A person stands on his two feet over a surface and experiences a press

A person stands on his two feet over a surface and experiences a pressure P. Now the person stands on only one foot. He would experience a pressure of magnitude

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4 P

$ rac{1}{2}$ P

2 P

Answer is Right!

Answer is Wrong!

This question was previously asked in

UPSC NDA-1 – 2015

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Pressure is defined as force per unit area (P = F/A). When a person stands on a surface, the force exerted is their weight (F = W). When standing on two feet, the weight is distributed over the total area of two feet ($A_{two}$). So the pressure is $P = W / A_{two}$. When standing on one foot, the weight is distributed over the area of one foot ($A_{one}$). Assuming the area of one foot is approximately half the area of two feet ($A_{one} \approx A_{two}/2$), the new pressure is $P’ = W / A_{one} \approx W / (A_{two}/2) = 2W/A_{two} = 2P$. Thus, the pressure is doubled.

Pressure is inversely proportional to the area over which the force is applied, assuming the force remains constant.

This is a simple application of the definition of pressure. By reducing the contact area by half (from two feet to one foot), the pressure exerted on the surface doubles for the same force (weight).

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