Mechanical Properties of Fluids

F=TL

$$\begin{gathered} \mathrm{L}=\frac{\mathrm{F}}{\mathrm{T}}=\frac{75\times {10}^{-4}}{6\times {10}^{-2}} \\ =12.5\times {10}^{-2}\mathrm{m} \end{gathered}$$

The Bernoulli's equation can be considered to be a statement of the conservation of energy principle appropriate for flowing fluids. It is one of the most important/useful in fluid mechanics. It puts into a relation pressure and velocity in an inviscid ( negligible viscosity ) incompressible flow.

The general energy equation is simplified to

     p₁ + $\frac{1}{2}\mathrm{p} {\mathrm{v}}_1^{ 2}$ + ρgh₁ = p₂ +  $\frac{1}{2}{\mathrm{\rho v}}_2^2$ + ρgh₂

A scent sprayer is an example of fall in pressure due to increase in velocity. Also as per Bernoulli's theorem that datum head, pressure head and velocity head of a flowing liquid is constant. Therefore scent sprayer is based on Bernoulli's theorem.

Radius of one drop of mercury is R

∴  The volume of one drop = $\frac{4}{3}\mathrm{\pi } {\mathrm{R}}^3$

∴  Total volume of the two drops,

  V =  2 × $\frac{4}{3}\mathrm{\pi } {\mathrm{R}}^3$

   V = $\frac{8}{3} {\mathrm{\pi R}}^3$

Let the radius of the large drop formed be R'

The total volume of the large is also V

    $\frac{4}{3}\mathrm{\pi } \mathrm{R}{\text{'}}^2 = \frac{8}{3}\mathrm{\pi } {\mathrm{R}}^3$

     R'³ = 2R³

⇒   R' = 2^1/3 R

Now the surface area of the resultant drop is 

   S₁ = 2 × 4$\mathrm{\pi }$R²

   S₁ = 8$\mathrm{\pi }$R² 

and the surface area of the resultant drop is 

    S₂ = 4$\mathrm{\pi }$R'² 

    S₂ = 4$\mathrm{\pi }$ 2^2/3 R²

Let T be the surface tension of the mercury. Therefore the surface energy of the two drops before coalescing is

     U₁ = S₁ T

     U₁ = 8$\mathrm{\pi }$R²T

and the surface energy after coalescing,

    U₂ = S₂T 

     U₂ = 2^2/3 × 4$\mathrm{\pi }$R²T

∴   $\frac{{\mathrm{U}}_1}{{\mathrm{U}}_2} = \frac{8\mathrm{\pi } {\mathrm{R}}^2\mathrm{T}}{2^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}} \times 4\mathrm{\pi } {\mathrm{R}}^2\mathrm{T}}$

           = $\frac{2}{2^{{\displaystyle \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}}$

    $\frac{{\mathrm{U}}_1}{{\mathrm{U}}_2}$ = 2^1/3

When impure water or kerosene is taken in a glass vessel, it is found that the surface near the walls is curved concave upwards.

Consider a molecule of water M on the free surface close to the wall of the glass vessel. The magnitude of  adhesive force A is greater than the magnitude of cohesive force C and resultant of the two molecular forces of attraction R is or outside the liquid. Hence the molecule A is attracted towards the walls of glass vessel. The free surface of water adjust itself at right angle to the resultant R. Therefore molecules  like M creep upwards and the angle of contact is acute.

Pressure and stress both have the dimensions Force/area. Strain and angle are both dimensionless. Energy and work have the dimension force × distance.

Tension and surface tension refer to two different physical quantites and their dimensions are different. Tension is a force and surface tension is force per unit length.

The beautiful colours are seen on account of interference of light reflected from the upper and the lower surfaces of the thin film. As conditions for constructive and destructive interference depend upon the wavelength of light, therefore coloured interference fringes are  observed.

The property utilized in the manufacture of lead shots is surface tension of liquid lead. In this process, molten lead is made to pass through a sieve from a high tower and allowed to fall in water. The molten lead particles, while descending, assume a spherical shape and solidify in this form, before falling into water.

$$\begin{gathered} \mathrm{P}\propto \frac{1}{\mathrm{r}} \\ \mathrm{hence} ,\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}=\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${\mathrm{r}}_1$}\right.}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{${\mathrm{r}}_2$}\right.}}=\frac{{\mathrm{r}}_2}{{\mathrm{r}}_1}=\frac{1}{2} ...\mathrm{since} {\mathrm{r}}_1:{\mathrm{r}}_2=2:1 \\ {\mathrm{P}}_1:{\mathrm{P}}_2=1:2 \end{gathered}$$

Given:- weight =6.24$\times$10^-4N,

             W=T.$2\mathrm{\pi r}$

         r=$\frac{\mathrm{W}}{2\mathrm{\pi T}}$=$\frac{6.28\times {10}^{-4}}{2\times 3.14\times 5\times {10}^{-2}}$

          r = 2$\times {10}^{-3}$m