Question

In the previous problem, if 15.0 cm of water and spirit each are further poured into the respective arms of the tube, what is the difference in the levels of mercury in the two arms ? (Specific gravity of mercury = 13.6)

Answer 1

Problem Context

U-tube with water (15.0 cm) in one arm, spirit (15.0 cm) in other, above mercury. Find mercury level difference.

Specific gravity mercury = 13.6, water = 1.0. Spirit SG ≈ 0.8 (typical value).

Left Arm: Water (15 cm) | Hg
Right Arm: Spirit (15 cm) | Hg
Find: \(\Delta h_\text{Hg}\)

Pressure Balance Principle

At mercury interface level, pressures equal:

$$P_\text{water side} = P_\text{spirit side}$$ $$\rho_w g h_w + \rho_\text{Hg} g h_1 = \rho_s g h_s + \rho_\text{Hg} g h_2$$

Difference in mercury levels: \(h = h_2 - h_1\)

$$\rho_w h_w = \rho_s h_s + \rho_\text{Hg} (h_2 - h_1)$$ $$h = h_1 - h_2 = \frac{\rho_w h_w - \rho_s h_s}{\rho_\text{Hg}}$$

Calculation

Given: \(h_w = h_s = 15.0\) cm = 0.15 m

$$\rho_w = 1000 \, \text{kg/m}^3, \quad \rho_s = 800 \, \text{kg/m}^3, \quad \rho_\text{Hg} = 13{,}600 \, \text{kg/m}^3$$ $$h = \frac{(1000 \times 0.15) - (800 \times 0.15)}{13{,}600}$$

Compute pressure difference:

$$\Delta P = 1000 \times 0.15 - 800 \times 0.15 = 150 - 120 = 30 \, \text{Pa·m}$$ $$h = \frac{30}{13{,}600} = 0.00221 \, \text{m} = 0.221 \, \text{cm}$$

Mercury Level Difference

\(\Delta h = \textbf{0.22 cm}\) (mercury higher on spirit side)

Physical Insight

• Water column heavier (150 kg/m² equivalent)
• Spirit column lighter (120 kg/m² equivalent)
• Mercury displaces upward on lighter side by 0.22 cm to balance