JEE PYQ: Units & Measurements - Question ID 152922fe2636 (JEE Main 2003)

Core Formula & Concept:

In electromagnetism, the two fundamental constants are:

• $\mu_0$ – the permeability of free space, whose SI unit is henry per metre (H/m).
• $\varepsilon_0$ – the permittivity of free space, whose SI unit is farad per metre (F/m).

From Maxwell’s equations we know that electromagnetic waves in vacuum travel at the speed of light $c$, and this speed is related to $\mu_0$ and $\varepsilon_0$ by the identity $c = \frac{1}{\sqrt{\mu_0\,\varepsilon_0}}.$ Consequently, $\frac{1}{\mu_0\,\varepsilon_0} = c^2.$ Since the speed of light $c$ has dimensions of length per time, $[c] = \mathrm{L\,T^{-1}}$, its square will have dimensions $\mathrm{L^2\,T^{-2}}$. This is the key to finding the dimensions of $1/(\mu_0\,\varepsilon_0)$.

Step-by-Step Derivation:

Step 1: Write down the SI units of $\mu_0$ and $\varepsilon_0$.

• $\mu_0$ has unit henry per metre (H/m).
• $\varepsilon_0$ has unit farad per metre (F/m).

Step 2: Express H and F in base SI units.

• 1 henry (H) = 1 $\mathrm{kg\,m^2\,s^{-2}\,A^{-2}}$.
• 1 farad (F) = 1 $\mathrm{kg^{-1}\,m^{-2}\,s^{4}\,A^{2}}$.

Step 3: Compute the unit of $\mu_0\,\varepsilon_0$.

$[\mu_0] = \frac{\mathrm{H}}{\mathrm{m}} = \frac{\mathrm{kg\,m^2\,s^{-2}\,A^{-2}}}{\mathrm{m}} = \mathrm{kg\,m\,s^{-2}\,A^{-2}}.$ $[\varepsilon_0] = \frac{\mathrm{F}}{\mathrm{m}} = \frac{\mathrm{kg^{-1}\,m^{-2}\,s^{4}\,A^{2}}}{\mathrm{m}} = \mathrm{kg^{-1}\,m^{-3}\,s^{4}\,A^{2}}.$ Multiply them: $[\mu_0\,\varepsilon_0] = \bigl(\mathrm{kg\,m\,s^{-2}\,A^{-2}}\bigr) \bigl(\mathrm{kg^{-1}\,m^{-3}\,s^{4}\,A^{2}}\bigr) = \mathrm{m^{-2}\,s^{2}}.$

Step 4: Find the unit of $1/(\mu_0\,\varepsilon_0)$.

$\Bigl[\frac{1}{\mu_0\,\varepsilon_0}\Bigr] = \frac{1}{[\mu_0\,\varepsilon_0]} = \frac{1}{\mathrm{m^{-2}\,s^{2}}} = \mathrm{m^{2}\,s^{-2}}.$

Step 5: Translate to dimensional notation.

The dimensions $\mathrm{m^2\,s^{-2}}$ correspond exactly to $\mathrm{L^2\,T^{-2}}$.

Step 6: Match with the given options.

Option C is $[\mathrm{L^2\,T^{-2}}]$, which agrees with our result. Common Traps & Exam Tip:

Many students mistakenly try to recall the dimensions of $\mu_0$ and $\varepsilon_0$ separately and then multiply them. A frequent error is to confuse the dimensions of $\mu_0$ (which are $\mathrm{M\,L\,T^{-2}\,A^{-2}}$) with those of $\varepsilon_0$ (which are $\mathrm{M^{-1}\,L^{-3}\,T^{4}\,A^{2}}$), leading to incorrect cancellation. Instead, remember the identity $\frac{1}{\mu_0\,\varepsilon_0} = c^2,$ and use the known dimensions of $c$ to arrive at the answer quickly and reliably.