The sap in trees, which consists mainly of water in summer, rises in a system of capillaries of radius $r=2.5 \times 10^{-5} \mathrm{~m}$. The surface tension of sap is $T=7.28 \times 10^{-2} \mathrm{Nm}^{-1}$ and the angle of contact is $0^{\circ}$. Does surface tension alone account for the supply of water to the top of all trees?
Given, radius $(r)=2.5 \times 10^{-5} \mathrm{~m}$
Surface tension $(S)=7.28 \times 10^{-2} \mathrm{~N} / \mathrm{m}$
Angle of contact $(\theta)=0^{\circ}$
The maximum height to which sap can rise in trees through capillarity action is given by
$$\begin{aligned} h= & \frac{2 S \cos \theta}{r \rho g} \text { where } S=\text { Surface tension, } \rho=\text { Density, } r=\text { Radius } \\ & =\frac{2 \times 7.28 \times 10^{-2} \times \cos 0^{\circ}}{2.5 \times 10^{-5} \times 1 \times 10^{-3} \times 9.8}=0.6 \mathrm{~m} \end{aligned}$$
This is the maximum height to which the sap can rise due to surface tension. Since, many trees have heights much more than this, capillary action alone cannot account for the rise of water in all trees.
The free surface of oil in a tanker, at rest, is horizontal. If the tanker starts accelerating the free surface will be titled by an angle $\theta$. If the acceleration is $a \mathrm{~ms}^{-2}$, what will be the slope of the free surface?
Consider the diagram where a tanker is accelerating with acceleration a.
Consider an elementary particle of the fluid of mass $d m$. The acting forces on the particle with respect to the tanker are shown above . Now, balancing forces (as the particle is in equilibrium) along the inclined direction component of weight $=$ component of pseudo force $d m g \sin \theta=d m a \cos \theta$ (we have assumed that the surface is inclined at an angle $\theta$ ) where, dma is pseudo force
$$\begin{array}{lrl} \Rightarrow & g \sin \theta & =a \cos \theta \\ \Rightarrow & a & =g \tan \theta \\ \Rightarrow & \tan \theta & =\frac{a}{g}=\text { slope } \end{array}$$
Two mercury droplets of radii 0.1 cm and 0.2 cm collapse into one single drop. What amount of energy is released? The surface tension of mercury $T=435.5 \times 10^{-3} \mathrm{Nm}^{-1}$.
Consider the diagram.
Radii of mercury droplets
$$\begin{aligned} & r_1=0.1 \mathrm{~cm}=1 \times 10^{-3} \mathrm{~m} \\ & r_2=0.2 \mathrm{~cm}=2 \times 10^{-3} \mathrm{~m} \end{aligned}$$
Surface tension $(T)=435.5 \times 10^{-3} \mathrm{~N} / \mathrm{m}$
Let the radius of the big drop formed by collapsing be $R$.
$\therefore$ Volume of big drop $=$ Volume of small droplets
$$\frac{4}{3} \pi R^3=\frac{4}{3} \pi r_1^3+\frac{4}{3} \pi r_2^2$$
$$\begin{aligned} \text{or}\quad R^3 & =r_1^3+r_2^3 \\ & =(0.1)^3+(0.2)^3 \\ & =0.001+0.008 \\ & =0.009 \end{aligned}$$
or $$\quad R=0.21 \mathrm{~cm}=2.1 \times 10^{-3} \mathrm{~m}$$
$$\begin{aligned} &\therefore \text { Change in surface area }\\ &\begin{aligned} \Delta A & =4 \pi R^2-\left(4 \pi r_1^2+4 \pi r_2^2\right) \\ & =4 \pi\left[R^2-\left(r_1^2+r_2^2\right)\right] \end{aligned} \end{aligned}$$
$\therefore \quad$ Energy released $=T \cdot \Delta A \quad$ (where $T$ is surface tension of mercury)
$$\begin{aligned} = & T \times 4 \pi\left[R^2-\left(r_1^2+r_2^2\right)\right] \\ = & 435.5 \times 10^{-3} \times 4 \times 3.14\left[\left(2.1 \times 10^{-3}\right)^2\right. \\ & \left.\quad-\left(1 \times 10^{-6}+4 \times 10^{-6}\right)\right] \\ = & 435.5 \times 4 \times 3.14[4.41-5] \times 10^{-6} \times 10^{-3} \\ = & -32.23 \times 10^{-7} \quad \text { (Negative sign shows absorption) } \end{aligned}$$
Therefore, $3.22 \times 10^{-6} \mathrm{~J}$ energy will be absorbed.
