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 $$(5x_{1}-2x_{2})$$,the tension in the spring will be,

A massless spring elongates by $$x_1$$ under 5N and $$x_2$$ under 7N. Find the tension for elongation $$(5x_1 - 2x_2)$$.

By Hooke's law:

$$F = kx$$, where k is the spring constant.

$$5 = kx_1 \Rightarrow x_1 = 5/k$$

$$7 = kx_2 \Rightarrow x_2 = 7/k$$

Find the elongation:

$$5x_1 - 2x_2 = 5 \times \frac{5}{k} - 2 \times \frac{7}{k} = \frac{25 - 14}{k} = \frac{11}{k}$$

Find the tension:

$$F = k \times \frac{11}{k} = 11 \text{ N}$$

The correct answer is Option 3: 11 N.