Calculate the depth of a well if the pressure at its bottom is 15 times that at a depth of 3 metres. Atmospheric pressure is 10 m column of water.

Options

A. 135m
B. 155m
C. 175m
D. 185m

Explanation

To calculate the depth of the well, we can follow these steps:

### Step 1: Understand the given information
We know that:
- The pressure at the bottom of the well (P_b) is 15 times the pressure at a depth of 3 meters (P_3).
- Atmospheric pressure corresponds to a 10-meter column of water.

### Step 2: Write the equations for pressure
The pressure at the bottom of the well can be expressed as:
\[ P_b = P_{atm} + H \cdot \rho \cdot g \]
Where:
- \( P_{atm} \) = Atmospheric pressure
- \( H \) = Depth of the well
- \( \rho \) = Density of water
- \( g \) = Acceleration due to gravity

The pressure at a depth of 3 meters is given by:
\[ P_3 = P_{atm} + 3 \cdot \rho \cdot g \]

### Step 3: Relate the pressures
According to the problem, we have:
\[ P_b = 15 \cdot P_3 \]

Substituting the expressions for \( P_b \) and \( P_3 \):
\[ P_{atm} + H \cdot \rho \cdot g = 15 \cdot (P_{atm} + 3 \cdot \rho \cdot g) \]

### Step 4: Expand and simplify the equation
Expanding the right side:
\[ P_{atm} + H \cdot \rho \cdot g = 15 \cdot P_{atm} + 45 \cdot \rho \cdot g \]

Rearranging gives:
\[ H \cdot \rho \cdot g = 15 \cdot P_{atm} - P_{atm} + 45 \cdot \rho \cdot g \]
\[ H \cdot \rho \cdot g = 14 \cdot P_{atm} + 45 \cdot \rho \cdot g \]

### Step 5: Substitute the value of atmospheric pressure
Given that atmospheric pressure corresponds to a 10-meter column of water:
\[ P_{atm} = 10 \cdot \rho \cdot g \]

Substituting this into the equation:
\[ H \cdot \rho \cdot g = 14 \cdot (10 \cdot \rho \cdot g) + 45 \cdot \rho \cdot g \]
\[ H \cdot \rho \cdot g = 140 \cdot \rho \cdot g + 45 \cdot \rho \cdot g \]
\[ H \cdot \rho \cdot g = 185 \cdot \rho \cdot g \]

### Step 6: Solve for H
Dividing both sides by \( \rho \cdot g \):
\[ H = 185 \text{ meters} \]

### Conclusion
The depth of the well is **185 meters**.

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