If energy (E), velocity (V) and force (F) are taken as fundamental quantity, then dimension of mass will be …..

Options

A. `E^(1)v^(2)F^(0)`
B. `E^(1)v^(-2)F^(0)`
C. `E^(1)v^(-1)F^(1)`
D. `E^(1)v^(-2)F^(1)`

Answer

A, B

Explanation

Energy `E=(1)/(2)mv^(2)`
`:.m=(2E)/(v^(2))` , 2 is constant
`:.[m]=[E^(1)][v^(-2)]`
`:.[m]=E^(1)v^(-2)F^(0)`
Second method :
Mass `m=KE^(a)v^(b)F^(c).......(i)`
where a, b, c `in` R and K is dimensionless constant.
Write dimension on both side,
`M^(1)L^(0)T^(0)=(M^(1)L^(2)T^(-2))^(a)(L^(1)T^(-1))^(b)(M^(1)L^(1)T^(-2))^(c)`
`=M^(a)L^(2b)T^(-2a)xxL^(b)T^(-b)xxM^(c)L^(c)T^(-2c)`
`M^(1)L^(0)T^(0)=M^(a+c)L^(2a+b+c)T^(-2a-b-2c)`
Comparing dimension of M, L, T
`a+c=1" ".......(i)`
`2a+b+c=0" ".......(ii)`
`-2a-b-2c=0" ".......(iii)`
Adding equation (ii) and (iii),
`-c=0`
`:.c=0`
`:.` From equation (i) `a+0=1`
`:.a=1`
From equation (ii) `2xx1+b+0=0`
`:.b=-2`
Substituting K=1, a=1, b=-1 and c=0
`m=E^(1)v^(-2)F^(0)`