Question

The two femurs each of 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 $(Take\, g = 10\, m\, s^{-2})$

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

Answer 1

Answer: (C)

Total cross-sectional area of the femurs is,
$A=2 \times 10\, cm^{2} =2 \times 10 \times 10^{-4} m^{2}=20 \times 10^{-4} \, m^{2}$
Force acting on them is
$F = mg = 40 \,kg \times 10\, m \,s^{-2} = 400 \,N$
$\therefore \quad$ Average pressure sustained by them is
$P=\frac{F}{A}=\frac{400 \,N}{20\times10^{-4} \,m^{2}}=2\times10^{5}\, N \, m^{-2}\quad$