Step 1: Understanding the Question:
The problem requires matching physical quantities with their correct dimensional formulas. Dimension of a quantity is the power to which fundamental units (Mass [M], Length [L], Time [T], Current [I]) are raised to represent it. Step 2: Key Formula or Approach:
We use basic physical definitions to derive dimensions:
Energy (\(E\)) = \([ML^{2}T^{-2}]\)
Electric Potential (\(V\)) = \(\frac{\text{Work}}{\text{Charge}}\)
Planck's constant (\(h\)) from \(E = hf\) Step 3: Detailed Explanation:
(A) Work function (f): It is a measure of energy required to remove an electron. Its dimensions are the same as energy:
\[ [f] = [ML^{2}T^{-2}] \Rightarrow \text{Matches (3)} \]
(B) Stopping potential (\(v_{s}\)): It is electric potential. Potential = \(\frac{\text{Work}}{\text{Charge}}\).
Charge (\(q\)) = \(Current \times Time = [IT]\).
\[ [v_{s}] = \frac{[ML^{2}T^{-2}]}{[IT]} = [ML^{2}T^{-3}I^{-1}] \Rightarrow \text{Matches (1)} \]
(C) Planck's constant (h): From the relation \(E = hf \Rightarrow h = \frac{E}{f}\).
Frequency (\(f\)) has dimensions \([T^{-1}]\).
\[ [h] = \frac{[ML^{2}T^{-2}]}{[T^{-1}]} = [ML^{2}T^{-1}] \Rightarrow \text{Matches (4)} \]
(D) Frequency (f): It is the number of oscillations per unit time.
\[ [f] = \frac{1}{[T]} = [T^{-1}] \Rightarrow \text{Matches (2)} \] Step 4: Final Answer:
The matching pairs are: A-3, B-1, C-4, and D-2. This corresponds to Option (B).
