Question 48: Properties of Matter

Question

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,

Accepted Answer

We can solve this using Hooke's Law, which states:
$$F = kx,$$
where:
Ã¢ÂÂ¢ $$F$$ is the force (tension),
Ã¢ÂÂ¢ $$k$$ is the spring constant, and
Ã¢ÂÂ¢ $$x$$ 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 $$\left(5x_1 - 2x_2ight)$$, use Hooke's Law again:
$$	ext{Tension, } F = k \left(5x_1 - 2x_2ight).$$
Substitute $$k = \frac{5}{x_1}$$:
$$F = \frac{5}{x_1} \left(5x_1 - 2x_2ight).$$
Substitute the expression for $$x_2$$:
$$F = \frac{5}{x_1} \left(5x_1 - 2\left(\frac{7}{5}x_1ight)ight) = \frac{5}{x_1} \left(5x_1 - \frac{14}{5}x_1ight).$$
Simplify the expression inside the parentheses:
$$5x_1 - \frac{14}{5}x_1 = \left(\frac{25}{5} - \frac{14}{5}ight)x_1 = \frac{11}{5}x_1.$$
Now substitute back:
$$F = \frac{5}{x_1} \cdot \frac{11}{5}x_1 = 11 	ext{ N}.$$
Thus, the tension in the spring for the extension $$\left(5x_1-2x_2ight)$$ is $$11 	ext{ N}.$$
The correct option is Option C.

Answer

20 N

Answer

39 N

Answer

11 N

Answer

15 N

Properties of Matter

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,

Choose the correct answer: