Question

On suspending a weight Mg, the length l of elastic wire and area of cross-section A its length becomes double the initial length. The instantaneous stress action on the wire is

• A. Mg/A
• B. Mg/2A
• C. 2Mg/A
• D. 4Mg/A

Answer 1

Answer: (C)

2Mg/A

Answer 2

When the length of wire becomes double, its area of cross section will become half because volume of wire is constant $(V = AL)$.

So the instantaneous stress = $\frac{\text{Force}}{\text{Area}} = \frac{Mg}{A/2} = \frac{2Mg}{A}$.