A massless spring gets elongated by amount $x_1$ under a tension of 5 N. Its elongation is $x_2$ under the…

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,
  1. 39 N
  2. 15 N
  3. 11 N
  4. 20 N

Solution

$\begin{aligned} & \mathrm{kx}_1=5 \mathrm{~N} \\ & \mathrm{kx}_2=7 \mathrm{~N} \\ & \mathrm{k}\left(5 \mathrm{x}_1-2 \mathrm{x}_2\right)=5 \mathrm{kx}_1-2 \mathrm{kx}_2 \\ & =5 \times 5-2 \times 7=11 \mathrm{~N}\end{aligned}$

Asked in: JEE Main 2025 (23 Jan Shift 2)

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