A wire whose cross-sectional area id $4 m m^{2}$ is stretched by $0.1$ mm by a certain load. If a similar wire of double the area of cross-section is under the sane load, then the elongation would be

0.5 mm

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0.05 mm

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0.005 mm

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5 mm

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Solution

The correct option is B 0.05 mm
$\begin{matrix} Given all parameter's are some except area of crosssection given A_{2} = 2 A_{1} \end{matrix}$

$\begin{matrix} Formula, \begin{matrix} Y & = \frac{F L}{A e} \Rightarrow & e \propto \frac{1}{A} \frac{e_{1}}{e_{2}} = \frac{A_{2}}{A_{1}} e_{2} & = (0.1) \times \frac{4}{8} e_{2} = 0.05 m m \end{matrix} \end{matrix}$

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