91Ó°ÊÓ

• 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$\mathrm{ÒÏ}=0.8\mathrm{g}/{\mathrm{cm}}^3$.
• 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, $\mathrm{ÒÏ}$is the density of the liquid, and ${\mathrm{ÒÏ}}_w$is the density of water. Then,

$\mathrm{ÒÏ}gh={\mathrm{ÒÏ}}_{w}g{h}_w$

$\mathrm{ÒÏ}h={\mathrm{ÒÏ}}_w{h}_w$

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

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

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

$=6.41\mathrm{cm}$

$=0.64\mathrm{m}$

The volume of the water column is

$V=\mathrm{Ï€}{r}^2{h}_w$

Substitute the values in the above equation.

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

$=4.525×{10}^{-5}{\mathrm{m}}^3$

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

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