# Pressure , Variation of Pressure with depth

### What is Pressure ?

When a surface is immersed within a fluid at rest, the fluid exerts a force perpendicular to the surface.

This normal force acting on a small surface is proportional to the surface area of the surface in contact with the fluid.

The magnitude of the normal force acting per unit area within a fluid is called pressure.

If a force $\vec{\Delta F}$  acts on any surface $\vec{\Delta A}$ such that $\vec{\Delta F}$  is acting towards $\vec{\Delta A}$  then the pressure is given by,

$\displaystyle P = \frac{|\vec{\Delta F}|}{|\vec{\Delta A}|}$

Pressure P is a scalar quantity. Its unit in S.I. system is N/m2. It is also known as Pascal. There are several other units of pressure like bar, atmosphere etc.

1 bar = 1.013 × 105 N/m2

= 1.013 × 105 Pascal = 1 atmosphere

= pressure exerted by 76 cm of Hg column.

Solved Example : What are the forces imparted by atmosphere on the walls of a room of dimension 6m × 5m × 4m ? (1 atm. = 105 N/m2)

Solution: The atmosphere imparts a force normal to any surface in contact with it.

We know that the atmospheric pressure $\large = \frac{Normal \; force}{Area \;of \;the \; Surface}$

$\displaystyle P = \frac{F}{A}$

⇒ F = P A

The area of the walls of the room are (6 × 4) m2 and (5 × 4) m2 respectively .

Putting P = 105 N/m2 and the area of the walls we obtain

F = 2.4 × 106 N and 2 × 106 N respectively.

### How to find Variation of Pressure with depth ?

(i)

Suppose that the pressure at A is P1 and the pressure at B is P2, then, considering the equilibrium of the fluid within the cylinder AB,

p2 ΔS = p1 ΔS + ρgΔS (y2 – y1)

or p2 = p1 + ρ g(y2 – y1)

Pressure increases with depth, then Δp = ρ g Δy and

$\large \lim_{\Delta y \rightarrow 0} \Delta p = \lim_{\Delta y \rightarrow 0} \rho g \Delta y$

$\large dp = \rho g dy$

Where y increases in the downward direction & Pressure increases.

(ii)

Pressure is same at two points in at the same horizontal level. As dy = 0; dp = 0

As the body is in equilibrium,

P1 ΔS = P2 ΔS

or, P1 = P2

Note: P is independent of shape of container.

Solved Example:  Assuming that the atmosphere has a uniform density of (1.3 kg/m3)  and an effective height of 10 km, find the force exerted on an area of dimensions 100m  × 80 m at the bottom of the atmosphere.

Solution :   F = PA

F = (ρ g h) A

F = (1.3) (9.8) (104) (100 × 80)

F = 12.74 × 8 × 107 N

F = 1.0192 × 109 N

Solved Example : For the arrangement shown in the figure, what is the density of oil ?

Solution : PB = Po + ρw g l

PA = Po + ρoil (l + d)g

PA = PB

Po + ρoil (l + d)g = Po + ρw g l

ρoil (l + d)g = ρw g l

$\large \rho_{oil} = \frac{\rho_w l}{l + d}$

$\large \rho_{oil} = \frac{1000 \times 135}{135 + 12.3}$

= 916.5 kg/m3

Solved Example: A water tank is 20 m deep. If the water barometer reads 10 m at that place, then what is the pressure at the bottom of the tank in atmosphere?

Solution: The pressure at the bottom is

P = P0 + hρg where P0 is the atmospheric pressure.

Therefore, $\large P = 1 atm + \frac{h \rho g}{P_0} atm$

$\large P = 1 + (\frac{20 \rho g}{10 \rho g})$

P = 3 atm

Solved Example: Two immiscible liquids of relative densities 1.5 and 2 are at rest in a rectangular container of dimensions 3 × 2 × 1 m. Find the force exerted on the bottom of the container due to the liquids.

Solution: The force exerted by the liquids at the bottom of the container is equal to the sum of the weights of the liquids. The weight of the liquid 1 = W1 = m1g

W1 = (ρ1V1) g.

Similarly, the weight of the liquid 2 = W2 = m2g

⇒ W2 = (ρ2V2)g.

The sum of the weights = W = W1 + W2

W = (ρ1V1 + ρ2V2)g

where V1 = volume of the liquid 1 = (3 × 2 × 1/2) = 3 m3

and V2 = volume of the liquid 2 = (3 × 2 × 1/2) = 3 m3.

W = (1.5 × 3 + 2 × 3) (9.8) × 103 N

= 7.35 × 104 N