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8.Mechanical Properties of Solids
medium
Longitudinal stress of $1\,kg/m{m^2}$ is applied on a wire. The percentage increase in length is $(Y = {10^{11}}\,N/{m^2})$
A
$0.002$
B
$0.01$
C
$0.003$
D
$0.001$
Solution
(b) Stress $=1\,kg/m{m^2}=10^6\;kg/m^2=10^7\;N/m^2$
Longitudinal strain $\frac{l}{L} = \frac{{{\rm{stress}}}}{Y} = \frac{{{{10}^7}}}{{{{10}^{11}}}} = {10^{ – 4}}$
Percentage increase in length $ = {10^{ – 4}} \times 100 = 0.01\% $
Standard 11
Physics