The potential energy of a particle varies with distance `x` from a fixed origin as `U = (A sqrt(x))/( x^(2) + B)`, where `A and B` are dimensional constants , then find the dimensional formula for `AB`.

Options

A. `[ML^(5//2)T^(-2)], [L]`
B. `MLT^(-2)],[L^2]`
C. `[L],[ML^(3//2)T^(-2)]`
D. `[L^2], [MLT^(-2)]`

Answer

To find the dimensional formula for the product \( AB \) where the potential energy \( U \) is given by the equation:

\[ U = \frac{A \sqrt{x}}{x^2 + B} \]

we will follow these steps:

### Step 1: Analyze the given equation
The potential energy \( U \) is expressed in terms of the distance \( x \) and the constants \( A \) and \( B \). We need to find the dimensions of \( A \) and \( B \).

### Step 2: Determine the dimensions of \( B \)
From the equation, we can observe that \( B \) must have the same dimensions as \( x^2 \) because \( x^2 + B \) needs to be dimensionally consistent.

Since \( x \) is a distance, its dimension is:
\[ [x] = L \]
Thus, the dimension of \( x^2 \) is:
\[ [B] = [x^2] = L^2 \]

### Step 3: Determine the dimensions of \( A \)
Rearranging the equation for \( A \):
\[ A = \frac{U (x^2 + B)}{\sqrt{x}} \]

We know that the dimension of potential energy \( U \) (energy) is:
\[ [U] = [\text{Energy}] = ML^2T^{-2} \]

Now substituting the dimensions we have:
\[ A = \frac{ML^2T^{-2} \cdot L^2}{L^{1/2}} \]

Calculating the dimensions:
\[ A = \frac{ML^4T^{-2}}{L^{1/2}} = ML^{4 - 1/2}T^{-2} = ML^{7/2}T^{-2} \]

### Step 4: Calculate the dimensions of \( AB \)
Now that we have the dimensions of \( A \) and \( B \):
\[ [A] = ML^{7/2}T^{-2} \]
\[ [B] = L^2 \]

The dimensions of the product \( AB \) are:
\[ [AB] = [A][B] = (ML^{7/2}T^{-2})(L^2) = ML^{7/2 + 2}T^{-2} = ML^{11/2}T^{-2} \]

### Final Answer
Thus, the dimensional formula for \( AB \) is:
\[ [AB] = ML^{11/2}T^{-2} \]