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The two femurs each of cross-sectional area `10 cm^(2)` support the upper part of a human body of mass `40 kg. the average pressure sustained by the femurs is (take `g=10 ms^(-2))`A. `2xx10^(3)Nm^(-2)`B. `2xx10^(4)Nm^(-2)`C. `2xx10^(5)Nm^(-2)`D. `2 xx10^(6)Nm^(-2)` |
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Answer» Correct Answer - C Total cross-sectional area of the femurs is , `A = 2 xx 10 cm^(2) = 2 xx 10 xx 10^(-4)m^(2)=20xx10^(-4)m^(2)` Force acting on them is `F = mg = 40 kg xx 10ms^(-2)=400N` `:.` Average pressure sustained by them is `P= F/A = (400N)/(20xx10^(-4)m^(2)) = 2 xx 10^(5)Nm^(-2)`. |
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