Question

A $45\text{kg}$ boy whose leg bones are $5{\text{cm}}^2$ in area and $50\text{cm}$ long fall through a height of $2\text{m}$ without breaking his leg bones. If the bones can stand a stress of $0.9\times {10}^8{\text{N/m}}^2$, calculate the Young's modulus for the material of the bones.
(Use $g=10{\text{m/s}}^2$)

• A. $1.05\times {10}^8{\text{N/m}}^2$
• B. $2.25\times {10}^8{\text{N/m}}^2$
• C. $1.05\times {10}^9{\text{N/m}}^2$
• D. $2.25\times {10}^9{\text{N/m}}^2$

Answer 1

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

Given:
$m=45\text{kg};h=2\text{m}$
$L=0.50\text{m};A=5\times {10}^{-4}{\text{m}}^2$
$\sigma =0.9\times {10}^8{\text{N/m}}^2$

Here, the gravitational potential energy will be converted into the form of elastic energy. So,

Loss in gravitational energy = gain in elastic energy in both leg bones

$\Rightarrow mgh=2\times \{\frac{1}{2}\times \text{stress}\times \text{strain}\times \text{volume}\}$

Hence,
Volume $=AL=2.5\times {10}^{-4}{\text{m}}^3$

Let $\epsilon$ be the strain produced in bones.

So we have

$mgh=45\times 10\times 2=90$

$(\frac{1}{2}\times \sigma \times \epsilon \times AL)=\frac{1}{2}\times 0.9\times {10}^8\times \epsilon \times 2.5\times 10$

So,

$mgh=(\frac{1}{2}\times \sigma \times \epsilon \times AL)\times 2$


$\Rightarrow 900=\frac{1}{2}\times 0.9\times {10}^8\times \epsilon \times 2.5\times {10}^{-4}\times 2$

$\Rightarrow \epsilon =\frac{900}{0.9\times 2.5\times {10}^4}=0.04$

Thus, we have the Young's modulus $Y$ as

$Y=\frac{\sigma }{\epsilon }=\frac{0.9\times {10}^8}{0.04}$

$\Rightarrow Y=2.25\times {10}^9{\text{N/m}}^2$

Hence, option $\left(d\right)$ is the correct answer.