The position of a particle moving on x -axis is given by x(t)=A t+B ^2 t+C t^2+D , where t is time. The dimension of (A B C)/(D) is

Correct answer: A. L^2 T^ -2 aligned & Dimension [x(t)]=[L] & [A]=[L] & [B]=[L] & [C]=[LT^ -2 ] & [D]=[L] & [ ABC D ]=[ L × L × LT^ -2 L ]=[L^2 T^ -2 ] aligned

JEE Main 202523 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 23 January
Shift: Morning Shift
Subject: Physics
Chapter: Units and Dimensions

JEE Main 2025 · 23 January · Morning Shift · Physics · Units and Dimensions

The position of a particle moving on x -axis is given by x(t)=A t+B...

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}

Options

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

Correct answer

A. $L^2 T^{-2}$

Step-by-step solution

\begin{aligned} & \text{ Dimension }[\mathrm{x}(\mathrm{t})]=[\mathrm{L}] \\ & {[\mathrm{A}]=[\mathrm{L}]} \\ & {[\mathbf{B}]=[\mathrm{L}]} \\ & {[\mathbf{C}]=\left[\mathrm{LT}^{-2}\right]} \\ & {[\mathrm{D}]=[\mathrm{L}]} \\ & {\left[\frac{\mathrm{\text{ ABC }}}{\mathrm{D}}\right]=\left[\frac{\mathrm{L} \times \mathrm{L} \times \mathrm{LT}^{-2}}{\mathrm{ L}}\right]=\left[\mathrm{L}^2 \mathrm{ T}^{-2}\right]}\end{aligned}

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