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 ........... $N / m ^2$
Medium
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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 $= mg =50 \times 10$ $=500 N$
Pressure, $P=\frac{\text { F orce }}{\text { Area }}=\frac{F}{A}=\frac{500}{20 \times 10^{-4}}$
$=2.5 \times 10^5 Pa$
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