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JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion

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. A 39 N
  2. B 15 N
  3. C 11 N
  4. D 20 N
Verified Solution

Answer & Solution

Correct Answer

(C) 11 N

Step-by-step Solution

Detailed explanation

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