The two thigh bones (femurs), each of cross-sectional area $10\; cm ^{2}$ support the upper part of a human body of mass $40\; kg$. Estimate the average pressure sustained by the femurs.
A$2 \times 10^{5} \;N m ^{-2}$
B$6 \times 10^{5} \;N m ^{-2}$
C$4 \times 10^{6} \;N m ^{-2}$
D$8 \times 10^{6} \;N m ^{-2}$
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A$2 \times 10^{5} \;N m ^{-2}$
a Answer Total cross-sectional area of the femurs is $A=2 \times 10 cm ^{2}=20 \times 10^{-4} m ^{2}$. The
force acting on them 1 s $F=40 kg$ wt $=400 N$ (taking $g=10 m s ^{-2}$ ). This force is acting vertically down and hence, normally on the femurs. Thus, the average pressure is
$P_{a v}=\frac{F}{A}=2 \times 10^{5} \;N m ^{-2}$
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