mg$$ $$\sqrt{3}\, T_{2} \left(\frac{\sqrt{3}}{2}\right) + T_{2} \left(\frac{1}{2}\right) = mg$$ $$\frac{3T_{2}}{2} + \frac{T_{2}}{2} = mg$$ $$2T_{2} = mg$$ $$T_{2} = \frac{mg}{2}$$

Thus, $\theta_{1} = 60^{\circ}$, $\theta_{2} = 30^{\circ}$, and $T_{2} = \frac{mg}{2}$. The correct option is B.