Question

Length of horizontal arm of a $U-$tube is $L=1m$ and ends of both the vertical arms are open to atmospheric pressure $P_0={10}^{5}N/m^2$. A liquid of density $\rho ={10}^{3}kg/m^3$ is poured in the tube such that liquid just fills the horizontal part of the tube as shown in Fig. 4.262. Now one end of the open end is sealed and the tube is than rotated about a vertical axis passing through the other vertical arm with angular speed ${\omega }_0=\sqrt{40/3}rad/sec$. If length of each vertical arm is $a=1m$ and in the sealed end liquid rises to a height $y=1/2m$, find pressure in the sealed tube during rotation. (in $\times {10}^{5}N/m^2$).

Answer 1

Let the cross-sectional area of the tube be $S$. Here element of length $dx$ is rotating with the tube. From the free-body diagram of the element, we can write
$(P+dP)S—PS=dm{\omega }^{2}x=\left(pSdx\right){\omega }^{2}x$
$dPS=\rho {\omega }^{2}Sxdx$
Here the pressure difference between points $A$ and $B$ can be given by integrating the pressure difference across an element of width $dx$, which is given as
$dP=\rho {\omega }^{2}xdx$
Now integrating from $A$ to $B$, we get
$P_B-P_A={\int }_y^L\rho {\omega }^{2}xdx=\frac{\rho {\omega }^2}{2}(L^2-y^2)$
$⟹P_B=P_0+\frac{\rho {\omega }^2}{2}(L^2-y^2)$
The pressure at point $C$ can be given as
$P_C=P_a-y_{p}g$
At point $A$, pressure is atmospheric. Thus, we have
$P_C=\frac{\rho {\omega }^2}{2}(L^2-y^2)+P_a-y_{p}g={10}^{5}N/m$

Hence, ans will be $1$