The kinetic energy of a particle moving along x-axis varies with the distance x of the particle from origin as `K=(A+x^3)/(Bx^(1//4)+C)` .Write the dimensional formula for `A^2B`

Answer

To find the dimensional formula for \( A^2B \) from the given expression for kinetic energy \( K = \frac{A + x^3}{B x^{1/4} + C} \), we will follow these steps:

### Step 1: Understand the given equation
The equation for kinetic energy is given as:
\[ K = \frac{A + x^3}{B x^{1/4} + C} \]
Where:
- \( K \) is kinetic energy.
- \( A \), \( B \), and \( C \) are constants.
- \( x \) is the distance from the origin.

### Step 2: Analyze the dimensions of kinetic energy
The dimensional formula for kinetic energy \( K \) is:
\[ [K] = M L^2 T^{-2} \]

### Step 3: Apply the principle of homogeneity
According to the principle of homogeneity, all terms in the equation must have the same dimensions. Therefore, we can equate the dimensions of the numerator and the denominator.

### Step 4: Analyze the numerator
The numerator is \( A + x^3 \). Since \( x^3 \) has dimensions:
\[ [x^3] = L^3 \]
This means that \( A \) must also have dimensions of \( L^3 \) for the terms to be dimensionally consistent.

### Step 5: Find the dimensions of \( A \)
Thus, we have:
\[ [A] = L^3 \]
Now, calculating \( A^2 \):
\[ [A^2] = (L^3)^2 = L^6 \]

### Step 6: Analyze the denominator
The denominator is \( B x^{1/4} + C \). The term \( x^{1/4} \) has dimensions:
\[ [x^{1/4}] = L^{1/4} \]
For dimensional consistency, the dimensions of \( B x^{1/4} \) must equal the dimensions of \( x^3 \) (which is \( L^3 \)). Thus:
\[ [B] \cdot [x^{1/4}] = L^3 \]
This implies:
\[ [B] \cdot L^{1/4} = L^3 \Rightarrow [B] = L^{3 - 1/4} = L^{11/4} \]

### Step 7: Find the dimensional formula for \( A^2B \)
Now we can find the dimensional formula for \( A^2B \):
\[ [A^2B] = [A^2] \cdot [B] = L^6 \cdot L^{11/4} = L^{6 + 11/4} = L^{24/4 + 11/4} = L^{35/4} \]

### Step 8: Consider the dimensions of kinetic energy in the denominator
The denominator also contains \( C \), which must have the same dimensions as \( A \) for consistency. Thus, \( [C] = L^3 \).

### Step 9: Combine the dimensions to find \( A^2B \)
Finally, we need to consider the dimensions of kinetic energy \( K \) in the denominator:
\[ [B x^{1/4}] \text{ must have dimensions } M L^2 T^{-2} \text{ as well.} \]
Thus:
\[ [B] = \frac{L^3}{L^{1/4}} = L^{11/4} \]

### Step 10: Final dimensional formula for \( A^2B \)
Now, we can combine everything:
\[ [A^2B] = L^6 \cdot L^{11/4} = L^{35/4} \]
However, we need to consider the kinetic energy's mass dimension:
\[ [A^2B] = \frac{L^{35/4}}{M L^2 T^{-2}} = M^{-1} L^{27/4} T^2 \]

### Final Answer
Thus, the dimensional formula for \( A^2B \) is:
\[ [A^2B] = M^{-1} L^{27/4} T^2 \]