Question

A 40 kg boy whose legs are 4 $cm^{2}$ in area and 50 cm long falls through a height of 2 m without breaking his leg bones. If the bones can withstand a stress of 0.9 $\times10^{8}N/m^{2}.$ The Young's modulus for the material of the bone is ________.

• A. $0.9\times10^{8}N/m^{2}$
• B. $5\times10^{8}N/m^{2}$
• C. $2.05\times10^{9}N/m^{2}$
• D. $2.05\times10^{8}N/m^{2}$

Hint:

Elastic potential energy per unit volume is $\frac{1}{2} \frac{(\text{stress})^2}{Y}$.

Answer 1

The Correct Option is C

Step 1: Concept
Energy conservation: Potential energy lost = Elastic energy gained. $mgh = \frac{1}{2} \times \text{Stress} \times \text{Strain} \times \text{Volume}$.
Step 2: Analysis
$m = 40$ kg, $h = 2$ m, $L = 0.5$ m, $A = 2 \times 4 \times 10^{-4}$ $m^2$ (two legs). Stress ($\sigma$) = $0.9 \times 10^8$ $N/m^2$. Volume ($V$) = $A \times L = 8 \times 10^{-4} \times 0.5 = 4 \times 10^{-4}$ $m^3$.
Step 3: Calculation
$mgh = (40)(10)(2) = 800$ J. Elastic energy = $\frac{1}{2} \frac{\sigma^2}{Y} V = 800$. $Y = \frac{\sigma^2 V}{2 \times 800} = \frac{(0.81 \times 10^{16}) \times (4 \times 10^{-4})}{1600} \approx 2.05 \times 10^9$ $N/m^2$.
Step 4: Conclusion
Hence, the Young's modulus is $2.05 \times 10^9$ $N/m^2$.
Final Answer:(C)