If the velocity `(v)` of a body in time `'t'` is given by `V = AT^(3) + BT^(2) + CT + D` then the dimensionas of `C` are...............

Options

A. `[LT^(-1)]`
B. `[LT^(-2)]`
C. `[LT^(-3)]`
D. `[LT^(-4)]`

Explanation

To find the dimensions of the constant \( C \) in the equation for velocity given by:

\[ V = AT^3 + BT^2 + CT + D \]

we need to analyze the dimensions of each term in the equation.

### Step-by-Step Solution:

1. **Identify the dimensions of velocity**:
The dimension of velocity \( V \) is given by:
\[ [V] = L T^{-1} \]
where \( L \) represents length and \( T \) represents time.

2. **Analyze each term in the equation**:
- The first term \( AT^3 \):
- The dimension of \( T^3 \) is \( [T^3] \).
- Therefore, the dimension of \( A \) must be such that:
\[ [A][T^3] = [V] \implies [A][T^3] = L T^{-1} \]
Rearranging gives:
\[ [A] = \frac{L T^{-1}}{T^3} = L T^{-4} \]

- The second term \( BT^2 \):
- The dimension of \( T^2 \) is \( [T^2] \).
- Therefore, the dimension of \( B \) must satisfy:
\[ [B][T^2] = [V] \implies [B][T^2] = L T^{-1} \]
Rearranging gives:
\[ [B] = \frac{L T^{-1}}{T^2} = L T^{-3} \]

- The third term \( CT \):
- The dimension of \( T \) is \( [T] \).
- Therefore, the dimension of \( C \) must satisfy:
\[ [C][T] = [V] \implies [C][T] = L T^{-1} \]
Rearranging gives:
\[ [C] = \frac{L T^{-1}}{T} = L T^{-2} \]

- The fourth term \( D \):
- The dimension of \( D \) must also match the dimension of velocity, which is \( L T^{-1} \).

3. **Conclusion**:
From our analysis, we found that the dimension of \( C \) is:
\[ [C] = L T^{-2} \]

### Final Answer:
The dimensions of \( C \) are \( L T^{-2} \).