A U-tube manometer with water and oil — find density of oil

The fundamental principle of U-tube manometers: the pressure at the same horizontal level in a connected, continuous fluid is equal.

Let’s identify a reference level: take the oil-water interface on the left side as the reference height, or better — take the level of the water surface on the right arm as our reference (let’s call this Point A).

The pressure at Point A (right arm, water surface level) must equal the pressure at the same height on the left side.

Let me set up the problem with clear geometry:

• Left arm: oil column of height $h_{oil} = 15$ cm
• Right arm: water column of height $h_{water} = 12$ cm
• The oil-water interface on the left is 3 cm ABOVE the water surface on the right

So the oil-water interface on the left is at height +3 cm relative to the water surface on the right.

The bottom of the oil column (oil-water interface, left) is 3 cm above the water level on the right.

The top of the oil column on the left is: $3 + 15 = 18$ cm above the water surface on the right.

Wait — both arms open to atmosphere, so both tops are at atmospheric pressure $P_0$.

Take the reference level as the oil-water interface on the left (the lowest point shared between the two fluid columns, noting that the left side has the denser fluid at the bottom connection).

Actually, let’s use the pressure at the U-tube bottom (the connection point) which is equal on both sides.

Pressure at bottom from left side: $P_{left} = P_0 + \rho_{oil} \cdot g \cdot h_{oil,total}$

where $h_{oil,total}$ is the total height of oil above the bottom.

Pressure at bottom from right side: $P_{right} = P_0 + \rho_{water} \cdot g \cdot h_{water,total}$

At equilibrium: $P_{left} = P_{right}$, so:

The oil-water interface on the left is 3 cm above the water surface on the right. The right arm has 12 cm of water above the bottom connection.

So: height of oil-water interface above bottom = height of water surface on right + 3 cm… wait, that’s not quite right either.

Let me restate with a cleaner diagram approach:

Let the bottom of the U-tube (the connection) = reference = 0.

Right side: water column of 12 cm → water surface is at height 12 cm.

Oil-water interface on the left is 3 cm ABOVE the right-side water surface → oil-water interface on left is at height $12 + 3 = 15$ cm.

Oil column of 15 cm on left → top of oil on left is at height $15 + 15 = 30$ cm.

Both tops are at atmospheric pressure.

Pressure at the bottom from the right:

Pressure at the bottom from the left: The left side has an oil-water interface at 15 cm height. Below the interface (from 0 to 15 cm) must also be water (the connecting fluid). Above the interface (from 15 to 30 cm = 15 cm of oil).

Setting them equal:

This gives a negative density, which means the geometry assumption needs revisiting.

Let me reconsider: the oil-water interface on LEFT is 3 cm ABOVE the water surface on RIGHT. This means the left level is higher, which makes sense if oil is less dense than water. Let me redo with the bottom as the connecting water column.

Right side water height above bottom = H. Left side: interface at H + 3, oil column of 15 cm above the interface, so top of oil at H + 18.

Setting pressures at the bottom:

This still gives a negative value. The issue is directionality. Let’s reconsider: if the oil side is higher overall, the oil side has MORE pressure contribution. Let me try the interface as the reference.

At the oil-water interface level (call it the datum):

• Left side: pressure = $P_0 + \rho_{oil} g (15)$ (15 cm of oil above this level)
• Right side: pressure = water pressure at this level = $P_0 + \rho_w g (12 + 3)$ (the water surface is 3 cm BELOW this level, so pressure at this level = $P_{atm} + \rho_w g \times 3$)

Wait — the water surface on the right is 3 cm BELOW the oil-water interface on the left. Pressure at the right water surface = $P_0$. Pressure in water 3 cm below = $P_0 + \rho_w g (3 \text{ cm})$… but going from right-surface downward 3 cm to the level of the interface.

Hmm — the interface is 3 cm ABOVE the water surface. So at the height of the interface, the pressure from the right side is:

$P_{right, at\ interface\ height} = P_0 - \rho_w g (3)$ (pressure decreases as you go up in a fluid)

Setting left and right pressures equal at the interface height:

This makes physical sense (oil is less dense than water, density < 1000 kg/m³).