A narrow pipe with a cross section area ‘A’ and length ‘L’ is half filled with liquid of density $\rho$ and is rotating with a constant angular velocity '$\omega$’ about vertical axis passing through its open end as shown in figure. A small hole of area ‘a’ (a << A) is made at its closed end at t = 0. The time in which entire liquid will come out of the pipe will be

 

$\text{ Pressure at }1,2\text{ and }3{\text{ are P}}_0,{\text{P and P}}_0 \text{respectively.}$

Consider an element of width dx located at distance x from axis. This element is in circular motion.

$\begin{array}{l}\left(P+dP\right)A-\left(P\right)A=dm{\omega }^{2}x\text{}\\ \Rightarrow \left(dP\right)A=\left(\rho Adx\right){\omega }^{2}x\text{}\\ \Rightarrow {\int }_{P_0}^{P}dP=\rho {\omega }^2{\int }_r^{L}xdx\end{array}$

$\begin{array}{l}\Rightarrow \text{P}-{\text{P}}_0=\frac{\text{ρ}}{2}\left({\text{L}}^2-{\text{r}}^2\right){\text{ω}}^2\\ \text{Since, a}<<\text{A} \\ \Rightarrow v_1<<\text{v}\end{array}$

Bernoulli’s equation for 2 and 3

$\begin{array}{l}\mathrm{P}+\frac{1}{2}{\mathrm{\rho v}}_1^2={\mathrm{P}}_0+\frac{1}{2}{\mathrm{\rho v}}^2\text{}\\ \Rightarrow \mathrm{P}-{\mathrm{P}}_0=\frac{1}{2}\mathrm{\rho }\left({\mathrm{v}}^2-{\mathrm{v}}_1^2\right) {\text{ where v}}_1<<\text{v},\\ {\text{therefore v}}^2-{\text{v}}_1^2\approx v^2\\ \therefore \frac{\text{ρ}}{2}\left({\text{L}}^2-{\text{r}}^2\right){\text{ω}}^2\text{=}\frac{1}{2}\mathrm{\rho }\left({\mathrm{v}}^2\right)\\ \Rightarrow \text{v}=\text{ω}\sqrt{{\text{L}}^2-{\text{r}}^2}\end{array}$

$\begin{array}{l}Equation of continuity between 2 \& 3,\text{}\\ A{v}_1=av\text{}\\ \Rightarrow v_1=\frac{av}{A}\end{array}$

$\begin{array}{l}{\text{⇒v}}_1=\frac{a\omega }{\text{A}}\sqrt{{\text{L}}^2-{\text{r}}^2}=\frac{dr}{dt}\\ \Rightarrow \text{t}=\frac{\text{A}}{a\omega }{\sin }^{-1}\frac{\text{r}}{\text{L}}{|}_{\text{L}/2}^{\text{L}}\text{}\\ \Rightarrow \text{t}=\frac{\text{A}}{a\omega }\frac{\text{π}}{3}\end{array}$