If force (F), velocity (v) and time (T) are taken as fundamental units, the dimension of mass are .....

Options

A. `[FV^(-1)]`
B. `[FVT^(-2)]`
C. `[FV^(-1)T^(-1)]`
D. `[FV^(-1)T]`

Explanation

`m=kF^(a)v^(b)T^(c) .....(1)`
Where k is dimensionless constant and a, b,CER comparing dimension of both sides,
`M^(1)L^(0)T^(0)=[M^(1)L^(1)T^(-2)]^(a)xx[M^(0)L^(1)T^(1)]^(c)`
`=M^(a)L^(a)T^(-2a)xxL^(b)T^(-b)xxT^(c)`
`M^(1)L^(0)T^(0)=M^(a)L^(a+b)T^(-2b-b+c)`
Comparing powers of M,L,T
`a=1,a+b=0 " " -2a+c=0`
`:.1+b=0, " " -2xx1-(-1)+c=0`
`:.b=-1 " " :. -2+1+c=0`
`:. c=1`
`:. m=F^(1)v^(-1)T^(1)` [From equ. (1)]