Chapter 14, Problem 14.22P

(a)

To determine

The length of the water column in the right arm of the U-tube.

Answer to Problem 14.22P

The length of the water column in the right arm of the U-tube is $20.0\text{cm}$.

Explanation of Solution

Given info: The cross sectional area $A_1$ of the left arm of the U-tube is $10.0{\text{cm}}^2$ and the cross sectional area $A_2$ of the right arm of the U-tube is $5.00{\text{cm}}^2$, the amount of water poured in the right arm is one hundred grams.

The formula to calculate the variation of pressure of the liquid with respect to the depth of the fluid is,

$P=P_0+h\rho g$

Here,

$P_0$ is the atmospheric pressure.

$h$ is the depth of the fluid.

$g$ is the acceleration due to gravity.

$P$ is the variation of pressure of the liquid with respect to the depth of the fluid.

Write the expression for the volume of the water column.

$V=A_2{l}_{\text{w}}$

Here,

$l_{\text{w}}$ is the length of the water column on the right side of the tube.

$A_2$ is the area of the right arm.

The formula to calculate the density of the water is,

$\begin{array}{c}{\rho }_{\text{w}}=\frac{{\text{m}}_{\text{water}}}{V}\\ =\frac{{\text{m}}_{\text{water}}}{A_2{l}_{\text{w}}}\end{array}$

Here,

${\text{m}}_{\text{water}}$ is the mass of the water.

$V$ is the volume of the water column.

Substitute $A_2{l}_{\text{w}}$ for $V$ in above equation.

${\rho }_{\text{w}}=\frac{{\text{m}}_{\text{water}}}{A_2{l}_{\text{w}}}$

Rearrange the above equation for the $l_{\text{w}}$.

$l_{\text{w}}=\frac{{\text{m}}_{\text{water}}}{A_2\rho }$

Substitute $5.00{\text{cm}}^2$ for $A_2$, $100\text{g}$ for ${\text{m}}_{\text{water}}$ and $1.0\text{g}/{\text{cm}}^3$ for ${\rho }_{\text{w}}$ in the above equation.

$\begin{array}{c}{l}_{\text{w}}=\frac{\left(100\text{g}\right)}{\left(5.00{\text{cm}}^2\right)\left(1.0\text{g}/{\text{cm}}^3\right)}\\ =20.0\text{cm}\end{array}$

Conclusion:

Therefore, the length of the water column in the right arm of the U-tube is $20.0\text{cm}$

(b)

To determine

The rise in the length of the mercury column in the left arm of the U-tube.

Answer to Problem 14.22P

The rise in the length of the mercury column in the left arm of the U-tube is $0.49\text{cm}$.

Explanation of Solution

Given info: The cross sectional area $A_1$ of the left arm of the U-tube is $10.0{\text{cm}}^2$ and the cross sectional area $A_2$ of the right arm of the U-tube is $5.00{\text{cm}}^2$, the amount of water poured in the right arm is one hundred grams. The density of the mercury is $13.6\text{g}/{\text{cm}}^3$.

Consider the height of the height of the left tube be $h_1$ and the height of the right tube is $h_2$.

The volume of mercury on the left arm of the tube is equal to the right arm of the tube.\

$A_1{h}_1=A_2{h}_2$

Here,

$A_1{h}_1$ is the volume of the mercury in the left arm of the tube.

$A_2{h}_2$ is the volume of the mercury in the right arm of the tube.

Rearrange the above equation.

$h_2=\frac{A_1{h}_1}{A_2}$

The pressure of the right tube is,

$P=P_0+{\rho }_{\text{w}}g{l}_{\text{w}}$ (1)

The pressure of the left tube is,

$P=P_0+{\rho }_{\text{m}}g\left(h_1+h_2\right)$ (2)

Here,

${\rho }_{\text{m}}$ is the density of mercury.

The pressure on the both the arm of the tube is same.

Compare the equation (1) and (2).

$\begin{array}{c}{P}_0+{\rho }_{\text{w}}g{l}_{\text{w}}=P_0+{\rho }_{\text{m}}g\left(h_1+h_2\right)\\ {\rho }_{\text{w}}g{l}_{\text{w}}={\rho }_{\text{m}}g\left(h_1+h_2\right)\\ \left(h_1+h_2\right)=\frac{{\rho }_{\text{w}}{l}_{\text{w}}}{{\rho }_{\text{m}}}\end{array}$

Substitute $\frac{A_1{h}_1}{A_2}$ for $h_2$ in the above equation.

$\begin{array}{c}\left(h_1+\frac{A_1{h}_1}{A_2}\right)=\frac{{\rho }_{\text{w}}{l}_{\text{w}}}{{\rho }_{\text{m}}}\\ h_1=\frac{{\rho }_{\text{w}}{l}_{\text{w}}}{{\rho }_{\text{m}}\left(1+\frac{A_1}{A_2}\right)}\end{array}$

Substitute $10.0{\text{cm}}^2$ for $A_1$, $5.0{\text{cm}}^2$ for $A_2$, $20.0\text{cm}$ for $l_{\text{w}}$, $1.0\text{g}/{\text{cm}}^3$ for ${\rho }_{\text{w}}$ and $13.6\text{g}/{\text{cm}}^3$ for ${\rho }_{\text{m}}$ in the above equation.

$\begin{array}{c}{h}_1=\frac{\left(1.0\text{g}/{\text{cm}}^3\right)\left(20.0\text{cm}\right)}{13.6\text{g}/{\text{cm}}^3\left(1+\frac{10.0{\text{cm}}^2}{5.0{\text{cm}}^2}\right)}\\ =0.49\text{cm}\end{array}$

Conclusion:

Therefore, the rise in the length of the mercury column in the left arm of the U-tube is $0.49\text{cm}$.