Two similar wires under the same load yield elongation of $0.1$ $mm$ and $0.05$ $mm$ respectively. If the area of cross- section of the first wire is $4m{m^2},$ then the area of cross section of the second wire is..... $mm^2$
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(b) $l = \frac{{FL}}{{AY}}\therefore l \propto \frac{1}{A}$ $(F,L$ and $Y$ are constant$)$
$\frac{{{A_2}}}{{{A_1}}} = \frac{{{l_1}}}{{{l_2}}} \Rightarrow {A_2} = {A_1}\left( {\frac{{0.1}}{{0.05}}} \right)$= $2{A_1} = 2 \times 4 = 8m{m^2}$
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