A container has a small hole at its bottom. Area of cross-section of the hole is `A_(1)` and that of the container is `A_(2)`. Liquid is poured in the container at a constant rate ` Q m^(3)s^(-1)`. The maximum level of liquid in the container will be

Options

A. `(Q^(2))/(2gA_(1)A_(2))`
B. `(Q^(2))/(2gA_(1)^(2))`
C. `(Q)/(2g A_(1)A_(2))`
D. `(Q^(2))/(2g A_(2_^(2))`

Explanation

To solve the problem, we need to find the maximum height of liquid in the container when the rate of inflow equals the rate of outflow. Here’s a step-by-step solution:

### Step 1: Understand the problem
We have a container with a hole at the bottom. Liquid is being poured into the container at a constant rate \( Q \) (in \( m^3/s \)). The cross-sectional area of the hole is \( A_1 \) and the cross-sectional area of the container is \( A_2 \). We need to find the maximum height \( h_{max} \) of the liquid in the container.

### Step 2: Set up the inflow and outflow rates
The inflow rate of the liquid into the container is given as \( Q \) (in \( m^3/s \)). The outflow rate through the hole can be expressed using Torricelli’s law, which states that the velocity of efflux \( v \) of a fluid under the influence of gravity is given by:
\[ v = \sqrt{2gh} \]
where \( g \) is the acceleration due to gravity and \( h \) is the height of the liquid above the hole.

### Step 3: Write the outflow rate
The outflow rate \( Q_{out} \) through the hole can be calculated as:
\[ Q_{out} = A_1 \cdot v = A_1 \cdot \sqrt{2gh} \]

### Step 4: Set inflow equal to outflow
At maximum height, the inflow rate equals the outflow rate:
\[ Q = A_1 \cdot \sqrt{2gh_{max}} \]

### Step 5: Solve for \( h_{max} \)
Rearranging the equation to solve for \( h_{max} \):
\[ \sqrt{2gh_{max}} = \frac{Q}{A_1} \]
Squaring both sides gives:
\[ 2gh_{max} = \left(\frac{Q}{A_1}\right)^2 \]
Now, isolate \( h_{max} \):
\[ h_{max} = \frac{Q^2}{2gA_1^2} \]

### Conclusion
The maximum height of the liquid in the container is given by:
\[ h_{max} = \frac{Q^2}{2gA_1^2} \]