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 follow these steps:

### Step 1: Understand the Concept of Gauge Pressure
Gauge pressure is defined as the pressure relative to the atmospheric pressure. It can be calculated using the formula:
\[ P_{\text{gauge}} = \rho g h \]
where:
- \( P_{\text{gauge}} \) 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 depth of the fluid column.

### Step 2: Identify Given Values
From the problem statement:
- 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 3: Substitute the Values into the Formula
Now, we can substitute the known values into the gauge pressure formula:
\[ P_{\text{gauge}} = \rho g h \]
\[ P_{\text{gauge}} = (1000 \, \text{kg/m}^3) \times (10 \, \text{m/s}^2) \times (10 \, \text{m}) \]

### Step 4: Calculate the Gauge Pressure
Now, we perform the multiplication:
\[ P_{\text{gauge}} = 1000 \times 10 \times 10 \]
\[ P_{\text{gauge}} = 100000 \, \text{Pa} \]

### Step 5: Convert the Pressure to Atmospheres
Since \( 1 \, \text{atm} = 10^5 \, \text{Pa} \), we can convert the gauge pressure into atmospheres:
\[ P_{\text{gauge}} = \frac{100000 \, \text{Pa}}{10^5 \, \text{Pa/atm}} = 1 \, \text{atm} \]

### Final Answer
Thus, the gauge pressure exerted below the column of water at a depth of 10 meters is:
\[ \text{Gauge Pressure} = 1 \, \text{atm} \]

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