8.Mechanical Properties of Solids
medium

A wire is stretched by $0.01$ $m$ by a certain force $F.$ Another wire of same material whose diameter and length are double to the original wire is stretched by the same force. Then its elongation will be

A

$0.005$ $m$

B

$0.01$ $m$

C

$0.02$ $m$

D

$0.002$ $m$

Solution

(a) $l = \frac{{FL}}{{\pi {r^2}Y}}$

$l \propto \frac{L}{{{r^2}}}$ $(Y$ and $F$ are constant$)$

$\frac{{{l_2}}}{{{l_1}}} = \frac{{{L_2}}}{{{L_1}}} \times {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2} = (2) \times {\left( {\frac{1}{2}} \right)^2} = \frac{1}{2}$

==> ${l_2} = \frac{{{l_1}}}{2} = \frac{{0.01m}}{2} = 0.005m$

Standard 11
Physics

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