Questions: Consider a U-tube filled with mercury as shown in the figure. The diameter of the right arm of the U-tube is D=1.5 cm, and the diameter of the left arm is twice that. Heavy oil with a specific gravity of 2.660 is poured into the left arm, forcing some mercury from the left arm into the right one. Determine the maximum amount of oil that can be added into the left arm. (Round the final answer to three decimal places.)

Solution Steps

We have a U-tube with mercury and oil. The left arm has a diameter of \(2D = 3.0 \, \text{cm}\) and the right arm has a diameter of \(D = 1.5 \, \text{cm}\). The specific gravity of the oil is 2.660. We need to find the maximum volume of oil that can be added to the left arm without causing mercury to spill out of the right arm.

The cross-sectional area \(A\) of a tube is given by:

\[ A = \pi \left(\frac{d}{2}\right)^2 \]

For the left arm:

\[ A_{\text{left}} = \pi \left(\frac{3.0}{2}\right)^2 = \pi \times 2.25 = 7.0686 \, \text{cm}^2 \]

For the right arm:

\[ A_{\text{right}} = \pi \left(\frac{1.5}{2}\right)^2 = \pi \times 0.5625 = 1.7671 \, \text{cm}^2 \]

The specific gravity of the oil is 2.660, which means its density is 2.660 times that of water (\(1000 \, \text{kg/m}^3\)), so:

\[ \rho_{\text{oil}} = 2.660 \times 1000 = 2660 \, \text{kg/m}^3 \]

The pressure exerted by the oil column must equal the pressure exerted by the displaced mercury column. The height \(h\) of the oil column is related to the height \(h_m\) of the mercury column by:

\[ \rho_{\text{oil}} \cdot g \cdot h = \rho_{\text{Hg}} \cdot g \cdot h_m \]

where \(\rho_{\text{Hg}} = 13600 \, \text{kg/m}^3\) is the density of mercury. Solving for \(h\):

\[ h = \frac{\rho_{\text{Hg}}}{\rho_{\text{oil}}} \cdot h_m = \frac{13600}{2660} \cdot h_m = 5.1128 \cdot h_m \]

The volume of oil \(V_{\text{oil}}\) is given by:

\[ V_{\text{oil}} = A_{\text{left}} \cdot h \]

Substituting \(h = 5.1128 \cdot h_m\):

\[ V_{\text{oil}} = A_{\text{left}} \cdot 5.1128 \cdot h_m \]

Since the mercury displaced in the right arm is equal to the mercury displaced in the left arm:

\[ A_{\text{right}} \cdot h_m = A_{\text{left}} \cdot h_m \]

Thus, the maximum height \(h_m\) is when the mercury level in the right arm reaches the top:

\[ h_m = \frac{A_{\text{left}}}{A_{\text{right}}} \cdot h_m \]

Finally, the maximum volume of oil is:

\[ V_{\text{oil}} = A_{\text{left}} \cdot 5.1128 \cdot \frac{A_{\text{right}}}{A_{\text{left}}} \cdot h_m \]

Simplifying:

\[ V_{\text{oil}} = 5.1128 \cdot A_{\text{right}} \cdot h_m \]

Final Answer