If force, acceleration and time are taken as fundamental quantities, then the dimensions of length will be:

Options

A. `FT^(2)`
B. `F^(-1)A^(2)T^(-1)`
C. `FA^(2)T`
D. `AT^(2)`

Explanation

To find the dimensions of length when force, acceleration, and time are taken as fundamental quantities, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the dimensions of the fundamental quantities:**
- Force (F) has the dimension:
\[ [F] = [M][L][T^{-2}] \]
- Acceleration (a) has the dimension:
\[ [a] = [L][T^{-2}] \]
- Time (t) has the dimension:
\[ [t] = [T] \]

2. **Express the dimensions of length (L) in terms of force, acceleration, and time:**
- Assume the dimensions of length can be expressed as:
\[ [L] = [F^a][a^b][t^c] \]
- Substitute the dimensions of force and acceleration:
\[ [L] = ([M][L][T^{-2}])^a \cdot ([L][T^{-2}])^b \cdot ([T])^c \]

3. **Expand the expression:**
- This gives:
\[ [L] = [M^a][L^{a+b}][T^{-2a - 2b + c}] \]

4. **Set up the equations based on the dimensions:**
- We want the dimensions of length to be:
\[ [L] = [L^1][M^0][T^0] \]
- This leads to the following equations:
- For mass (M): \( a = 0 \)
- For length (L): \( a + b = 1 \)
- For time (T): \( -2a - 2b + c = 0 \)

5. **Solve the equations:**
- From \( a = 0 \), substitute into \( a + b = 1 \):
\[ 0 + b = 1 \Rightarrow b = 1 \]
- Substitute \( a = 0 \) and \( b = 1 \) into \( -2a - 2b + c = 0 \):
\[ -2(0) - 2(1) + c = 0 \Rightarrow c = 2 \]

6. **Conclusion:**
- Now we have \( a = 0 \), \( b = 1 \), and \( c = 2 \).
- Therefore, the dimension of length can be expressed as:
\[ [L] = [F^0][a^1][t^2] = [a][t^2] \]
- Thus, the dimensions of length when force, acceleration, and time are taken as fundamental quantities is:
\[ [L] = [L][T^2] \]

### Final Answer:
The dimensions of length are given by:
\[ [L] = [a][t^2] \]

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