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

Explanation

To solve the problem, we will use the concept of pressure in a fluid at different depths. Let's break down the solution step by step.

### Step 1: Define the variables
Let:
- \( h \) = depth of the lake
- \( P_0 \) = atmospheric pressure
- \( \rho \) = density of water
- \( g \) = acceleration due to gravity

### Step 2: Write the expression for pressure at half the depth
The pressure at half the depth of the lake (\( h/2 \)) can be expressed as:
\[ P_{\text{half}} = P_0 + \left(\frac{h}{2}\right) \rho g \]

### Step 3: Write the expression for pressure at the bottom of the lake
The pressure at the bottom of the lake (at depth \( h \)) is given by:
\[ P_{\text{bottom}} = P_0 + h \rho g \]

### Step 4: Set up the equation based on the problem statement
According to the problem, the pressure at half the depth is equal to \( \frac{2}{3} \) of the pressure at the bottom of the lake:
\[ P_0 + \left(\frac{h}{2}\right) \rho g = \frac{2}{3} \left(P_0 + h \rho g\right) \]

### Step 5: Simplify the equation
Expanding the right side:
\[ P_0 + \frac{h}{2} \rho g = \frac{2}{3} P_0 + \frac{2}{3} h \rho g \]

### Step 6: Rearranging the equation
Now, let's rearrange the equation to isolate terms involving \( P_0 \) and \( h \):
\[ P_0 - \frac{2}{3} P_0 = \frac{2}{3} h \rho g - \frac{h}{2} \rho g \]
This simplifies to:
\[ \frac{1}{3} P_0 = \left(\frac{2}{3} - \frac{1}{2}\right) h \rho g \]

### Step 7: Find a common denominator and simplify
To combine the fractions on the right side:
\[ \frac{2}{3} - \frac{1}{2} = \frac{4}{6} - \frac{3}{6} = \frac{1}{6} \]
Thus, we have:
\[ \frac{1}{3} P_0 = \frac{1}{6} h \rho g \]

### Step 8: Solve for \( h \)
Now, we can solve for \( h \):
\[ h = \frac{2 P_0}{\rho g} \]

### Step 9: Substitute known values
Assuming standard values for \( P_0 \), \( \rho \), and \( g \):
- \( P_0 = 10^5 \, \text{Pa} \) (approximate atmospheric pressure)
- \( \rho = 1000 \, \text{kg/m}^3 \) (density of water)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)

Substituting these values:
\[ h = \frac{2 \times 10^5}{1000 \times 10} = \frac{200000}{10000} = 20 \, \text{m} \]

### Final Answer
The depth of the lake is \( h = 20 \, \text{meters} \).

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