A massless spring gets elongated by amount x₁ under a tension of 5N. Its elongation is x₂ under the tension of 7N. For the elongation of (5x₁ – 2x₂), the tension in the spring will be,

We are given that a massless spring stretches under different tensions, and we need to determine the tension for a specific elongation.

Step 1: Hooke’s Law

For a spring, Hooke’s law states:

$$F = k x$$

where:

$F$ is the force applied (tension in this case),

$x$ is the elongation,

$k$ is the spring constant.

Step 2: Express Tension in Terms of Elongation

From the given data:

For a force of 5N, the elongation is $x_1$:

$$5 = k x_1$$

For a force of 7N, the elongation is $x_2$:

$$7 = k x_2$$

Step 3: Express the Required Tension

We need to find the tension $F$ corresponding to an elongation of $(5x_1 - 2x_2)$. Using Hooke’s law:

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

Substituting $k = \frac{5}{x_1}$ from the first equation:

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

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

Using $x_2 = \frac{7}{k} = \frac{7x_1}{5}$, we substitute:

$$F = 25 - 5 \times \frac{2 \times 7x_1}{5x_1}$$

 $$F = 25 - 14$$ $$F = 11N$$

Final Answer:

The tension in the spring will be 11 N.