If force, mass and time are taken to be base physical quantities with dimensional formula [F], [M] and [T], then :
Dimension of mass in velocity is :

Options

A. `-1`
B. `-2`
C. `1`
D. `2`

Answer

a

Explanation

`E prop F^(x) M^(y) T^(z)`
`kg m^(2) s^(-2) = [kg m s^(-2)]^(x) (kg)^(y)(s)^(z)`
`x+y=1`
`x=2`
`-2x+z=-2`
`x=2, y=-1, =2`
Hence, (Dimensional formula of ) `E= [F^(2)M^(-1) T^(2)]`
since, `E=[ML^(2)T^(-2)]=M[LT^(-1)]^(2)`
and `E=M^(-1)[FT]^(2)`
Hence, dimension of mass in velocity is `=-1`
Unit of pressure kg `m^(-1) s^(-2)`
Hence, `[ML^(-1)T^(-2)]=[MLT^(-2)]^(x)[M]^(y)[T]^(z)`
So, `x+y=1` and `-2x+z=-2`
`x=-1`
`rArr z=-4`