Dimensions of universal gravitational constant \((G)\) in terms of Planck's constant \((h)\), distance \((L)\), mass \((M)\) and time \((T)\) are _______.

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: \([hT^{-1}LM^{-2}]\)