suppose the kinetic energy of a body oscillating with amplitude A and at a distance x is given by
`K=(Bx)/(x^(2)+A^(2))`
The dimension of B are the same as that of

Options

A. work/time
B. `"work"xx"distance"`
C. work/distance
D. `"work"xx"time"`

Explanation

To solve the problem, we need to determine the dimensions of the constant \( B \) in the given equation for kinetic energy:

\[ K = \frac{Bx}{x^2 + A^2} \]

### Step 1: Identify the dimensions of each term in the equation.

1. **Kinetic Energy (K)**: The dimension of kinetic energy is given by:
\[ [K] = [\text{Energy}] = [M][L^2][T^{-2}] = \text{ML}^2\text{T}^{-2} \]

2. **Distance (x)**: The dimension of distance is:
\[ [x] = [L] \]

3. **Amplitude (A)**: The amplitude is also a distance, so:
\[ [A] = [L] \]

### Step 2: Analyze the denominator \( x^2 + A^2 \).

Since both \( x \) and \( A \) have the same dimension, the dimension of \( x^2 + A^2 \) is:
\[ [x^2 + A^2] = [L^2] \]

### Step 3: Write the dimensions of the entire right-hand side of the equation.

The right-hand side of the equation is:
\[ \frac{Bx}{x^2 + A^2} \]

Substituting the dimensions we have:
\[ \left[\frac{Bx}{x^2 + A^2}\right] = \frac{[B][L]}{[L^2]} = \frac{[B]}{[L]} \]

### Step 4: Set the dimensions of both sides equal.

Since both sides of the equation represent kinetic energy, we can set their dimensions equal:
\[ \frac{[B]}{[L]} = [K] \implies \frac{[B]}{[L]} = \text{ML}^2\text{T}^{-2} \]

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

To find the dimensions of \( B \), we multiply both sides by \( [L] \):
\[ [B] = [K] \cdot [L] = \text{ML}^2\text{T}^{-2} \cdot [L] = \text{ML}^3\text{T}^{-2} \]

### Conclusion

The dimensions of \( B \) are:
\[ [B] = \text{ML}^3\text{T}^{-2} \]