Step 1: Understanding the Concept:
We need to find the dimensional formula of the universal gravitational constant \(G\) and match it with an expression given in terms of \(h\), \(L\), \(M\), and \(T\). Step 2: Key Formula or Approach:
From Newton's law of gravitation, \(F = \frac{G M_1 M_2}{r^2} \implies G = \frac{F r^2}{M^2}\).
From the energy of a photon, \(E = \frac{h c}{\lambda} \implies h = E \cdot T\).
Find the base dimensions of \(G\) and \(h\) in terms of standard M, L, T.
Then substitute \(h\) into the given options to see which one equals the dimensions of \(G\). Step 3: Detailed Explanation:
First, find the dimensions of \(G\):
\[ [G] = \frac{[Force] \times [Distance]^2}{[Mass]^2} = \frac{(M L T^{-2}) (L^2)}{M^2} = [M^{-1} L^3 T^{-2}] \]
Next, find the dimensions of \(h\):
\[ [h] = [Energy] \times [Time] = (M L^2 T^{-2}) (T) = [M L^2 T^{-1}] \]
Now, let's test the options to see which one simplifies to \([M^{-1} L^3 T^{-2}]\).
Let's test Option (B): \([h T^{-1} L M^{-2}]\).
Substitute the dimensional formula for \(h\):
\[ [h T^{-1} L M^{-2}] = (M L^2 T^{-1}) \cdot T^{-1} \cdot L \cdot M^{-2} \]
Group the like terms:
\[ = (M \cdot M^{-2}) \cdot (L^2 \cdot L) \cdot (T^{-1} \cdot T^{-1}) \]
\[ = M^{-1} L^3 T^{-2} \]
This perfectly matches the dimensional formula for \(G\). Step 4: Final Answer:
The dimensions of \(G\) are \([h T^{-1} L M^{-2}]\).
