If force `F`, acceleration `a, and time `T` are taken as the fundamental physical quantities , the dimensions of length on this systemof units are

Options

A. `FAT^(2)`
B. `FAT`
C. `FT`
D. `AT^(2)`

Explanation

To find the dimensions of length in terms of the fundamental physical quantities force (F), acceleration (a), and time (T), we can follow these steps:

### Step 1: Write the dimensions of the quantities involved.
- The dimension of force (F) is given by:
\[ [F] = [M][L][T^{-2}] \]
- The dimension of acceleration (a) is given by:
\[ [a] = [L][T^{-2}] \]
- The dimension of time (T) is:
\[ [T] = [T] \]

### Step 2: Assume the dimensions of length (L) in terms of F, a, and T.
Let the dimensions of length (L) be expressed as:
\[ [L] = [F^a][a^b][T^c] \]

### Step 3: Substitute the dimensions of F, a, and T into the equation.
Substituting the dimensions we have:
\[ [L] = [F^a][a^b][T^c] = ([M][L][T^{-2}])^a \cdot ([L][T^{-2}])^b \cdot [T^c] \]

### Step 4: Expand the equation.
Expanding this gives:
\[ [L] = [M^a][L^{a+b}][T^{-2a-2b+c}] \]

### Step 5: Set up the equations based on dimensional analysis.
For the dimensions of length, we can equate the powers of M, L, and T:
1. For mass (M): \( a = 0 \)
2. For length (L): \( a + b = 1 \)
3. For time (T): \( -2a - 2b + c = 0 \)

### Step 6: Solve the equations.
From the first equation, we have:
\[ a = 0 \]
Substituting \( a = 0 \) into the second equation:
\[ 0 + b = 1 \implies b = 1 \]
Now substituting \( a = 0 \) and \( b = 1 \) into the third equation:
\[ -2(0) - 2(1) + c = 0 \implies c = 2 \]

### Step 7: Write the final expression for the dimensions of length.
Now substituting the values of a, b, and c back into the expression for length:
\[ [L] = [F^0][a^1][T^2] = [a][T^2] \]

Thus, the dimensions of length in this system of units are:
\[ [L] = a T^2 \]