If a drop of liquid breaks into smaller droplets, it results in lowering of temperature of the droplets. Let a drop of radius $R$, break into $N$ small droplets each of radius $r$. Estimate the drop in temperature.
When a big drop of radius R, breaks into N droplets each of radius r, the volume remains constant.
$$\begin{aligned} \therefore \quad &\text { Volume of big drop }=N \times \text { Volume of each small drop }\\ &\begin{aligned} \frac{4}{3} \pi R^3 & =N \times \frac{4}{3} \pi r^3 \\ \text{or}\quad R^3 & =N r^3 \\ \text{or}\quad N & =\frac{R^3}{r^3} \end{aligned} \end{aligned}$$
$$\begin{aligned} \text { Now, } \quad \quad \text { change in surface area } & =4 \pi R^2-N 4 \pi r^2 \\ & =4 \pi\left(R^2-N r^2\right) \\ \text { Energy released }=T \times \Delta A & =S \times 4 \pi\left(R^2-N r^2\right) \quad \text{[T = Surface tension]} \end{aligned}$$
Due to releasing of this energy, the temperature is lowered. If $\rho$ is the density and $s$ is specific heat of liquid and its temperature is lowered by $\Delta \theta$, then energy released $=m s \Delta \theta$ [ $s=$ specific heat $\Delta \theta=$ change in temperature]
$$T \times 4 \pi\left(R^2-N r^2\right)=\left(\frac{4}{3} \times R^3 \times \rho\right) s \Delta \theta \quad\left[\therefore m=v \rho=\frac{4}{3} \pi R^3 \rho\right]$$
$$\begin{aligned} \Rightarrow \quad \Delta \theta & =\frac{T \times 4 \pi\left(R^2-N r^2\right)}{\frac{4}{3} \pi R^3 \rho \times s} \\ & =\frac{3 T}{\rho s}\left[\frac{R^2}{R^3}-\frac{N r^2}{R^3}\right] \\ & =\frac{3 T}{\rho s}\left[\frac{1}{R}-\frac{\left(R^3 / r^3\right) \times r^2}{R^3}\right] \\ & =\frac{3 T}{\rho s}\left[\frac{1}{R}-\frac{1}{r}\right] \end{aligned}$$
The surface tension and vapour pressure of water at $20^{\circ} \mathrm{C}$ is $7.28 \times 10^{-2} \mathrm{Nm}^{-1}$ and $2.33 \times 10^3 \mathrm{~Pa}$, respectively. What is the radius of the smallest spherical water droplet which can form without evaporating at $20^{\circ} \mathrm{C}$ ?
Given, surface tension of water
$$(S)=7.28 \times 10^{-2} \mathrm{~N} / \mathrm{m}$$
Vapour pressure $(p)=2.33 \times 10^3 \mathrm{~Pa}$
The drop will evaporate, if the water pressure is greater than the vapour pressure. Let a water droplet or radius $R$ can be formed without evaporating.
Vapour pressure $=$ Excess pressure in drop.
$\therefore p=\frac{2 S}{R}$
$$\begin{aligned} \text{or}\quad R & =\frac{2 S}{p}=\frac{2 \times 7.28 \times 10^{-2}}{2.33 \times 10^3} \\ & =6.25 \times 10^{-5} \mathrm{~m} \end{aligned}$$