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^(11//2)T^(-1)]`
B. `[ML^(11)T^(-2)]`
C. `[ML^(7//2)T^(-2)]`
D. `[ML^(11//2)T^(-2)]`

Explanation

To find the dimensional formula for the product \( AB \) given the potential energy function \( U = \frac{A \sqrt{x}}{x^2 + B} \), we will follow these steps:

### Step 1: Identify the dimensions of potential energy \( U \)
The dimensional formula for potential energy \( U \) is given by:
\[ [U] = [\text{Energy}] = [\text{Force}] \times [\text{Distance}] = [M L^2 T^{-2}] \]

### Step 2: Analyze the expression for \( U \)
The expression for potential energy is:
\[ U = \frac{A \sqrt{x}}{x^2 + B} \]
Here, \( x \) represents distance, which has the dimension of length:
\[ [x] = [L] \]

### Step 3: Determine the dimension of \( B \)
Since \( B \) is added to \( x^2 \), it must have the same dimension as \( x^2 \):
\[ [B] = [x^2] = [L^2] \]

### Step 4: Determine the dimension of \( A \)
Rearranging the equation for \( U \):
\[ U = \frac{A \sqrt{x}}{x^2 + B} \implies U(x^2 + B) = A \sqrt{x} \]
To find the dimension of \( A \), we can express it as:
\[ A = \frac{U (x^2 + B)}{\sqrt{x}} \]
Since \( B \) has the dimension \( [L^2] \), we can approximate \( x^2 + B \) as having the dimension \( [L^2] \) as well. Thus:
\[ [x^2 + B] = [L^2] \]
Now substituting the dimensions:
\[ [A] = \frac{[U] \cdot [L^2]}{[\sqrt{x}]} = \frac{[M L^2 T^{-2}] \cdot [L^2]}{[L^{1/2}]} = \frac{[M L^4 T^{-2}]}{[L^{1/2}]} = [M L^{4 - 1/2} T^{-2}] = [M L^{7/2} T^{-2}] \]

### Step 5: Calculate the dimensions of \( AB \)
Now we can find the dimensions of the product \( AB \):
\[ [AB] = [A] \cdot [B] = [M L^{7/2} T^{-2}] \cdot [L^2] = [M L^{7/2 + 2} T^{-2}] = [M L^{11/2} T^{-2}] \]

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