Question

The thigh bones of each cross-sectional area 10 $c{m^2}$ support the upper part of the human body of mass 40kg. Average pressure sustained by the femurs is:
A) $10\dfrac{N}{{{m^2}}}$
B) $2 \times {10^5}\dfrac{N}{{{m^2}}}$
C) $6 \times {10^{ - 5}}\dfrac{N}{{{m^2}}}$
D) ${10^{ - 3}}\dfrac{N}{{{m^2}}}$

Answer 1

Average pressure is nothing but normal pressure in this case but on two thigh bones. Pressure is the ratio of force to the area. Force is the product of mass and acceleration due to gravity. Area is twice that of a thigh bone as the force is applied on both thighs. The ratio of the force and area calculated gives the pressure sustained by the femurs.

Complete Step-by-step solution
Pressure on a point is defined as force applied over a surface or area. Therefore,
Pressure $= \dfrac{{Force}}{{Area}}$
In this case, force is generated by gravity i.e. it is a gravitational force and it is given by
$F = mg$
M=mass of the body considered
g= acceleration due to gravity
Therefore,
$F = 40 \times 9.81$
F=392.4M
This force is balanced by 2 thigh bones or femurs. The cross-sectional area of 1 femur is 10 $c{m^2}$ . As this force is balanced on 2 femurs, the total area becomes
$A = 2 \times 10 = 20c{m^2}$
As this area is given in $c{m^2}$ , we need to convert it into ${m^2}$ . This is done because the units of force are in Newton which is in the mks system.
$A = 20 \times {10^{ - 4}}{m^2}$
Therefore, the average pressure applied on the femur bone or the thigh bone is given as:
Pressure $= \dfrac{{392.4}}{{20 \times {{10}^{ - 4}}}}$
$P = 19.6 \times {10^4}N/{m^2}$
Average pressure $P = 1.96 \times {10^5}\dfrac{N}{{{m^2}}}$

So, the correct answer is option B.

Note: Average pressure is the same thing as stress applied on a body. Both of these quantities even have the same dimensions, so we have to follow the same approach when we have to find stress developed on the legs.