A spherical tank of `1.2 m` radius is half filled with oil of relative density `0.8`. If the tank is given a horizontal acceleration of `10 m//s^(2)`, the maximum pressure on the tank is `sqrt(2)P` pascal. Find the value of `P`.

Options

A. 8600
B. 9600
C. 5800
D. 6500

Explanation

To solve the problem step by step, we will analyze the situation involving the spherical tank filled with oil and subjected to horizontal acceleration.

### Step 1: Identify the parameters
- Radius of the spherical tank, \( R = 1.2 \, \text{m} \)
- Relative density of oil, \( \rho_r = 0.8 \)
- Acceleration of the tank, \( a = 10 \, \text{m/s}^2 \)
- Density of water, \( \rho_w = 1000 \, \text{kg/m}^3 \)

### Step 2: Calculate the density of oil
The density of oil can be calculated using the relative density:
\[ \rho = \rho_r \times \rho_w = 0.8 \times 1000 \, \text{kg/m}^3 = 800 \, \text{kg/m}^3 \]

### Step 3: Determine the angle of tilt (θ) due to acceleration
The angle of tilt can be determined using the formula:
\[ \tan \theta = \frac{a}{g} \]
where \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity).
\[ \tan \theta = \frac{10}{10} = 1 \implies \theta = 45^\circ \]

### Step 4: Calculate the maximum height difference (h_max)
When the tank is accelerated, the oil will tilt, causing a height difference across the tank. The maximum height difference can be calculated as follows:
\[ h_{\text{max}} = R \sin \theta \]
Since \( \theta = 45^\circ \), we have:
\[ \sin 45^\circ = \frac{1}{\sqrt{2}} \]
Thus,
\[ h_{\text{max}} = R \cdot \frac{1}{\sqrt{2}} = 1.2 \cdot \frac{1}{\sqrt{2}} = \frac{1.2}{\sqrt{2}} = \frac{1.2 \sqrt{2}}{2} = 0.6 \sqrt{2} \, \text{m} \]

### Step 5: Calculate the maximum pressure (P_max)
The pressure at the bottom of the oil column can be calculated using the formula:
\[ P = h \cdot \rho \cdot g \]
Substituting \( h_{\text{max}} \):
\[ P_{\text{max}} = h_{\text{max}} \cdot \rho \cdot g = (0.6 \sqrt{2}) \cdot (800) \cdot (10) \]
Calculating this gives:
\[ P_{\text{max}} = 0.6 \sqrt{2} \cdot 8000 = 4800 \sqrt{2} \, \text{Pa} \]

### Step 6: Relate P_max to the given condition
According to the problem, the maximum pressure is given as:
\[ P_{\text{max}} = \sqrt{2} P \]
Setting the two expressions for \( P_{\text{max}} \) equal:
\[ 4800 \sqrt{2} = \sqrt{2} P \]
Dividing both sides by \( \sqrt{2} \):
\[ P = 4800 \, \text{Pa} \]

### Final Answer
Thus, the value of \( P \) is:
\[ \boxed{4800} \]