Physics 2000 JAMB Past Questions - Question 6: At a fixed point below a liquid surface, the pressure downward is P1 and the pressure upward is P2. It can be deduced that

Correct Answer

D

Explanation

At a fixed point below a liquid surface, the pressure difference between the downward pressure (P1) and the upward pressure (P2) can be deduced using the hydrostatic pressure equation. The hydrostatic pressure equation states that the pressure at a point in a fluid is directly proportional to the depth of the point below the surface of the fluid and the density of the fluid.the pressure difference between P1 and P2 is equal to the pressure exerted by the liquid column above that point. This is known as Pascal's law, which states that the pressure in a fluid at rest is transmitted equally in all directions.

Mathematically, we can express this relationship as:

P1 - P2 = ρgh

Where:
- P1 is the pressure downward
- P2 is the pressure upward
- ρ is the density of the liquid
- g is the acceleration due to gravity
- h is the height of the liquid column above the point

So, the pressure difference between P1 and P2 is equal to the product of the density of the liquid, the acceleration due to gravity, and the height of the liquid column.

Mathematically, it is expressed as:

\[P = P_0 + \rho \cdot g \cdot h\]

Where:
- P is the total pressure at the point.
- P0 is the pressure at the surface of the liquid (which is often atmospheric pressure).
- ρ (rho) is the density of the liquid.
- g is the acceleration due to gravity.
- h is the depth of the point below the liquid surface.

Now, let's apply this equation to your scenario. P1 and P2 represent the pressures at the same point below the liquid surface but in different directions.

- P1 is the pressure acting downward.
- P2 is the pressure acting upward.

For P1:
\[P_1 = P_0 + \rho \cdot g \cdot h\]

For P2:
\[P_2 = P_0 - \rho \cdot g \cdot h\]

Now, to find the pressure difference between P1 and P2, subtract P2 from P1:

\[P1 - P2 = (P_0 + \rho \cdot g \cdot h) - (P_0 - \rho \cdot g \cdot h)\]

When you subtract these equations, the P0 terms cancel out:

\[P1 - P2 = \rho \cdot g \cdot h + \rho \cdot g \cdot h\]

Simplify:

\[P1 - P2 = 2 \cdot \rho \cdot g \cdot h\]

So, the pressure difference (P1 - P2) is equal to twice the product of the density of the liquid (ρ), the acceleration due to gravity (g), and the depth below the liquid surface (h).

In summary, the pressure difference between P1 and P2 is equal to 2 times the product of the density of the liquid, the acceleration due to gravity, and the depth below the liquid surface.