The dimensions of `a/b` in the equation `P=(a-t^(2))/(bx)` where `P` is pressure, `x` is distance and `t` is time are

Options

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

Answer

To find the dimensions of \( \frac{a}{b} \) in the equation \( P = \frac{a - t^2}{bx} \), where \( P \) is pressure, \( x \) is distance, and \( t \) is time, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the dimensions of each variable**:
- Pressure \( P \) is defined as force per unit area. The dimensions of force \( F \) are given by:
\[ [F] = [m][a] = [m][L][T^{-2}] = [M][L][T^{-2}] \]
Therefore, the dimensions of pressure \( P \) are:
\[ [P] = \frac{[F]}{[A]} = \frac{[M][L][T^{-2}]}{[L^2]} = [M][L^{-1}][T^{-2}] \]

2. **Identify the dimensions of \( x \) and \( t \)**:
- Distance \( x \) has dimensions:
\[ [x] = [L] \]
- Time \( t \) has dimensions:
\[ [t] = [T] \]

3. **Determine the dimensions of \( a \)**:
- In the equation \( P = \frac{a - t^2}{bx} \), for \( a \) and \( t^2 \) to be added, they must have the same dimensions. The dimensions of \( t^2 \) are:
\[ [t^2] = [T^2] \]
- Therefore, the dimensions of \( a \) must also be:
\[ [a] = [T^2] \]

4. **Determine the dimensions of \( b \)**:
- Rearranging the equation gives us:
\[ b = \frac{a - t^2}{Px} \]
- Since \( a \) and \( t^2 \) have the same dimensions, we can simplify to:
\[ b = \frac{[T^2]}{[P][L]} \]
- Substituting the dimensions of \( P \):
\[ b = \frac{[T^2]}{[M][L^{-1}][T^{-2}][L]} = \frac{[T^2]}{[M][L^0][T^{-2}]} = \frac{[T^2]}{[M][T^{-2}]} = [M^{-1}][T^4] \]

5. **Calculate the dimensions of \( \frac{a}{b} \)**:
- Now we find the dimensions of \( \frac{a}{b} \):
\[ \frac{a}{b} = \frac{[T^2]}{[M^{-1}][T^4]} = [M^1][T^{-2}] \]

### Final Result:
The dimensions of \( \frac{a}{b} \) are:
\[ [M^1][L^0][T^{-2}] \]