In the relation `V = ( pi)/( 8) ( P r^(4))/( nl)` , where the letters have their usual meanings , the dimensions of `V` are

Options

A. `M^(0) L^(3)T^(0)`
B. `M^(0)L^(3)T^(-1)`
C. `M^(0)L^(-3)T^(-1)`
D. `M^(1)L^(3)T^(0)`

Explanation

To find the dimensions of \( V \) in the relation

\[ V = \frac{\pi}{8} \frac{P r^4}{n l} \]

we need to analyze the dimensions of each variable involved in the equation.

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

1. **Pressure (P)**:
Pressure is defined as force per unit area. The dimension of force is \( [M L T^{-2}] \) (mass times acceleration). Area has the dimension \( [L^2] \). Therefore, the dimension of pressure is:
\[ [P] = \frac{[F]}{[A]} = \frac{[M L T^{-2}]}{[L^2]} = [M L^{-1} T^{-2}] \]

2. **Radius (r)**:
The radius is a length, so its dimension is:
\[ [r] = [L] \]

3. **Viscosity (n)**:
The dimension of viscosity is given as \( [M^1 L^{-1} T^{-1}] \).

4. **Length (l)**:
Length has the dimension:
\[ [l] = [L] \]

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

Now we substitute the dimensions into the equation for \( V \):

\[ V = \frac{P r^4}{n l} \]

Substituting the dimensions we found:

\[ [V] = \frac{[M L^{-1} T^{-2}] \cdot [L^4]}{[M^1 L^{-1} T^{-1}] \cdot [L]} \]

### Step 3: Simplify the expression

Now we simplify the expression step by step:

1. **Numerator**:
\[ [M L^{-1} T^{-2}] \cdot [L^4] = [M L^{4-1} T^{-2}] = [M L^{3} T^{-2}] \]

2. **Denominator**:
\[ [M^1 L^{-1} T^{-1}] \cdot [L] = [M^1 L^{-1 + 1} T^{-1}] = [M^1 L^{0} T^{-1}] = [M^1 T^{-1}] \]

Now, substituting back into the equation for \( V \):

\[ [V] = \frac{[M L^{3} T^{-2}]}{[M^1 T^{-1}]} = [M^{1-1} L^{3} T^{-2 - (-1)}] = [L^{3} T^{-1}] \]

### Final Result

Thus, the dimensions of \( V \) are:

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

### Conclusion

The correct option for the dimensions of \( V \) is \( [L^3 T^{-1}] \).

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