Force is given by the expression, F = A cos(B | vedclass

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(A) In the expression $F = A \cos(Bx) + C \cos(Dt)$, the argument of the trigonometric function must be dimensionless. Therefore, the dimensions of $B

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velocity

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(A) In the expression $F = A \cos(Bx) + C \cos(Dt)$, the argument of the trigonometric function must be dimensionless. Therefore, the dimensions of $Bx$ and $Dt$ must be equal to the dimension of a constant (dimensionless). $[Bx] = [M^0 L^0 T^0] \implies [B] = [x^{-1}] = [L^{-1}]$. $[Dt] = [M^0 L^0 T^0] \implies [D] = [t^{-1}] = [T^{-1}]$. Now, we find the dimension of $\left(\frac{D}{B}\right)$: $\left[\frac{D}{B}\right] = \frac{[T^{-1}]}{[L^{-1}]} = [L T^{-1}]$. Since $[L T^{-1}]$ represents the dimension of velocity, the correct option is $A$.

Force is given by the expression, $F = A \cos(Bx) + C \cos(Dt)$, where $x$ is displacement and $t$ is time. The dimension of $\left(\frac{D}{B}\right)$ is the same as that of

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