JEE Physics 2024 Question with Solution

Given: A mass of $1 \, \text{kg}$ is suspended by a rope, and a horizontal force $F$ is applied at the midpoint so that the rope makes an angle of $45^\circ$ with the vertical.

Find: The magnitude of $F$.

Let $T_1$ be the tension in the inclined part of the rope. The vertical component of tension balances the weight of the mass:

$$T_1 \cos 45^\circ = mg$$

Given,

$$m = 1 \, \text{kg}, \quad g = 10 \, \text{m/s}^2$$

So,

$$T_1 \times \frac{1}{\sqrt{2}} = 1 \times 10$$ $$T_1 = 10\sqrt{2} \, \text{N}$$

The horizontal component of tension gives the applied force:

$$T_1 \sin 45^\circ = F$$

Substituting $T_1 = 10\sqrt{2} \, \text{N}$,

$$10\sqrt{2} \times \frac{1}{\sqrt{2}} = F$$ $$F = 10 \, \text{N}$$

Therefore, the magnitude of the force is $10 \, \text{N}$. The correct option is D.

Given: A horizontal force $F$ is applied at the midpoint of the rope. Let $T_1$ be the tension in the upper inclined section and $T_2$ be the tension in the lower vertical section.

Find: The value of $F$.

At equilibrium, the lower vertical section supports the weight of the mass, so

$$T_2 = mg = 10 \, \text{N}$$

Now resolve $T_1$ into components. Its vertical component balances $T_2$:

$$T_1 \cos 45^\circ = T_2$$ $$T_1 \cos 45^\circ = 10 \, \text{N}$$

Using

$$\cos 45^\circ = \frac{1}{\sqrt{2}}$$

we get

$$T_1 \times \frac{1}{\sqrt{2}} = 10$$ $$T_1 = 10\sqrt{2} \, \text{N}$$

The horizontal component of $T_1$ balances the applied force:

$$T_1 \sin 45^\circ = F$$

Using

$$\sin 45^\circ = \frac{1}{\sqrt{2}}$$ $$F = 10\sqrt{2} \times \frac{1}{\sqrt{2}} = 10 \, \text{N}$$

Hence, the force is $10 \, \text{N}$, so the correct option is D.