Increase in length of a wire is $1\, mm$ when suspended by a weight. If the same weight is suspended on a wire of double its length and double its radius, the increase in length will be ........ $mm$
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(b) $l = \frac{{FL}}{{AY}} \Rightarrow l \propto \frac{L}{{{r^2}}}$ ($F$ and $Y$ are same)
$\frac{{{l_2}}}{{{l_1}}} = \frac{{{L_2}}}{{{L_1}}}{\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2} = 2 \times {\left( {\frac{1}{2}} \right)^2} = \frac{1}{2}$==>${l_2} = \frac{{{l_1}}}{2} = \frac{l}{2} = 0.5mm.$
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