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

Answer

A

Explanation

Given, `F=AsinCt+BcosDx`
where, `t`= time and `x` = distance
As, we know that trigonometric ratios are dimensionless.
This implies
`sin Ct` = dimensionless and `cosDx` = dimensionless
Also,`" "[C]=[(1)/(t)]=["T"^(-1)]and [D]=[(1)/(x)]=["L"^(-1)]` As, Eq. (i) represents the force. So, `A//B` is dimensionless i.e. `["M"^(0)"L"^(0)"T"^(0)]`
While, `" "[(C)/(D)]=[("T"^(-1))/("L"^(-1))]=["M"^(0)"LT"^(-1)]`