Question 27

The potential energy of a particle changes with distance $$x$$ from a fixed origin as $$V = \frac{A\sqrt{x}}{x + B}$$, where $$A$$ and $$B$$ are constants with appropriate dimensions. The dimensions of $$AB$$ are _______.

$$V = \frac{A\sqrt{x}}{x + B}$$

$$[B] = [x] = [L]$$

$$[V] = \frac{[A]\sqrt{[x]}}{[x + B]}$$

$$[M^1 L^2 T^{-2}] = \frac{[A] [L^{1/2}]}{[L]}$$

$$[A] = [M^1 L^{5/2} T^{-2}]$$

$$[AB] = [M^1 L^{5/2} T^{-2}] \times [L^1]$$

$$[AB] = [M^1 L^{7/2} T^{-2}]$$

Create a FREE account and get:

Download resources