JEE Main 2025 Jan 23 Shift 1, Physics Q8: Units And Dimensions

JEE Main2025Jan 23, Shift 1Physics

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

A
$L$
B
$L^{3} T^{- 2}$
C
$L^{2} T^{- 2}$
D
$L^{2}$

Solution

Every term added to give $x (t)$ must itself have dimensions of length. The arguments of $sin$ and $cos$ are dimensionless, so:

$[A] = [B] = [D] = L$ , and $[C] [T]^{2} = L$ gives $[C] = L T^{- 2}$ .

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

Concept behind this question

Units And DimensionsNotes, formulas and examples →

Practice more Physics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2025 Jan 23 Shift 1, Physics Q8: Units And Dimensions

Concept behind this question

More previous year questions on Units And Dimensions

Practice more Physics