In a new system of units, the units of mass, length, time and...

The dimensional formula for permeability of free space $\mu_0$ is $[M L T^{-2} A^{-2}]$.

The dimensional formula for permittivity of free space $\epsilon_0$ is $[M^{-1} L^{-3} T^4 A^2]$.

The dimensional formula for $\sqrt{\dfrac{\mu_0}{\epsilon_0}}$ is:
\left( \dfrac{[M L T^{-2} A^{-2}]}{[M^{-1} L^{-3} T^4 A^2]} \right)^{1/2} = \left( [M^2 L^4 T^{-6} A^{-4}] \right)^{1/2} = [M L^2 T^{-3} A^{-2}]$<br><br>Let the magnitude of$1$SI unit in the new system be$n$. Using the principle of dimensional homogeneity:<br>$n_1 u_1 = n_2 u_2$<br>$1 \times [M_1 L_1^2 T_1^{-3} A_1^{-2}] = n \times [M_2 L_2^2 T_2^{-3} A_2^{-2}]$<br><br>Given the new units are$M_2 = 5$kg,$L_2 = 5$m,$T_2 = 5$s, and$A_2 = 5$A, while the SI units are$M_1 = 1$kg,$L_1 = 1$m,$T_1 = 1$s, and$A_1 = 1$A.<br><br>$n = \left( \dfrac{M_1}{M_2} \right) \left( \dfrac{L_1}{L_2} \right)^2 \left( \dfrac{T_1}{T_2} \right)^{-3} \left( \dfrac{A_1}{A_2} \right)^{-2}$<br><br>$n = \left( \dfrac{1}{5} \right) \left( \dfrac{1}{5} \right)^2 \left( \dfrac{1}{5} \right)^{-3} \left( \dfrac{1}{5} \right)^{-2}$<br><br>$n = 5^{-1} \times 5^{-2} \times 5^3 \times 5^2$<br><br>$n = 5^{-1 - 2 + 3 + 2} = 5^2 = 25$<br><br>Thus, the magnitude of one SI unit in the new system of units is$25$.<br><br>Answer:$25