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,
- A 39 N
- B 15 N
- C 11 N
- D 20 N
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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