Q97 / 125 · BITSAT 2011 · Physics

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

  • A
    $19.6 \times 10^{20} \mathrm{~N} / \mathrm{m}^{2}$
  • B
    $19.6 \times 10^{18} \mathrm{~N} / \mathrm{m}^{2}$
  • Correct answer
    C
    $19.6 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}$
  • D
    $19.6 \times 10^{15} \mathrm{~N} / \mathrm{m}^{2}$

Solution

$Y=\frac{\mathrm{F} / \mathrm{A}}{\Delta \ell / \ell}=\frac{\frac{250 \times 9.8}{50 \times 10^{-6}}}{\frac{0.5 \times 10^{-3}}{2}}$ $=\frac{250 \times 9.8}{50 \times 10^{-6}} \times \frac{2}{0.5 \times 10^{-3}} \Rightarrow 19.6 \times 10^{10} \mathrm{~N} / \mathrm{m}^{2}$

Hence, the correct answer is option: 'C'.

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