(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.

When a body is stretched, work is done against the restoring force (internal force). This work done is stored in the body in the form of elastic energy. Consider a wire whose un-stretch length is L and area of cross section is A. Let a force produce an extension I and further assume that the elastic limit of the wire has not been exceeded and there is no loss in energy. Then, the work done by the force F is equal to the energy gained by the wire.
The work done in stretching the wire by dl, dW = Fdl
The total work done in stretching the wire from 0 to l is
\(W=\int^{l}_{o}F dl\)   --- (1)
From Young's modulus of elasticity,
\(\mathrm{Y}=\frac{F}{A} \times \frac{L}{l} \Rightarrow \frac{Y A l}{L}\)  ......(2)
Substituting equation (2) in equation (1), we get
\(\mathrm{W}=\int_{0}^{1} \frac{Y A l}{L} d l\)
Since, l is the dummy variable in the integration, we can change l to l' (not in limits), therefore
\(\mathrm{W}=\int_{0}^{l} \frac{Y A l^{\prime}}{L} d l^{\prime} =\frac{Y A}{L}\left(\frac{l^{\prime 2}}{2}\right)_{0}^{i}=\frac{Y A}{I} \frac{l^{2}}{2}=\frac{1}{2}\left(\frac{Y A l}{L}\right) l=\frac{1}{2} F l \)
\(W =\frac{1}{2} F l=\text { Elastic potential energy }\)
Energy per unit volume is called energy density,
\(u =\frac{\text { Elastic potential energy }}{\text { Volume }}=\frac{\frac{1}{2} F l}{A L}  \)
\(\frac{1}{2} \frac{F}{A} \frac{l}{L} =\frac{1}{2}(\text { Stress } \times \text { Strain })\)  ......(3)

Hooke's law states that within the elastic limit, the strain produced in a body is directly proportional to the stress applied.
It can be verified in a simple way by stretching a thin straight wire (stretches like spring) of length L and uniform cross-sectional area A suspended from a fixed point O. A pan and a pointer are attached at the free end of the wire as shown in Figure. The extension produced on the wire is measured using a vernier scale arrangement. The experiment shows that for a given load, the corresponding stretching force is F and the elongation produced on the wire is ΔL. It is directly proportional to the original length L and inversely proportional to the area of cross section A. A graph is plotted using F on the X-axis and ΔL on the Y-axis. This graph is a straight line passing through the origin as shown in Figure.


Therefore,
ΔL = (slope)F
Multiplying and dividing by volume,
V = AL,
F (slope) = \(\frac{AL}{AL} \Delta L\)
Rearranging, we get
\(\frac{F}{A}=[\frac{L}{A(Slope)}]\frac{\Delta L}{L}\)
Therefore, \(\frac{F}{A} \alpha [\frac{\Delta L}{L}]\)
Comparing with equations stress and strain \(\sigma=\frac{\text { Force }}{\text { Area }}=\frac{F}{A}, \varepsilon=\frac{\text { Change in size }}{\text { Original size }}=\frac{\Delta l}{l}\),
we get volume strain, \(\varepsilon_{v}=\frac{\Delta V}{V}\) equation as
\(\sigma \propto \varepsilon\)
i.e., the stress is proportional to the strain in the elastic limit.

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.