The thrust developed by a rocket-motor is given by `F = mv + A(P_1 -P_2)` where m is the mass of the gas ejected per unit time, v is velocity of the gas, A is area of cross-section of the nozzle, `P_1 and P_2` are the pressures of the exhaust gas and surrounding atmosphere. The formula is dimensionally

Options

A. Correct
B. Wrong
C. Some time wrong, sometimes correct
D. Data is not adequate

Answer

To verify the dimensional consistency of the thrust formula \( F = mv + A(P_1 - P_2) \), we will analyze the dimensions of each term in the equation.

### Step-by-Step Solution:

1. **Identify the dimensions of force (F)**:
- The dimension of force \( F \) is given by \( [F] = M L T^{-2} \) where:
- \( M \) = mass
- \( L \) = length
- \( T \) = time

2. **Analyze the first term \( mv \)**:
- Here, \( m \) is the mass flow rate (mass per unit time), and \( v \) is the velocity.
- The dimension of \( m \) (mass flow rate) is \( [m] = M T^{-1} \).
- The dimension of \( v \) (velocity) is \( [v] = L T^{-1} \).
- Therefore, the dimension of \( mv \) is:
\[ [mv] = [m][v] = (M T^{-1})(L T^{-1}) = M L T^{-2} \]

3. **Analyze the second term \( A(P_1 - P_2) \)**:
- \( A \) is the area of the nozzle, and its dimension is \( [A] = L^2 \).
- \( P_1 \) and \( P_2 \) are pressures, and the dimension of pressure is \( [P] = M L^{-1} T^{-2} \).
- The difference \( (P_1 - P_2) \) has the same dimension as pressure, which is \( [P] = M L^{-1} T^{-2} \).
- Therefore, the dimension of \( A(P_1 - P_2) \) is:
\[ [A(P_1 - P_2)] = [A][P] = (L^2)(M L^{-1} T^{-2}) = M L T^{-2} \]

4. **Combine the dimensions of both terms**:
- Now we have:
\[ [F] = [mv] + [A(P_1 - P_2)] = (M L T^{-2}) + (M L T^{-2}) = M L T^{-2} \]
- Since both terms have the same dimension, we can conclude that:
\[ [F] = M L T^{-2} \]

5. **Conclusion**:
- The dimensions of both sides of the equation are equal, confirming that the thrust formula \( F = mv + A(P_1 - P_2) \) is dimensionally consistent.