A wire of length $2\, m$ is made from $10\;c{m^3}$ of copper. A force $F$ is applied so that its length increases by $2\, mm.$ Another wire of length 8 m is made from the same volume of copper. If the force $F$ is applied to it, its length will increase by......... $cm$
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(d) $l = \frac{{FL}}{{AY}} = \frac{{F{L^2}}}{{(AL)Y}} = \frac{{F{L^2}}}{{VY}}$
$l \propto {L^2}$ If volume of the wire remains constant
$\frac{{{l_2}}}{{{l_1}}} = {\left( {\frac{{{L_2}}}{{{L_1}}}} \right)^2} = {\left( {\frac{8}{2}} \right)^2} = 16$
${l_2} = 16 \times {l_1} = 16 \times 2 = 32mm = 3.2cm$
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