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. $\frac{Mg}{A}$
• B. $\frac{Mg}{2A}$
• C. $\frac{2Mg}{A}$
• D. $\frac{4Mg}{A}$

Answer 1

The correct option is C $\frac{2Mg}{A}$

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 $=\frac{force}{Area}=\frac{Mg}{\frac{A}{2}}=\frac{2Mg}{A}$