There is a small hole in the bottom of a fixed container containing a liquid upto height `'h'`. The top of the liquid as well as the hole at the bottom are exposed to atmosphere. As the liquid come out of the hole. (Area of the hole is `'a'` and that of the top surface is `'A'`) :

Options

A. upwards at g
B. upwards at ` g (a ^ 2 )/( A^2 ) `
C. downwards at ` g (a)/(A)`
D. downwards at ` g (a ^ 2 )/(A^2 ) `

Explanation

To solve the problem step by step, we will analyze the situation using principles from fluid dynamics, specifically Torricelli's law and the continuity equation.

### Step 1: Understand the Setup
We have a container filled with liquid up to a height \( h \). There is a small hole at the bottom of the container, and both the top surface of the liquid and the hole are exposed to atmospheric pressure.

### Step 2: Determine the Velocity of the Liquid Exiting the Hole
According to Torricelli's law, the velocity \( v_2 \) of the liquid flowing out of the hole can be determined using the formula:
\[ v_2 = \sqrt{2gh} \]
where \( g \) is the acceleration due to gravity and \( h \) is the height of the liquid column above the hole.

### Step 3: Apply the Continuity Equation
The continuity equation states that the product of the cross-sectional area and the velocity of the fluid must remain constant along a streamline. If \( A \) is the area of the top surface of the liquid and \( a \) is the area of the hole, we can write:
\[ A v_1 = a v_2 \]
where \( v_1 \) is the velocity of the liquid surface.

### Step 4: Express \( v_1 \) in Terms of \( v_2 \)
From the continuity equation, we can express \( v_1 \) as:
\[ v_1 = \frac{a}{A} v_2 \]
Substituting \( v_2 \) from Step 2:
\[ v_1 = \frac{a}{A} \sqrt{2gh} \]

### Step 5: Analyze the Change in Height of the Liquid
As the liquid flows out, the height \( h \) of the liquid in the container decreases. We need to find the acceleration of the liquid surface, which can be expressed as the change in velocity with respect to the change in height:
\[ a_1 = \frac{dv_1}{dt} = \frac{dv_1}{dh} \cdot \frac{dh}{dt} \]

### Step 6: Differentiate \( v_1 \) with Respect to \( h \)
To find \( \frac{dv_1}{dh} \), we differentiate \( v_1 \):
\[ v_1 = \frac{a}{A} \sqrt{2gh} \]
Differentiating with respect to \( h \):
\[ \frac{dv_1}{dh} = \frac{a}{A} \cdot \frac{1}{2\sqrt{2gh}} \cdot 2g = \frac{ag}{A\sqrt{2gh}} \]

### Step 7: Substitute Back into the Acceleration Equation
Now substituting \( \frac{dv_1}{dh} \) back into the acceleration equation:
\[ a_1 = \frac{ag}{A\sqrt{2gh}} \cdot \frac{dh}{dt} \]
Since \( \frac{dh}{dt} \) is negative (as height decreases), we can denote it as:
\[ a_1 = -\frac{ag}{A\sqrt{2gh}} \cdot \frac{dh}{dt} \]

### Step 8: Conclusion
The negative sign indicates that the acceleration of the liquid surface is directed downward, which is consistent with the fact that the height of the liquid is decreasing as it flows out.

### Final Result
The acceleration of the liquid surface is given by:
\[ a_1 = -\frac{a^2 g}{A^2} \]