In a new system of units, the units of mass, length, time and current are 5 kg, 5 m, 5 s and 5 A, respectively. If μ₀ and ε₀ are the permeability and permittivity of free space, respectively, then in this new system of units, the magnitude of one SI unit of $\sqrt{\frac{{\mu }_0}{{\epsilon }_0}}$, is:

Correct Answer: 25

Step 1: Understand the physical quantity and its dimensions

The given physical quantity is $\sqrt{\frac{{\mu }_0}{{\epsilon }_0}}$, which represents the characteristic impedance of free space.

Let's determine its dimensional formula in terms of mass ($M$), length ($L$), time ($T$), and electric current ($A$).

The speed of light in vacuum is given by:

$c=1\sqrt{{\mu }_0{\epsilon }_0}$

Thus, we can rewrite the expression as:

$\sqrt{\frac{{\mu }_0}{{\epsilon }_0}}={\mu }_{0}c$

The formula for magnetic permeability ${\mu }_0$ comes from the force between two parallel current-carrying conductors: $F=\frac{{\mu }_0{I}^{2}l}{2\pi r}$.

Therefore, the dimensions of ${\mu }_0$ are:

$\left[{\mu }_0\right]=\left[ML{T}^{-2}{A}^{-2}\right]$

Since the dimensions of velocity $c$ are $\left[L{T}^{-1}\right]$, the dimensional formula for $\sqrt{\frac{{\mu }_0}{{\epsilon }_0}}$ is:

$\left[Q\right]=\left[ML{T}^{-2}{A}^{-2}\right]\times \left[L{T}^{-1}\right]=\left[M{L}^2{T}^{-3}{A}^{-2}\right]$

Step 2: Conversion of units between the two systems

In the SI system:

Mass unit, $M_1=1\text{kg}$

Length unit, $L_1=1\text{m}$

Time unit, $T_1=1\text{s}$

Current unit, $A_1=1\text{A}$

In the new system of units:

Mass unit, $M_2=5\text{kg}$

Length unit, $L_2=5\text{m}$

Time unit, $T_2=5\text{s}$

Current unit, $A_2=5\text{A}$

Step 3: Calculate the magnitude in the new system

Using the principle of homogeneity, $n_1{U}_1=n_2{U}_2$, where $n_1=1$ (magnitude of 1 SI unit):

$n_2=n_1{\left(\frac{M_1}{M_2}\right)}^1{\left(\frac{L_1}{L_2}\right)}^2{\left(\frac{T_1}{T_2}\right)}^{-3}{\left(\frac{A_1}{A_2}\right)}^{-2}$

Substitute the given unit values into the formula:

$n_2=1\times {\left(\frac{1}{5}\right)}^1\times {\left(\frac{1}{5}\right)}^2\times {\left(\frac{1}{5}\right)}^{-3}\times {\left(\frac{1}{5}\right)}^{-2}$

Simplify the terms:

$n_2=5^{-1}\times 5^{-2}\times 5^3\times 5^2$

Combine the exponents of 5:

$n_2=5^{-1-2+3+2}=5^2=25$

Thus, the magnitude of 1 SI unit of $\sqrt{\frac{{\mu }_0}{{\epsilon }_0}}$ in the new system of units is 25.