When a weight of $10\, kg$ is suspended from a copper wire of length $3$ metres and diameter $0.4\, mm,$ its length increases by $2.4\, cm$. If the diameter of the wire is doubled, then the extension in its length will be ........ $cm$
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(d) $l \propto \frac{1}{{{r^2}}}$ $(F,L$ and $Y$ are constant$)$
$\frac{{{l_2}}}{{{l_1}}} = {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2} = {\left( {\frac{1}{2}} \right)^2} \Rightarrow {l_2} = \frac{{{l_1}}}{4} = \frac{{2.4}}{4}$$ \Rightarrow {l_2} = 0.6\;cm$
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