A massless spring gets elongated by amount $x_1$ under a tension of 5 N . Its elongation is $x_2$ under the tension of 7 N . For the elongation of $\left(5 x_1-2 x_2\right)$, the tension in the spring will be,

<p>We can solve this using Hooke's Law, which states:</p> <p>$F = kx,$</p> <p>where:</p> <p>• $F$ is the force (tension),</p> <p>• $k$ is the spring constant, and</p> <p>• $x$ is the extension of the spring.</p> <p>Follow these steps:</p> <p><p>For the first scenario (extension $x_1$ under 5 N):</p> <p>$5 = kx_1 \quad \Rightarrow \quad k = \frac{5}{x_1}.$</p></p> <p><p>For the second scenario (extension $x_2$ under 7 N):</p> <p>$7 = kx_2 \quad \Rightarrow \quad k = \frac{7}{x_2}.$</p></p> <p><p>Equate the two expressions for $k$:</p> <p>$$\frac{5}{x_1} = \frac{7}{x_2} \quad \Rightarrow \quad \frac{x_1}{x_2} = \frac{5}{7} \quad \Rightarrow \quad x_2 = \frac{7}{5}x_1.$$</p></p> <p><p>To find the tension for the extension of $\left(5x_1 - 2x_2\right)$, use Hooke's Law again:</p> <p>$\text{Tension, } F = k \left(5x_1 - 2x_2\right).$</p> <p>Substitute $k = \frac{5}{x_1}$:</p> <p>$F = \frac{5}{x_1} \left(5x_1 - 2x_2\right).$</p></p> <p><p>Substitute the expression for $x_2$:</p> <p>$$F = \frac{5}{x_1} \left(5x_1 - 2\left(\frac{7}{5}x_1\right)\right) = \frac{5}{x_1} \left(5x_1 - \frac{14}{5}x_1\right).$$</p></p> <p><p>Simplify the expression inside the parentheses:</p> <p>$$5x_1 - \frac{14}{5}x_1 = \left(\frac{25}{5} - \frac{14}{5}\right)x_1 = \frac{11}{5}x_1.$$</p></p> <p><p>Now substitute back:</p> <p>$F = \frac{5}{x_1} \cdot \frac{11}{5}x_1 = 11 \text{ N}.$</p></p> <p>Thus, the tension in the spring for the extension $\left(5x_1-2x_2\right)$ is $11 \text{ N}.$ </p> <p>The correct option is Option C.</p>