JEE Mains · Physics · STD 11 - Units,dimensions and measurement
- JEE Mains 2025
- Easy
- MCQ
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
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(L^2 T^{-2}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Dimension }[\mathrm{x}(\mathrm{t})]=[\mathrm{L}] \\ & {[\mathrm{A}]=[\mathrm{L}]} \\ & {[\mathbf{B}]=[\mathrm{L}]} \\ & {[\mathbf{C}]=\left[\mathrm{LT}^{-2}\right]} \\ & {[\mathrm{D}]=[\mathrm{L}]} \\ & {\left[\frac{\mathrm{ABC}}{\mathrm{D}}\right]=\left[\frac{\mathrm{L} \times \mathrm{L} \times \mathrm{LT}^{-2}}{\mathrm{~L}}\right]=\left[\mathrm{L}^2 \mathrm{~T}^{-2}\right]}\end{aligned}\)
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