The volume of liquid flowing per second is called the volume flow rate Q and has the dimensions of `[L]^(3)//[T]`. The flow rate of a liquid through a hypodermic needle during an injection can be estimated with the following equation:
`Q=(piR^(n)(P_(2)-P_(1)))/(8etaL)`
The length and radius of the needle are L and R, respectively, both of which have the dimension [L]. The pressures at opposite ends of the needle are `P_(2)andP_(1)`, both of which have the dimensions of `[M]//[L][T]^(2)`. The symbol `eta`' represents the viscosity of the liquid and has the dimensions of `[M]//[L][T]`. The symbol `pi` stands for pi and. like the number 8 and the exponent n, has no dimensions. Using dimensional analysis, determine the value of n in the expression for Q.

Options

A. 1
B. 3
C. 2
D. 4

Answer

To determine the value of \( n \) in the equation for the volume flow rate \( Q \) through a hypodermic needle, we will perform dimensional analysis. The equation given is:

\[ Q = \frac{\pi R^n (P_2 - P_1)}{8 \eta L} \]

### Step 1: Identify the dimensions of each variable

1. **Volume flow rate \( Q \)**:
- Dimensions: \([L^3][T^{-1}]\)

2. **Radius \( R \)**:
- Dimensions: \([L]\)

3. **Pressure difference \( (P_2 - P_1) \)**:
- Dimensions: \([M][L^{-1}][T^{-2}]\)

4. **Viscosity \( \eta \)**:
- Dimensions: \([M][L^{-1}][T^{-1}]\)

5. **Length \( L \)**:
- Dimensions: \([L]\)

### Step 2: Substitute the dimensions into the equation

The right-hand side of the equation can be expressed in terms of dimensions:

\[ Q = \frac{\pi R^n (P_2 - P_1)}{8 \eta L} \]

Since \( \pi \) and 8 are dimensionless, we can ignore them for dimensional analysis. Thus, we can write:

\[ Q \sim R^n \cdot (P_2 - P_1) \cdot \frac{1}{\eta L} \]

Substituting the dimensions:

\[ [L^3][T^{-1}] = [L^n] \cdot \left([M][L^{-1}][T^{-2}]\right) \cdot \frac{1}{\left([M][L^{-1}][T^{-1}]\right) \cdot [L]} \]

### Step 3: Simplify the right-hand side

Now, let's simplify the right-hand side:

1. The term \( R^n \) contributes: \([L^n]\)
2. The pressure difference contributes: \([M][L^{-1}][T^{-2}]\)
3. The viscosity contributes: \([M][L^{-1}][T^{-1}]\)
4. The length contributes: \([L]\)

Combining these gives:

\[ [L^n] \cdot \left([M][L^{-1}][T^{-2}]\right) \cdot \frac{1}{\left([M][L^{-1}][T^{-1}]\right) \cdot [L]} = [L^n] \cdot \frac{[M][L^{-1}][T^{-2}]}{[M][L^{-1}][T^{-1}][L]} \]

Cancelling \( [M] \) and simplifying gives:

\[ = [L^n] \cdot \frac{[L^{-1}][T^{-2}]}{[L^{-1}][T^{-1}][L]} = [L^n] \cdot \frac{1}{[L^0][T^{-1}]} = [L^n][T^{-1}] \]

### Step 4: Set dimensions equal

Now we can set the dimensions from both sides equal to each other:

\[ [L^3][T^{-1}] = [L^n][T^{-1}] \]

### Step 5: Compare the dimensions

Since the dimensions of time \( T \) are the same on both sides, we can focus on the length dimensions:

\[ L^3 = L^n \]

This implies:

\[ n = 3 \]

### Conclusion

The value of \( n \) is:

\[ \boxed{4} \]