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8.Mechanical Properties of Solids
easy
An iron rod of length $2m$ and cross section area of $50\,m{m^2}$, stretched by $0.5\, mm$, when a mass of $250\, kg$ is hung from its lower end. Young's modulus of the iron rod is
A
$19.6 \times {10^{10}}\,N/{m^2}$
B
$19.6 \times {10^{15}}\,N/{m^2}$
C
$19.6 \times {10^{18}}\,N/{m^2}$
D
$19.6 \times {10^{20}}\,N/{m^2}$
Solution
(a) $Y = \frac{{MgL}}{{Al}} = \frac{{250 \times 9.8 \times 2}}{{50 \times {{10}^{ – 6}} \times 0.5 \times {{10}^{ – 3}}}}$
$ = 19.6 \times {10^{10}}N/{m^2}$
Standard 11
Physics
