If the energy ( E) ,velocity (v) and force (F) be taken as fundamental quantities,then the dimension of mass will be

Options

A. `Ev^(2)`
B. `Ev^(-2)`
C. `Fv^(-1)`
D. `Fv^(-2)`

Answer

B

Explanation

Let us assume that `mprop E^(a)v^(b)F^(c)`
`m=KE^(a)v^(b)f^(c)`…….i
where K is the constnat of proportional and is a dimensionaless constant
The dimensions of mass `m=[M]`
Energy `(E)=[ML^(2)T^(-2)],` velocity `(v)=[LT^(-1)]`, Force `(F)=[MLT^(-2)]`
Putting these dimensionless is eqn (i) we get `[M]=[ML^(2)T^(-2)]^(a)[LT^(-1)]^(b)[MLT^(-2)]^(c)`
equation the dimensions on both side we get `a+c=1` (ii)
`2a+b+c=0` (iii) `-2a-b-2c=0` (iv)
Solving ii, iii and iv we get
`a=1,b=-2,c=0" ":.[m]=K[E][v]^(-2)`