The pressure at a point in water is 10 `N//m^(2)`. The depth below this point where the pressure becomes double is (Given density of water = `10^(3)` kg `m^(-3)`, g = 10 m `s^(-2)`)

Options

A. 1 mm
B. 1 cm
C. 1 m
D. 10 cm

Answer

To solve the problem, we need to find the depth below a point in water where the pressure becomes double. Let's break down the solution step by step.

### Step 1: Understand the given data
We know:
- Initial pressure at point A, \( P_a = 10 \, \text{N/m}^2 \)
- The pressure at point B (which is double the pressure at point A), \( P_b = 2 \times P_a = 20 \, \text{N/m}^2 \)
- Density of water, \( \rho = 10^3 \, \text{kg/m}^3 \)
- Acceleration due to gravity, \( g = 10 \, \text{m/s}^2 \)

### Step 2: Calculate the change in pressure
The change in pressure \( \Delta P \) as we move from point A to point B is:
\[ \Delta P = P_b - P_a = 20 \, \text{N/m}^2 - 10 \, \text{N/m}^2 = 10 \, \text{N/m}^2 \]

### Step 3: Use the hydrostatic pressure formula
The change in pressure in a fluid column can be expressed as:
\[ \Delta P = \rho g h \]
where:
- \( \Delta P \) is the change in pressure,
- \( \rho \) is the density of the fluid,
- \( g \) is the acceleration due to gravity,
- \( h \) is the height (or depth) of the fluid column.

### Step 4: Rearrange the formula to solve for height \( h \)
We can rearrange the equation to find \( h \):
\[ h = \frac{\Delta P}{\rho g} \]

### Step 5: Substitute the known values
Substituting the values we have:
\[ h = \frac{10 \, \text{N/m}^2}{(10^3 \, \text{kg/m}^3)(10 \, \text{m/s}^2)} \]
\[ h = \frac{10}{10^4} = 10^{-3} \, \text{m} \]

### Step 6: Convert the height to millimeters
To convert meters to millimeters:
\[ h = 10^{-3} \, \text{m} = 1 \, \text{mm} \]

### Final Answer
The depth below the point where the pressure becomes double is \( 1 \, \text{mm} \).

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