Force `F` is given in terms of time `t` and distance `x` by `F = A sin Ct + B cos Dx`. Then the dimensions of `A//B` and `C//D` are

Options

A. `[M^(0) L^(0) T^(0)] , [M^(0) L^(0) T^(-1)]`
B. `[M L T^(-2)] , [M^(0) L^(-1) T^(0)]`
C. `[M^(0) L^(0) T^(0)] , [M^(0) L T^(-1)]`
D. `[M^(0) L^(1) T^(-1)] , [M^(0) L^(0) T^(0)]`

Answer

C

Explanation

`(A)/(B) = ("Force")/("Force") = [M^(0) L^(0) T^(0)]`
`Ct = angle rArr C = ("Angle")/("Time") = (1)/( T) = T^(-1)`
` Dx = angle rArr D = ("Angle")/("Distance") = (1)/( L) = L^(-1)`
` :. (C )/( D) = ( T^(-1))/(L^(-1)) = [ M^(0)LT^(-1)]`