A tank 5 m high is half filled with water and then is filled to top with oil of density `0.85 g//cm^3` The pressure at the bottom of the tank, due to these liquids is

Options

A. `1.85 g(wt)/(cm^(2))`
B. `89.25 g(wt)/(cm^(2))`
C. `462.5 g(wt)/(cm^(2))`
D. `500 g(wt)/(cm^(2))`

Answer

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

### Step 1: Understand the setup
The tank has a total height of 5 meters. It is half-filled with water and then topped off with oil. The density of the oil is given as \(0.85 \, \text{g/cm}^3\).

### Step 2: Determine the heights of the liquids
Since the tank is 5 meters high and half-filled with water, the height of the water column (\(h_1\)) is:
\[ h_1 = \frac{5}{2} = 2.5 \, \text{m} \]
The height of the oil column (\(h_2\)) is also:
\[ h_2 = 5 - 2.5 = 2.5 \, \text{m} \]

### Step 3: Convert the density of oil to SI units
The density of oil is given as \(0.85 \, \text{g/cm}^3\). To convert this to SI units (kg/m³):
\[ \rho_{\text{oil}} = 0.85 \, \text{g/cm}^3 = 0.85 \times 1000 \, \text{kg/m}^3 = 850 \, \text{kg/m}^3 \]

### Step 4: Use the formula for pressure
The pressure at the bottom of the tank (\(P\)) is given by:
\[ P = P_0 + P_{\text{water}} + P_{\text{oil}} \]
Where:
- \(P_0\) is the atmospheric pressure (approximately \(10^5 \, \text{Pa}\)),
- \(P_{\text{water}} = \rho_{\text{water}} \cdot g \cdot h_1\),
- \(P_{\text{oil}} = \rho_{\text{oil}} \cdot g \cdot h_2\).

### Step 5: Calculate the pressures due to water and oil
The density of water (\(\rho_{\text{water}}\)) is \(1000 \, \text{kg/m}^3\) and \(g\) (acceleration due to gravity) is approximately \(9.81 \, \text{m/s}^2\).

Calculating the pressure due to water:
\[ P_{\text{water}} = \rho_{\text{water}} \cdot g \cdot h_1 = 1000 \cdot 9.81 \cdot 2.5 = 24525 \, \text{Pa} \]

Calculating the pressure due to oil:
\[ P_{\text{oil}} = \rho_{\text{oil}} \cdot g \cdot h_2 = 850 \cdot 9.81 \cdot 2.5 = 20896.25 \, \text{Pa} \]

### Step 6: Calculate total pressure at the bottom of the tank
Now, substituting the values into the pressure formula:
\[ P = P_0 + P_{\text{water}} + P_{\text{oil}} = 10^5 + 24525 + 20896.25 \]
Calculating this gives:
\[ P = 100000 + 24525 + 20896.25 = 145421.25 \, \text{Pa} \]

### Step 7: Convert to more convenient units if needed
If we want to express this pressure in terms of grams per square centimeter:
\[ P = \frac{145421.25 \, \text{Pa}}{9.81 \, \text{m/s}^2} \approx 14800 \, \text{g/cm}^2 \]

### Final Answer
The pressure at the bottom of the tank due to the liquids is approximately \(145421.25 \, \text{Pa}\) or \(14800 \, \text{g/cm}^2\).

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