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

Answer: (C)

$\frac{2Mg}{A}$

$\text{When the length of the wire becomes double its area of cross-section}$

$\text{becomes half as volume of wire remains constant}$

$\begin{array}{c}A=\text{Area}\\ L=\text{Length}\\ \text{Let}{A}^{\prime }\text{be new area.}\end{array}$

$\begin{array}{cc}& V=AL\\ \Rightarrow AL& =A^{\prime }\left(2L\right)\\ \Rightarrow A^{\prime }& =\frac{A}{2}\end{array}$

$\begin{array}{c}\text{So, instantaneous stress}=\frac{\text{Force}}{\text{Area}}=\frac{Mg}{A/2}=\frac{2Mg}{A}\\ \text{Hence (c) option is correct}\end{array}$