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

[E] `=ML^(2)T^(-2)rArr [v]=M^(0)L^(1)T^(-1)rArr [F] =M^(1)L^(1)T^(-2)`
Let dimension of mass in E is =x
Let dimension of mass in V is =y
Let dimension of mass in F is =z
Then , `M=E^(x)V^(y)F^(x)`
`M^(1)L^(0)T^(0)=[ML^(2)T^(-2)]^(x)[M^(0)L^(1)T^(-1)]^(y)[M^(1)L^(1)T^(-2)]^(z)`
`M^(1)L^(0)T^(0)rArr M^(x+2)L^(2x+y+z)T^(-2x-y-2x)`
x+y+z=1..........(i)
2x+y+z=0.........(ii)
2x+y+2z=0..........(iii)
By equation (i), (ii) and (iii)
`x=1`
`y=-2`
`z=0`
So, [M]`=E^(1)V^(-2)`