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

If energy (E), velocity (V) and force (F) are taken as fundamental qua

1.	If energy (E), velocity (V) and force (F) are taken as fundamental quantity, then dimension of mass will be …..
Answer» `E^(1)v^(2)F^(0)` `E^(1)v^(-2)F^(0)` `E^(1)v^(-1)F^(1)` `E^(1)v^(-2)F^(1)` Solution :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 Kis 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)`