If force `(F)` velocity `(V)` and time `(T)` are taken as fundamental units, then the dimensions of mass are

Options

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

Answer

d

Explanation

Let `m prop F^(x)V^(y)T^(z)`
By substituting the following dimension
`[F] = [MLT^(-2)],[V] = [LT^(-1)],[T]=[T]`
`M^(1)L^(0)T^(0)] = [MLT^(-1) [LT^(-1)][T]`
`rArr [M^(1)L^(0)T^(0)] = [M^(x)L^(x + y) T^(-2x -y +z)]`
`x = 1`
`x + y =0 , rArr y = -1`
` rArr -2 xx 1 - (-1) + z = 0`
`rArr z = 1`
Hence `[M] = [F^(1)V^(-1)T^(1)] = [FV^(1)T]`