Physics Mechanical Properties of Fluids

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

$Now,Q=ms\Delta \theta$

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

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

Thermal strain $=$ Longitudinal strain $=\alpha \Delta T$

$\Rightarrow$ Longitudinal strain, $\delta =10^{-5}*10^2=10^{-3}$

$\Rightarrow$ Compressive stress $=\delta *$  Young's Modulus

$=10^{-3}*0.5*10^{11}$

$=0.5*10^8$

$\Rightarrow$ Compressive force $=0.5*10^8*10^{-3}=0.5*10^5$

$=5*10^4*\frac{10}{10}$

$=50*10^3 N$

In the case for maximum elongation,

Stress = Elastic limit

${\delta }_{max}=\frac{{\sigma }_{elastic }*L}{ Young\text{'}s modulus }=\frac{8*10^8*1}{2*10^{11}}=4*10^{-3}$

$=4 mm$

i.e. maximum elongation is 4 mm

Excess force  $=T*2\pi R$

$=\frac{7}{100}*2*3.14*\frac{4.5}{100}$

$=197.82*10^{-4}$

$=19.8*10^{-3} N$

$=19.8mN$

 $\frac{4}{3}\pi R^3=729\frac{4}{3}\pi r^3$

R³ = 729r³

$R=\left(729\right)\frac{1}{3}\left(r\right)\left(\frac{1}{3}\right)$

R = 9r

$\Delta u=T\left(4\pi r^2\right)*729-T*4\pi R^2$

$=T*4\pi *8R^2$

$\begin{array}{l}=75.39*10^{-5}J\\ \Delta u=7.5*10^{-4}J\end{array}$

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

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