JEE PYQ: Units & Measurements - Question ID 48919af26bc7 (JEE Main 2020)

In dimensional analysis, every physical quantity can be expressed in terms of the fundamental dimensions: mass ($M$), length ($L$), time ($T$), electric current ($I$), etc. The key formulas and concepts we use here are:

• The speed of light in vacuum is given by $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$. This means the quantity $x = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$ has the dimension of speed, i.e., $[L T^{-1}]$.
• The ratio of electric field ($E$) to magnetic field ($B$) in an electromagnetic wave is also the speed of light: $\frac{E}{B} = c$. Thus, $y = \frac{E}{B}$ also has the dimension of speed, $[L T^{-1}]$.
• The time constant of an $RC$ circuit is $\tau = RC$, where $R$ is resistance and $C$ is capacitance. The quantity $z = \frac{l}{CR}$ can be rewritten as $z = \frac{l}{\tau}$, which is length divided by time, i.e., speed. Hence, $z$ also has the dimension $[L T^{-1}]$.

Step 1: Dimension of $x = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$

We know: $\mu_0$ (permeability of free space) has dimension $[M L T^{-2} I^{-2}]$, $\varepsilon_0$ (permittivity of free space) has dimension $[M^{-1} L^{-3} T^{4} I^{2}]$.

Compute the product: $\mu_0 \varepsilon_0$ has dimension $[M L T^{-2} I^{-2}] \cdot [M^{-1} L^{-3} T^{4} I^{2}] = [L^{-2} T^{2}]$.

Taking the square root: $\sqrt{\mu_0 \varepsilon_0}$ has dimension $[L^{-1} T]$.

Taking the reciprocal: $x = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$ has dimension $[L T^{-1}]$.

Step 2: Dimension of $y = \frac{E}{B}$

Electric field $E$ has dimension $[M L T^{-3} I^{-1}]$, Magnetic field $B$ has dimension $[M T^{-2} I^{-1}]$.

Compute the ratio: $y = \frac{E}{B}$ has dimension $\frac{[M L T^{-3} I^{-1}]}{[M T^{-2} I^{-1}]} = [L T^{-1}]$.

Step 3: Dimension of $z = \frac{l}{CR}$

Length $l$ has dimension $[L]$, Capacitance $C$ has dimension $[M^{-1} L^{-2} T^{4} I^{2}]$, Resistance $R$ has dimension $[M L^{2} T^{-3} I^{-2}]$.

Compute the product $CR$: $CR$ has dimension $[M^{-1} L^{-2} T^{4} I^{2}] \cdot [M L^{2} T^{-3} I^{-2}] = [T]$.

Taking the ratio: $z = \frac{l}{CR}$ has dimension $\frac{[L]}{[T]} = [L T^{-1}]$.

Step 4: Compare dimensions of $x$, $y$, and $z$

We have shown: $[x] = [y] = [z] = [L T^{-1}]$.

Therefore, all three quantities $x$, $y$, and $z$ have the same dimension.

Students often make the following mistakes:

• Forgetting the dimension of $\mu_0$ and $\varepsilon_0$, leading to incorrect dimensional analysis of $x$.
• Assuming $\frac{E}{B}$ has a different dimension because they confuse electric and magnetic field units.
• Misapplying the dimensions of $C$ and $R$, especially the powers of $T$ and $I$, leading to wrong dimension for $z$.
• Overlooking that $\frac{l}{CR}$ is actually a speed (length over time), not just a ratio of length to resistance.

Exam Tip: Always write down the dimensions of each physical quantity involved before performing dimensional analysis. This systematic approach minimizes errors.

Based on the above derivation, the correct option is B: x, y and z have the same dimension.