Physics Mechanical Properties of Fluids

In this case,

$F=\eta \frac{v}{h}\mathcal{l}\left(\mathcal{l}-x\right)$            

$\therefore v=\frac{Fh}{\eta \mathcal{l}\left(\mathcal{l}-x\right)}=\frac{dx}{dt}$            

So,     $t=\frac{3\eta {\mathcal{l}}^3}{8Fh}=5sec$

  $\tau ={\displaystyle \int F.r={\displaystyle \underset{0}{\overset{R}{\int }}\eta \left(2\pi rdr\right)\left(\frac{rw}{t}\right).r}}$  

  $=\frac{2\pi \eta W{R}^4}{4t}=\frac{2*3.14*1*8*10^{-4}}{4*2*10^{-3}}$          f

  $=0.625N-m?0.63N-m$            

Let $2x$ $$ length of rod is immersed in water.

$$ ${\tau }_{\text{Hinge}}($  net $)=0$ $$

$\Rightarrow \mathrm{}\mathrm{m}\mathrm{g}?\frac{b}{2}\mathrm{s}\mathrm{i}\mathrm{n}?\theta -F_B(b-x)\mathrm{s}\mathrm{i}\mathrm{n}?\theta =0$

$\Rightarrow \mathrm{}mgb=2(b-x)F_B$

$$ $\therefore \mathrm{}{F}_B=\frac{mgb}{2(b-x)}=\frac{\left(Abd\right)b}{2(b-x)}g$  , where $d=$ $$ density of material of rod

${}_{}$  $F_B=$ Buoyant force $\mathrm{}\frac{}{}\frac{}{}$

Equate $F_B,\frac{A{b}^{2}d}{2(b-x)}=2Ax\rho$ ${}_{}\frac{{}^{}}{}$

$\Rightarrow \mathrm{}{b}^2=4x(b-x)\frac{9}{5}$

$\mathrm{}{}^{}\frac{{}^{}}{}\frac{}{}$  $\Rightarrow \mathrm{}{x}^2-bx+\frac{5b^2}{36}=0\Rightarrow x=\frac{b}{6}\therefore$ immersed $=\frac{b}{3}$

Fact. Due to decrease in intermolecular forces

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

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