Variation of Pressure in Liquids with depth

Concept Explanation

Variation of Pressure in Liquids with depth

Variation of Pressure in Liquids With Depth: When we swim in a river, we feel the water pressure acting against our eardrums. The deeper we swim, the greater the pressure. The cause of this pressure is the weight of the fluid, water plus air directly above us. As we swim deeper, there is more water above us. Thereofore, there is more pressure. If we swim twice as deep, there is twice the weight of water above us, so the water contribution to the pressure felt by us is doubled. The pressure due to a liquid also depends upon its density.

Mathematical expression for fluid pressure

Let us consider a liquid of density $large rho$ in a beaker of base area A. If h be the height of the liquid column then

Volume of liquid in beaker, $large V=Atimes h$

Force (or thrust) acting at the bottom of the beaker is given by

F = Weight of liquid = mg

$large =Vrho g;;;;(because m=Vrho ;and;V=Atimes h)$

$large =(Atimes h)rho g=rho; g;h;A$

$large therefore$ Pressure at the bottom of the beaker $large P=frac{F}{A}=frac{rho ghA}{A}=rho gh,$

Thus, we observe that pressure at a depth h due to a liquid column is directly proportional to

(i) the density $large rho$ of the liquid.

(ii) the height of the liquid column and

(iii) the acceleration due to gravity g

Q. Calculate the average density of the atmosphere, given that it extends to an altitude of 120 km. Compare this density with that of the air.

Solution : If we solve $rho :=:frac{P}{hg}$

Here, we can take P = atmospheric pressure, h (height) is known and g(accelaeration due to garvity) also known.

Entering the value in the value of $rho$ ,

$rho:=;frac{1.01times10^5: N/m^2}{(120times10^3 m)(9.8 m/s^2)}$ $rho:=;frac{1.01times10^5: N/m^2}{(120times10^3 m)(9.8 m/s^2)}:=;8.59times10^-^2:kg/m^3$

Q. Calculate the depth below the surface of water at which the pressure due to the weight of water equals 1.00 atm.

Solution: As we know that $P:=:hrho g$

$h:=;frac{P}{rho g}$

$h:=;frac{1.01times10^5:N/m^2}{(1.00times10^3:m)(9.8:m/s^2)}:=;10.3;m$

Sample Questions

(More Questions for each concept available in Login)

Question : 1

The pressure due to a liquid depends upon its ________________.
A. Force	B. Area	C. Density	D. None of these.
Right Option : C
View Explanation

Question : 2

Which of the following are correct : (a)The pressure due to a liquid also depends its area. (b) The greater the height, the greater the pressure. (c) The greater the density, the greater the pressure.
A. Only (a) and (b) are correct.		B. Only (b) and (c) are correct.
C. Only (a) and (c) are correct.		D. None of these.
Right Option : B
View Explanation

Question : 3

Which of the following are correct : (a) The deeper we swim, the greater the pressure. (b) The deeper we swim, the smaller the pressure. (c) The pressure due to a liquid depends upon its density.
A. Only (a) and (b) are correct.		B. Only (a) and (c) are correct.
C. Only (b) and (c) are correct.		D. All of these.
Right Option : C
View Explanation

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Variation of Pressure in Liquids with depth

Variation of Pressure in Liquids with depth

Students / Parents Reviews [10]

Prisha Gupta

Shreya Shrivastava

Manish Kumar

Cheshta

Eshan Arora

Om Umang

Aman Kumar Shrivastava

Barkha Arora

Archit Segal

Hiya Gupta