A tank 4 m high is half fi lled with water and then filled to the top with oil of density 0.60 g/cc. What is the pressure at the bottom of the tank due to these liquids ? (Take `g = 10 m s^(-2)`)

Options

A. `1.6xx10^3Nm^(-2)`
B. `3.2xx10^3Nm^(-2)`
C. `1.6xx10^4Nm^(-2)`
D. `3.2xx10^4Nm^(-2)`

Explanation

To find the pressure at the bottom of a tank filled with water and oil, we can follow these steps:

### Step 1: Understand the tank setup
The tank is 4 meters high. It is half-filled with water (2 meters) and the remaining half (2 meters) is filled with oil.

### Step 2: Identify the densities
- Density of water (ρ_water) = 1 g/cm³ = 1000 kg/m³
- Density of oil (ρ_oil) = 0.60 g/cm³ = 600 kg/m³

### Step 3: Calculate the pressure due to oil
The pressure at the bottom of the tank due to the oil can be calculated using the formula:
\[ P_{\text{oil}} = h_{\text{oil}} \times \rho_{\text{oil}} \times g \]
Where:
- \( h_{\text{oil}} \) = height of the oil = 2 m
- \( \rho_{\text{oil}} \) = density of oil = 600 kg/m³
- \( g \) = acceleration due to gravity = 10 m/s²

Substituting the values:
\[ P_{\text{oil}} = 2 \, \text{m} \times 600 \, \text{kg/m}^3 \times 10 \, \text{m/s}^2 \]
\[ P_{\text{oil}} = 12000 \, \text{Pa} \, \text{(or N/m²)} \]

### Step 4: Calculate the pressure due to water
The pressure at the bottom of the tank due to the water can be calculated similarly:
\[ P_{\text{water}} = h_{\text{water}} \times \rho_{\text{water}} \times g \]
Where:
- \( h_{\text{water}} \) = height of the water = 2 m
- \( \rho_{\text{water}} \) = density of water = 1000 kg/m³

Substituting the values:
\[ P_{\text{water}} = 2 \, \text{m} \times 1000 \, \text{kg/m}^3 \times 10 \, \text{m/s}^2 \]
\[ P_{\text{water}} = 20000 \, \text{Pa} \, \text{(or N/m²)} \]

### Step 5: Calculate the total pressure at the bottom of the tank
The total pressure at the bottom of the tank is the sum of the pressure due to oil and the pressure due to water:
\[ P_{\text{total}} = P_{\text{oil}} + P_{\text{water}} \]
\[ P_{\text{total}} = 12000 \, \text{Pa} + 20000 \, \text{Pa} \]
\[ P_{\text{total}} = 32000 \, \text{Pa} \]

### Step 6: Convert to standard form
Expressing the total pressure in standard form:
\[ P_{\text{total}} = 3.2 \times 10^4 \, \text{Pa} \]

### Final Answer
The pressure at the bottom of the tank is \( 3.2 \times 10^4 \, \text{N/m}^2 \).

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