Dimensions of universal gravitational constant (G) in terms of...

The dimensional formula of the universal gravitational constant $G$ is obtained from Newton's law of gravitation $F = \dfrac{G m_1 m_2}{r^2}$.

$[G] = \dfrac{[F][r^2]}{[m_1 m_2]} = \dfrac{[MLT^{-2}][L^2]}{[M^2]} = [M^{-1}L^3T^{-2}]$

The dimensional formula of Planck's constant $h$ is obtained from the energy of a photon $E = h\nu$.

$[h] = \dfrac{[E]}{[\nu]} = \dfrac{[ML^2T^{-2}]}{[T^{-1}]} = [ML^2T^{-1}]$

Let the dimensions of $G$ be expressed as $[G] = [h]^a [L]^b [M]^c [T]^d$.

Substituting the dimensions, we get:

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

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

Equating the powers of $M$, $L$, and $T$ on both sides:

$a + c = -1$

$2a + b = 3$

$-a + d = -2$

Assuming $a = 1$ (as per the given options), we get:

$c = -1 - 1 = -2$

$b = 3 - 2(1) = 1$

$d = -2 + 1 = -1$

Therefore, the dimensions of $G$ are $[h^1 L^1 M^{-2} T^{-1}]$, which can be written as $[hT^{-1}LM^{-2}]$.

Answer: $[hT^{-1}LM^{-2}]$