A massless spring gets elongated by amount x₁ under a tension of 5 N. Its elongation is x₂ under the tension of 7 N. For the elongation of (5x₁ - 2x₂), the tension in the spring will be, (a) 20 N (b) 39 N (c) 11 N (d) 15 N

Answer: (c). Hooke's law T = kx: x₁ = 5/k, x₂ = 7/k 5x₁ - 2x₂ = (25 - 14)/(k) = 11/k T = k × 11/k = 11 N. So the answer is (c) 11 N.

A massless spring gets elongated by amount x₁ under a tension of 5 N… | JEE Main 2025 (23 Jan, Shift 2)

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Inertia, First & Second Laws

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 $(5x_1 - 2x_2)$ , the tension in the spring will be,

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Answer: (c)

Hooke's law $T = kx$ : $x_1 = \frac{5}{k},\ x_2 = \frac{7}{k}$
$5x_1 - 2x_2 = \frac{25 - 14}{k} = \frac{11}{k}$
$T = k \times \frac{11}{k} = 11\text{ N}$ . So the answer is (c) 11 N.

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