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

This problem involves dimensional analysis and the principle of homogeneity. The principle of homogeneity states that the dimensions of each term in a physical equation must be identical.
Step 1: Determine the dimensions of potential energy $V$ and distance $x$.
Potential energy has dimensions of work: $[V] = [M L^2 T^{-2}]$.
Distance $x$ has dimensions: $[x] = [L]$.
Step 2: Find the dimension of $B$.
In the expression $V = \frac{A\sqrt{x}}{x + B}$, the denominator is $x + B$. According to the principle of homogeneity, we can only add quantities with the same dimensions. Therefore, dimensions of $B$ must be the same as dimensions of $x$.
$$[B] = [x] = [L]$$
Step 3: Find the dimension of $A$.
Equating dimensions on both sides of the original equation:
$$[V] = \frac{[A][x]^{1/2}}{[x + B]}$$
$$[ML^2T^{-2}] = \frac{[A][L]^{1/2}}{[L]}$$
Solving for $[A]$:
$$[A] = [ML^2T^{-2}] \times [L] \times [L]^{-1/2}$$
$$[A] = [ML^2T^{-2}] \times [L]^{1/2} = [ML^{5/2}T^{-2}]$$
Step 4: Calculate dimensions of the product $AB$.
$$[AB] = [A] \times [B] = [ML^{5/2}T^{-2}] \times [L] = [ML^{7/2}T^{-2}]$$
Thus, the dimensions of $AB$ are $[M^1 L^{7/2} T^{-2}]$.