Question

The two femurs each of the cross-sectional area 10 cm${}^2$ support the upper part of a human body of mass 40 kg. The average pressure sustained by the femurs is then (Takes g= 10 m s${}^{-2}$)

• A. 2 x 10${}^3$ N m${}^{-2}$
• B. 2 x 10${}^4$ N m${}^{-2}$
• C. 2 x 10${}^5$ N m${}^{-2}$
• D. 2 x 10${}^6$ N m${}^{-2}$.

Answer 1

The correct option is C 2 x 10${}^5$ N m${}^{-2}$

Given that,

$Area=10c{m}^2$

$Mass=40kg$

$g=10m/s^2$

Total cross-sectional area of the femurs is,

$A=2\times 10c{m}^2$

$=2\times 10\times {10}^{-4}{m}^2$

$=20\times {10}^{-4}{m}^2$

Force acting on them is

$F=mg=40kg\times 10m{s}^{-2}=400N$

$\therefore$ Average pressure sustained by them is
$P=\frac{F}{A}=\frac{400N}{20\times {10}^{-4}{m}^2}$

$=2\times {10}^{5}N{m}^{-2}$