Mechanical Properties of Fluids

√2gh = (2r²g/9η) (ρ_t - ρ)
⇒ h = (2/81) (r? g (ρ_t - ρ)²/η²)
⇒ h ∝ r?
After falling through h, the velocity be equal to terminal velocity.

Range R = v*t = √ (2gh) * √ (2 (H-h)/g) = 2√ (h (H-h).
For R to be max, dR/dh = 0.
h (H-h) must be max. d/dh (Hh-h²)=H-2h=0.
h=H/2 = 12/2=6m.

At terminal speed
Mg = Fv = 6πηRv
⇒ V = mg / 6πηR
V = (4/3)πR³ρg / 6πηR
⇒ V = (2/9) * (ρR²g/η)
= (2/9) * (1000 * (0.2 * 10? ³)² * 10) / (1.8 * 10? )
= 4.94 m/s

v = √2gh velocity of efflux.
F = v ( dm/dt ) = v (aρv) = aρv² = 2aρgh
fr = µR = µAhρg
For just sliding, for = F
µAhρg = 2aρgh
or µ = 2a/A

Thermal stress is developed on heating when expansion of rod is hindered.

Total surface energy before coalesce = $U_i=2*4\pi R^2*\sigma =8\pi R^2\sigma ..........\left(i\right)$ ${}_{}{}^{}{}^{}$

Let new radius becomes r, so according to conservation energy we can write

$2*\frac{4\pi R^3}{3}=\frac{4\pi r^3}{3}\Rightarrow r=2^{\frac{1}{3}}R$

Total surface energy after coalesce = $U_f=4\pi r^2*\sigma =2^{\frac{2}{3}}*4\pi R^2\sigma ..........\left(ii\right)$ ${}_{}{}^{}{}^{\frac{}{}}{}^{}$

$\Rightarrow \frac{U_i}{U_f}=\frac{8\pi R^2\sigma }{2^{\frac{2}{3}}*4\pi R^2\sigma }=2^{\frac{1}{3}}:1$

P_B – P_A = $\rho gh$

$\Rightarrow \left(P_0-\frac{2T}{r^2}\right)-\left(P_0-\frac{2T}{r_1}\right)=\rho gh$

$h=\frac{2T\left(\frac{1}{r_1}-\frac{1}{r_2}\right)}{\rho g}$

$=\frac{2*7.3*10^{-2}*\left(\frac{1}{2.5*10^{-3}}-\frac{1}{4*10^{-3}}\right)}{1000*10}$

$\begin{array}{l}=2.19*10^{-3}m\\ =2.19mm\end{array}$  

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

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

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

$F_b+F_v=mg$

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

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

$v_T\propto r^2$

Bernoulli's equation between (1) & (2)

$\frac{P_1}{\rho g}+\frac{v_1^2}{2g}+z_1=\frac{P_2}{\rho g}+\frac{v_2^2}{zg}+z_2$

$\frac{P_{atm}+mg/A}{\rho g}+0+40+10^{-2}$

$\frac{P_{atm}}{\rho g}+\frac{v_2^2}{2g}+0\left[v_1\approx 0\right]$

$\frac{mg}{A\rho g}+40*10^{-2}=\frac{v_2^2}{2g}$

$\frac{m}{A\rho }+40*10^{-2}=\frac{v_2^2}{2g}$

$\Rightarrow \frac{25}{0.5*10^3}+40*10^2=\frac{V_2^2}{2*10}$

$\Rightarrow V_2^2=\left(5*10^{-2}+40*10^{-2}\right)20$

$V_2^2=45*10^{-2}*20$

$V_2^2=90*10^{-1}$

$V_2=3m/sec$

$=300cm/sec$

mg = pvg

mg = d₁vg           ……. (i)

F_b = weight of fluid displaced = d₂vg         …… (ii)

by NLM 1            F_b + F_v = mg

F_V = mg - F_b

$F_v=mg\left(1-\frac{d_2}{d_1}\right)$