Physics Mechanical Properties of Fluids

Ratio of masses on two pistons of the hydraulic lift equals to that of their cross- section area.

$\frac{A_1}{A_2}=\frac{100}{m}$            

Now,   $\frac{4^2{A}_1}{A_2/4^2}=\frac{M}{m}\Rightarrow M=256\frac{A_1}{A_2}.m=25600kg.$  

$\tau =\eta \frac{v}{h}$  

              Given

              $\tau =10^{-3}N/m^2$  

              (shear stress)

              h =?

              v = 36 km/hr = 10 m/sec

              $10^{-3}=\frac{10^{-2}*10}{h}$

h = 100 m

Let, Rain drop moving with terminal velocity v_t in the air F_b (Bnoyancy force) = $\frac{4}{3}\pi r^3{\rho }_{air}g$  

$mg=\frac{4}{3}\pi r^3{\rho }_{air}g$

$F_v=\left(visconsforce\right)=6\pi \eta r{v}_T$

$F_b+F_v=mg$

$\Rightarrow \frac{4}{3}\pi r^3{\rho }_{air}g+6\pi \eta r{v}_T=\frac{4}{3}\pi r^3\rho g$

$v_T=\frac{2}{9}\frac{r^2}{\eta }\left(\rho -{\rho }_{air}\right)$

$v_T\propto r^2$

Deformation energy per unit volume,
U = ∫dU = ∫ (p²/2B)Ady
= ∫ (ρ²g²y/2B)Ady
U = (ρ²g²h³/2B) (A/3)
= (ρ²g²h²/6B) (volume)
= 1.5 μJ

Factual (theory based)

Surface energy of bubble $=2*$ charge in surface area x surface tension

$\begin{array}{rr}& =8?{\mathrm{R}}^2*\mathrm{T}\\ & =8*3.142*4*{10}^{-4}*3*{10}^{-2}\mathrm{J}\\ & =30.1*{10}^{-5}\mathrm{J}\end{array}$

Mass $=\mathrm{M}$

Density of ball $=\mathrm{d}$

Density of glycerine $=\frac{\mathrm{d}}{2}$

${\mathrm{F}}_{\mathrm{B}}={\mathrm{V}}_{\mathrm{s}}{\rho }_{\mathcal{l}}\mathrm{g}=\mathrm{V}\frac{\mathrm{d}}{2}\mathrm{g}$

${\mathrm{F}}_{\mathrm{g}}=\mathrm{M}\mathrm{g}=\mathrm{v}\mathrm{d}\mathrm{g}$

For constant velocity, ${\mathrm{F}}_{\text{net}}=0$

$\therefore {\mathrm{F}}_{\mathrm{B}}+{\mathrm{F}}_{\mathrm{v}}=\mathrm{M}\mathrm{g}$

${\mathrm{F}}_{\mathrm{V}}=\mathrm{M}\mathrm{g}-{\mathrm{F}}_{\mathrm{B}}=\mathrm{V}\mathrm{d}\mathrm{g}-\frac{\mathrm{V}\mathrm{d}\mathrm{g}}{2}=\frac{\mathrm{V}\mathrm{d}\mathrm{g}}{2}=\frac{\mathrm{M}\mathrm{g}}{2}$

$\mathrm{P}={\mathrm{P}}_0+\frac{4\mathrm{T}}{\mathrm{R}}$

on expansion $\mathrm{R}$ increase, as a result, $\mathrm{P}$  decreases.

Initially speed is zero, then increases & after some time it becomes constant.

$\eta =1-\frac{T_{\text{sin}k}}{T_{\text{source}}}=0.5$

$\begin{array}{rr}& \frac{{\mathrm{T}}_{\text{sin}}}{{\mathrm{T}}_{\text{source}}}=0.5\\ & \Rightarrow {\mathrm{T}}_{\text{sin}}=\frac{1}{2}*(273+327)\\ & =\frac{1}{2}*600\\ & =300\mathrm{K}\\ & ={27}^{?}\mathrm{C}\end{array}$