The force is given


The force is given in terms of time $t$ and displacement $x$ by the equation $F=A \cos B x+C \sin D t$

$\mathrm{F}=\mathrm{A} \cos \mathrm{Bx}+\mathrm{C} \sin \mathrm{Dt}$

The dimensional formula of $\frac{\mathrm{AD}}{\mathrm{B}}$ is :

  1. $\left[\mathrm{M}^{0} \mathrm{~L} \mathrm{~T}^{-1}\right]$

  2. $\left[\mathrm{M} \mathrm{L}^{2} \mathrm{~T}^{-3}\right]$

  3. $\left[\mathrm{M}^{1} \mathrm{~L}^{1} \mathrm{~T}^{-2}\right]$

  4. $\left[\mathrm{M}^{2} \mathrm{~L}^{2} \mathrm{~T}^{-3}\right]$

Correct Option: , 2


${[\mathrm{A}]=\left[\mathrm{MLT}^{-2}\right] }$

${[\mathrm{B}]=\left[\mathrm{L}^{-1}\right] }$

${[\mathrm{D}]=\left[\mathrm{T}^{-1}\right] }$

${\left[\frac{\mathrm{AD}}{\mathrm{B}}\right]=\frac{\left[\mathrm{MLT}^{-2}\right]\left[\mathrm{T}^{-1}\right]}{\left[\mathrm{L}^{-1}\right]} }$

${\left[\frac{\mathrm{AD}}{\mathrm{B}}\right]=\left[\mathrm{ML}^{2} \mathrm{~T}^{-3}\right] }$

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