A large open top container of negligible mass and uniform cross sectional area `A` has a small uniform cross sectional area `a` in its side wall near the bottom. The container is kept over a smooth horizontal floor and contains a liquid of density `rho` and mass `m_(0)`. Assuming that the liqud starts flowing through the hole A the acceleration of the container will be

Options

A. g
B. `g//50 `
C. `g//100 `
D. `g//10`

Explanation

To find the acceleration of the container when the liquid starts flowing through the hole, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the velocity of efflux**:
The velocity of the liquid flowing out through the hole at the bottom of the container can be determined using Torricelli's theorem. The velocity \( v \) of efflux is given by:
\[ v = \sqrt{2gh} \]
where \( h \) is the height of the liquid column above the hole.

2. **Calculate the mass flow rate**:
The mass flow rate \( \frac{dm}{dt} \) can be expressed in terms of the density \( \rho \) and the volume flow rate. The volume flow rate \( Q \) through the hole is given by:
\[ Q = a \cdot v = a \cdot \sqrt{2gh} \]
where \( a \) is the cross-sectional area of the hole. Thus, the mass flow rate becomes:
\[ \frac{dm}{dt} = \rho Q = \rho a \sqrt{2gh} \]

3. **Determine the force due to the efflux**:
The force \( F \) exerted on the container due to the liquid flowing out can be calculated using the rate of change of momentum. The force is given by:
\[ F = \frac{dm}{dt} \cdot v = \left(\rho a \sqrt{2gh}\right) \cdot \sqrt{2gh} = \rho a (2gh) \]

4. **Calculate the mass of the liquid in the container**:
The mass \( m \) of the liquid in the container at any instant can be expressed as:
\[ m = \rho V = \rho A h \]
where \( A \) is the cross-sectional area of the container and \( h \) is the height of the liquid.

5. **Find the acceleration of the container**:
The acceleration \( a \) of the container can be found using Newton's second law, \( F = ma \). Thus, the acceleration is:
\[ a = \frac{F}{m} = \frac{\rho a (2gh)}{\rho A h} \]
Simplifying this expression, we find:
\[ a = \frac{2a g}{A} \]

### Final Result:
The acceleration of the container is:
\[ a = \frac{2a g}{A} \]