QJEEM25Ph9f9e: JEE Main 2025 Physics - Class 11 JEEM25Ph9f9e Solution: A massless spring gets elongated by amount under a tension of 5 …

We can solve this using Hooke's Law, which states:

$F = kx,$

where:

• $F$ is the force (tension),

• $k$ is the spring constant, and

• $x$ is the extension of the spring.

Follow these steps:

For the first scenario (extension $x_1$ under 5 N):

$5 = kx_1 \quad \Rightarrow \quad k = \frac{5}{x_1}.$

For the second scenario (extension $x_2$ under 7 N):

$7 = kx_2 \quad \Rightarrow \quad k = \frac{7}{x_2}.$

Equate the two expressions for $k$ :

$\frac{5}{x_1} = \frac{7}{x_2} \quad \Rightarrow \quad \frac{x_1}{x_2} = \frac{5}{7} \quad \Rightarrow \quad x_2 = \frac{7}{5}x_1.$

To find the tension for the extension of $\left(5x_1 - 2x_2\right)$ , use Hooke's Law again:

$\text{Tension, } F = k \left(5x_1 - 2x_2\right).$

Substitute $k = \frac{5}{x_1}$ :

$F = \frac{5}{x_1} \left(5x_1 - 2x_2\right).$

Substitute the expression for $x_2$ :

$F = \frac{5}{x_1} \left(5x_1 - 2\left(\frac{7}{5}x_1\right)\right) = \frac{5}{x_1} \left(5x_1 - \frac{14}{5}x_1\right).$

Simplify the expression inside the parentheses:

$5x_1 - \frac{14}{5}x_1 = \left(\frac{25}{5} - \frac{14}{5}\right)x_1 = \frac{11}{5}x_1.$

Now substitute back:

$F = \frac{5}{x_1} \cdot \frac{11}{5}x_1 = 11 \text{ N}.$

Thus, the tension in the spring for the extension $\left(5x_1-2x_2\right)$ is $11 \text{ N}.$

The correct option is Option C.