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8.Mechanical Properties of Solids
medium
A wire of area of cross-section ${10^{ - 6}}{m^2}$ is increased in length by $0.1\%$. The tension produced is $1000 N$. The Young's modulus of wire is
A
${10^{12}}\,N/{m^2}$
B
${10^{11}}\,N/{m^2}$
C
${10^{10}}\,N/{m^2}$
D
${10^9}\,N/{m^2}$
Solution
(a) $Y = \frac{{FL}}{{Al}} = \frac{{1000 \times 100}}{{{{10}^{ – 6}} \times 0.1}} = {10^{12}}N/{m^2}$
Standard 11
Physics