Force F is given in terms of time t and distance x by `F= A sinCt+ B cosDx`. Then dimensions of `(A)/(B)` and `(C)/(D)` are

Options

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

Answer

C

Explanation

As `[A]= [F]` and `[B]= [F]`
`:. [(A)/(B)]= ([F])/([F])= [M^(0)L^(0)T^(0)]`
As `sintheta` is dimensionless,
`:. [Ct]= 1` or `[C]=(1)/([t])= (1)/([T])= [T^(-1)]`
`[Dx]=1` or `[D]= (1)/([x])= (1)/([L])= [L^(-1)]`
`:. [(C)/(D)]= ([T^(-1)])/([L^(-1)])= [M^(0)LT^(-1)]`