JEE Mains · Physics · STD 11 - 1. units,dimensions and measurement
The potential energy of a particle changes with distance \(x\) from a fixed origin as \(V = \dfrac{A\sqrt{x}}{x + B}\), where \(A\) and \(B\) are constant with appropriate dimensions. The dimensions of \(AB\) are _______.
- A \([M^1 L^{5/2} T^{-2}]\)
- B \([M^{3/2} L^{5/2} T^{-2}]\)
- C \([M^1 L^2 T^{-2}]\)
- D \([M^1 L^{7/2} T^{-2}]\)
Answer & Solution
Correct Answer
(D) \([M^1 L^{7/2} T^{-2}]\)
Step-by-step Solution
Detailed explanation
By the principle of dimensional homogeneity, quantities added or subtracted must have the same dimensions. In the denominator, \(B\) is added to \(x\) (distance). \([B] = [x] = [L]\) The dimensions of potential energy \(V\) are \([M^1 L^2 T^{-2}]\). From the given equation…
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