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 find the dimensions of \( \frac{A}{B} \) given the force equation \( F = A\sqrt{x} + Bt^2 \), we will follow these steps:

### Step 1: Identify the dimensions of force \( F \)
The dimension of force \( F \) is given by:
\[ [F] = MLT^{-2} \]

### Step 2: Analyze the first term \( A\sqrt{x} \)
The term \( \sqrt{x} \) represents the square root of distance \( x \). The dimension of distance \( x \) is:
\[ [x] = L \]
Thus, the dimension of \( \sqrt{x} \) is:
\[ [\sqrt{x}] = L^{1/2} \]
Now, for the term \( A\sqrt{x} \) to have the same dimension as force \( F \), we have:
\[ [A\sqrt{x}] = [F] \implies [A][\sqrt{x}] = MLT^{-2} \]
Substituting the dimension of \( \sqrt{x} \):
\[ [A][L^{1/2}] = MLT^{-2} \]
From this, we can solve for the dimension of \( A \):
\[ [A] = \frac{MLT^{-2}}{L^{1/2}} = ML^{1/2}T^{-2} \]

### Step 3: Analyze the second term \( Bt^2 \)
The dimension of \( t^2 \) is:
\[ [t^2] = T^2 \]
Thus, for the term \( Bt^2 \) to also have the dimension of force \( F \):
\[ [Bt^2] = [F] \implies [B][t^2] = MLT^{-2} \]
Substituting the dimension of \( t^2 \):
\[ [B][T^2] = MLT^{-2} \]
From this, we can solve for the dimension of \( B \):
\[ [B] = \frac{MLT^{-2}}{T^2} = ML T^{-4} \]

### Step 4: Calculate the dimensions of \( \frac{A}{B} \)
Now we can find the ratio \( \frac{A}{B} \):
\[ \frac{A}{B} = \frac{ML^{1/2}T^{-2}}{MLT^{-4}} = \frac{L^{1/2}}{L} \cdot \frac{T^{-2}}{T^{-4}} = L^{-1/2}T^{2} \]

### Final Result
Thus, the dimensions of \( \frac{A}{B} \) are:
\[ \frac{A}{B} = L^{-1/2}T^{2} \]