Match List-I with List-II.

	List-I		List-II
A	Permeability of free space	I	[M L2 T–2]
B	Magnetic field	II	[M T–2 A–1]
C	Magnetic moment	III	[M L T–2 A–2]
D	Torsional constant	IV	[L2 A]

Choose the correct answer from the options given below :

We will determine the dimensional formulas for each physical quantity and match them correctly.

Step 1: Dimensional Formulas

Permeability of free space ($\mu_0$)

From $\mathrm{B}=\frac{{\mathrm{\mu }}_0\mathrm{I}}{2\mathrm{\pi r}}\Rightarrow \left[{\mathrm{\mu }}_0\right]=\left[\frac{\mathrm{B}\times \mathrm{r}}{\mathrm{I}}\right]=\left[\frac{{\mathrm{MT}}^{-2}{\mathrm{A}}^{-1}\times \mathrm{L}}{\mathrm{A}}\right]=\left[{\mathrm{MLT}}^{-2}{\mathrm{A}}^{-2}\right]$

So, $\mu_0$ matches with III.

Magnetic field ($B$)

In terms of fundamental quantities: $$B = \frac{\text{Force}}{\text{Current} \times \text{Length}} = \frac{M L T^{-2}}{A L} = [M T^{-2} A^{-1}]$$

So, $B$ matches with II.

Magnetic moment ($\mu$)

Magnetic moment is given by $\text{current} \times \text{area}$, or $$\mu = I \cdot L^2$$

Since current $I$ has the dimension $[A]$ and area has $[L^2]$, we get: $$[L^2 A]$$

So, magnetic moment matches with IV.

Torsional constant ($K$)

Torsional constant has the same dimension as torque per unit angle. Torque is force $\times$ distance: 

$\mathrm{\tau }=\mathrm{c\theta }\Rightarrow \left[\mathrm{c}\right]=\left[\frac{\mathrm{\tau }}{\mathrm{\theta }}\right]=\left[{\mathrm{ML}}^2{\mathrm{T}}^{-2}\right]$

So, torsional constant matches with I.

Step 2: Matching the Columns

Final Answer:

A - III, B - II, C - IV, D - I