Q 11-04-065JEE MainJEE Main 2025 (23 Jan, Shift 2)Easy
A massless spring gets elongated by amount $x_1$ under a tension of $5\ \text{N}$. Its elongation is $x_2$ under the tension of $7\ \text{N}$. For the elongation of $(5x_1 - 2x_2)$, the tension in the spring will be
Answer: (C) $11\ \text{N}$
By Hooke's law $x = T/k$: $x_1 = \dfrac{5}{k}$, $x_2 = \dfrac{7}{k}$.
$$5x_1 - 2x_2 = \frac{25 - 14}{k} = \frac{11}{k}$$
The tension for this elongation is $k\cdot\dfrac{11}{k} = 11\ \text{N}$.
Solution by Sreeraj P, M.Sc Physics