The set-up illustrated above shows a capillary tube of uniform cross-sectional area in two different arrangements. Using the data in the diagrams. Calculate...

Answer Details

From the two diagrams, we can see that the heights of the mercury columns in both the open and closed ends of the capillary tube are different. This difference in height is due to the atmospheric pressure acting on the surface of the mercury in the dish. Let h₁ and h₂ be the heights of the mercury columns in the two arrangements, and let d be the density of mercury. For the first arrangement, we have: h₁ = 7 cm h₂ = 6 cm The pressure difference between the two ends of the capillary tube is given by the difference in height of the mercury columns: ΔP = dgh where g is the acceleration due to gravity. ΔP = (13.6 x 10³ kg/m³) x (10.0 m/s²) x (0.01 m) x (h₁ - h₂) ΔP = 1.36 x 10³ (h₁ - h₂) Pa For the second arrangement, we have: h₁ = 75 cm h₂ = 74 cm Using the same formula, we get: ΔP = 1.36 x 10³ (h₁ - h₂) Pa ΔP = 1.36 x 10³ (75 - 74) Pa ΔP = 1.36 x 10³ Pa To find the pressure of the atmosphere, we need to add this pressure difference to the atmospheric pressure: P_atm = P₀ + ΔP where P₀ is the atmospheric pressure at the surface of the mercury in the dish. Since the height of the mercury column in the dish is 76 cm, we have: P₀ = dgh P₀ = (13.6 x 10³ kg/m³) x (10.0 m/s²) x (0.01 m) x (76 cm) P₀ = 1.032 x 10⁵ Pa Therefore, the atmospheric pressure is: P_atm = P₀ + ΔP P_atm = 1.032 x 10⁵ Pa + 1.36 x 10³ Pa P_atm = 1.046 x 10⁵ Pa Converting this to cm of Hg, we get: P_atm = 1.046 x 10⁵ Pa x (1 cm/133.322 Pa) P_atm = 784.8 cm of Hg Therefore, the pressure of the atmosphere is approximately 784.8 cm of Hg. The answer is closest to 75 cm