Animated Solution for Physics - Physics and Measurement: A length-scale ($l$) depends on the permittivity ($\epsilon$) of a dielectric material, Boltzmann's constant ($k_B$), the absolute temperature ($T$), the number per unit volume ($n$) of certain charged particles, and the charge ($q$) carried by each of the particles. Which of the following expression (s) for $l$ is (are) dimensionally correct?

Dimensional analysis is one of the most powerful tools in a physicist's arsenal. It acts as a universal truth serum for equations—if the dimensions on the left side don't match the dimensions on the right side, the equation is fundamentally flawed. In this classic JEE Advanced problem, we are tasked with finding a dimensionally correct expression for a length scale $l$ using a specific set of physical constants and variables.

At first glance, the options look intimidating. They are packed with square roots, fractional powers, and a jumble of constants. However, the secret to conquering this problem lies not in brute-force calculation, but in strategic grouping.

Before we can test any of the options, we must establish the dimensional foundation of every variable involved. Let's start with the straightforward ones:

Number density ($n$): This is the number of particles per unit volume. Since 'number' is dimensionless, $[n]=[L^{-3}]$. Charge ($q$): Charge is current multiplied by time, so $[q]=[AT]$. Absolute Temperature ($T$):* The dimension of temperature is simply $[K]$.

Now, we face the slightly more complex constants: permittivity ($\epsilon$) and Boltzmann's constant ($k_B$). Instead of memorizing their dimensions, we can quickly derive them from fundamental laws.

From Coulomb's Law, the electrostatic force is $F=\frac{1}{4\pi \epsilon }\frac{q^2}{r^2}$. Rearranging for permittivity gives us:

For Boltzmann's constant, we recall the kinetic energy of a gas molecule, $E=\frac{3}{2}{k}_{B}T$. Rearranging for $k_B$ yields:

If we were to substitute these massive dimensional formulas directly into each of the four options, we would be inviting algebraic disaster. A keen eye will notice a pattern: the term $\epsilon k_{B}T$ appears in the numerator or denominator of every single option.

This is our golden ticket. Let's evaluate the combined dimension of this block:

Notice how beautifully the mass ($M$) and temperature ($K$) dimensions cancel out. We are left with a much cleaner expression:

We can take this simplification one step further. The term $q^2$ also appears frequently. Its dimension is $[A^2{T}^2]$. If we evaluate the ratio of our grouped term to $q^2$, the time and current dimensions vanish entirely!

This single realization transforms a tedious calculation into a rapid mental check.

Armed with our simplified ratio, testing the options becomes trivial. We are looking for an expression that yields the dimension of length, $[L]$.

Testing Option (a): This is the square root of $n$ multiplied by the inverse of our ratio. This is not length. Option (a) is incorrect.

Testing Option (b): This is the square root of our ratio divided by $n$. We have a match! Option (b) is dimensionally correct.

Testing Option (c): This is incorrect.

Testing Option (d): We have another match! Option (d) is also dimensionally correct.

By strategically grouping terms, we bypassed the chaos of raw substitution and elegantly arrived at the correct answers: (b) and (d).