A beaker containing a liquid of density `rho` moves up with an acceleration `a`. The pressure due to the liquid at a depth h below the free surface of the liquid is.

Options

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

Explanation

To solve the problem of finding the pressure due to a liquid at a depth \( h \) below the free surface when the beaker is moving upwards with an acceleration \( a \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Basic Concept of Pressure in Fluids**:
The pressure at a certain depth \( h \) in a fluid at rest is given by the formula:
\[ P = \rho g h \]
where \( \rho \) is the density of the liquid, \( g \) is the acceleration due to gravity, and \( h \) is the depth below the free surface.

2. **Consider the Effect of Acceleration**:
When the beaker is accelerating upwards with an acceleration \( a \), we need to consider the effect of this acceleration on the pressure at depth \( h \). In this case, we introduce a pseudo force acting downwards due to the upward acceleration of the beaker.

3. **Combine the Effects of Gravity and Acceleration**:
The effective acceleration acting on the liquid at depth \( h \) will be the sum of the gravitational acceleration \( g \) and the upward acceleration \( a \). Therefore, the total effective acceleration acting downwards is:
\[ g + a \]

4. **Write the Modified Pressure Formula**:
Now, substituting the effective acceleration into the pressure formula, we get:
\[ P = \rho (g + a) h \]

5. **Final Expression for Pressure**:
Thus, the pressure at a depth \( h \) below the free surface of the liquid in the moving beaker is:
\[ P = \rho h (g + a) \]

### Final Answer:
The pressure due to the liquid at a depth \( h \) below the free surface of the liquid is given by:
\[ P = \rho h (g + a) \]