When a uniform wire of radius r is stretched | vedclass

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(d) $l = \frac{{FL}}{{\pi {r^2}Y}}	herefore l \propto \frac{1}{{{r^2}}}$ $(F,L$ and $Y$ are constant$)$

$\frac{{{l_2}}}{{{l_1}}} = {\left( {\frac{

Vedclass · Accepted Answer

$8.00$

Vedclass · Answer

(d) $l = \frac{{FL}}{{\pi {r^2}Y}}	herefore l \propto \frac{1}{{{r^2}}}$ $(F,L$ and $Y$ are constant$)$

$\frac{{{l_2}}}{{{l_1}}} = {\left( {\frac{{{r_1}}}{{{r_2}}}} ight)^2} = {(2)^2} \Rightarrow {l_2} = 4{l_1} = 4 	imes 2 = 8\;mm.$

When a uniform wire of radius $r$ is stretched by a $2kg$ weight, the increase in its length is $2.00\, mm$. If the radius of the wire is $r/2$ and other conditions remain the same, the increase in its length is .......... $mm$

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