JEE Physics 2025 Question with Solution

Given: A massless spring has elongation $x_1$ under tension $5 \, \text{N}$ and elongation $x_2$ under tension $7 \, \text{N}$.

Find: The tension corresponding to elongation $5x_1 - 2x_2$.

Using Hooke's law, tension is directly proportional to elongation:

$$F = kx$$

For the first case,

$$5 = kx_1$$

For the second case,

$$7 = kx_2$$

The required tension is

$$T = k(5x_1 - 2x_2)$$

From the above relations,

$$x_1 = \frac{5}{k}, \qquad x_2 = \frac{7}{k}$$

Substituting,

$$T = k\left(5 \cdot \frac{5}{k} - 2 \cdot \frac{7}{k}\right)$$ $$T = k\left(\frac{25}{k} - \frac{14}{k}\right)$$ $$T = k \cdot \frac{11}{k} = 11$$

Therefore, the tension in the spring is $11 \, \text{N}$. The correct option is C.

Given: Elongation is directly proportional to tension for the spring.

Find: Tension for elongation $5x_1 - 2x_2$.

Since $x \propto T$, any linear combination of elongations corresponds to the same linear combination of tensions.

So for elongation

$$5x_1 - 2x_2$$

the corresponding tension is

$$5 \times 5 - 2 \times 7 = 25 - 14 = 11$$

Hence, the required tension is $11 \, \text{N}$. The correct option is C.