Question 6: Units , Dimensions and Error Analysis

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 $A B$ are $\_\_\_\_$

Accepted Answer

According to the principle of homogeneity, dimensions of each term on both sides of a equation must be the same. Also, only physical quantities of the same dimensions can be added or subtracted.
The potential energy of the particle is given as $V=\frac{A \sqrt{x}}{x+B}$, where $A$ and $B$ are constant.
In the denominator of the expression $\mathrm{x}+\mathrm{B}$, the constant B is added to distance x . Therefore, the dimensions of $B$ must be the same as the dimensions of $x$.
$$ [\mathrm{B}]=[\mathrm{x}]=\left[\mathrm{L}^1ight] $$
The potential energy $V$ has the same dimensions as work or energy.
$$ [\mathrm{V}]=[\mathrm{F} 	imes \mathrm{x}]=\left[\mathrm{M}^1 \mathrm{~L}^1 \mathrm{~T}^{-2} 	imes \mathrm{L}^1ight]=\left[\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-2}ight] $$
On rearranging the formula of potential energy for A :
$$ A=\frac{V(x+B)}{\sqrt{x}} $$
Substituting the dimensions:
$$ [\mathrm{A}]=\frac{\left[\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-2}ight]\left[\mathrm{L}^1ight]}{\left[\mathrm{L}^{1 / 2}ight]} $$
$$\Rightarrow $$ $[\mathrm{A}]=\frac{\left[\mathrm{M}^1 \mathrm{~L}^3 \mathrm{~T}^{-2}ight]}{\left[\mathrm{L}^{1 / 2}ight]}$
$$\Rightarrow $$$$ [\mathrm{A}]=\left[\mathrm{M}^1 \mathrm{~L}^{3-\frac{1}{2}} \mathrm{~T}^{-2}ight]=\left[\mathrm{M}^1 \mathrm{~L}^{5 / 2} \mathrm{~T}^{-2}ight] $$
To find the dimensions of the product AB :
$$ [\mathrm{AB}]=[\mathrm{A}] 	imes[\mathrm{B}] $$
$$\Rightarrow $$ $[\mathrm{AB}]=\left[\mathrm{M}^1 \mathrm{~L}^{5 / 2} \mathrm{~T}^{-2}ight] 	imes\left[\mathrm{L}^1ight]$
$$\Rightarrow $$ $[\mathrm{AB}]=\left[\mathrm{M}^1 \mathrm{~L}^{5 / 2+1} \mathrm{~T}^{-2}ight]$
$$\Rightarrow $$ $$ [\mathrm{AB}]=\left[\mathrm{M}^1 \mathrm{~L}^{7 / 2} \mathrm{~T}^{-2}ight] $$
Therefore, the dimensions of $A B$ is $\left[M^1 L^{7 / 2} T^{-2}ight]$.
Thus, the correct option is (D).

Answer

$$ \left[\mathrm{M}^1 \mathrm{~L}^{5 / 2} \mathrm{~T}^{-2}\right] $$

Answer

$$ \left[\mathrm{M}^{3 / 2} \mathrm{~L}^{5 / 2} \mathrm{~T}^{-2}\right] $$

Answer

$$ \left[\mathrm{M}^1 \mathrm{~L}^2 \mathrm{~T}^{-2}\right] $$

Answer

$$ \left[\mathrm{M}^1 \mathrm{~L}^{7 / 2} \mathrm{~T}^{-2}\right] $$

Units , Dimensions and Error Analysis

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 A B are \_\_\_\_

Choose the correct answer: