JEE Main · 2025 · 23 Jan 2025 · Shift 2 · 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

(5 x_{1} - 2 x_{2})

, the tension in the spring will be,

(A)15 N
(B)20 N
(C)11 N✓ Correct
(D)39 N

Answer: (C)

Solution

For an ideal, massless spring, tension is proportional to elongation:

F = k x

.

From the given data,

k x_{1} = 5 N, k x_{2} = 7 N

The tension corresponding to an elongation of

(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

The tension in the spring is

11 N

JEE Main · 2025 · 23 Jan 2025 · Shift 2 · Single correct option

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