The rate of flow of liquid ina tube of radius r, length l, whose ends are maintained at a pressure difference P is `V = (piQPr^(4))/(etal)` where `eta` is coefficient of the viscosity and Q is

Options

A. `8`
B. `1/8`
C. `16`
D. `1/16`

Answer

To solve the problem, we need to derive the value of \( Q \) from the given equation for the rate of flow of liquid in a tube. The equation provided is:

\[ V = \frac{\pi Q P r^4}{\eta l} \]

where:
- \( V \) is the volume flow rate,
- \( P \) is the pressure difference,
- \( r \) is the radius of the tube,
- \( \eta \) is the coefficient of viscosity,
- \( l \) is the length of the tube.

### Step 1: Rearranging the Equation
We want to isolate \( Q \) in the equation. To do this, we can rearrange the equation as follows:

\[ Q = \frac{V \eta l}{\pi P r^4} \]

### Step 2: Understanding the Poiseuille's Law
According to Poiseuille's law, the volume flow rate \( V \) can also be expressed as:

\[ V = \frac{\pi r^4 (P_1 - P_2)}{8 \eta l} \]

where \( P_1 - P_2 \) is the pressure difference across the length of the tube. In our case, we can denote this pressure difference as \( P \).

### Step 3: Substituting Poiseuille's Law into the Rearranged Equation
Now, substituting the expression for \( V \) from Poiseuille's law into our rearranged equation gives:

\[ Q = \frac{\left(\frac{\pi r^4 P}{8 \eta l}\right) \eta l}{\pi P r^4} \]

### Step 4: Simplifying the Expression
Now, we can simplify the expression:

1. The \( \pi \) in the numerator and denominator cancels out.
2. The \( r^4 \) in the numerator and denominator cancels out.
3. The \( \eta l \) in the numerator cancels with \( \eta l \) in the denominator.

This results in:

\[ Q = \frac{1}{8} \]

### Conclusion
Thus, the value of \( Q \) is:

\[ Q = \frac{1}{8} \]

### Final Answer
The answer is \( \frac{1}{8} \), which corresponds to option B.

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