Question

A boy of mass $45\text{kg}$, whose leg bones have a cross-sectional area of $5{\text{cm}}^2$ and length $50\text{cm}$, falls through a height of $2\text{m}$ without breaking his leg bones. If the bones can withstand a stress of $0.9\times {10}^8{\text{Nm}}^{-2}$, calculate the Young's modulus for the material of the bone. Use, $g=10{\text{ms}}^{-2}$

• A. $9\times {10}^9{\text{N m}}^{-2}$
• B. $4.5\times {10}^9{\text{N m}}^{-2}$
• C. $2.25\times {10}^9{\text{N m}}^{-2}$
• D. $1.2\times {10}^9{\text{N m}}^{-2}$

Answer 1

The correct option is C $2.25\times {10}^9{\text{N m}}^{-2}$

Here,
$m=45\text{kg},h=2\text{m};$
$L=0.50\text{m},A=5\times {10}^{-4}{\text{m}}^2$

Using conservation of energy,
$\text{Loss in gravitational energy = gain in elastic energy in both leg bones}$
i.e $mgh=2\times [\frac{1}{2}\times \text{stress}\times \text{strain}\times \text{volume}]$
$\text{Here, volume}=A\times L=5\times {10}^{-4}\times 0.50=2.5\times {10}^{-4}{\text{m}}^3$

$\therefore 45\times 10\times 2=2\times [\frac{1}{2}\times 0.9\times {10}^8\times \text{strain}\times 2.5\times {10}^{-4}]$
[using maximum permissible stress]
or Strain $=\frac{45\times 10\times 2}{0.9\times 2.5\times {10}^4}=0.04$

$\therefore Y=\frac{\text{stress}}{\text{strain}}=\frac{0.9\times {10}^8}{0.04}$

$Y=2.25\times {10}^9{\text{Nm}}^{-2}$