Question

A wire of length 1m and radius 1 mm is subjected to a load. The extension is x. The wire is melted and then drawn into a wire of square cross-section of side 2 mm. Its extension under the same load will be

• A. $\frac{{\pi }^{2}x}{8}$
• B. $\frac{{\pi }^{2}x}{16}$
• C. $\frac{{\pi }^{2}x}{2}$
• D. $\frac{x}{2\pi }$

Answer 1

The correct option is B $\frac{{\pi }^{2}x}{16}$

$\begin{array}{c}l=1m\\ r=1mm\\ \Delta l=x\end{array}$

$\begin{array}{cc}& \frac{F}{A}=Y\frac{\Delta l}{l}\\ \Rightarrow F& =AY\frac{\Delta l}{l}\\ \Rightarrow F& =\frac{Yx\left(\pi \right){(1\times {10}^{-3})}^2}{1}\\ =& 4x\pi \times {10}^{-6}N\end{array}$

$\begin{array}{c}{V}_1\text{(volume)}=\pi r^{2}l=\pi \times {10}^{-6}\times 1={10}^{-6}\pi \\ \text{now new wire's volume ,}{v}_2=a^2\times l_2=4\times {10}^{-6}{\ell }_2\\ {10}^{-6}=4\times {10}^{-6}{l}_2\\ \Rightarrow l_2=\pi /4\end{array}$

$\begin{array}{c}\Delta l=\frac{Fl}{AY}=\frac{Yx\pi \times {10}^{-8}\times \pi }{4\times 4\times {10}^{-6}\times Y}=\frac{{\pi }^{2}x}{16}\\ \text{Ans:}B\text{.}\end{array}$