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 ms^(-2))`

Options

A. `2xx10^(3)Nm^(-2)`
B. `2xx10^(4)Nm^(-2)`
C. `2xx10^(5)Nm^(-2)`
D. `2 xx10^(6)Nm^(-2)`

Answer

C

Explanation

Total cross-sectional area of the femurs is ,
`A = 2 xx 10 cm^(2) = 2 xx 10 xx 10^(-4)m^(2)=20xx10^(-4)m^(2)`
Force acting on them is
`F = mg = 40 kg xx 10ms^(-2)=400N`
`:.` Average pressure sustained by them is
`P= F/A = (400N)/(20xx10^(-4)m^(2)) = 2 xx 10^(5)Nm^(-2)`.