Question

Potential energy of a particle is given as \( u = \dfrac{A\sqrt{x}}{B+x} \). Find the dimensions of \(A\) and \(B\).

• A. \( M^{3}L^{3/2}T^{-2},\, L \)
• B. \( ML^{5/2}T^{-1},\, L^{2} \)
• C. \( ML^{5/2}T^{-2},\, L \)
• D. \( M^{7/2}T^{-3},\, L \)

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
According to the principle of dimensional homogeneity, physical quantities can only be added or subtracted if they have the same dimensions. Both sides of a physical equation must also possess identical dimensions.
Step 2: Key Formula or Approach:
The given formula is: \[ U = \frac{A\sqrt{x}}{B+x} \] The dimension of potential energy is \([U] = \text{ML}^2\text{T}^{-2}\), and for distance, it is \([x] = \text{L}\). We equate dimensions: \([B] = [x]\) and \([U] = \frac{[A]\sqrt{[x]}}{[B+x]}\).
Step 3: Detailed Explanation:
Since \(B\) is added to \(x\) in the denominator, they must have the same dimension. \[ [B] = [x] = \text{L} \] Now substitute the known dimensions into the overall equation: \[ [U] = \frac{[A][x]^{1/2}}{[x]} \] \[ \text{ML}^2\text{T}^{-2} = \frac{[A]\text{L}^{1/2}}{\text{L}} \] Solving for \([A]\): \[ [A] = \text{ML}^2\text{T}^{-2} \times \frac{\text{L}}{\text{L}^{1/2}} \] \[ [A] = \text{ML}^2\text{T}^{-2} \times \text{L}^{1/2} = \text{ML}^{5/2}\text{T}^{-2} \]
Step 4: Final Answer:
The dimensions of A and B are \(\text{ML}^{5/2}\text{T}^{-2}\) and \(\text{L}\) respectively.