An iron rod of length $2 m$ and cross-sectional area of $50 m m^{2}$ stretched by $0.5 m m$ , when a mass of $250 k g$ is hung from its lower end. Young's modulus of iron rod is

19.6 \times 10^{20} N / m^{2}

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19.6 \times 10^{18} N / m^{2}

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19.6 \times 10^{10} N / m^{2}

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19.6 \times 10^{15} N / m^{2}

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Solution

The correct option is A $19.6 \times 10^{10} N / m^{2}$
$Y = \frac{(\frac{F}{A})}{(\frac{Δ l}{l})} = \frac{\frac{250 \times 9.8}{50 \times 10^{- 6}}}{\frac{0.5 \times 10^{- 3}}{2}}$
$Y = \frac{250 \times 9.8}{50 \times 10^{- 6}} \times \frac{2}{0.5 \times 10^{- 3}} = 19.6 \times 10^{10} N / m^{2}$

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An iron rod of length 2 m and cross-sectional area of 50 mm2 stretched by 0.5 mm, when a mass of 250 kg is hung from its lower end. Young's modulus of iron rod is

The correct option is A 19.6×1010 N/m2Y=(FA)(Δll)=250×9.850×10−60.5×10−32Y=250×9.850×10−6×20.5×10−3=19.6×1010 N/m2

An iron rod of length $2 m$ and cross-sectional area of $50 m m^{2}$ stretched by $0.5 m m$ , when a mass of $250 k g$ is hung from its lower end. Young's modulus of iron rod is

The correct option is A $19.6 \times 10^{10} N / m^{2}$
$Y = \frac{(\frac{F}{A})}{(\frac{Δ l}{l})} = \frac{\frac{250 \times 9.8}{50 \times 10^{- 6}}}{\frac{0.5 \times 10^{- 3}}{2}}$
$Y = \frac{250 \times 9.8}{50 \times 10^{- 6}} \times \frac{2}{0.5 \times 10^{- 3}} = 19.6 \times 10^{10} N / m^{2}$