Laws of Motion

JEE Main level, single correct, medium

112 of 210

Question 112 of 210

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 (5x1−2x2)\left(5 x_1-2 x_2\right), the tension in the spring will be,

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Answer: (C) 11 N

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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