If pressure at half the depth of a lake is equal to `2//3` pressure at the bottom of the lake then what is the depth of the lake ?

Options

A. 10 m
B. 20 m
C. 60 m
D. 30 m

Answer

To solve the problem, we need to find the depth of the lake (H) given that the pressure at half the depth of the lake is equal to \( \frac{2}{3} \) of the pressure at the bottom of the lake.

### Step-by-Step Solution:

1. **Define the Variables:**
Let:
- \( H \) = depth of the lake
- \( P_0 \) = atmospheric pressure
- \( \rho \) = density of water (approximately \( 1000 \, \text{kg/m}^3 \))
- \( g \) = acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \))

2. **Pressure at Half Depth:**
The pressure at half the depth of the lake (i.e., at \( \frac{H}{2} \)) can be expressed as:
\[ P\left(\frac{H}{2}\right) = P_0 + \rho g \left(\frac{H}{2}\right) \]

3. **Pressure at the Bottom:**
The pressure at the bottom of the lake (i.e., at depth \( H \)) can be expressed as:
\[ P(H) = P_0 + \rho g H \]

4. **Set Up the Equation:**
According to the problem, the pressure at half the depth is equal to \( \frac{2}{3} \) of the pressure at the bottom:
\[ P\left(\frac{H}{2}\right) = \frac{2}{3} P(H) \]
Substituting the expressions for pressure:
\[ P_0 + \rho g \left(\frac{H}{2}\right) = \frac{2}{3} \left(P_0 + \rho g H\right) \]

5. **Simplify the Equation:**
Expanding the right side:
\[ P_0 + \frac{\rho g H}{2} = \frac{2}{3} P_0 + \frac{2}{3} \rho g H \]
Rearranging gives:
\[ P_0 - \frac{2}{3} P_0 + \frac{\rho g H}{2} - \frac{2}{3} \rho g H = 0 \]

6. **Combine Like Terms:**
This simplifies to:
\[ \frac{1}{3} P_0 = \left(\frac{2}{3} - \frac{1}{2}\right) \rho g H \]
Finding a common denominator for the fractions:
\[ \frac{2}{3} - \frac{1}{2} = \frac{4 - 3}{6} = \frac{1}{6} \]
Thus, we have:
\[ \frac{1}{3} P_0 = \frac{1}{6} \rho g H \]

7. **Solve for H:**
Rearranging gives:
\[ H = \frac{2 P_0}{\rho g} \]

8. **Substitute Values:**
Substitute \( P_0 = 1.01 \times 10^5 \, \text{Pa} \), \( \rho = 1000 \, \text{kg/m}^3 \), and \( g = 10 \, \text{m/s}^2 \):
\[ H = \frac{2 \times 1.01 \times 10^5}{1000 \times 10} \]
Simplifying:
\[ H = \frac{2.02 \times 10^5}{10000} = 20.2 \, \text{m} \]

9. **Final Result:**
The depth of the lake is approximately \( 20 \, \text{m} \).