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The two thigh bones (femurs) each of cross-sectional area $10 c m^{2}$ support the upper part of a human body of mass $40 k g$ . The average pressure sustained by the femurs is

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$2 \times 10^{5} N / m^{2}$

$6 \times 10^{5} N / m^{2}$

$3 \times 10^{5} N / m^{2}$

$5 \times 10^{5} N / m^{2}$

answer is B.

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Detailed Solution

$A = 10 c m^{2} = 10 \times 10^{- 4} m^{2}; m = 40 k g$

$P = \frac{F}{2 A} = \frac{m g}{2 A} = \frac{40 \times 10}{2 \times 10^{- 3}} = 2 \times 10^{5} P a$

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