Physics Mechanical Properties of Fluids

$\rho ={\rho }_0\left(1-\frac{r^2}{R^2}\right)\mathrm{}0<r\le R$

$\begin{array}{rr}& \mathrm{m}\mathrm{g}=\mathrm{B}\\ & \int \rho \left(4\pi r^{2}dr\right)={\rho }_{\mathrm{L}}\frac{4}{3}\pi {\mathrm{R}}^3;\mathrm{}{\int }_0^{\mathrm{R}}??{\rho }_0\left(1-\frac{{\mathrm{r}}^2}{{\mathrm{R}}^2}\right)4\pi r^{2}dr={\rho }_{\mathrm{L}}\frac{4}{3}\pi {\mathrm{R}}^3\\ & {\int }_0^{\mathrm{R}}??{\rho }_{0}4\pi \left({\mathrm{r}}^2-\frac{{\mathrm{r}}^4}{{\mathrm{R}}^2}\right)\mathrm{d}\mathrm{r}={\rho }_{0}4\pi {\left(\frac{{\mathrm{r}}^3}{3}-\frac{{\mathrm{r}}^5}{5{\mathrm{R}}^2}\right)}_0^{\mathrm{R}}={\rho }_{\mathrm{L}}\frac{4}{3}\pi {\mathrm{R}}^3;\mathrm{}\frac{2}{5}{\rho }_0={\rho }_{\mathrm{L}}\end{array}$

Using equation of continuity

$\begin{array}{rr}& A_1{V}_1=A_2{V}_2\\ & \frac{V_1}{V_2}=\frac{A_2}{A_1}={\left(\frac{d_2}{d_1}\right)}^2={\left(\frac{4.8}{6.4}\right)}^2=\frac{9}{16}\end{array}$

Kindly consider the following Image

As we know that Reynolds's number R = ρvD/η
In First case: v? = (0.18*10? ³)/ (π (0.5*10? ²)²*60) = 0.03822 m/s
R = (10³ * 0.03822 * 0.01)/10? ³ = 382.2 < 2000 (Laminar/Steady)
In Second case: v? = (0.48*10? ³)/ (π (0.5*10? ²)²*60) = 0.10191 m/s
R? = (10³ * 0.10191 * 0.01)/10? ³ = 1019.1 < 2000 (Laminar/Steady)
The provided solution has a different calculation for R, leading to a different conclusion.

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

Pressure outside is 0.
Here, Pin = 4T/r
By, P? V? + P? V? = PV (isothermal process, Boyle's law applied to the mass of gas inside)
(4T/r? ) (4/3 πr? ³) + (4T/r? ) (4/3 πr? ³) = (4T/r) (4/3 πr³)
r? ² + r? ² = r²
r = √r? ² + r? ²

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

From volume conservation

$n*\frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3\Rightarrow R^3=n{r}^3.........\left(1\right)$

Decrease in surface area = $n*4\pi r^2-4\pi R^2$

$\left(\Delta A\right)=4\pi \left[n{r}^2-R^2\right]=4\pi \left[\frac{n*r^3}{r}-R^2\right]=4\pi \left[\frac{R^3}{r}-R^2\right]=4\pi R^3\left[\frac{1}{r}-\frac{1}{R}\right]$

Energy released (W) = $T*\Delta A=4\pi R^{3}T\left[\frac{1}{r}-\frac{1}{R}\right]$

Heat produced (Q) = $\frac{W}{J}=\frac{4\pi T{R}^3}{J}\left[\frac{1}{r}-\frac{1}{R}\right]$

$Now,Q=ms\Delta \theta$

$\Rightarrow \frac{4\pi R^{3}T}{J}\left[\frac{1}{r}-\frac{1}{R}\right]=\left(\frac{4}{3}\pi R^3\right)*\left|x\right|*\Delta \theta =\frac{3T}{J}\left[\frac{1}{r}-\frac{1}{R}\right]$