Q53 / 75 · JEE Main 2025 (23 Jan Shift 1) · Physics

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

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

Solution

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

Hence, the correct answer is option: 'A'.

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