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. `2 xx 10^(2) N m^(-2)`
B. `2 xx 10^(4) N m^(-2)`
C. `2 xx 10^(5) N m^(-2)`
D. `2 xx 10^(6) N m^(-2)`

Answer

C

Explanation

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