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 1: Understand the Problem
We have a container filled with liquid, and it is being accelerated horizontally. The liquid's surface will tilt due to this acceleration. We need to find the angle \( \theta \) that the surface makes with the horizontal.

### Step 2: Analyze Forces Acting on the Liquid
When the container accelerates, the effective forces acting on the liquid can be considered. The gravitational force acts downwards, while the inertial force due to the acceleration acts horizontally.

### Step 3: Set Up the Pressure Equations
1. Let \( P_A \) be the pressure at point A (the surface of the liquid) and \( P_B \) be the pressure at point B (some depth in the liquid).
2. The pressure difference between points A and B due to the height \( h \) of the liquid column can be expressed as:
\[ P_B - P_A = \rho g h \]
(where \( \rho \) is the density of the liquid and \( g \) is the acceleration due to gravity).

### Step 4: Consider the Effect of Horizontal Acceleration
1. The pressure difference due to horizontal acceleration \( a = \frac{g}{\sqrt{3}} \) can be expressed as:
\[ P_B - P_C = \rho a x \]
where \( x \) is the horizontal distance from point A to point B.

### Step 5: Relate the Two Pressure Differences
Since the pressure at the surface (point A) is the same for both equations, we can equate them:
\[ \rho g h = \rho \left(\frac{g}{\sqrt{3}}\right) x \]
Cancelling \( \rho \) from both sides gives:
\[ g h = \frac{g}{\sqrt{3}} x \]
This simplifies to:
\[ h = \frac{x}{\sqrt{3}} \]

### Step 6: Find the Angle \( \theta \)
The angle \( \theta \) can be found using the tangent function:
\[ \tan \theta = \frac{h}{x} \]
Substituting \( h \) from the previous step:
\[ \tan \theta = \frac{\frac{x}{\sqrt{3}}}{x} = \frac{1}{\sqrt{3}} \]
Thus, we find:
\[ \theta = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) \]

### Step 7: Determine the Angle
The angle \( \theta \) corresponds to:
\[ \theta = 30^\circ \]

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

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