The quantity $\sqrt{\frac{\mu_0}{\epsilon_0}}$ is known as the characteristic impedance of free space ($Z_0$).

Using the relationship between the speed of light $c$, permeability $\mu_0$, and permittivity $\epsilon_0$:

$$c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \rightarrow \mu_0\,\epsilon_0 = \frac{1}{c^2} \rightarrow \mu_0 = \frac{1}{c^2\,\epsilon_0}$$

Substituting this for $\mu_0$ into $\sqrt{\frac{\mu_0}{\epsilon_0}}$

$$\sqrt{\frac{\mu_0}{\epsilon_0}} = \sqrt{\frac{1}{c^2\,\epsilon_0\epsilon_0}} = \sqrt{\frac{1}{c^2\,\epsilon_0^2}} = \frac{1}{c\,\epsilon_0}$$

The dimensional formula of permittivity $\epsilon_0$ can be determined using Coulomb's Law.

The electrostatic force $F$ between two point charges $q_1$ and $q_2$ separated by a distance $r$ is:

$$F = \frac{1}{4\pi\epsilon_0}\frac{q_1\,q_2}{r^2} \rightarrow \epsilon_0 = \frac{1}{4\pi F}\frac{q_1\,q_2}{r^2}$$

Electric charge $q$ is defined as current $(I)$ multiplied by time $(T)$, so its dimensional formula is $[A\ T]$.

Force has dimensions of mass times acceleration, $[M\ L\ T^{-2}]$.

Distance $r$ has the dimension of length $[L]$.

Substituting these definitions gives the dimensional formula for $\epsilon_0$:

$$[\epsilon_0] = \frac{[A\ T][A\ T]}{[M\ L\ T^{-2}][L^2]} = \frac{[A^2\ T^2]}{[M\ L^3\ T^{-2}]} = [M^{-1}\ L^{-3}\ T^4\ A^2]$$

The speed of light $c$ is a velocity, which has the dimensional formula $[L\ T^{-1}]$.

Now, substituting the dimensions of $c$ and $\epsilon_0$ into our expression $Z_0 = \frac{1}{c\epsilon_0}$:

$$Z_0 = \left[\sqrt{\frac{\mu_0}{\epsilon_0}}\right] = \frac{1}{[L\ T^{-1}][M^{-1}\ L^{-3}\ T^4\ A^2]} = \frac{1}{[M^{-1}\ L^{-2}\ T^3\ A^2]} = [M^1\ L^2\ T^{-3}\ A^{-2}]$$

Let the unit of the quantity in the standard SI system be $U_1$ and its numerical magnitude be $N_1$.

Let its unit in the new system be $U_2$ and its numerical magnitude be $N_2$.

Since the physical value remains identical regardless of the unit system chosen:

$$N_1\,U_1 = N_2\,U_2 \Rightarrow N_2 = N_1\left(\frac{U_1}{U_2}\right)$$

Using dimensional formula $[M^1\ L^2\ T^{-3}\ A^{-2}]$, we can express the ratio of the units as:

$$N_2 = N_1\left[\frac{M_1}{M_2}\right]^1\left[\frac{L_1}{L_2}\right]^2\left[\frac{T_1}{T_2}\right]^{-3}\left[\frac{A_1}{A_2}\right]^{-2}$$

Where $M_1, L_1, T_1, A_1$ represent the base units in the standard SI system $(1 { kg}, 1 { m}, 1 { s}, 1 { A})$. And $M_2, L_2, T_2, A_2$ represent the base units in the new system $(5 { kg}, 5 { m}, 5 { s}, 5 { A})$.

The magnitude in the new system that corresponds to exactly one SI unit of the quantity.

$$N_1 = 1$$

Substituting the standard SI values and the given new system values into the conversion equation:

$$N_2 = 1 \times \left[\frac{1\text{ kg}}{5\text{kg}}\right]^1 \left[\frac{1\text{ m}}{5\text{ m}}\right]^2 \left[\frac{1\text{ s}}{5\text{ s}}\right]^{-3} \left[\frac{1\text{ A}}{5\text{ A}}\right]^{-2}$$

$$N_2 = \left(\frac{1}{5}\right)^1 \times \left(\frac{1}{5}\right)^2 \times \left(\frac{1}{5}\right)^{-3} \times \left(\frac{1}{5}\right)^{-2}$$

$$N_2 = \left(\frac{1}{5}\right)^{1+2-3-2}$$

$$N_2 = \left(\frac{1}{5}\right)^{-2}$$

$$N_2 = (5)^2 = 25$$

Therefore, the correct answer is 25.

Units And Measurements

Continue this paper

Similar Units And Measurements PYQs

More practice from this question