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PART - 2 CH - 9 Mechanical Properties of Fluids — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsPART - 2 CH - 9 Mechanical Properties of Fluids2 Marks
Question
The mass of a person is 30 kg and the area of the sole of his foot is $0.01 m^2$. If $g =10 m /$ $s^2$ then calculate the pressure exerted by the person on the floor.
(i) When the person is standing on one leg.
(ii) When the person is standing on both legs.
✓

Answer

(i) Given : Area of the sole A $=0.01 m^2$
$\begin{aligned} \text { Force applied on the floor } & =\text { Weight of the person } \\ & = mg =30 kg \times 10 m / s ^2 \\ & =300 \text { Newton }\end{aligned}$
Pressure exerted on the floor $P =\frac{\text { Force }( F )}{\operatorname{Area}( A )}$
$P=\frac{300}{0.01}=3 \times 10^4 N / m ^2$
(ii) Area of both soles of a man standing on both feet
$A=2 \times 0.01=0.02 m^2$
$\begin{aligned} \text { Force applied on the floor } & =\text { Weight of the floor } \\ & =300 \text { Newton }\end{aligned}$
Pressure exerted on the floor $P =\frac{\text { Force }( F )}{\text { Area }( A )}$
$P=\frac{300 \text { Newton }}{0.02 m^2}=1.5 \times 10^4 N / m ^2$

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