Question

A beaker contains a liquid of density ‘ρ’ up to height ‘h’ such that ‘P_A’ is atmospheric pressure and ‘g’ is the acceleration due to gravity. Answer the following questions:

1. What is the pressure on the free surface of the liquid?
2. What is the pressure on the base of the beaker?
3. What is the lateral pressure at the base on the inner walls of the beaker?

Answer 1

a. Pressure on the free surface of the liquid is equal to the atmospheric pressure (P_a).

b. Consider a liquid contained in a beaker, such that ‘ p’ is the density of the liquid.
Consider a point B at the base of liquid and the liquid column of the area of cross-section ‘a’ around it, such that ‘h’ is the height of the liquid column as shown in the figure

∴ Volume of the imaginary column of liquids = area of cross-section × length = ah

∴ Mass of liquid column = Volume × density
= V × ρ = a.h.ρ

∴ Weight of liquid column =mass × g = mg
= a.h.ρ.g.

∴Thrust exerted by liquid column on the base of the breaker
= a.h.ρ.g.

∴ Pressure due to the liquid column

P = `"Force"/"Area"="F"/"a"="ahρg"/"a"`
P = h.ρ.g.
So, pressure on the base of beaker = hρg
∴ Total pressure at the base of beaker = Atmospheric pressure + hρg
= P_a + hρg

c. Also lateral pressure at the base on the inner walls of beaker = P_a + hρg