Question

Mercury is poured into a U-tube as shown in Figure (a). The left arm of the tube has cross-sectional area ${\mathit{A}}_{\mathbf{1}}$of $\mathbf{10}\mathbf{.}\mathbf{0}\mathit{c}{\mathit{m}}^{\mathbf{2}}$, and the right arm has a cross-sectional area${\mathit{A}}_{\mathbf{2}}$ of $\mathbf{5}\mathbf{.}\mathbf{0}\mathit{c}{\mathit{m}}^{\mathbf{2}}$. One hundred grams of water are then poured into the right arm as shown in Figure (b).

(a) Determine the length of the water column in the right arm of the U-tube.

(b) Given that the density of mercury is $\mathbf{13}\mathbf{.}\mathbf{6}\mathit{g}\mathbf{/}\mathit{c}{\mathit{m}}^{\mathbf{3}}$, what distance h does the mercury rise in the left arm?

Short answer

(a) The length of the water column in the right arm of the U-tube is$h_w=20cm$

(b) The distance h does the mercury rise in the left arm$h=1.49cm$

Step-by-step solution

Step 1:

The pressure in a fluid at rest varies with depth h in the fluid according to the expression

$\mathit{P}\mathbf{=}{\mathit{P}}_{\mathbf{0}}\mathbf{+}\mathit{\rho }\mathit{g}\mathit{h}$

Where ${\mathit{P}}_{\mathbf{0}}$is the pressure at $\mathit{h}\mathbf{=}\mathbf{0}$and$\mathit{\rho }$ is the density of the fluid, assumed uniform. Pascal’s law states that when pressure is applied to an enclosed fluid, the pressure is transmitted undiminished to every point in the fluid and to every point on the walls of the container

Step 2:

Part(a):

Given:

Area of left tube$A_L=10c{m}^2$

Area of right tube$A_R=5c{m}^2$

Mass of water poured$m=100g$

Density of mercury${\rho }_w=13.6g/c{m}^3$

Let the height of water column$h_w$

So, the volume V,

$\begin{array}{rcl}V& =& A_R\times h_w\\ \frac{m}{{\rho }_w}& =& 5\times h_w\\ h_w& =& \frac{100}{5\times 1}\\ h_w& =& 20cm\end{array}$

Step 3:

Part(b):

Now if we observed, the volume of water poured of $h_1$is equals to volume of mercury Hg column of height$h$

$\begin{array}{rcl}{A}_L\times h& =& A_R\times h_1\\ \because h_1& =& 2h\\ & & \end{array}$

Now balancing the pressure in two columns at

level $XX\text{'}$

$\begin{array}{rcl}{P}_1& =& P_2\\ P_0+{\rho }_{Hg}g\left(h+h_1\right)& =& P_0+{\rho }_{w}g{h}_w\\ 13.6\times 9.8\times \left(h+2h\right)& =& 1\times 9.8\times 20\\ h& =& \frac{20}{13.6\times 3}\\ h& =& 1.49cm\\ & & \end{array}$