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8.Mechanical Properties of Solids
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A wire of cross section $4 \;mm^2$ is stretched by $0.1\, mm$ by a certain weight. How far (length) will be wire of same material and length but of area $8 \;mm^2$ stretch under the action of same force......... $mm$
A
$0.05$
B
$0.10$
C
$0.15$
D
$0.20$
Solution
(a) $l = \frac{{FL}}{{AY}}$
$l \propto \frac{1}{A}$ $(F,L$ and $Y$ are constant$)$
$\frac{{{l_2}}}{{{l_1}}} = \frac{{{A_1}}}{{{A_2}}} = \frac{4}{8} = \frac{1}{2}$$ \Rightarrow {l_2} = \frac{{{l_1}}}{2} = \frac{{0.1}}{2} = 0.05mm$
Standard 11
Physics
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