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 (5 x₁-2 x₂), the tension in the spring will be,

Solution

Physics sir · 3 steps

Tension is proportional to stretch, so the tension follows the same combination as the elongations.

1. Hooke's law. For a spring of constant k, the tension is T = kx. So x₁ = 5/k and x₂ = 7/k.
2. Build the elongation. Put these into the required elongation and keep the subtraction.

   5x₁ - 2x₂ = 25/k - 14/k = 11/k

3. So. The tension is T = k × 11/k = 11 N. The answer is option (c).

The idea to keep. Since T = kx is linear, any combination of elongations gives the same combination of tensions: 5x₁ - 2x₂ needs 5(5) - 2(7) newtons.

Answer from the NTA final answer key. Solution by PrepWiser.