Question

If the velocity of light = c, acceleration due to gravity =g and mass = m are taken as fundamental quantity then find out dimensional formula for universal gravitational constant = G.

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

For these sums we need to know the Dimension formula,

of the given Quantities,

$\large{\bold{G = M^{-1} L^3 T^{-2}}}$

$\large{\bold{c = M^0 L^1 T^{-1}}}$

$\large{\bold{g = M^0 L^1 T^{-2}}}$

$\large{\bold{m = M^1 L^0 T^0}}$

$\huge{\bold{\underline{Explanation:-}}}$

Let be the fundamental Quantities be,

$\large{\bold{[G] = [c]^a [g]^b [m]^c} -----(1)}$

Now, Substituting the Dimensional formula,

$\large{M^{-1} L^3 T^{-2} = [ M^0 L^1 T^{-1}]^a [ M^0 L^1 T^{-2}]^b [ M^1 L^0 T^0]^c}$

Now,

$\large{M^{-1} L^3 T^{-2} = [L^a T^{-a}] [L^b T^{-2b}] [M^c]}$

Rearranging,

$\large{M^{-1} L^3 T^{-2} = L^{(a + b)} T^{(-a - 2b)} M^c}$

Now, Comparing Quantities,

Case-1

$\large{M^{-1} = M^c}$

$\large{\bold{c = - 1}}$

Case-2

$\large{L^3 = L^{(a + b)}}$

$\large{a + b = 3 -----(2)}$

Case-3.

$\large{T^{-2} = T^{ (- a - 2b)}}$

$\large{ - a - 2b = -2}$

Multiplying by minus(-) on both sides,

$\large{a + 2b = 2 -----(3)}$

Now from equation 2 and 3.

Simplifying we get

(I.e Subtract equation (2) from equation(3))

$\large{\bold{b = -1}}$

and,

$\large{\bold{a = 4}}$

Substituting the values in equation (1).

$\large{\bold{[G] = [c]^a [g]^b [m]^c}}$

Now,

$\large{G = [c]^4 [g]^{-1} [m]^{-1}}$

It can be written as,

$\huge{\boxed{\boxed{G = \frac{c^4}{gm}}}}$

Hence derived.

Note:-

"g" denotes acceleration due to gravity.

"c" denotes Speed of light.

"m" denotes Mass.