Volume flow rate through a capillary tube is given by `V=(K(r_2 -r_1)^(4))/((x_(2)-x_(1)))`, where `r_(1)" and "r_(2)` are the inner and outer radii and `x_(1)" and "x_(2)` are the positions crossed by the liquid. What will be the dimensional formula for K?

Options

A. `[MLT^(-1)]`
B. `[M^(0)L^(0)T^(-2)]`
C. `[M^(0)L^(0)T^(-1)]`
D. None of these

Explanation

To find the dimensional formula for \( K \) in the given equation for volume flow rate through a capillary tube, we start with the equation:

\[ V = \frac{K (r_2 - r_1)^4}{(x_2 - x_1)} \]

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

1. **Volume Flow Rate (V)**:
Volume flow rate is defined as the volume of fluid passing through a section per unit time. Its dimensional formula is:
\[ [V] = \frac{[L^3]}{[T]} = L^3 T^{-1} \]

2. **Inner and Outer Radii (r1 and r2)**:
Both \( r_1 \) and \( r_2 \) are lengths, so their dimensional formula is:
\[ [r_1] = [r_2] = [L] \]

3. **Position Difference (x2 - x1)**:
The difference in positions \( (x_2 - x_1) \) is also a length, so its dimensional formula is:
\[ [x_2 - x_1] = [L] \]

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

Substituting the dimensions into the equation gives:

\[ [L^3 T^{-1}] = \frac{[K] \cdot [L^4]}{[L]} \]

### Step 3: Simplify the right side

The right side simplifies as follows:

\[ \frac{[K] \cdot [L^4]}{[L]} = [K] \cdot [L^3] \]

### Step 4: Set the dimensions equal

Now we equate the dimensions from both sides:

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

### Step 5: Solve for the dimensional formula of K

To isolate \( [K] \), we divide both sides by \( [L^3] \):

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

### Step 6: Write the final dimensional formula

Thus, the dimensional formula for \( K \) is:

\[ [K] = M^0 L^0 T^{-1} \]

### Conclusion

The final answer is:

\[ \text{Dimensional formula for } K = M^0 L^0 T^{-1} \]