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The mass of the liquid flowing per secon...

The mass of the liquid flowing per second per unit area of cross-section of the tube is proportional to (pressure difference across the ends)^(n) and (average velocity of the liquid)^(m) . Which of the following relations between m and n is correct?

A

`m=n`

B

`m=-n`

C

`m^(2)=n`

D

`m=-n^(2)`

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The correct Answer is:
To solve the problem, we need to establish a relationship between the mass flow rate of a liquid (per second per unit area) and the pressure difference and average velocity of the liquid using dimensional analysis. ### Step-by-Step Solution: 1. **Understanding the Problem**: We are given that the mass flow rate \( \dot{m} \) (mass per unit time per unit area) is proportional to the pressure difference \( P \) raised to the power \( n \) and the average velocity \( v \) raised to the power \( m \). This can be expressed as: \[ \dot{m} \propto P^n v^m \] 2. **Writing the Equation**: We can write the equation as: \[ \dot{m} = k P^n v^m \] where \( k \) is a proportionality constant. 3. **Dimensional Analysis**: The dimensions of the quantities involved are: - Mass flow rate \( \dot{m} \): \([M][T^{-1}][L^{-2}]\) - Pressure \( P \): \([M][L^{-1}][T^{-2}]\) - Velocity \( v \): \([L][T^{-1}]\) 4. **Expressing Dimensions**: The dimensional formula for each term can be expressed as: \[ \text{Dimensions of } \dot{m} = [M][T^{-1}][L^{-2}] \] \[ \text{Dimensions of } P^n = [M^n][L^{-n}][T^{-2n}] \] \[ \text{Dimensions of } v^m = [L^m][T^{-m}] \] 5. **Combining Dimensions**: Now, we combine the dimensions of pressure and velocity: \[ P^n v^m = [M^n][L^{-n}][T^{-2n}] \cdot [L^m][T^{-m}] = [M^n][L^{m-n}][T^{-2n-m}] \] 6. **Setting Up the Equation**: Now, we equate the dimensions of both sides: \[ [M][T^{-1}][L^{-2}] = [M^n][L^{m-n}][T^{-2n-m}] \] 7. **Comparing Dimensions**: - For mass (\( M \)): \[ 1 = n \quad \text{(1)} \] - For length (\( L \)): \[ -2 = m - n \quad \text{(2)} \] - For time (\( T \)): \[ -1 = -2n - m \quad \text{(3)} \] 8. **Solving the Equations**: From equation (1), we have: \[ n = 1 \] Substituting \( n = 1 \) into equation (2): \[ -2 = m - 1 \implies m = -1 \] 9. **Finding the Relationship**: Thus, we have: \[ m = -n \] 10. **Conclusion**: The correct relationship between \( m \) and \( n \) is: \[ m = -n \] Therefore, the answer is option 2.

To solve the problem, we need to establish a relationship between the mass flow rate of a liquid (per second per unit area) and the pressure difference and average velocity of the liquid using dimensional analysis. ### Step-by-Step Solution: 1. **Understanding the Problem**: We are given that the mass flow rate \( \dot{m} \) (mass per unit time per unit area) is proportional to the pressure difference \( P \) raised to the power \( n \) and the average velocity \( v \) raised to the power \( m \). This can be expressed as: \[ \dot{m} \propto P^n v^m ...
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