If speed of light (c), acceleration due to gravity (g) and pressure (P) are taken as fundamental units, the dimensions of gravitational constant (G) are

Options

A. `[c^(0)gP^(-3)]`
B. `[c^(2)g^(3)P^(-2)]`
C. `c^(0)g^(2)P^(-1)]`
D. `c^(2)g^(2)P^(-2)]`

Answer

C

Explanation

Let `G= kc^(x)g^(y)P^(z)`
Where k is a dimensionless constant.
`:. [M^(-1)L^(3)T^(-2)]= [LT^(-1)]^(x)[LT^(-2)]^(y)[ML^(-1)T^(-2)]^(z)= [M^(z)L^(x+y-z)T^(-x-2y-2z)]`
Applying principle of homogeneity of dimensions we get,
z=- 1...(i)
x+y-z= 3 ...(ii)
-x-2y-2z= -2 ,..(iii)
On solving (i), (ii) and (iii) we get,
`x=0, y=2, z=-1 :. [G]= [c^(0)g^(2)P^(-1)]`