Is viscosity a vector?
Viscosity is a property of liquid it does not have any direction, hence it is a scalar quantity.
Is surface tension a vector?
No, surface tension is a scalar quantity.
Surface tension $=\frac{\text { Work done }}{\text { Surface area }}$, where work done and surface area both are scalar quantities.
Iceberg floats in water with part of it submerged. What is the fraction of the volume of iceberg submerged, if the density of ice is $\rho_i=0.917 \mathrm{~g} \mathrm{~cm}^{-3} ?$
Given, density of ice $\left(\rho_{\text {ice }}\right)=0.917 \mathrm{~g} / \mathrm{cm}^3$
Density of water $\left(\rho_w\right)=1 \mathrm{~g} / \mathrm{cm}^3$
Let $V$ be the total volume of the iceberg and $V^{\prime}$ of its volume be submerged in water.
In floating condition.
Weight of the iceberg = Weight of the water displaced by the submerged part by ice
$$V_{\rho_{\text {ice }}} g=V^{\prime} \rho_w g$$
or $$\frac{V^{\prime}}{V}=\frac{\rho_{c e}}{\rho_w}=\frac{0.917}{1}=0.917 \quad(\because \text { Weight }=m g=v \rho g)$$
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.