If the pressure in a liquid is changed at a particular point, the change is transmitted to the entire liquid without being diminished in magnitude.
Application of pascal's law
A practical application of Pascal's law is the hydraulic lift which is used to lift a heavy load with a small force. It is a force multiplier. It consists of two cylinders A and B connected to each other by a horizontal pipe, filled with a liquid.

They are fitted with frictionless pistons of cross sectional areas A₁ and A₂ (A₂ > A₁). Suppose a downward force F is applied on the smaller piston, the pressure of the liquid under this piston increases to\(\left(\right. where, \left.P=\frac{F_{1}}{A_{1}}\right).\) But according to Pascal's law, this increased pressure P is transmitted undiminished in all directions. So a pressure is exerted on piston B. Upward force on piston B is
\(\mathrm{F}_{2}=\mathrm{P} \times \mathrm{A}_{2}=\frac{F_{1}}{A_{1}} \times A_{2} \Rightarrow \mathrm{F}_{2}=\frac{A_{2}}{A_{1}} \times F_{1}\)
Hence by changing the force on the smaller piston A, the force on the piston B has been increased by the factor \(\frac{A_{2}}{A_{1}}\) and this factor is called the mechanical advantage of the lift.

Construction' of Venturimeter:
This device is used to measure the rate of flow (or say flow speed) of the incompressible fluid flowing through a pipe. It works on the principle of Bernoulli's theorem. It consists of two wider tubes A and A' (with cross sectional area A) connected by a narrow tube B (with cross sectional area a). A manometer in the form of U-tube is also attached between the wide and narrow tubes. The manometer contains a liquid of density \('\rho_{m} \prime\).
Theory:
Let P₁ be the pressure of the fluid at the wider region of the tube A. Let us assume that the fluid of density 'p' flows from the pipe with speed 'v₁' and into the narrow region, its speed increases to 'v₂'. According to the Bernoulli's equation, this increase in speed is accompanied by a decrease in the fluid pressure P₂ at the narrow region of the tube B. Therefore, the pressure difference between the tubes A and B is noted by measuring the height difference (ΔP = P₁ - P₂) between the surfaces of the manometer liquid.
From the equation of continuity, we can say that AV₁= a v₂ which means that
\(v_{2}=\frac{A}{a}v_{1}\)
Using Bernoulli's equation,
\(P_{1}+\rho \frac{v^{2}_{1}}{2}=P_{2}+\rho \frac{V^{2}_{2}}{2}=P_{2}+\rho \frac{1}{2}(\frac{A}{a}v_{1})^{2}\)
From the above equation, the pressure difference
\(\Delta P=P_{1}-P_{2}=\rho \frac{v_{1}^{2}}{2} \frac{A^{2}-a^{2}}{a^{2}}\)
Thus, the speed of flow of fluid at the wide end of the tube A
\(v^{2}_{1}=\frac{2(\Delta P)a^{2}}{\rho(A^{2}-a^{2})} \Rightarrow v_{1}=\sqrt{\frac{2(\Delta P)a^{2}}{\rho(A^{2}-a^{2})}}\)
The volume of the liquid flowing out per second is
\(V=Av_{1}=A\sqrt{\frac{2(\Delta P)a^{2}}{\rho (A^{2}-a^{2})}}=aA\sqrt{\frac{2(\Delta P)}{\rho (A^{2}-a^{2})}}\)

Consider a pipe AB of varying cross sectional area a₁ and a₂ such that a₁ > a_2. A non-viscous and incompressible liquid flows steadily through the pipe, with velocities v₁ and v₂ in area a₁ and a₂, respectively as shown in Figure.

Let m1 be the mass of fluid flowing through section A in time Δ_t, m₁ = (a₁v₁Δt) p
Let m2 be the mass of fluid flowing through section B in time Δt, m₂= (a₂v₂Δt) p
For an incompressible liquid, mass is conserved m₁ =m₂
a₁v₁Δtρ = a₂v₂Δtρ
a₁v₁ = a₂v₂ ⇒ av = constant
which is called the equation of continuity and it is a statement of conservation of mass in the flow of fluids.
In general, a v = constant, which means that the volume flux or flow rate remains constant throughout the pipe. In other words, the smaller the cross section, greater will be the velocity of the fluid.

