A load $W$ produces an extension of $1mm$ in a thread of radius $r.$ Now if the load is made $4W$ and radius is made $2r$ all other things remaining same, the extension will become..... $mm$
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(c) $l = \frac{{FL}}{{AY}}\therefore l \propto \frac{F}{{{r^2}}}$
$\frac{{{l_1}}}{{{l_2}}} = \frac{{{F_2}}}{{{F_1}}}{\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2} = (4) \times {\left( {\frac{1}{2}} \right)^2} = 1$
${l_2} = {l_1} = 1mm$
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