Force $F$ is given in terms of time $t$ and distance $x$ by $F = A s i n C t + B c o s D x$ .Then the dimensions of $\frac{A}{B}$ and $\frac{C}{D}$ are given by :

M L T^{- 2}, M^{\circ} L^{\circ} T^{- 1}

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M L T^{- 2}, M^{\circ} L^{- 1} T^{\circ}

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M^{\circ} L^{\circ} T^{\circ}, M^{\circ} L^{1} T^{- 1}

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M^{\circ} L^{1} T^{- 1}, M^{\circ} L^{\circ} T^{\circ}

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Solution

The correct option is D $M^{\circ} L^{\circ} T^{\circ}, M^{\circ} L^{1} T^{- 1}$
Given force, $F = A s i n C t + B c o s D x$
We know that the trigonometric functions are dimensionless because they are ratio of length.
Thus, $s i n C t$ and $c o s D x$ will be dimensionless. So dimension of C should be $[T^{- 1}]$ and dimension of D should be $[L^{- 1}]$
So, $C / D = [T^{- 1}] / [L^{- 1}] = [L T^{- 1}] = [M^{0} L^{1} T^{- 1}]$
Since the dimension on each side of a equation is same so dimension of A and B should be equal to dimension of force.
Thus, $A / B = 1 = [M^{0} L^{0} T^{0}]$

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