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A U-shaped tube of constant cross-sectional area contains water of density 1000 kg / m3 . Bothsides of the U-tube are open to the atmosphere.Fig. 3.1 shows that the water levels in the two sides of the tube are equal.stopper0.200 mrubber tubingconnected to gassupplyFig. 3.1 Fig. 3.2The atmospheric pressure is 1.00 × 10 5 Pa.The left-hand side of the tube is now connected to a gas supply using a length of rubber tubing.This causes the level of the water in the left-hand side of the tube to drop by 0.200 m, as shown inFig. 3.2.(a) Calculate the pressure of the gas supply. Give your answer to 3 significant figures.pressure = ......................................................... [3

Question

A U-shaped tube of constant cross-sectional area contains water of density 1000 kg / m3 . Bothsides of the U-tube are open to the atmosphere.Fig. 3.1 shows that the water levels in the two sides of the tube are equal.stopper0.200 mrubber tubingconnected to gassupplyFig. 3.1 Fig. 3.2The atmospheric pressure is 1.00 × 10 5 Pa.The left-hand side of the tube is now connected to a gas supply using a length of rubber tubing.This causes the level of the water in the left-hand side of the tube to drop by 0.200 m, as shown inFig. 3.2.(a) Calculate the pressure of the gas supply. Give your answer to 3 significant figures.pressure = ......................................................... [3

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Solution

The pressure of the gas supply can be calculated using the formula for pressure due to a fluid column, which is P = ρgh, where ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height of the fluid column.

In this case, the density of the water (ρ) is 1000 kg/m^3, the acceleration due to gravity (g) is approximately 9.81 m/s^2, and the change in height of the water column (h) is 0.200 m.

So, the pressure due to the water column is P = (1000 kg/m^3) * (9.81 m/s^2) * (0.200 m) = 1962 Pa.

However, this is the pressure difference between the atmospheric pressure and the gas pressure. Since the atmospheric pressure is 1.00 × 10^5 Pa, the pressure of the gas supply is the atmospheric pressure minus the pressure due to the water column.

Therefore, the pressure of the gas supply is (1.00 × 10^5 Pa) - 1962 Pa = 9.80 × 10^4 Pa or 98000 Pa to 3 significant figures.

This problem has been solved

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