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.
$\therefore p=\frac{2 S}{R}$
$$\begin{aligned} \text{or}\quad R & =\frac{2 S}{p}=\frac{2 \times 7.28 \times 10^{-2}}{2.33 \times 10^3} \\ & =6.25 \times 10^{-5} \mathrm{~m} \end{aligned}$$
(a) Pressure decreases as one ascends the atmosphere. If the density of air is $\rho$, what is the change in pressure $d p$ over a differential height $d h$ ?
(b) Considering the pressure $p$ to be proportional to the density, find the pressure $p$ at a height $h$ if the pressure on the surface of the earth is $p_0$.
(c) If $p_0=1.03 \times 10^5 \mathrm{Nm}^{-2}, \rho_0=1.29 \mathrm{~kg} \mathrm{~m}^{-3}$ and $g=9.8 \mathrm{~ms}^{-2}$, at what height will be pressure drop to $(1 / 10)$ the value at the surface of the earth?
(d) This model of the atmosphere works for relatively small distances. Identify the underlying assumption that limits the model.
(a) Consider a horizontal parcel of air with cross-section A and height dh.
Let the pressure on the top surface and bottom surface be $p$ and $p+d p$. If the parcel is in equilibrium, then the net upward force must be balanced by the weight.
$$\begin{aligned} & \text { i.e., } \\ & (p+d p) A-p A=-p g A d h \quad(\because \text { Weight }=\text { Density } \times \text { Volume } \times g) \\ & =-\rho \times A d h \times g \\ & \Rightarrow \quad d p=-\rho g d h . \quad(\rho=\text { density of air) } \end{aligned}$$
Negative sign shows that pressure decreases with height.
(b) Let $p_0$ be the density of air on the surface of the earth.
As per question, pressure $\propto$ density
$$\begin{aligned} \Rightarrow \quad \frac{p}{\rho_0} & =\frac{\rho}{\rho_0} \\ \Rightarrow \quad \rho & =\frac{\rho_0}{\rho_0} p \\ \therefore \quad d p & =-\frac{\rho_0 g}{P_0} p d h \quad[\because d p=-\rho g d h] \end{aligned}$$
$$\Rightarrow \quad \frac{d p}{p}=-\frac{\rho_0 g}{\rho_0} d h$$
$$\Rightarrow \quad \int_\limits{\rho_0}^p \frac{d p}{p}=-\frac{\rho_o g}{\rho_0} \int_\limits0^h d h \quad\left[\begin{array}{l} \because \text { at } h=0, r=p_0 \\ \text { and } \quad \text { at } h=h, p=p \end{array}\right]$$
$$\Rightarrow \quad \ln \frac{\rho}{p_0}=-\frac{\rho_0 g}{p_0} h$$
By removing log, $$p=p_0 e\left(-\frac{\rho_0 g h}{p_0}\right)$$
$$\begin{aligned} &\text { (c) As } p=p_0 e^{-\frac{p_0 g h}{p_o}} \text {, }\\ &\Rightarrow \quad \ln \frac{P}{P_0}=-\frac{\rho_0 g h}{P_0} \end{aligned}$$
By question,
$$p=\frac{1}{10} p_0$$
$$ \begin{array}{ll} \Rightarrow & \ln \left(\frac{\frac{1}{10} P_0}{P_0}\right)=-\frac{\rho_0 g}{P_0} h \\ \Rightarrow & \ln \frac{1}{10}=-\frac{\rho_o g}{P_0} h \rho_{\circ} \end{array}$$
$$\begin{aligned} \therefore \quad h & =-\frac{p_0}{\rho_o g} \ln \frac{1}{10}=-\frac{p_0}{p_0 g} \ln (10)^{-1}=\frac{p_0}{p_0 g} \ln 10 \\ & =\frac{p_0}{\rho_o g} \times 2.303 \quad \left[\because \ln (x)=2.303 \log _{10}(\mathrm{x})\right]\\ & =\frac{1.013 \times 10^5}{1.22 \times 9.8} \times 2.303=0.16 \times 10^5 \mathrm{~m} \\ & =16 \times 10^3 \mathrm{~m} \end{aligned}$$
(d) We know that $\quad \rho \propto \rho \quad$ (when $T=$ constant i.e., isothermal pressure) Temperature $(T)$ remains constant only near the surface of the earth, not at greater heights.