JEE Main 2025 Jan 23 Shift 2, Physics Q20: Properties of Solids

JEE Main2025Jan 23, Shift 2Physics

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 $(5 x_{1} - 2 x_{2})$ , the tension in the spring will be,

A
$15$ N
B
$20$ N
C
$11$ N
D
$39$ N

Solution

Hooke's law: tension is $k$ times the elongation. Given $k x_{1} = 5$ N and $k x_{2} = 7$ N.

Tension for elongation $5 x_{1} - 2 x_{2}$ :

$F = k (5 x_{1} - 2 x_{2}) = 5 (k x_{1}) - 2 (k x_{2}) = 5 (5) - 2 (7) = 25 - 14 = 11 N .$

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JEE Main 2025 Jan 23 Shift 2, Physics Q20: Properties of Solids

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