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8.Mechanical Properties of Solids
easy
A force $F$ is applied on the wire of radius $r$ and length $L$ and change in the length of wire is $l.$ If the same force $F$ is applied on the wire of the same material and radius $2r$ and length $2L,$ Then the change in length of the other wire is
A
$l$
B
$2l$
C
$l/2$
D
$4l$
Solution
(c) $l = \frac{{FL}}{{AY}} \Rightarrow l \propto \frac{L}{{{r^2}}}$ $(F$ and $Y$ are constant$)$
$\frac{{{l_2}}}{{{l_1}}} = \frac{{{L_2}}}{{{L_1}}} \times {\left( {\frac{{{r_1}}}{{{r_2}}}} \right)^2} = 2 \times {\left( {\frac{1}{2}} \right)^2} = \frac{1}{2}$
$\therefore {l_2} = \frac{{{l_1}}}{2}$
i.e. the change in the length of other wire is $\frac{l}{2}$
Standard 11
Physics
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