Question

Derive an expression for the pressure exerted by the liquid having density d, at a depth h.

Answer 1

• Consider a liquid at rest in a container. In a liquid, all points at the same depth are at the same pressure.
• To show it, let us consider a cylindrical column of the liquid of height h, and cross-sectional area a, as shown in (Fig) If d is the density of the liquid.

1. Mass of the liquid in the column = $axhxd$
2. If g is the acceleration due to gravity, then Weight of the liquid in the column = Mass x Acceleration due to gravity=$ahdxg=ahdg$
3. Therefore, Thrust at the base of the cylindrical column of liquid =$ahdg$
4. Then, The pressure exerted by the liquid at a depth of h (at point X) $=\frac{Thrust}{Area}$$=\frac{ahdg}{a}=hdg$
5. $P=hdg$

i.e. pressure at a point in a liquid is proportional to its depth, density, and acceleration due to gravity.

This shows that at any place, inside a liquid, pressure at all the points at the same depth is the same.