Figure 14-30 shows a modified U-tube: the right arm is shorter than the left arm. The open end of the right arm is height $\mathit{d}\mathbf{=}\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{}\mathit{c}\mathit{m}$above the laboratory bench. The radius throughout the tube is $\mathbf{1}\mathbf{.}\mathbf{50}\mathbf{}\mathit{c}\mathit{m}$. Water is gradually poured into the open end of the left arm until the water begins to flow out the open end of the right arm. Then a liquid of density $\mathbf{0}\mathbf{.}\mathbf{80}\mathbf{}\mathbf{g}\mathbf{/}\mathit{c}{\mathit{m}}^{\mathbf{3}}$is gradually added to the left arm until its height in that arm is$\mathbf{8}\mathbf{.}\mathbf{0}\mathbf{}\mathit{c}\mathit{m}$(it does not mix with the water). How much water flows out of the right arm?

• The height of the right arm of the tube from the laboratory bench is, $d=10\mathrm{cm}=0.1\mathrm{m}$.
• The radius of the tube is, $r=1.5\mathrm{cm}=0.015\mathrm{m}$.
• The density of the liquid is$\rho =0.8\mathrm{g}/{\mathrm{cm}}^3$.
• The height of the liquid column in the left arm is, $h=8\mathrm{cm}=0.08\mathrm{m}$.

Pascal's Principle is stated as A change in the pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and to the walls of the containing vessel.

There are two types of liquid on both sides of the U-tube, which are in equilibrium. So, the pressure on the sides should be the same. We can find the height of the water column in the right arm of the tube using Pascal's principle and by equating the pressure at the left and right arms of the tube. From this, we can find the amount of water that flows out of the right arm.

Formula:

Pascal's principle gives$P_{\text{in}}=P_{\text{out}}$

According to Pascal's principle, the pressure at the left arm of the tube should be equal to the pressure at the right arm of the tube. Suppose $\mathrm{h}$is the height of the liquid column, $h_w$is the height of the water column, and, $\rho$is the density of the liquid, and ${\rho }_w$is the density of water. Then,

$\rho gh={\rho }_{w}g{h}_w$

$\rho h={\rho }_w{h}_w$

$0.8(0.08)=0.998\left(h_w\right)$

$0.8\mathrm{g}/{\mathrm{cm}}^3(0.08)=0.998\left(h_w\right)$

$h_w=8.0\mathrm{cm}\frac{0.8\mathrm{g}/{\mathrm{cm}}^3}{0.998\mathrm{g}/{\mathrm{cm}}^3}$

$=6.41\mathrm{cm}$

$=0.64\mathrm{m}$

The volume of the water column is

$V=\pi r^2{h}_w$

Substitute the values in the above equation.

$=(3.142)(0.015\mathrm{m}{)}^2(0.064\mathrm{m})$

$=4.525\times {10}^{-5}{\mathrm{m}}^3$

$=45.3{\mathrm{cm}}^3$

Therefore, the amount of water that flows out of the right arm is $45.3{\mathrm{cm}}^3$.