Dimensions of Universal Gravitational Constant in Terms of Planck Constant Distance Mass and Time

To find the dimensions of the universal gravitational constant ($G$) in terms of Planck's constant ($h$), distance ($L$), mass ($M$), and time ($T$), we first express the dimensional formulas of $G$ and $h$ in terms of fundamental dimensions of mass ($M$), length ($L$), and time ($T$).

Step 1: Dimensions of Universal Gravitational Constant ($G$)

From Newton's Law of Gravitation: $F = \frac{G m_1 m_2}{r^2} \implies G = \frac{F r^2}{m_1 m_2}$

Substituting the dimensions of force ($[F] = [M L T^{-2}]$), distance ($[r] = [L]$), and mass ($[m] = [M]$): $[G] = \frac{[M L T^{-2}] [L^2]}{[M^2]} = [M^{-1} L^3 T^{-2}]$

Step 2: Dimensions of Planck's Constant ($h$)

From Planck's equation: $E = h \nu \implies h = \frac{E}{\nu}$

Substituting the dimensions of energy ($[E] = [M L^2 T^{-2}]$) and frequency ($[\nu] = [T^{-1}]$): $[h] = \frac{[M L^2 T^{-2}]}{[T^{-1}]} = [M L^2 T^{-1}]$

Step 3: Dimensional Analysis for $G$

Let the dimension of $G$ in terms of $h$, $T$, $L$, and $M$ be represented as: $[G] = [h]^a [T]^b [L]^c [M]^d$

Substitute the fundamental dimensions into this equation: $[M^{-1} L^3 T^{-2}] = [M L^2 T^{-1}]^a [T]^b [L]^c [M]^d$ $[M^{-1} L^3 T^{-2}] = M^{a+d} L^{2a+c} T^{-a+b}$

Step 4: Comparing Exponents

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

1. For $M$: $a + d = -1$
2. For $L$: $2a + c = 3$
3. For $T$: $-a + b = -2$

Assuming $a = 1$ (as present in the given options):

• From $a + d = -1 \implies 1 + d = -1 \implies d = -2$
• From $2a + c = 3 \implies 2(1) + c = 3 \implies c = 1$
• From $-a + b = -2 \implies -1 + b = -2 \implies b = -1$

Conclusion

Substituting $a = 1, b = -1, c = 1, d = -2$ gives: $[G] = [h T^{-1} L M^{-2}]$

This corresponds to Option B.