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JEE Main 202523 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Evening Shift
Subject
Physics
Chapter
Laws of Motion

JEE Main 2025 · 23 January · Evening Shift · Physics · Laws of Motion

A massless spring gets elongated by amount x_1 under a tension of 5...

A massless spring gets elongated by amount x1x_1 under a tension of 5 N . Its elongation is x2x_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,

Options

  1. A39 N
  2. B15 N
  3. C11 N
  4. D20 N

Correct answer

C. 11 N

Step-by-step solution

kx1=5 Nkx2=7 Nk(5x1−2x2)=5kx1−2kx2=5×5−2×7=11 N\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}
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