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8.Mechanical Properties of Solids
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A steel wire of lm long and $1\,m{m^2}$ cross section area is hang from rigid end. When weight of $1\,kg$ is hung from it then change in length will be given ..... $mm$ $(Y = 2 \times {10^{11}}N/{m^2})$
A
$0.5$
B
$0.25$
C
$0.05$
D
$5$
Solution
(c) $l = \frac{{MgL}}{{YA}} = \frac{{1 \times 10 \times 1}}{{2 \times {{10}^{11}} \times {{10}^{ – 6}}}} = 0.05\;mm$
Standard 11
Physics
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