The two thigh bones (femures), each of cross-sectional area $10 \,cm ^2$ support the upper part of a person of mass $50 \,kg$. The average pressure sustained by the femures is ............. $N / m ^2$
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(a)
Given, mass of body, $m =50\,kg ; g =10\,ms ^{-2}$;
area, $A =2 \times$ area of each thigh bone
$=2 \times 10\,cm ^2=20 \times 10^{-4}\,m ^2$
(weight of the body is supported by two thigh bones)
Force, $F=$ weight of the body $= mg =40 \times 10$ $=400\, N$
Pressure, $P=\frac{\text { Force }}{\text { Area }}=\frac{F}{A}=\frac{400}{20 \times 10^{-4}}$ $=2 \times 10^5\,Pa$
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