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Physics Question on Pressure

The two femurs each of cross-sectional area 10 cm210 \,cm^{2} support the upper part of a human body of mass 40 kg40\, kg. The average pressure sustained by the femurs is (Take g=10 m s−2)(Take\, g = 10\, m\, s^{-2})

A

2×103N m−22 \times 10^{3} N \, m^{-2}

B

2×104 N m−22 \times 10^{4} \, N \, m^{-2}

C

2×105 N m−22 \times 10^{5} \, N\, m^{-2}

D

2×106 N m−22 \times 10^{6} \, N \, m^{-2}

Answer

2×105 N m−22 \times 10^{5} \, N\, m^{-2}

Explanation

Solution

Total cross-sectional area of the femurs is, A=2×10 cm2=2×10×10−4m2=20×10−4 m2A=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×10 m s−2=400 NF = mg = 40 \,kg \times 10\, m \,s^{-2} = 400 \,N ∴\therefore\quad Average pressure sustained by them is P=FA=400 N20×10−4 m2=2×105 N m−2P=\frac{F}{A}=\frac{400 \,N}{20\times10^{-4} \,m^{2}}=2\times10^{5}\, N \, m^{-2}\quad