The position of a particle moving on x-axis is given by $$x(t)=A\sin t+B\cos^{2}t+Ct^{2}+D$$, where is time. The dimension of $$\frac{ABC}{D}$$ is

To find the dimension of $$\frac{ABC}{D}$$, observe that $$x(t) = A\sin t + B\cos^2 t + Ct^2 + D$$ represents a position, so each term on the right must have dimension of length $$[L]$$.

Since $$\sin t$$ and $$\cos^2 t$$ are dimensionless and $$[t^2] = [T^2]$$, we have $$[A\sin t] = [L]$$ implying $$[A] = [L]$$, $$[B\cos^2 t] = [L]$$ implying $$[B] = [L]$$, and $$[Ct^2] = [L]$$ so $$[C][T^2] = [L]$$ and thus $$[C] = [LT^{-2}]$$. The constant $$D$$ must satisfy $$[D] = [L].$$

Therefore,

$$\left[\frac{ABC}{D}\right] = \frac{[L][L][LT^{-2}]}{[L]} = \frac{L^3 T^{-2}}{L} = L^2 T^{-2}$$

The correct answer is Option (1): $$L^2T^{-2}$$.