A wire of length L is hanging from a fixed support. The length changes to L₁ and L₂ when masses 1 kg and 2 kg are suspended respectively from its free end. Then the value of L is equal to :

Solution

Physics sir · 3 steps

Extension grows in step with the load, so the 2 kg stretch is exactly twice the 1 kg stretch.

1. Hooke's law. Within the elastic limit, Y = (F/A)/(Δ L/L) gives Δ L = FL/AY. For the same wire, L, A and Y are fixed, so the extension is proportional to the hanging weight.
2. Write both extensions. With 1 kg the extension is L₁ - L. With 2 kg the load doubles, so the extension doubles: L₂ - L = 2(L₁ - L).

   L₂ - L = 2L₁ - 2L ⇒ L = 2L₁ - L₂

3. So. The natural length is L = 2L₁ - L₂, which is option (c).

The idea to keep. Whenever a wire is loaded twice, write each extension as (stretched length minus original length) and set their ratio equal to the ratio of the loads. The unknown original length drops out in one line.

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