Question

The force $F$ is given by expression $F=A \cos (B x)+C \sin (D t)$, where $x$ is the displacement and $t$ is the time. Then dimension of $\frac{D}{B}$ are same as that of

• A. velocity $[LT^{-1}]$
• B. angular velocity $[T^{-1}]$
• C. angular momentum $[ML^2T^{-1}]$
• D. velocity gradient $[T^{-1}]$

Answer 1

Answer: (A)

velocity $[LT^{-1}]$

Answer 2

In trigonometric function like $\sin \,\theta, \cos \,\theta$ etc $\theta$ is dimensionless So, $\left.[B x]= M ^{0} L^{0} T^{0}\right]$ $[B]=\left[ L ^{-1}\right]$ Also, $[Dt] = [M^0L^0T^0]$ $[D] = [T^{-1}]$ Now $\left[\frac{D}{B}\right]=\left[ LT ^{-1}\right]$