40. If the velocity of light c, Plancks constant h, and gravitational constant G are taken as fundamental quantities, then express mass, length, and time in terms of dimensions of these quantities.

Hint: Use the principle of the derivation of formula through dimensional analysis.
Step 1: Find the dimensions of individual quantities.

We know that dimensions of (h)=[ML2T1]Dimensionsof(c)=[LT1]Dimensions of gravitational constant (G)=[M1L3T2]
i Step 2: Define the dependence of m with C, h and G.
 Let mCxhyGzm=kCxhyG                                         ...(i)

where k is a dimensionless constant of proportionality.

Step 3: Put the dimensions of each quantity.
Substituting dimensions of each term in Eq (i), we get,

[ML0T0]=[LT1]x×[ML2T1]γ[M1L3T2]2=[MyzLx+2y+3zTxy2z]

Comparing powers of the same terms on both sides, we get,

y - z = 1                            ...(I)
x + 2y + 3z = 0                 ...(iii)
- x - y - 2z = 0                   ...(iii)

Adding Eqs. (i). (i) and (v), we get,
2y=1y=12
Substituting the value of y in Eq. (i), we get,
z=12
From Eq. (iv)
x = - y - 2z
Substituting values of y and z, we get
x=122(12)=12
Putting values of x, y, and z in Eq. (i), we get

m=kc1/2h1/2G1/2 m=kchG

(ii) Step 4: Define the dependence of L with C, h and G.

 Let LcxhyGzL=kcxhyGz

where K is a dimensionless constant.

Step 3: Put the dimensions of each quantity.

Substituting dimensions of each term in Eq. (v), we get

[M0LT0] = [LT1]x×[ML2T1]γ×[M1L3T2]z
              = [Myzx+2y+3zTxy2z]

On comparing powers of same terms, we get
y - z = 0                                  ...(vi)
x + 2y + 3z = 1                       ...(vii)
- x - y - 2z = 0                         ...(viii)

(iii) Let TcahbGc

T=cahbGc            ...........(ix)

where, k is a dimensionless constant.

Substituting dimensions of each term in Eq. (ix), we get

[M0L0T1]=[LT-1]a×[ML2T-1]b×[M-1L3T-2]c
=[Mb-cLa+2b+3cT-a-b-2c]

On comparing powers of the same terms, we get

b-c=0               ...(x)
a+2b+3c=1     ...(xi)
-a-b-2c=1   ...(xii)

Adding Eqs. (x), (xi) and (xii), we get

2b=1b=12 

Substituting the value of b in Eq. (x), we get

c=b=12

From Eq. (xii),

a=-b-2c-1

Substituting values of b and c, we get

a=-12-2(12)-1=-52

Putting values of a, b and c in Eq. (ix), we get

T=kc-5/2h1/2G1/2=khGc5