If velocity of light c, Planck's constant h ,Gravitational constant G are taken as fundamental quantites, then express mass, length and time in terms of dimensions of these quantites.

Explanation

We know that, dimensions of `(h)=[ML^(2)T^(-1)]`
`"Dimensions of "(c)=[LT^(-1)]`
Dimensions of gravitational constant `(G)=[M^(_1)L^(3)T^(-2)]`
`(i)"Let"" "mproptoc^(x)h^(y)G^(z)`
`Rightarrow m=kc^(x)h^(y)G^(z)" "...(i)`
where, k is a dimensionless constant of proportionality.
Substituting dimensions of each term in Eq. (i), we get
`[ML^(0)T^(0)]=[LT^(-1)]^(x)xx[ML^(2)T^(-1)]^(y)[M^(-1)L^(3)T^(-2)]^(z)`
`=[M^(y-z)L^(x+2y+2y+3z)T^(-x-y-2z)]`
Comparing powers of same terms on both sides, we get
`y-z=1" "...(ii)`
`x+2y+3z=0" "...(iii)`
`-x-y-2z=0" "...(iv)`
Adding Eqs. (ii),(iii) and (iv), we get
`2y=1 Rightarrow y=1/2`
Substituting value of y in Eq. (ii), we get
`z=-1/2`
From Eq. (iv)
`x=-y-2z`
Substituting values of y and z, we get
`x=-1/2-2(-1/2)=1/2`
Putting values of x, y and z in Eq. (i), we get
`m=kc^(1//2)h^(1//2)G^(-1//2)`
`Rightarrow m=ksqrt((ch)/G)`
`(ii)"Let "" "Lproptoc^(x)h^(y)G^(z)`
`Rightarrow L=kc^(x)h^(y)G^(z)`
where, k is a dimensionless constant.
Substituting dimensions of each term in Eq. (v), we get
`[M^(0)LT^(0)]=[LT^(-1)]^(x)xx[ML^(2)T^(-1)]^(y)xx[M^(-1)L^(3)T^(-2)]^(z)`
`=[M^(y-z)L^(x+2y+3z)T^(-x-y-2z)]`
On comparing powers of same terms, we get
`y-z=0" "...(vi)`
`x+2y+3z=1" "...(vii)`
`-x-y-2z=0" "...(viii)`
Adding Eqs. (vi), (vii), and (viii), we get
`2y=1`
`Rightarrow y=1/2`
Substituting value of y in Eq. (vi), we get
`z=1/2`
From Eq. (viii),
`x=-y-2z`
Substituting values of y and z, we get
`x=-1/2-2(1/2)=-3/2`
Putting values of x,y and z in Eq. (v), we get
`L=kc^(-3//2)h^(1//2)G^(1//2)`
`L=ksqrt(hG)/(c^(3))`
(iii)`"Let "Tproptoc^(x)h^(y)G^(z)`
`Rightarrow T=kc^(x)h^(y)G^(z)" "...(ix)`
where, k is a dimensionless constant.
Substituting dimensions of each term in Eq. (ix), we get
`[M^(0)L^(0)T^(0)=[LT^(-1)]^(x)xx[ML^(2)T^(-1)]^(y)xx[M^(-1)L^(3)T^(-2)]^(z)`
`=[M^(y-z)L^(x+2y+3z)T^(-x-y-2z)]`
On comparing powers of same terms, we get
`y-z=0" "...(x)`
`x+2y+3z=0" "...(x i)`
`-x-y-2z=1" "...(x ii)`
Adding Eqs. (x), (x i) and (xii), we get
`2y=1`
`Rightarrow y=1/2`
Substituting value of y in Eq. (x), we get
`z=y=1/2`
Form Eq. (xii),
`x=-y-2 z-1`
Substituting values of y and z, we get
`x=-1/2-2(1/2)-1=-5/2`
Putting values of x, y and z in Eq. (i x), we get
`T=kc^(-5//2)h^(1//2)G^(1//2)`
`T=ksqrt((hG)/c^(5))`