Question

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

Hint:

Use the principle of homogeneity of dimensions. Terms added together must have the same dimensions. Potential energy $V$ has dimensions $[ML^2T^{-2}]$.

Answer 1

The Correct Option is D

We use dimensional analysis to solve for the dimensions of constants in a physical formula. The given formula is $V = \frac{A\sqrt{x}}{x + B}$.
1. From the denominator $(x + B)$, we know that $B$ must have the same units as $x$ because only quantities of the same dimension can be added. Since $x$ is distance, $[B] = [L]$.
2. The whole term $(x + B)$ then has the dimension of length, $[L]$.
3. Now we analyze the formula for $V$:
$$V = \frac{A x^{1/2}}{\text{Length}}$$
The dimensions of potential energy $V$ are $[ML^2T^{-2}]$.
$$[ML^2T^{-2}] = \frac{[A] [L]^{1/2}}{[L]}$$
$$[A] = \frac{[ML^2T^{-2}] [L]}{[L]^{1/2}} = [ML^2T^{-2}] [L]^{1/2} = [ML^{2 + 0.5} T^{-2}] = [ML^{2.5} T^{-2}]$$
4. Now, find the product $AB$:
$$[AB] = [A][B] = [ML^{2.5}T^{-2}] [L^1] = [ML^{2.5 + 1} T^{-2}] = [ML^{3.5} T^{-2}]$$
Representing $3.5$ as a fraction:
$$[AB] = [M^1 L^{7/2} T^{-2}]$$
Comparing this with the given options, we find it matches option 4.