Question: SAE 30W oil flows steadily upward through the circular tube depicted below. The manometer fluid is mercury, and it indicates a pressure difference of $P_1 - P_2 = 8.5$ kPa across the two points. The length of the tube is $L = 25$ cm. Determine the angle $\theta$ in degrees.

1. **Problem Statement:** We need to find the angle $\theta$ of the tube given the pressure difference $P_1 - P_2 = 8500$ Pa (converted from kPa) and the tube length $L = 0.25$ m (converted from cm). 2. **Known Data:** - Pressure difference: $P_1 - P_2 = 8500$ Pa - Tube length: $L = 0.25$ m - Manometer fluid: mercury with density $\rho_m = 13600$ kg/m$^3$ - Oil density (SAE 30W oil) approximately $\rho = 870$ kg/m$^3$ - Gravitational acceleration: $g = 9.81$ m/s$^2$ 3. **Relevant Formula:** The pressure difference due to the height difference in the manometer is related to the hydrostatic pressure difference: $$P_1 - P_2 = (\rho_m - \rho) g h$$ where $h$ is the vertical height difference of the mercury column. 4. **Relating $h$ and $\theta$:** The vertical height $h$ is related to the tube length $L$ and angle $\theta$ by: $$h = L \sin \theta$$ 5. **Combine equations:** Substitute $h$ into the pressure difference equation: $$P_1 - P_2 = (\rho_m - \rho) g L \sin \theta$$ 6. **Solve for $\sin \theta$:** $$\sin \theta = \frac{P_1 - P_2}{(\rho_m - \rho) g L}$$ 7. **Calculate $\sin \theta$:** $$\sin \theta = \frac{8500}{(13600 - 870) \times 9.81 \times 0.25}$$ 8. **Calculate denominator:** $$13600 - 870 = 12730$$ $$12730 \times 9.81 \times 0.25 = 12730 \times 2.4525 = 31227.33$$ 9. **Calculate $\sin \theta$ value:** $$\sin \theta = \frac{8500}{31227.33} \approx 0.272$$ 10. **Find $\theta$:** $$\theta = \arcsin(0.272) \approx 15.8^\circ$$ **Final answer:** The angle $\theta$ is approximately $15.8^\circ$.