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,

[JEE Main 2025, 23 Jan (Shift 2)]

Question

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,

[JEE Main 2025, 23 Jan (Shift 2)]

YESPYQ · Accepted Answer

Correct answer: d) 11 N. ${F}_{1}=k{x}_{1}\ 5=k{x}_{1}\ {x}_{1}=\frac{5}{k}\ 7=k{x}_{2}\ {x}_{2}=\frac{7}{k}$ $x=5{x}_{1}-2{x}_{2}\ x=\frac{25}{k}-\frac{14}{k}\Rightarrow x=\frac{11}{k}\ F=kx\ F=k\times \frac{11}{k}\ F=11N$

A massless spring gets elongated by amount \({x}_{1}\) under a tension of 5 N . Its elongation is \({x}_{2}\) under the …

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