The angle which the free surface of a liquid filled in a container will make with horizontal if the container is accelerated horizontally with acceleration `(g)/(sqrt(3))` is

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `15^(@)`

Answer

To solve the problem of finding the angle that the free surface of a liquid makes with the horizontal when a container is accelerated horizontally with an acceleration of \( \frac{g}{\sqrt{3}} \), we can follow these steps:

### Step-by-Step Solution

1. **Understand the Situation**:
- We have a container filled with liquid, which is being accelerated horizontally with an acceleration \( a = \frac{g}{\sqrt{3}} \).
- We need to determine the angle \( \theta \) that the free surface of the liquid makes with the horizontal.

2. **Identify Forces Acting on the Liquid**:
- The liquid experiences two accelerations:
- The gravitational acceleration \( g \) acting downwards.
- The horizontal acceleration \( a = \frac{g}{\sqrt{3}} \).

3. **Use Pressure Difference**:
- Consider two points in the liquid: point B (at height \( h \)) and point C (at height 0).
- The pressure difference between these two points can be expressed as:
\[ P_B - P_A = \rho \cdot a \cdot x \]
where \( x \) is the horizontal distance between points A and B.

4. **Vertical Pressure Difference**:
- The vertical pressure difference can be expressed as:
\[ P_B - P_C = \rho \cdot g \cdot h \]

5. **Set Up the Equations**:
- From the first equation:
\[ P_B - P_A = \rho \cdot \frac{g}{\sqrt{3}} \cdot x \]
- From the second equation:
\[ P_B - P_C = \rho \cdot g \cdot h \]

6. **Relate the Pressures**:
- Since \( P_A \) and \( P_C \) are both atmospheric pressure, we can set the two expressions for \( P_B \) equal to each other:
\[ P_A + \rho \cdot \frac{g}{\sqrt{3}} \cdot x = P_A + \rho \cdot g \cdot h \]
- Cancel \( P_A \) from both sides:
\[ \rho \cdot \frac{g}{\sqrt{3}} \cdot x = \rho \cdot g \cdot h \]

7. **Simplify**:
- Cancel \( \rho \) and \( g \) from both sides:
\[ \frac{x}{\sqrt{3}} = h \]
- Rearranging gives:
\[ \frac{h}{x} = \frac{1}{\sqrt{3}} \]

8. **Use Trigonometry**:
- From the triangle formed by points A, B, and C, we have:
\[ \tan(\theta) = \frac{h}{x} \]
- Substitute the value we found:
\[ \tan(\theta) = \frac{1}{\sqrt{3}} \]

9. **Find the Angle**:
- The angle \( \theta \) whose tangent is \( \frac{1}{\sqrt{3}} \) is:
\[ \theta = 30^\circ \]

### Final Answer:
The angle which the free surface of the liquid makes with the horizontal is \( 30^\circ \).