Correct Answer - C
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)`