Q 11-01-077JEE MainJEE Main 2025 (23 Jan, Shift 1)Easy
The position of a particle moving on the $x$-axis is given by $x(t) = A\sin t + B\cos^2 t + Ct^2 + D$, where $t$ is time. The dimension of $\dfrac{ABC}{D}$ is
Answer: (A) $[L^2T^{-2}]$
Every term must have the dimension of length. $\sin t$ and $\cos^2 t$ are dimensionless, so
$$[A] = [B] = [D] = [L], \qquad [C] = \frac{[L]}{[T^2]} = [LT^{-2}]$$
$$\left[\frac{ABC}{D}\right] = \frac{[L][L][LT^{-2}]}{[L]} = [L^2T^{-2}]$$
Solution by Sreeraj P, M.Sc Physics