Physics · Mechanical Properties Of Fluids · NEET
Both arms are open to air, so the air pushes the same on both tops. At the common bottom level the pressure must be equal (same fluid, same height = same pressure). Pressure of a column is rho x g x h. If a liquid is lighter (small rho), it needs a taller column h to make the same pressure. So the lighter liquid rises higher and the heavier liquid stays lower. Equal push, but the lighter one must be tall to match.
Pick a level that lies inside the SAME single liquid on both sides - usually the horizontal level of the oil-water interface (the touching surface), or the bottom of the U-tube. Below this chosen level the fluid is continuous and connected, so pressure is equal there. Never balance at a level where one arm has oil and the other has water at different heights unless you count both columns correctly.
Both open ends have the same atmospheric pressure P0 pushing down. When you write the pressure balance, P0 appears on both sides and cancels out. That is why the working formula is simply rho_water x h_water = rho_oil x h_oil for the liquid columns above the common level. You only keep P0 if one side is closed or sealed.
Measure each liquid column height from the common balancing level up to the free surface of that liquid. For the water-oil interface method, water height is measured from the interface up, and oil height from the same interface up. Using the top of the tube gives wrong numbers because the tube length is not the liquid column.
Liquid is conserved. If water goes UP by 65 mm in one arm, an equal 65 mm of water must leave the other arm, so it goes DOWN by 65 mm there. Up 65 plus down 65 gives a total water level difference of 130 mm between the two arms. This is a very common NEET trap - the shift on one side is only half the total difference.
A U-tube with both ends open to the atmosphere is partially filled with water. Oil, which is immiscible with water, is poured into one side until it stands at a distance of 10 mm above the water level on the other side. Meanwhile the water rises by 65 mm from its original level. The density of the oil is:
In a U-tube, water and oil are in the left and right arms respectively. The heights of the water and oil columns (measured from the bottom) are 15 cm and 20 cm respectively. The density of the oil is: (take rho_water = 1000 kg m^-3)
Try the real previous-year questions from this chapter — each with the answer and a full solution.
At a common level below both liquids the pressure is equal, so rho1 x g x h1 = rho2 x g x h2. The g cancels, giving rho1 x h1 = rho2 x h2. Heights are measured from the common level to each free surface.
Immiscible means they do not mix. They form two separate columns with a clear boundary (interface). If they mixed, there would be one blended liquid of one density and no separate heights to compare.
The denser (heavier) liquid ends up as the shorter column, and it settles at the bottom of the bend. The lighter liquid floats above and stands as a taller column in its arm.
No, when both ends are open. The atmospheric pressure P0 acts equally on both free surfaces and cancels in the balance. It only matters when one arm is closed or connected to a different pressure.
The physics is the same - equal pressure at the same connected level. A manometer uses this idea to measure an unknown gas or fluid pressure by reading a height difference of a known liquid, while the two-liquid U-tube usually asks you to find an unknown density.
