1. Find dimensions of each quantity in List-I

(A) Permeability of free space, $\mu_0$

Using the relation for magnetic force between two parallel currents,

$$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi r}$$

So,

$$[\mu_0] = \frac{[F/L]\,[r]}{[I]^2}$$

Now,

$$[F] = [MLT^{-2}], \quad [F/L] = [MT^{-2}], \quad [r]=[L], \quad [I]=[A]$$

Therefore,

$$[\mu_0] = \frac{[MT^{-2}][L]}{[A^2]} = [MLT^{-2}A^{-2}]$$

So,

$$(A) \to (III)$$

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(B) Magnetic field, $B$

Using Lorentz force,

$$F = qvB$$

Hence,

$$[B] = \frac{[F]}{[q][v]}$$

Now,

$$[q]=[AT], \quad [v]=[LT^{-1}]$$

Thus,

$$[B] = \frac{[MLT^{-2}]}{[AT][LT^{-1}]} = [MT^{-2}A^{-1}]$$

So,

$$(B) \to (II)$$

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(C) Magnetic moment

Magnetic moment of a current loop is

$$m = IA$$

where area has dimension $[L^2]$. Therefore,

$$[m] = [A][L^2] = [L^2A]$$

So,

$$(C) \to (IV)$$

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(D) Torsional constant

Torsional constant is torque per unit angular twist:

$$C = \frac{\tau}{\theta}$$

Since angle $\theta$ is dimensionless,

$$[C] = [\tau]$$

And torque has dimensions

$$[\tau] = [\text{force} \times \text{distance}] = [MLT^{-2}]\,[L] = [ML^2T^{-2}]$$

So,

$$(D) \to (I)$$

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2. Final matching

$$(A)-(III), \quad (B)-(II), \quad (C)-(IV), \quad (D)-(I)$$

This corresponds to Option A.

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3. Comparison with stored answer

Stored correct answer = A

Our derived answer = A

So, the answer agrees with the stored answer.