Question

A horizontal oriented tube AB of length $5m$ rotates with a constant angular velocity $0.5rad/s$ about a stationary vertical axis OO' passing through the end A. The tube is filled with ideal fluid. The end A of the tube is open, the closed end B has a very small orifice. The velocity with which the liquid comes out from the hole (in $m/s$) is

Answer 1

Let us consider a small elemental length $dx$ at a distance $x$ from the end $A$.
For a rotating frame with constant angular velocity, the change in pressure for this elemental length is given by,

$\frac{dP}{dx}=\rho {\omega }^{2}x$
$dP=\rho {\omega }^{2}xdx$

Since we need to calculate the pressure at the end B (just before the orifice) the limits of $x$ goes from $x=5-2$ to $x=5$ and limits of $P$ from $P_0$ to $P\left(B\right)$
Integrating both sides,
${\int }_{P_0}^{P\left(B\right)}dP=\rho {\omega }^2{\int }_{5-2}^{5}xdx$
$P\left(B\right)-P_0=\rho {\omega }^2{\int }_3^{5}xdx$
$P\left(B\right)=P_0+\rho {\omega }^2{\int }_3^{5}xdx$

Applying the Bernoulli's theorem,
$P_0+\rho {\omega }^2{\int }_3^{5}xdx=P_0+\frac{1}{2}\rho v^2$
$\Rightarrow \frac{\rho {\omega }^2}{2}[5^2-3^2]=\frac{1}{2}\rho v^2$
$\Rightarrow (0.5{)}^2\times 16=v^2$
$\Rightarrow v=2m/s$