Units & Dimensions

JEE Main level, single correct, medium

49 of 149

Question 49 of 149

The position of a particle moving on xx-axis is given by x(t)=Asin⁡t+Bcos⁡2t+Ct2+Dx(t)=A \sin t+B \cos ^2 t+C t^2+D, where tt is time. The dimension of ABCD\frac{A B C}{D} is

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Answer: (A) L2T−2L^2 T^{-2}

 Dimension [x(t)]=[L][A]=[L][B]=[L][C]=[LT−2][D]=[L][ABCD]=[L×L×LT−2 L]=[L2 T−2]\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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