Question

Wire of length $L$ is stretched by length l when a force $F$ is applied at one end. If elastic limit is not exceeded, the amount of energy stored in wire is

• A. $Fl$
• B. $\frac{1}{2}Fl$
• C. $\frac{F{l}^2}{L}$
• D. $\frac{1}{2}\frac{E{l}^2}{L}$

Answer 1

The correct option is B $\frac{1}{2}Fl$

$\begin{array}{c}\text{Given,}\\ \text{length of wire}=L\\ \text{length streched}=b\\ \text{Force applied}=F\end{array}$

$\begin{array}{c}\text{if elastic limit is not exceeded then}\\ \text{stress is directly proportional to strain}\\ \text{and}\end{array}$

$\begin{array}{c}\text{Energy stored}=\frac{1}{2}\times \text{stress}\times \text{strain}x\text{volume}\\ =\frac{1}{2}\times \frac{F}{A}\times \frac{l}{L}\times A\times L\\ E=\frac{1}{2}Fl\end{array}$