Q141 / 150 · MHT CET 2020 (16 Oct Shift 1) · Physics

Let force $F=A \sin (C t)+B \cos (D x)$ where $x$ and $t$ are displacement and time respectively. The dimensions of $\frac{\mathrm{C}}{\mathrm{D}}$ are same as dimensions of

  • A
    angular Velocity
  • B
    angular monnentun
  • C
    Velocity gradient
  • Correct answer
    D
    Velocity

Solution

$$ \mathrm{F}=\mathrm{A} \sin \mathrm{Ct}+\mathrm{B} \cos \mathrm{Dx} $$ Since the argument of trigonometric functions is the angle, which is dimensionless $$ \begin{aligned} & \therefore[\mathrm{Ct}]=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0} \\ & \Rightarrow\left[\mathrm{CT}^{1}\right]=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0} \\ & \Rightarrow[\mathrm{C}]=\mathrm{T}^{-1}...(1) \end{aligned} $$ Similarly $$ \begin{aligned} &[\mathrm{Dx}]=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0} \\ \Rightarrow &\left[\mathrm{DL}^{1}\right]=\mathrm{M}^{0} \mathrm{~L}^{0} \mathrm{~T}^{0} \\ \Rightarrow &[\mathrm{D}]=\mathrm{L}^{-1}...(2) \end{aligned} $$

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

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