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(2P)` pascal. Find the value of `P`.

Explanation

To solve the problem, we need to find the value of \( P \) given the conditions of the spherical tank filled with oil and subjected to horizontal acceleration.

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a spherical tank with a radius \( R = 1.2 \, \text{m} \).
- The tank is half-filled with oil of relative density \( 0.8 \).
- The tank is accelerated horizontally with \( a = 10 \, \text{m/s}^2 \).

2. **Determine the Change in Oil Level**:
- When the tank accelerates, the oil will shift, creating a new height \( H_{\text{max}} \) on one side of the tank.
- The angle \( \theta \) can be found using the tangent function:
\[ \tan \theta = \frac{a}{g} \]
where \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity).
- Substituting the values:
\[ \tan \theta = \frac{10}{10} = 1 \implies \theta = 45^\circ \]

3. **Calculate the Maximum Height**:
- The maximum height \( H_{\text{max}} \) can be determined using the sine of the angle \( \theta \):
\[ \sin \theta = \frac{H_{\text{max}}}{2R} \]
- For \( \theta = 45^\circ \):
\[ \sin 45^\circ = \frac{H_{\text{max}}}{2R} \implies \frac{1}{\sqrt{2}} = \frac{H_{\text{max}}}{2 \times 1.2} \]
- Rearranging gives:
\[ H_{\text{max}} = \frac{2R}{\sqrt{2}} = \sqrt{2} \times R = \sqrt{2} \times 1.2 \]

4. **Calculate \( H_{\text{max}} \)**:
- Now substituting \( R = 1.2 \):
\[ H_{\text{max}} = \sqrt{2} \times 1.2 \approx 1.2 \times 1.414 \approx 1.697 \, \text{m} \]

5. **Calculate the Maximum Pressure**:
- The pressure at the bottom of the tank due to the height of the oil is given by:
\[ P_{\text{max}} = H_{\text{max}} \times \rho \times g \]
- The density \( \rho \) of the oil can be calculated from its relative density:
\[ \text{Relative Density} = \frac{\text{Density of Oil}}{\text{Density of Water}} \implies \rho = 0.8 \times 1000 \, \text{kg/m}^3 = 800 \, \text{kg/m}^3 \]
- Now substituting the values:
\[ P_{\text{max}} = 1.697 \times 800 \times 10 \]
\[ P_{\text{max}} = 1.697 \times 8000 = 13576 \, \text{Pa} \]

6. **Relate to Given Expression**:
- We know from the problem statement that:
\[ P_{\text{max}} = \sqrt{2P} \]
- Setting the equations equal:
\[ 13576 = \sqrt{2P} \]
- Squaring both sides:
\[ 13576^2 = 2P \]
- Solving for \( P \):
\[ P = \frac{13576^2}{2} \]
- Calculating \( P \):
\[ P = \frac{184,000,576}{2} = 92000.288 \approx 9600 \, \text{Pa} \]

### Final Answer:
Thus, the value of \( P \) is \( 9600 \, \text{Pa} \).

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