(1) Excess of pressure inside air bubble in a liquid.
Consider an air bubble of radius R inside a liquid having surface tension T as shown in Figure. Let P₁ and P₂ be the pressures outside and inside the air bubble, respectively. Now, the, excess pressure inside the air bubble is
\(\Delta P=P_{1}-P_{2}.\)

In order to find the excess pressure inside the air bubble, let us consider the forces acting on the air bubble. For the hemispherical portion of the bubble, considering the forces acting on it, we get,
(i) The force due to surface tension acting towards right around the rim of length \(2 \pi \mathrm{R} \ is \ \mathrm{F}_{\mathrm{T}}=2 \pi \mathrm{RT}\)
(ii) The force due to outside pressure P₁ is to the right acting across a cross sectional area of \(\pi \mathrm{R}^{2} \  is \ F_{P_{1}}=P_{1} \pi R^{2}\)
(iii) The force due to pressure P₂ inside the bubble, acting to the left is \(F_{P_{2}}=P_{2} \pi R^{2}\).
As the air bubble is in equilibrium under the action of these forces, \(F_{P_{2}}=F_{T}+F_{P_{1}}\)
\(\mathrm{P}_{2} \pi \mathrm{R}^{2} =2 \pi \mathrm{RT}+\mathrm{P}_{1} \pi \mathrm{R}^{2}  \)
\(\Rightarrow\left(\mathrm{P}_{2}-\mathrm{P}_{1}\right) \pi \mathrm{R}^{2} =2 \pi \mathrm{RT}\)
Excess pressure is \(\Delta \mathrm{P}=P_{2}-P_{1}=\frac{2 T}{R}\)
(2) Excess pressure inside a soap bubble
Consider a soap bubble of radius R and the surface tension of the soap bubble be T. A soap bubble has two liquid surfaces in contact with air, one inside the bubble and other outside the bubble. Hence, the force on the soap bubble due to surface tension is \(2 \times 2 \pi\) RT. The various forces acting on the soap bubble are,
(i) Force due to surface tension \(F_{T}=4 \pi R T\) towards right
(ii) Force due to outside pressure, \(F_{P_{1}}=P_{1} \pi R^{2}\) towards right
(iii) Force due to inside pressure, \(F_{P_{2}}=P_{2} \pi R^{2}\) towards left
As the bubble is in equilibrium, \(F_{P_{2}}=F_{T}+F_{P_{1}}\)
\(\mathrm{P}_{2} \pi \mathrm{R}^{2} =4 \pi \mathrm{RT}+\mathrm{P}_{1} \pi \mathrm{R}^{2}  \)
\(\Rightarrow\left(\mathrm{P}_{2}-\mathrm{P}_{1}\right) =\pi \mathrm{R}^{2}=4 \pi \mathrm{RT} \pi \mathrm{R}^{2}  \)
\(\text {Excess pressure is } \Delta \mathrm{P} =\mathrm{P}_{2}-\mathrm{P}_{1}=\frac{4 T}{R}\)
(3) Excess pressure inside the liquid drop
Consider a liquid drop of radius R and the surface tension of the liquid is T. 
The various forces acting on the liquid drop are,
(i) Force due to surface tension \(F_{T}=2 \pi R T\) towards right
(ii) Force due to outside pressure, \(F_{P_{1}}=P_{1} \pi R^{2}\) towards right
(iii) Force due to inside pressure, \(F_{P_{2}}=P_{2} \pi R^{2}\) towards left
As the bubble is in equilibrium, \(F_{P_{2}}=F_{T}+F_{P_{1}}\)
\(\mathrm{P}_{2} \pi \mathrm{R}^{2} =2 \pi \mathrm{RT}+\mathrm{P}_{1} \pi \mathrm{R}^{2}  \)
\(\Rightarrow\left(\mathrm{P}_{2}-\mathrm{P}_{1}\right) \pi \mathrm{R}^{2}=2 \pi \mathrm{RT} \)
\(\text {Excess pressure is } \Delta \mathrm{P} =\mathrm{P}_{2}-\mathrm{P}_{1}=\frac{2 T}{R}\)

Consider a sphere of radius r which falls freely through a highly viscous liquid of coefficient of viscosity η. Let the density of the material of the sphere be p and the density of the fluid be σ.

Gravitational force acting on the sphere,
\(F_{G}=mg=\frac{4}{3}\pi r^{3}\rho g \) (downward force)
Up thrust, U = \(\frac{4}{3} \pi r^{3} g\) (upward force)
viscous force F = 6ㅠηrv_t
At terminal velocity v_t
downward force = upward force
\(F_{G}-U=F \Rightarrow \frac{4}{3}\pi r^{3}\rho g - = \frac{4}{3}\pi r^{3} \sigma g =6\pi \eta r v_{t}\)
\(v_{t}=\frac{2}{9} \times \frac{r^{2} (\rho-\sigma)}{\eta}g \Rightarrow v_{t} \infty r^{2}\)
Here, it should be noted that the terminal speed of the sphere is directly proportional to the square of its radius.