Hydrostatic pressure 'P' varies with displacement 'x' as `P = (A)/(B)log(Bx^(2) +C)` where A,B, and C are constants. The dimensional formula for 'A' is

Options

A. `[M^(1)L^(-1)T^(-2)]`
B. `MLT^(-2)]`
C. `[ML^(-2)T^(-2)]`
D. `ML^(-3)T^(-2)]`

Explanation

To find the dimensional formula for the constant \( A \) in the equation \( P = \frac{A}{B} \log(Bx^2 + C) \), we will follow these steps:

### Step 1: Understand the given equation
The equation relates hydrostatic pressure \( P \) to displacement \( x \). The term \( \log(Bx^2 + C) \) is dimensionless because the logarithm of a quantity must be dimensionless.

### Step 2: Identify the dimensions of pressure \( P \)
Hydrostatic pressure \( P \) is defined as force per unit area. The dimensional formula for force is given by:
\[ \text{Force} = \text{mass} \times \text{acceleration} = MLT^{-2} \]
The dimensional formula for area is:
\[ \text{Area} = L^2 \]
Thus, the dimensional formula for pressure \( P \) is:
\[ P = \frac{\text{Force}}{\text{Area}} = \frac{MLT^{-2}}{L^2} = ML^{-1}T^{-2} \]

### Step 3: Analyze the logarithmic term
Since \( \log(Bx^2 + C) \) is dimensionless, the term \( Bx^2 + C \) must also be dimensionless. This implies that \( Bx^2 \) must have the same dimensions as \( C \).

### Step 4: Determine the dimensions of \( x \)
The displacement \( x \) has dimensions of length:
\[ [x] = L \]
Thus, \( x^2 \) has dimensions:
\[ [x^2] = L^2 \]

### Step 5: Find the dimensions of \( B \)
For \( Bx^2 \) to be dimensionless:
\[ [B] \cdot [x^2] = [B] \cdot L^2 \implies [B] = L^{-2} \]

### Step 6: Relate dimensions of \( A \) and \( B \) to \( P \)
From the equation \( P = \frac{A}{B} \log(Bx^2 + C) \), since \( \log(Bx^2 + C) \) is dimensionless, we have:
\[ [P] = \frac{[A]}{[B]} \]
Substituting the dimensions we found:
\[ ML^{-1}T^{-2} = \frac{[A]}{L^{-2}} \]
This implies:
\[ [A] = P \cdot [B] = (ML^{-1}T^{-2}) \cdot (L^{-2}) = ML^{-3}T^{-2} \]

### Step 7: Conclusion
The dimensional formula for \( A \) is:
\[ [A] = ML^{-3}T^{-2} \]