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 (5 x 1 -2 x 2 ) , the tension in the spring will be,

Physics · Newton's Laws of Motion

JEE Main 2025 — 23 January, Evening Shift — Question 66

A massless spring gets elongated by amount $x_{1}$ under a tension of 5 N . Its elongation is $\mathrm{x}_{2}$ 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,

Option A:
15N
Option B:
20N
Option C:
11N
Correct
Option D:
39N

Answer: C

Step-by-step solution

$\mathrm{kx}_{1}=5 \mathrm{~N}$

$\mathrm{kx}_{2}=7 \mathrm{~N}$

$\mathrm{k}\left(5 \mathrm{x}_{1}-2 \mathrm{x}_{2}\right)=5 \mathrm{kx}_{1}-2 \mathrm{kx}_{2}$

$=5 \times 5-2 \times 7=11 \mathrm{~N}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 23 January, Evening Shift
Subject: Physics
Chapter: Newton's Laws of Motion
Topic: Spring Force and Combination of Springs