Find the dimension of universal gravitational constant taking the units of density D velocity V and work W as base units instead of those of mass, distance and time.

Explanation

From Newton.s law of gravitation F = `G(M_(1)M_(2))/r^(2)` we
[G] = `([F][r^(2)])/[[m^(2)])=(MLT^(-2)xxL^(2))/(M^(2))=M^(-1)L^(3)T^(-2)`
Let `M^(-1)L^(3)T^(-2)=D^(x)V^(y)W^(z)`
`=(ML^(-3))^(x)xx(LT^(-1))^(y)xx(ML^(2)T^(-2))^(z)`
`= M^(x+z)xxL^(-3x+y+2z)xxT^(-y-2z)`
Equating powers of same bases,
x+z = -1
-3x+y+2z=3
and `" "` -y-2z = -2
Solving x,y,z, we get x = `-(1)/(3),y=(10)/(3) "and"z=-(2)/(3)`
`:." "[G]="D"^(-(1)/(3))"V"^(10/(3))"W"^(-(2)/(3))`