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

Concept: Dimensional analysis is used to determine the relationship between physical quantities. Key ideas used here:

• Dimension of potential energy: \( [U] = ML^2T^{-2} \)
• Quantities added together must have the same dimensions
• If \(x\) represents displacement, then \( [x] = L \)

Step 1: Determine the dimension of \(B\).} Given expression: \[ u=\frac{A\sqrt{x}}{B+x} \] Since \(B\) is added to \(x\), \[ [B] = [x] \] Because \(x\) is displacement, \[ [B] = L \]
Step 2: Use dimensional consistency of the equation.} Dimension of potential energy: \[ [U] = ML^2T^{-2} \] Now taking dimensions of the given equation: \[ [U] = \left[\frac{A\sqrt{x}}{B+x}\right] \] Since \(B+x\) has dimension \(L\), \[ [U] = \frac{[A][x]^{1/2}}{L} \] Substitute \( [x]=L \): \[ ML^2T^{-2} = \frac{[A]L^{1/2}}{L} \] \[ ML^2T^{-2} = [A]L^{-1/2} \]
Step 3: Solve for the dimension of \(A\).} \[ [A] = ML^2T^{-2}\times L^{1/2} \] \[ [A] = ML^{5/2}T^{-2} \] Final Result \[ [A] = ML^{5/2}T^{-2}, \qquad [B] = L \]