Question

A person of mass 45 kg having leg bones each of length 50 cm and area of cross-section of 5 cm$^2$ jumps safely from a height of 2 m. If his leg bones can withstand a stress of $0.9 \times 10^8$ N m$^{-2}$, then the Young’s modulus of the leg bones is
(Acceleration due to gravity $g = 10$ m s$^{-2}$)

• A. $45 \times 10^8$ N m$^{-2}$
• B. $2.25 \times 10^8$ N m$^{-2}$
• C. $2 \times 10^8$ N m$^{-2}$
• D. $1.125 \times 10^8$ N m$^{-2}$

Hint:

Young’s modulus relates stress and strain in elastic materials, often requiring careful unit conversions in biomechanical problems.

Answer 1

The Correct Option is A

The person jumps from 2 m, so the potential energy is converted into kinetic energy: $mgh = \frac{1}{2}mv^2$. 
With $m = 45$ kg, $g = 10$ m s$^{-2}$, $h = 2$ m: \[ v^2 = 2gh = 2 \times 10 \times 2 = 40 \implies v = \sqrt{40} { m s}^{-1} \] Upon landing, the kinetic energy is absorbed by the leg bones, which compress under the stress. 
Stress is given as $0.9 \times 10^8$ N m$^{-2}$. 
Stress = $\frac{F}{A}$, where $A = 5$ cm$^2 = 5 \times 10^{-4}$ m$^2$ (per leg, assuming two legs): \[ F = {stress} \times A = 0.9 \times 10^8 \times 5 \times 10^{-4} = 4.5 \times 10^4 { N (per leg)} \] Total force for two legs: $F_{{total}} = 2 \times 4.5 \times 10^4 = 9 \times 10^4$ N. 
This force decelerates the person to rest. Using $F = ma$, the deceleration $a = \frac{F}{m} = \frac{9 \times 10^4}{45} = 2000$ m s$^{-2}$. 
The stopping distance $s$ is found using $v^2 = 2as$: \[ s = \frac{v^2}{2a} = \frac{40}{2 \times 2000} = 0.01 { m} = 1 { cm} \] Strain = $\frac{\Delta L}{L} = \frac{1}{50} = 0.02$ (since $L = 50$ cm). 
Young’s modulus $Y = \frac{{stress}}{{strain}} = \frac{0.9 \times 10^8}{0.02} = 4.5 \times 10^9 = 45 \times 10^8$ N m$^{-2}$.