If force F is related with distance x and time t as `F=Asqrtx+Bt^(2),` the dimensions of `(A)/(B)` is

Options

A. `M^(0)L^(-1//2)T`
B. `ML^(-1//2)T^(-2)`
C. `M^(0)L^(-1//2)T^(2)`
D. `M^(0)LT^(-2)`

Answer

To solve the problem, we need to find the dimensions of \( \frac{A}{B} \) given the equation \( F = A \sqrt{x} + B t^2 \).

### Step-by-Step Solution:

1. **Identify the dimensions of force \( F \)**:
- The dimension of force \( F \) is given by \( [F] = M L T^{-2} \).

2. **Analyze the equation**:
- The equation is \( F = A \sqrt{x} + B t^2 \).
- Since both terms on the right side are being added, their dimensions must be equal to the dimension of \( F \).

3. **Find the dimension of \( \sqrt{x} \)**:
- Let \( x \) be a distance, so \( [x] = L \).
- Therefore, \( [\sqrt{x}] = [x]^{1/2} = L^{1/2} \).

4. **Set up the equation for \( A \)**:
- From the term \( A \sqrt{x} \), we have:
\[ [F] = [A] [\sqrt{x}] \implies M L T^{-2} = [A] L^{1/2} \]
- Rearranging gives:
\[ [A] = \frac{M L T^{-2}}{L^{1/2}} = M L^{1/2} T^{-2} \]

5. **Find the dimension of \( t^2 \)**:
- The dimension of time \( t \) is \( [t] = T \).
- Thus, \( [t^2] = T^2 \).

6. **Set up the equation for \( B \)**:
- From the term \( B t^2 \), we have:
\[ [F] = [B] [t^2] \implies M L T^{-2} = [B] T^2 \]
- Rearranging gives:
\[ [B] = \frac{M L T^{-2}}{T^2} = M L T^{-4} \]

7. **Find the dimension of \( \frac{A}{B} \)**:
- Now we can find the ratio \( \frac{A}{B} \):
\[ \frac{[A]}{[B]} = \frac{M L^{1/2} T^{-2}}{M L T^{-4}} \]
- The \( M \) cancels out:
\[ = \frac{L^{1/2} T^{-2}}{L T^{-4}} = \frac{L^{1/2}}{L} \cdot \frac{T^{-2}}{T^{-4}} = L^{-1/2} T^{2} \]

8. **Final answer**:
- The dimensions of \( \frac{A}{B} \) are:
\[ [\frac{A}{B}] = L^{-1/2} T^{2} \]