The guage pressure exerted below a column of water, open to the earth's atmosphere at depth of 10 m is (density of water = 1000 kg/`m^3`, g = 10 m/`s^2` and 1 atm pressure = `10^5` Pa)

Options

A. 1 atm
B. 2 atm
C. 3 atm
D. 4 atm

Answer

To find the gauge pressure exerted below a column of water at a depth of 10 meters, we can use the formula for pressure in a fluid column, which is given by:

\[ P = \rho g h \]

Where:
- \( P \) is the gauge pressure,
- \( \rho \) is the density of the fluid (water in this case),
- \( g \) is the acceleration due to gravity,
- \( h \) is the height (or depth) of the fluid column.

### Step 1: Identify the given values
- Density of water, \( \rho = 1000 \, \text{kg/m}^3 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)
- Depth of water column, \( h = 10 \, \text{m} \)

### Step 2: Substitute the values into the formula
Now, substitute the values into the pressure formula:

\[ P = \rho g h = (1000 \, \text{kg/m}^3) \times (10 \, \text{m/s}^2) \times (10 \, \text{m}) \]

### Step 3: Perform the multiplication
Calculating the above expression:

\[ P = 1000 \times 10 \times 10 = 100000 \, \text{Pa} \]

### Step 4: Convert the pressure to standard units
The pressure we calculated is in Pascals (Pa). Since \( 1 \, \text{Pa} = 1 \, \text{N/m}^2 \), we can express this as:

\[ P = 10^5 \, \text{Pa} \]

### Step 5: Relate to atmospheric pressure
We know that \( 1 \, \text{atm} = 10^5 \, \text{Pa} \). Therefore, we can conclude:

\[ P = 1 \, \text{atm} \]

### Final Answer
The gauge pressure exerted below the column of water at a depth of 10 meters is \( 1 \, \text{atm} \).

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