Archimedes principle states that when a body is partially or wholly immersed in a fluid, it experiences an upward thrust equal to the weight of the fluid displaced by it and its up thrust acts through the centre of gravity of the liquid displaced.

Poisson's ratio is defined as the ratio of relative contraction (lateral strain) to relative expansion (longitudinal strain).

Capillary descent, cos140 = cos(90+50)–sin50 = –0.7660
\(h=\frac { 2Tcos\theta }{ r\rho g } =\frac { 2\times ({ 0.465N \ m }^{ -1 })({ cos \ 140 }^{ 0 }) }{ \left( 2\times { 10 }^{ -3 }m \right) \left( 13.6\times { 10 }^{ 3 } \right) \left( { 9.8 }ms^{ -2 } \right) } \)
\(=\frac{2 \times 0.456 \times(-0.7660)}{2 \times 13.6 \times 9.8} \)
\(=\frac{-0.6986}{266.56}=-2.62 \times 10^{-3} \mathrm{~m}\)
where, negative sign indicates that there is fall of mercury (mercury is depressed) in glass tube.

From equation (7.34), we have 
h∝\(\frac { 1 }{ r } \Rightarrow hr=\)constant
Consider two capillary tubes with radius r₁ and r₂ which on placing in a liquid, capillary rises to height h₁ and h₂, respectively. Then,
h₁r₁ = h₂r₂ = constant
\(\Rightarrow { h }_{ 2 }=\frac { { h }_{ 1 }{ r }_{ 1 } }{ { r }_{ 2 } } =\frac { \left( 2\times { 10 }^{ -2 }m \right) }{ \frac { r }{ 3 } } \Rightarrow { h }_{ 2 }=6{ \times 10 }^{ 2 }m\)

A soap bubble has two free surfaces,
therefore increase in surface area \(\Delta\)A = A₂ − A₁ = 2(100 - 50)\(\times\)10^-4m² = 100\(\times\)10^-4m².
Since, work done W = T\(\times\) \(\Delta\)A \(\Rightarrow\) T \(=\frac { W }{ \Delta A }\)
\( =\frac { 2.4\times 10^{ -4 }J }{ 100\times { 10 }^{ -4 }{ m }^{ 2 } } =2.4\times { 10 }^{ -2 }Nm^{ -1 }\)

We know that \(\frac{F}{A}\) = Y\(\times \frac{\Delta L}{L}\)
\(\frac{\Delta{L}}{L}\)=\(\left( \frac { F }{ A } \right) \left( \frac { L }{ Y } \right) \)
\(=\left( \frac { 50 }{ 1.25\times { 10 }^{ -4 } } \right) \left( \frac { 10 }{ 4\times { 10 }^{ 10 } } \right) ={ 10 }^{ -4 }m\)

\(\text { Strain }=\frac{\text { Change in length }}{\text { Original length }}\)
\(=\frac{2 l-l}{l}=\frac{l}{l}=1\)

\(\text { Stress, } \ \frac{F}{A}=Y \frac{\Delta L}{L}\)
\(\text { Stress } \mathrm{S}=\frac{\text { Force }}{\pi r^{2}}\)
\(S_{1} \propto \frac{1}{r_{1}^{2}} \ S_{2} \propto \frac{1}{r_{2}^{2}} \ F \text { is same }\)
\(\frac{S_{1}}{S_{2}}=\frac{r_{2}^{2}}{r_{1}^{2}} \quad r_{1}=3 r_{2}\)
\(\text { Let } S_{1}=S_{x} \text { and } S_{2}=S_{y}\)
\( \frac{S_{x}}{S_{y}}=\frac{r_{2}^{2}}{\left(3 r_{2}\right)^{2}}=\frac{r_{2}^{2}}{9 r_{2}^{2}}=\frac{1}{9} \)
\( \frac{S_{x}}{S_{y}}=\frac{1}{9} \)
\(\therefore S_{y}=9 S_{x} \)

It states that when a body is partially or wholly immersed in a fluid, it experiences an upward thrust equal to the weight of the fluid displaced by it and its upthrust acts through the centre of gravity of the liquid displaced.
upthrust or buoyant force = weight of liquid displaced
Proof:
Consider a body of height h lying inside a liquid of density ρ at a depth x below the free surface of the liquid.

Let the area of cross section be a.
The forces on the sides of the body cancel out. Pressure at the upper face of the body
\(\mathrm{P}_{1}=x \rho g\)
Pressure at the lower surface of the body
\(\mathrm{P}_{2}=(x+h) \rho g\)
Thrust acting on the upper face of the body is
\(\mathrm{F}_{1}=\mathrm{P}_{1} \mathrm{a}=x \rho g a\)
acting vertically downwards.
Thrust acting on the lower face of the body is
\(\mathrm{F}_{2}=\mathrm{P}_{2} \mathrm{a}=(x+h) \rho g a\)
acting vertically upwards.
The resultant force \(\left(\mathrm{F}_{2}-\mathrm{F}_{1}\right)\) is acting on the body is the upward direction and is called upthrust (U)
\(\therefore \mathrm{U} =\mathrm{F}_{2}-\mathrm{F}_{1}=(x+h) \rho g a-x \rho g a \)
\(=\mathrm{a} \rho g h  \)
\(\text {But ah } =\mathrm{V}=\text { Volume of the body }  \)
\(=\text {Volume of the displaced liquid }  \)
\(\mathrm{U} =\mathrm{V} \rho \mathrm{g}=\mathrm{Mg}\)
\(\because M=V \rho=\) mass of liquid displaced i.e., upthrust or buoyant force = Weight of liquid displaced.
This proves the Archimedes principle.

(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, \(F_{P_{2}}=F_{T}+F_{P_{1}}\)
\(\mathrm{P}_{2} \pi \mathrm{R}^{2} =2 \pi \mathrm{RT}+\mathrm{P}_{1} \pi \mathrm{R}^{2}  \)
\(\Rightarrow\left(\mathrm{P}_{2}-\mathrm{P}_{1}\right) \pi \mathrm{R}^{2}=2 \pi \mathrm{RT} \)
\(\text {Excess pressure is } \Delta \mathrm{P} =\mathrm{P}_{2}-\mathrm{P}_{1}=\frac{2 T}{R}\)