Force acting on a particle is given by `F=(A-x)/(Bt)`, where x is in metre and t is in seconds. The dimensions of B is -

Options

A. `MLT^(-2)`
B. `M^(-1)T^(-3)`
C. `M^(-1)T`
D. `MT^(-1)`

Answer

To find the dimensions of \( B \) in the equation \( F = \frac{A - x}{Bt} \), we will follow a step-by-step approach:

### Step 1: Understand the given equation
The force \( F \) is expressed as:
\[ F = \frac{A - x}{Bt} \]
where:
- \( F \) is the force,
- \( A \) is a constant (we will assume it has dimensions of length),
- \( x \) is the displacement (in meters),
- \( B \) is the unknown we need to find the dimensions for,
- \( t \) is time (in seconds).

### Step 2: Identify the dimensions of known quantities
- The dimension of force \( F \) is given by:
\[ [F] = MLT^{-2} \]
- The dimension of displacement \( x \) is:
\[ [x] = L \]
- The dimension of time \( t \) is:
\[ [t] = T \]

### Step 3: Set up the equation based on dimensional homogeneity
According to the principle of dimensional homogeneity, the dimensions on both sides of the equation must be equal. Thus, we can write:
\[ [F] = \frac{[A] - [x]}{[B][t]} \]
Since \( A \) is assumed to have the same dimension as \( x \), we have:
\[ [A] = L \]
So, substituting the dimensions we have:
\[ MLT^{-2} = \frac{L - L}{[B][T]} \]

### Step 4: Simplify the equation
The term \( A - x \) simplifies to \( 0 \) when \( A = x \). However, we need to analyze the dimensions of \( B \) with respect to the remaining terms. We can express the equation as:
\[ F = \frac{0}{Bt} \]
This does not help us find \( B \). Instead, we should consider the general form of the equation without assuming \( A = x \).

### Step 5: Rearranging the equation
We can express the equation as:
\[ F = \frac{A}{Bt} - \frac{x}{Bt} \]
This means both terms must have the same dimensions as \( F \).

### Step 6: Analyze one term for dimensions
Considering the term \( \frac{x}{Bt} \):
\[ \text{Dimensions of } \frac{x}{Bt} = \frac{L}{[B][T]} \]
Setting this equal to the dimensions of force:
\[ MLT^{-2} = \frac{L}{[B][T]} \]

### Step 7: Solve for the dimensions of \( B \)
Rearranging gives:
\[ [B] = \frac{L}{MT^{-2} \cdot T} = \frac{L}{MT^{-1}} = \frac{L T^{-1}}{M} \]
Thus, the dimensions of \( B \) are:
\[ [B] = M^{-1} L T \]

### Final Answer
The dimensions of \( B \) are:
\[ [B] = M^{-1} T \]