The two femurs each of cross-sectional area l0 cm2 support the upper part of a human body of mass 40 kg. The average pressure sustained by the femurs is (Take g : l0 ms-2)

Total cross-sectional area of the femurs is,A = 2 x lO cm2 : 2 x l0x 10-4 m2 = 20 x l0-4 m2Force acting on them isF = mg = 40 kg x l0 ms-2 = 400 N.'. Average pressure sustained by them isP = FA = 400 N20 × 10-4m2 = 2×105Nm-2