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
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(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}$
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