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

Answer

C

Explanation

All the terms in the equations must have the dimension of force
`:. [A sin (C t)] -= MLT^(-2)`
And `[B cos Dx] -= MLT^(-2)`
Since `sin Ct` and `cos Dx` are dimensionless
`[A]=MLT^(-2)`
`[B]=MLT^(-2)`
`rArr (A)/(B)=M^(0)L^(0)T^(0)`
Since Ct and Dx are dimensionless
So, `[C]=M^(0)L^(0)T^(-1)`
`[D]=M^(0)L^(-1)T^(0)`
`rArr (C)/(D) = M^(0)LT^(-1)`