# Problem Analysis Given: - A U-shaped tube open to air at both ends. - Mercury is initially at the same level in both arms. - Water is poured the left arm, forming a column **10.6 cm** high. We are to find **how much the mercury in the right arm rises** (let's call this rise \( h \)). # Step-by-Step Solution ## 1. Initial Setup - The air pressure at the top of both arms is atmospheric and equal. - The water column pushes down on the mercury in the left arm, causing mercury to rise in the right arm. ## 2. Pressure Relationships **At the same horizontal level at the bottom of the U-tube:** - The pressure must be equal in both arms. Let: - \( \rho_{Hg} \) = density of mercury (\( 13,600 \, \text{kg/m}^3 \)) - \( \rho_{w} \) = density of water (\( 1,000 \, \text{kg/m}^3 \)) - \( g \) = acceleration due to gravity (\( 9.8 \, \text{m/s}^2 \)) - \( h_w = 10.6 \, \text{cm} = .106 \, \text{m} \) (water column height) - \( h \) = the rise of mercury in the right arm (\( \text{m} \)) After pouring water: - The mercury surface in the left arm drops by \( h \). - The mercury surface in the right arm rises by \( h \). - The difference in mercury levels between the two arms is \( 2h \). ## 3. Applying Hydrostatic Pressure **Left arm (from top to bottom):** - Atmospheric pressure (\( P_{atm} \)) - Water column: \( \rho_w g h_w \) - Mercury column: \( \rho_{Hg} g (H - h) \) (if \( H \) is initial height, but cancels out) **Right arm (from top to bottom):** - Atmospheric pressure (\( P_{atm} \)) - Mercury column: \( \rho_{Hg} g (H + h) \) Set the pressures at the bottom equal (at the same horizontal level): \[ P_{atm} + \rho_w g h_w + \rho_{Hg} g (H - h) = P_{atm} + \rho_{Hg} g (H + h) \] ## 4. Simplify Equation Cancel \( P_{atm} \) and \( \rho_{Hg} g H \): \[ \rho_w g h_w - \rho_{Hg} g h = \rho_{Hg} g h \] \[ \rho_w g h_w = 2\rho_{Hg} g h \] \[ h = \frac{\rho_w h_w}{2\rho_{Hg}} \] ## 5. Plug in Values \[ h = \frac{(1,000 \, \text{kg/m}^3)(.106 \, \text{m})}{2 \times 13,600 \, \text{kg/m}^3} \] \[ h = \frac{106}{27,200} \] \[ h \approx .003897 \, \text{m} \] \[ h \approx .39 \, \text{cm} \] # Final Answer **The mercury in the right arm rises about \( \boxed{.39 \ \text{cm}} \) from its initial position.** --- ## Summary Table | Quantity | Value | |----------------|------------------| | Water column | \( 10.6 \) cm | | Mercury rise | \( .39 \) cm | --- **Diagram (text only):** ``` Left: water (10.6 cm) + mercury (lowered) Right: mercury (raised) Difference in mercury levels: 2h h ≈ .39 cm ```