a modified U-tube: the right arm is shorter than the left arm. The open end of the right arm is height d = 10.0 cm above the laboratory bench. The radius throughout the tube is 1.60 cm. Water is gradually poured into the open end of the left arm until the water begins to flow out the open end of the right arm. Then a liquid of density 0.770 g/cm3 is gradually added to the left arm until its height in that arm is 6.70 cm (it does not mix with the water). How much water (in cm3) flows out of the right arm?
To solve this problem, we need to use the principles of hydrostatics, which involves the pressure equilibrium in fluids. Here's the step-by-step explanation of how to find the answer:
1. Calculate the height difference between the two arms: Since the right arm is shorter, the difference in height can be found by subtracting the height d from the height in the left arm: h_diff = 6.70 cm - 10.0 cm = -3.30 cm. Note that the negative sign indicates that the right arm is at a lower height.
2. Calculate the pressure at the surface of the water in the left arm: The pressure at a certain height in a fluid is given by the equation P = ρgh, where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height. Converting the height into meters: h = -3.30 cm = -0.033 m. Plugging in the values, we find P_left = (0.770 g/cm^3)(9.8 m/s^2)(-0.033 m) = -0.237 N/m^2.
3. Calculate the pressure at the open end of the right arm: The pressure at the open end is atmospheric pressure, which we'll assume to be approximately 1.013 x 10^5 N/m^2.
4. Find the difference in pressure between the two arms: ΔP = P_right - P_left. Since P_right is the atmospheric pressure and P_left is negative, we add the absolute values: ΔP = |-0.237 N/m^2| + 1.013 x 10^5 N/m^2 = 1.013 x 10^5 N/m^2 + 0.237 N/m^2 = 1.013 x 10^5 N/m^2.
5. Apply the equation ΔP = ρgh_diff: Rearranging the equation, we can solve for the unknown height h_diff: h_diff = ΔP / (ρg). Calculating the value: h_diff = (1.013 x 10^5 N/m^2) / ((0.770 g/cm^3)(9.8 m/s^2)) = 14.27 cm.
6. Calculate the volume of the water that flows out of the right arm: To find the volume, we need to multiply the difference in heights by the cross-sectional area of the right arm. The cross-sectional area can be calculated using the formula for the area of a cylinder: A = πr^2. Converting the radius to meters: r = 1.60 cm = 0.016 m. The volume of water is then V = A * h_diff = π(0.016 m)^2 * 0.1427 m = 0.000036 m^3. Converting to cubic centimeters (1 m^3 = 10^6 cm^3), the volume is approximately 36 cm^3.
Therefore, approximately 36 cm^3 of water flows out of the right arm.