An equation is given as `(p +(a)/(V^(2))) = b(theta)/(V)`,where `p=` pressure `V=` volumen and `theta =` absolute temperature. If a and b are constants, then dimensions of a will be

Options

A. `[ML^(5)T^(-2)]`
B. `[M^(-1)L^(5)T^(-2)]`
C. `[ML^(-5)T^(-1)]`
D. `[ML^(5)T]`

Explanation

To find the dimensions of the constant \( a \) in the equation

\[ p + \frac{a}{V^2} = \frac{b \theta}{V} \]

where \( p \) is pressure, \( V \) is volume, and \( \theta \) is absolute temperature, we will follow these steps:

### Step 1: Write down the dimensions of the known quantities.

1. **Pressure \( p \)**: The dimension of pressure is given by the formula:
\[ [p] = \frac{\text{Force}}{\text{Area}} = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2} \]

2. **Volume \( V \)**: The dimension of volume is:
\[ [V] = L^3 \]

3. **Absolute Temperature \( \theta \)**: The dimension of temperature is:
\[ [\theta] = K \]

### Step 2: Analyze the equation.

The equation can be rewritten as:
\[ p + \frac{a}{V^2} = \frac{b \theta}{V} \]

For the left-hand side \( p + \frac{a}{V^2} \) to be dimensionally consistent, the dimensions of \( \frac{a}{V^2} \) must equal the dimensions of \( p \).

### Step 3: Set up the dimensional equation.

From the equation, we can write:
\[ [p] = \left[\frac{a}{V^2}\right] \]

This implies:
\[ ML^{-1}T^{-2} = \frac{[a]}{[V^2]} \]

### Step 4: Substitute the dimensions of volume.

Now substituting the dimension of volume:
\[ [V^2] = (L^3)^2 = L^6 \]

Thus, we can rewrite the equation as:
\[ ML^{-1}T^{-2} = \frac{[a]}{L^6} \]

### Step 5: Solve for the dimensions of \( a \).

Rearranging gives:
\[ [a] = ML^{-1}T^{-2} \cdot L^6 \]
\[ [a] = ML^{5}T^{-2} \]

### Conclusion:

The dimensions of \( a \) are:
\[ [a] = ML^{5}T^{-2} \]