The absolute pressure at a depth h below the surface of a liquid of density `rho` is [Given `P_(a)` = atmospheric pressure, g = acceleration due to gravity]

Options

A. `rho gh`
B. `P_(a) + rho gh `
C. `P_(a) - rho gh`
D. `P_(a) 2 rho g h`

Answer

To find the absolute pressure at a depth \( h \) below the surface of a liquid with density \( \rho \), we can follow these steps:

### Step 1: Understand the Concept of Pressure in Fluids
Pressure in a fluid increases with depth due to the weight of the fluid above. The pressure at a certain depth is the sum of the atmospheric pressure and the pressure due to the column of liquid above that point.

### Step 2: Identify the Variables
- Let \( P_a \) be the atmospheric pressure at the surface of the liquid.
- Let \( \rho \) be the density of the liquid.
- Let \( g \) be the acceleration due to gravity.
- Let \( h \) be the depth below the surface of the liquid.

### Step 3: Calculate the Pressure Due to the Liquid Column
The pressure exerted by a column of liquid of height \( h \) is given by the formula:
\[ P_{\text{liquid}} = \rho g h \]
This represents the additional pressure due to the weight of the liquid above the point at depth \( h \).

### Step 4: Calculate the Absolute Pressure at Depth \( h \)
The absolute pressure at depth \( h \) is the sum of the atmospheric pressure and the pressure due to the liquid column:
\[ P = P_a + P_{\text{liquid}} \]
Substituting the expression for \( P_{\text{liquid}} \):
\[ P = P_a + \rho g h \]

### Final Answer
Thus, the absolute pressure at a depth \( h \) below the surface of a liquid of density \( \rho \) is:
\[ P = P_a + \rho g h \]

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