A large container (with open top) of negligible mass and uniform cross-sectional area A has a small hole of cross-sectional area a in its side wall near the bottom. The container contains a liquid of density ` rho` and mass m. If the liquid starts flowing out of the hole at time t = 0, the velocity of efflux of the liquid when 75% of the liquid has drained out is:

Options

A. ` sqrt((3m g )/( 4 A rho )) `
B. ` sqrt ((2m g )/(A rho )) `
C. `2 sqrt ((mg )/(A rho)) `
D. `sqrt((mg )/(2 A rho )) `

Explanation

To find the velocity of efflux of the liquid when 75% of the liquid has drained out of the container, we can follow these steps:

### Step 1: Understand the Problem
We have a large container with a small hole at the bottom. The container is filled with a liquid of density \( \rho \) and mass \( m \). We need to determine the velocity of the liquid flowing out of the hole when 75% of the liquid has drained out.

### Step 2: Determine the Remaining Volume of Liquid
When 75% of the liquid has drained out, 25% of the liquid remains. If the initial volume of the liquid is \( V \), the remaining volume \( V' \) can be calculated as:
\[ V' = \frac{1}{4} V \]

### Step 3: Relate Volume to Height
The volume of liquid in the container can be expressed in terms of the cross-sectional area \( A \) and the height \( h \) of the liquid:
\[ V = A \cdot h \]
Thus, when 75% of the liquid is drained, the height of the remaining liquid \( h' \) is:
\[ V' = A \cdot h' = \frac{1}{4} (A \cdot h) \]
From this, we can derive:
\[ h' = \frac{h}{4} \]

### Step 4: Apply Torricelli's Law
According to Torricelli's Law, the velocity \( v \) of efflux of a fluid under the force of gravity through a small hole is given by:
\[ v = \sqrt{2gh'} \]
Substituting \( h' \) from the previous step:
\[ v = \sqrt{2g \cdot \frac{h}{4}} = \sqrt{\frac{2gh}{4}} = \sqrt{\frac{gh}{2}} \]

### Step 5: Relate Height to Mass and Density
We know that the mass \( m \) of the liquid is related to its density \( \rho \) and volume \( V \):
\[ m = \rho \cdot V = \rho \cdot (A \cdot h) \]
From this, we can express height \( h \) as:
\[ h = \frac{m}{\rho A} \]

### Step 6: Substitute Height into Velocity Equation
Substituting \( h \) back into the velocity equation:
\[ v = \sqrt{\frac{g \cdot \frac{m}{\rho A}}{2}} = \sqrt{\frac{gm}{2\rho A}} \]

### Conclusion
Thus, the velocity of efflux of the liquid when 75% has drained out is:
\[ v = \sqrt{\frac{gm}{2\rho A}} \]