A rocket ejects the fuel (hot gases) of density `rho` from the chamber which is maintained at a pressure `P_(1)`, say. The atmospheric pressure is `P_(0)`. If the base area of the cylindrical chamber is `A_(1)` and the area of the hole through which hot gases escape from the rocket is `A_(2)` find (a) velocity of efflux of the fuel, (b) thrust exerted on the rocket given by the escaping fuel.

Explanation

To solve the problem step by step, we will break it down into two parts: (a) finding the velocity of efflux of the fuel and (b) calculating the thrust exerted on the rocket.

### Part (a): Velocity of Efflux of the Fuel

1. **Understanding the System**:
We have a cylindrical chamber with base area \( A_1 \) and a hole with area \( A_2 \) through which the fuel (hot gases) is ejected. The pressure inside the chamber is \( P_1 \) and the atmospheric pressure outside is \( P_0 \).

2. **Applying the Continuity Equation**:
The volume flow rate must be conserved. Therefore, we can write:
\[ A_1 V_1 = A_2 V_2 \]
where \( V_1 \) is the velocity of the fluid inside the chamber and \( V_2 \) is the velocity of the fluid as it exits through the hole.

3. **Applying Bernoulli's Equation**:
According to Bernoulli's theorem, we can relate the pressures and velocities at two points (inside the chamber and at the exit). We can express this as:
\[ P_1 + \frac{1}{2} \rho V_1^2 = P_0 + \frac{1}{2} \rho V_2^2 \]
Rearranging gives:
\[ P_1 - P_0 = \frac{1}{2} \rho V_2^2 - \frac{1}{2} \rho V_1^2 \]
This simplifies to:
\[ P_1 - P_0 = \frac{\rho}{2} (V_2^2 - V_1^2) \]

4. **Assuming \( V_1 \) is Small**:
If the hole area \( A_2 \) is much smaller than the base area \( A_1 \), we can assume that \( V_1 \) is much smaller than \( V_2 \) (i.e., \( V_1 \approx 0 \)). Thus, the equation simplifies to:
\[ P_1 - P_0 = \frac{\rho}{2} V_2^2 \]

5. **Solving for \( V_2 \)**:
Rearranging the equation gives:
\[ V_2^2 = \frac{2(P_1 - P_0)}{\rho} \]
Taking the square root:
\[ V_2 = \sqrt{\frac{2(P_1 - P_0)}{\rho}} \]

### Part (b): Thrust Exerted on the Rocket

1. **Understanding Thrust**:
The thrust \( F \) exerted on the rocket is due to the change in momentum of the escaping fuel. It can be expressed as:
\[ F = \text{(mass flow rate)} \times V_2 \]

2. **Calculating Mass Flow Rate**:
The mass flow rate \( \dot{m} \) can be calculated using the density and the volume flow rate:
\[ \dot{m} = \rho \cdot \text{(volume flow rate)} = \rho A_2 V_2 \]

3. **Substituting into the Thrust Equation**:
Therefore, the thrust can be expressed as:
\[ F = \dot{m} \cdot V_2 = (\rho A_2 V_2) V_2 = \rho A_2 V_2^2 \]

4. **Substituting \( V_2 \)**:
Now substituting \( V_2^2 \) from part (a):
\[ F = \rho A_2 \left(\frac{2(P_1 - P_0)}{\rho}\right) = 2 A_2 (P_1 - P_0) \]

### Final Answers
- (a) The velocity of efflux of the fuel is:
\[ V_2 = \sqrt{\frac{2(P_1 - P_0)}{\rho}} \]
- (b) The thrust exerted on the rocket is:
\[ F = 2 A_2 (P_1 - P_0) \]