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^(@)L^(@)T^(@),M^(@)L^(1)T^(-1)`
B. `MLT^(-1),M^(@)L^(-1)T`
C. `M^(@)L^(@)T^(@),M^(@)L^(-1)T^(1)`
D. `M^(@)LT^(-1),M^(@)L^(@)T^(@)`

Answer

A

Explanation

`F=A sin Ct=B cos Dx`
As `sin theta` and `cos theta` are dimensionless therefore A and B both, dimensions of force `[MLT^(-2)]`
`:.A/B=([MLT^(-2)])/([MLT^(-2)])=[M^(@)L^(@)T^(@)]," "` Also `C=1/t=1/T` and `D=1/x=1/L`
`:.C/D=L/T=[M^(@)L^(-1)T^(-1)]`