Variation of force with displacement ‘x’ and time ‘t’ is given by $F = A \cos B x + C sin Dt$ .
Then the dimensions of $\frac{A}{C}$ and $\frac{D}{B}$ are

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$[M^{0} L^{0} T^{0}], [M^{0} {LT}^{- 1}]$

$[M^{0} L^{1} T^{- 1}], [M^{0} L^{0} T^{0}]$

$[M^{0} L^{- 1} T^{- 1}], [M^{0} L^{0} T^{0}]$

$[M^{0} L^{0} T^{0}], [{ML}^{0} T^{- 1}]$

answer is D.

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Detailed Solution

Dimensions of A and C are same and equal to force

$[A] = [C] = [M^{1} L^{1} T^{- 2}] [B] = [L^{- 1}] [D] = [T^{- 1}] \frac{A}{C} = M^{0} L^{0} T^{0}, \frac{D}{B} = \frac{T^{- 1}}{L^{- 1}} = [{LT}^{- 1}]$

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