A vessel filled with water is kept on a weighing pan and the scale adjusted to zero. A block of mass $M$ and density $\rho$ is suspended by a massless spring of spring constant $k$. This block is submerged inside into the water in the vessel. What is the reading of the scale?
Consider the diagram,
The scale is adjusted to zero, therefore, when the block suspended to a spring is immersed in water, then the reading of the scale will be equal to the thrust on the block due to water.
$$\begin{aligned} \text { Thrust } & =\text { weight of water displaced } \\ & =V_\rho g \text { (where } V \text { is volume of the block and } \rho_w \text { is density of water) } \\ & =\frac{m}{\rho} \rho_w g=\left(\frac{\rho_w}{\rho}\right) m g \end{aligned}$$
$$\left(\because \text { Density of the block } \rho=\frac{\text { mass }}{\text { volume }}=\frac{m}{V}\right)$$
A cubical block of density $\rho$ is floating on the surface of water. Out of its height $L$, fraction $x$ is submerged in water. The vessel is in an elevator accelerating upward with acceleration $a$. What is the fraction immersed?
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.
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.