Physics Mechanical Properties of Fluids

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

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

  $p_1-p_2=\frac{4T}{r_1}.......\left(1\right)$

$p_2-p_0=\frac{4T}{r_2}.......\left(2\right)$

(1) + (2), p₁ – p₀ = 4T $\left(\frac{1}{r_1}+\frac{1}{r_2}\right)$

$\Rightarrow \frac{4T}{r_{eq}}=4T\left(\frac{1}{r_1}+\frac{1}{r_2}\right)$

$r_{eg}=\frac{r_1{r}_2}{r_1+r_2}=\frac{3*6}{3+6}=2cm$

P_B – P_A = $\rho gh$

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

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

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

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

  $\rho =10^{3}kg/m^3$

g = 10 m/s²

Final height in both vessels

$=\frac{100+150}{2}=125cm$

So, less in U =    ${\displaystyle \int A*0.25*g*0.25}$

= $10^3*625*10^{-4}*10*16*10^{-4}$  

= 625 * 16 * 10^-4

= 1J.

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

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

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

$F_b+F_v=mg$

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

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

$v_T\propto r^2$