The two femurs each of cross-sectional area l0 ${\mathrm{cm}}^2$ support the upper part of a human body of mass 40 kg. The average pressure sustained by the femurs is (Take g : l0 ${\mathrm{ms}}^{-2}$)

Total cross-sectional area of the femurs is,
A = 2 x lO ${\mathrm{cm}}^2$ : 2 x l0x ${10}^{-4}$ ${\mathrm{m}}^2$ = 20 x $\mathrm{l}{0}^{-4} {\mathrm{m}}^2$
Force acting on them is
F = mg = 40 kg x l0 ${\mathrm{ms}}^{-2}$ = 400 N
.'. Average pressure sustained by them is
$\mathrm{P} = \frac{\mathrm{F}}{\mathrm{A}} = \frac{400 \mathrm{N}}{20 \times {10}^{-4}{\mathrm{m}}^2} = 2\times {10}^5{\mathrm{Nm}}^{-2}$