Write the dimensions of `a//b` in the relation `P = ( a - t^(2))/( bx)` , where `P` is the pressure , `x` is the distance , and `t` is the time .

Options

A. `M^(-1) L^(0)T^(-2)`
B. `M L^(0)T^(-2)`
C. `M L^(0)T^(2)`
D. `M LT^(-2)`

Explanation

To find the dimensions of \( \frac{a}{b} \) in the relation \( P = \frac{a - t^2}{bx} \), we will follow these steps:

### Step 1: Identify the dimensions of pressure \( P \)
Pressure \( P \) is defined as force per unit area. The dimensions of force are given by:
\[ \text{Force} = \text{mass} \times \text{acceleration} = M \cdot [L \cdot T^{-2}] = M L T^{-2} \]
Therefore, the dimensions of pressure are:
\[ [P] = \frac{M L T^{-2}}{L^2} = M L^{-1} T^{-2} \]

### Step 2: Analyze the expression \( a - t^2 \)
Since \( a \) and \( t^2 \) are being subtracted, they must have the same dimensions. The dimension of \( t^2 \) (where \( t \) is time) is:
\[ [t^2] = [T^2] \]
Thus, we can conclude:
\[ [a] = [T^2] \]

### Step 3: Analyze the term \( bx \)
In the expression \( bx \), \( x \) represents distance, which has the dimension:
\[ [x] = [L] \]
Let’s denote the dimension of \( b \) as \( [b] \). Therefore, the dimension of \( bx \) becomes:
\[ [bx] = [b][L] \]

### Step 4: Set the dimensions of the right-hand side equal to those of the left-hand side
From the equation \( P = \frac{a - t^2}{bx} \), we can write:
\[ [P] = \frac{[a]}{[bx]} \]
Substituting the dimensions we have:
\[ M L^{-1} T^{-2} = \frac{[T^2]}{[b][L]} \]

### Step 5: Rearranging to find the dimension of \( b \)
Rearranging gives:
\[ [b][L] = \frac{[T^2]}{[P]} = \frac{[T^2]}{M L^{-1} T^{-2}} = \frac{[T^2] \cdot [L]}{[M]} \cdot [T^2] = \frac{[L T^4]}{[M]} \]
Thus, we can express \( [b] \):
\[ [b] = \frac{[T^2]}{[P][L]} = \frac{[T^2]}{M L^{-1} T^{-2} \cdot [L]} = \frac{[T^4]}{[M]} \]

### Step 6: Find the dimensions of \( \frac{a}{b} \)
Now we can find the dimensions of \( \frac{a}{b} \):
\[ \frac{a}{b} = \frac{[T^2]}{[b]} = \frac{[T^2]}{\frac{[T^4]}{[M]}} = [T^2] \cdot \frac{[M]}{[T^4]} = [M T^{-2}] \]

### Final Answer
Thus, the dimensions of \( \frac{a}{b} \) are:
\[ [M T^{-2}] \]