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

(C) $\mathrm{L}^{2} \mathrm{~T}^{-2}$

(B) $\mathrm{L}^{3} \mathrm{~T}^{-2}$

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Units & Measurement

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 t is time. The dimension of

\frac{ABC}{D}

is-

Answer: (C)

For

x (t) = A sin t + B cos^{2} t + C t^{2} + D

, the argument of the sine and cosine must be dimensionless, and each additive term must carry the dimension of

x

, which is

[L]

.

From

A sin t

[A] = [L]

.
From

B cos^{2} t

[B] = [L]

.
From

C t^{2}

[C] [T^{2}] = [L] \Rightarrow [C] = [L T^{- 2}]

.
From the constant term:

[D] = [L]

.

So

[\frac{A B C}{D}] = \frac{[ L ] [ L ] [ L T ^{- 2} ]}{[ L ]} = [L^{2} T^{- 2}]

JEE Main · 2025 · 23 Jan 2025 · Shift 1 · Single correct option