The cross-sectional area of the U-tube shown below is uniform and is equal to one square centimeter (1.00cm^2). One end is open to the atmosphere. Atmosphere pressure may be taken as P1=1.01 * 10^5 Pa. The other end of the U-tube is connected to a pipe in which a constant gas pressure of P2=1.25 atmospheres is maintained. The U-tube contains a liquid which has a density of 1.0*10^4kg/m^3.

QUESTION: What the second liquid still in the arm that is open to the atmosphere, the pressure p2 is now reduced until it is equal to atmospheric pressure. What now is the difference (if any) between the liquid levels in the two arms of the U-tube?

When the liquid is in equilibrium, the pressure difference between the two arms of the U-tube is equal to the difference in the heights of the liquid columns. This can be calculated using the equation:

ΔP = ρgh

where:
ΔP is the pressure difference,
ρ is the density of the liquid, and
h is the height difference.

Initially, the pressure at the pipe end (P2) is greater than the atmospheric pressure (P1), so the liquid will be higher in the arm connected to the pipe. Let's assume that the height of the liquid in the arm connected to the pipe is h1, and the height in the arm open to the atmosphere is h2.

Initially, the pressure difference (ΔP) is given by:
ΔP = P2 - P1

Since P1 is given as the atmospheric pressure (1.01 *10^5 Pa), and P2 is given as 1.25 atmospheres, we need to convert P2 to Pa:
P2 = 1.25 * 1.01 * 10^5 Pa

Now, using the equation ΔP = ρgh, we can express the heights h1 and h2 in terms of the pressure difference:

ΔP = ρgh1
ΔP = ρgh2

Since the cross-sectional area of the U-tube is uniform, we can cancel out the area from both equations:

gh1 = gh2

Now, let's consider the situation when P2 is reduced until it is equal to atmospheric pressure (P1). In this case, the pressure difference becomes zero:
P2 - P1 = 0

Substituting the values:
(1.25 * 1.01 * 10^5) - (1.01 * 10^5) = 0

Simplifying, we find:
0.2425 * 10^5 = 0

Since the equation is not true, it means that the pressure difference cannot be zero. Therefore, when the pressure at the pipe end is reduced to atmospheric pressure, there will still be a pressure difference between the two arms of the U-tube.

As a result, there will still be a difference in the heights of the liquid in the two arms of the U-tube.

To find the difference in liquid levels in the two arms of the U-tube when the pressure in the open arm is reduced to atmospheric pressure, we need to consider the equilibrium of forces acting on the liquid in the U-tube.

Let's analyze the forces acting on the liquid in each arm of the U-tube separately:

1. Arm connected to the pipe (closed arm):
- The pressure exerted by the gas in the pipe is P2=1.25 atmospheres.
- The cross-sectional area of the U-tube is 1 square centimeter.
- The height difference between the liquid levels in the two arms is denoted as 'h'.

Since the pressure inside the U-tube is greater in the closed arm, the liquid will move down in this arm until the resultant force acting on the liquid is balanced.

The resultant force acting on the liquid in the closed arm is the difference between the downward force due to the liquid column and the upward force due to the gas pressure:

Downward force = pressure x area = P2 x A = 1.25 atmospheres x 1 cm^2
Upward force = pressure x area = P1 x A = 1.01 * 10^5 Pa x 1 cm^2

2. Arm open to the atmosphere (open arm):
- The pressure at the open end of the U-tube is atmospheric pressure P1=1.01 * 10^5 Pa.
- The cross-sectional area of the U-tube is 1 square centimeter.
- The height of the liquid column in the open arm is denoted as 'H'.

The resultant force acting on the liquid in the open arm is the difference between the downward force due to the liquid column and the upward force due to atmospheric pressure:

Downward force = density x height x gravity = ρ x H x g = 1.0*10^4 kg/m^3 x H x 9.81 m/s^2
Upward force = pressure x area = P1 x A = 1.01 * 10^5 Pa x 1 cm^2

When the pressure in the open arm is reduced to atmospheric pressure (P2 = P1), the liquid levels in the two arms will equilibrate. This means that the resultant forces in both arms will be balanced, leading to the expression:

1.25 atmospheres x 1 cm^2 - 1.01 * 10^5 Pa x 1 cm^2 = 1.0*10^4 kg/m^3 x H x 9.81 m/s^2

Simplifying the equation, we have:

1.25 atmospheres - 1.01 * 10^5 Pa = 1.0*10^4 kg/m^3 x H x 9.81 m/s^2

Now we can solve for 'H' to find the difference in liquid levels in the two arms of the U-tube when the pressure is reduced to atmospheric pressure.