JEE Physics 2025 Question with Solution

Given: A body of mass $1 \, \text{kg}$ is in equilibrium under tensions $T_1$ and $T_2$. The left string makes $60^\circ$ with the horizontal and the right string makes $30^\circ$ with the horizontal.

Find: The magnitudes of $T_1$ and $T_2$.

For equilibrium, resolve the tensions into horizontal and vertical components and apply

$$\sum F_x = 0, \qquad \sum F_y = 0$$

Using $g = 10 \, \text{m/s}^2$, the weight is

$$mg = 10 \, \text{N}$$

Horizontal equilibrium gives

$$T_2 \cos 30^\circ = T_1 \cos 60^\circ$$ $$T_2 \left(\frac{\sqrt{3}}{2}\right) = T_1 \left(\frac{1}{2}\right)$$ $$T_1 = \sqrt{3} \, T_2$$

Vertical equilibrium gives

$$T_1 \sin 60^\circ + T_2 \sin 30^\circ = 10$$ $$T_1 \left(\frac{\sqrt{3}}{2}\right) + T_2 \left(\frac{1}{2}\right) = 10$$

Substitute $T_1 = \sqrt{3} \, T_2$:

$$\frac{\sqrt{3}}{2}(\sqrt{3}T_2) + \frac{1}{2}T_2 = 10$$ $$\left(\frac{3}{2} + \frac{1}{2}\right)T_2 = 10$$ $$2T_2 = 10$$ $$T_2 = 5 \, \text{N}$$

Then

$$T_1 = \sqrt{3} \times 5 = 5\sqrt{3} \, \text{N}$$

Therefore, the magnitudes of tensions are $T_1 = 5\sqrt{3} \, \text{N}$ and $T_2 = 5 \, \text{N}$. The correct option is B.

Given: The mass is suspended by two strings and remains at rest.

Find: The pair $\left(T_1, T_2\right)$.

The solution contains an earlier inconsistent working using equal angles, but the detailed approach clearly states the figure angles as $60^\circ$ and $30^\circ$ with the horizontal. Using the figure-consistent equilibrium equations:

$$\begin{aligned} \sum F_x &= 0 \\ \sum F_y &= 0 \end{aligned}$$

Write the components:

$$\begin{aligned} T_1 &: \text{ horizontal } = T_1\cos 60^\circ, \quad \text{vertical } = T_1\sin 60^\circ \\ T_2 &: \text{ horizontal } = T_2\cos 30^\circ, \quad \text{vertical } = T_2\sin 30^\circ \end{aligned}$$

Horizontal balance:

$$T_2\cos 30^\circ = T_1\cos 60^\circ$$ $$T_2\left(\frac{\sqrt{3}}{2}\right) = T_1\left(\frac{1}{2}\right)$$ $$T_1 = \sqrt{3}T_2$$

Vertical balance:

$$T_1\sin 60^\circ + T_2\sin 30^\circ = 10$$ $$T_1\left(\frac{\sqrt{3}}{2}\right) + T_2\left(\frac{1}{2}\right) = 10$$

Substitute the relation from horizontal balance:

$$\frac{\sqrt{3}}{2}(\sqrt{3}T_2) + \frac{1}{2}T_2 = 10$$ $$\frac{3}{2}T_2 + \frac{1}{2}T_2 = 10$$ $$2T_2 = 10$$ $$T_2 = 5 \, \text{N}$$

Hence

$$T_1 = \sqrt{3}\times 5 = 5\sqrt{3} \, \text{N}$$

So the required pair is $(5\sqrt{3}, 5)$, which matches option B.