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प्रश्न
A beaker contains a liquid of density ‘ρ’ up to height ‘h’ such that ‘PA’ is atmospheric pressure and ‘g’ is the acceleration due to gravity. Answer the following questions:
- What is the pressure on the free surface of the liquid?
- What is the pressure on the base of the beaker?
- What is the lateral pressure at the base on the inner walls of the beaker?
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उत्तर
a. Pressure on the free surface of the liquid is equal to the atmospheric pressure (Pa).
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
= Pa + hρg
c. Also lateral pressure at the base on the inner walls of beaker = Pa + hρg
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