Consider the diagram. Let the density of water be $\rho_w$ and a cubical block of ice of side $L$ be floating in water with $x$ of its height $(L)$ submerged in water.

$$\begin{aligned} \text { Volume of the block }(V) & =L^3 \\ \text { Mass of the block }(m) & =V \rho=L^3 \rho \\ \text { Weight of the block } & =m g=L^3 \rho g \end{aligned}$$

1st case

Volume of the water displaced by the submerged part of the block $=x L^2$

$\therefore$ Weight of the water displaced by the block

In floating condition, $x L^2 \rho_w g$

Weight of the block = Weight of the water displaced by the block

$$\begin{gathered} L^3 \rho g=x L^2 \rho_w g \\ \text{or}\quad\frac{x}{L}=\frac{\rho}{\rho_w}=x \end{gathered}$$

2nd case

When elevator is accelerating upward with an acceleration $a$, then effective acceleration

Then, weight of the block

$$\begin{aligned} & =(g+a) \quad(\because \text { Pseudo force is downward }) \\ & =m(g+a) \\ & =L^3 \rho(g+a) \end{aligned}$$

Let $x_1$ fraction be submerged in water when elevator is accelerating upwards.

Now, in the floating condition, weight of the block = weight of the displaced water

$$\begin{aligned} L^3 \rho(g+a) & =\left(x_1 L^2\right) \rho_w(g+a) \\ \text{or}\quad\frac{x_1}{L} & =\frac{\rho}{\rho_w}=x \end{aligned}$$

From 1st and 2nd case,

We see that, the fraction of the block submerged in water is independent of the acceleration of the elevator.

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.

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

$$\begin{array}{lrl} \Rightarrow & g \sin \theta & =a \cos \theta \\ \Rightarrow & a & =g \tan \theta \\ \Rightarrow & \tan \theta & =\frac{a}{g}=\text { slope } \end{array}$$

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.

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]

$$T \times 4 \pi\left(R^2-N r^2\right)=\left(\frac{4}{3} \times R^3 \times \rho\right) s \Delta \theta \quad\left[\therefore m=v \rho=\frac{4}{3} \pi R^3 \rho\right]$$

$$\begin{aligned} \Rightarrow \quad \Delta \theta & =\frac{T \times 4 \pi\left(R^2-N r^2\right)}{\frac{4}{3} \pi R^3 \rho \times s} \\ & =\frac{3 T}{\rho s}\left[\frac{R^2}{R^3}-\frac{N r^2}{R^3}\right] \\ & =\frac{3 T}{\rho s}\left[\frac{1}{R}-\frac{\left(R^3 / r^3\right) \times r^2}{R^3}\right] \\ & =\frac{3 T}{\rho s}\left[\frac{1}{R}-\frac{1}{r}\right] \end{aligned}$$