Write the dimensions of `a xx b `in the relation `E = ( b - x^(2))/( at)`, where `E` is the energy , `x` is the displacement , and `t` is the time.

Options

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

Explanation

To find the dimensions of \( A \times B \) in the relation \( E = \frac{B - x^2}{At} \), we will follow these steps:

### Step 1: Identify the dimensions of each variable
- **Energy (E)**: The dimension of energy is given by \( [E] = [M][L^2][T^{-2}] \).
- **Displacement (x)**: The dimension of displacement is \( [x] = [L] \).
- **Time (t)**: The dimension of time is \( [t] = [T] \).

### Step 2: Analyze the equation
The equation is given as:
\[ E = \frac{B - x^2}{At} \]
From dimensional homogeneity, we know that the dimensions of \( B \) must match the dimensions of \( x^2 \) since they are being subtracted. Thus:
\[ [B] = [x^2] = [L^2] \]

### Step 3: Find the dimension of A
Rearranging the equation gives:
\[ At = B - x^2 \]
From this, we can express \( A \):
\[ A = \frac{B - x^2}{t} \]
Since \( B \) and \( x^2 \) both have dimensions of \( [L^2] \), we can write:
\[ [B - x^2] = [L^2] \]
Thus, the dimension of \( A \) becomes:
\[ [A] = \frac{[L^2]}{[T]} = [L^2][T^{-1}] \]

### Step 4: Calculate the dimensions of \( A \times B \)
Now we can find the dimensions of \( A \times B \):
\[ [A \times B] = [A] \times [B] = [L^2][T^{-1}] \times [L^2] \]
Calculating this gives:
\[ [A \times B] = [L^4][T^{-1}] \]

### Final Answer
The dimensions of \( A \times B \) are:
\[ [A \times B] = [L^4][T^{-1}] \]