Question

If the work done in stretching a wire by $1mm$ is $2J$ , the work necessary for stretching another wire of same material but with double radius of cross-section and half the length by $1mm$ is

• A. $16J$
• B. $8J$
• C. $4J$
• D. $1/4J$

Answer 1

Answer: (A)

$16J$

$\begin{array}{cc}w& =\frac{1}{2}\times F\times \Delta LY=\frac{\left(\frac{F}{A}\right)}{\left(\frac{\Delta L}{L}\right)}\\ & =\frac{1}{2}\times \frac{Y\Delta LA}{L}\times \Delta L\\ \omega & =\frac{1}{2}\frac{YA}{L}(\Delta L{)}^2\end{array}$

we can write

$\text{w}\alpha \cdot \frac{\Delta }{L}$

$\omega \alpha \frac{\pi R^2}{L}\Rightarrow \omega \alpha \frac{R^2}{L}$

$\begin{array}{cc}& w_1=\frac{R^2}{L}{w}_2=\frac{(2R{)}^2}{\left(\frac{L}{2}\right)}\text{(from data)}\\ & \frac{w_1}{w_2}=\frac{R^2\cdot K}{L\cdot 8R^2}\\ & \frac{2}{w_2}=\frac{1}{8}\\ & w_2=16J\end{array}$