Question:
In a new system of units, the units of mass, length, time and current are \(5\,\text{kg}\), \(5\,\text{m}\), \(5\,\text{s}\) and \(5\,\text{A}\), respectively. If \(\mu_0\) and \(\epsilon_0\) are the permeability and permittivity of free space, respectively, then in this new system of units, the magnitude of one SI unit of \(\sqrt{\mu_0/\epsilon_0}\) is:
In a new system of units, the units of mass, length, time and current are \(5\,\text{kg}\), \(5\,\text{m}\), \(5\,\text{s}\) and \(5\,\text{A}\), respectively. If \(\mu_0\) and \(\epsilon_0\) are the permeability and permittivity of free space, respectively, then in this new system of units, the magnitude of one SI unit of \(\sqrt{\mu_0/\epsilon_0}\) is:
Show Hint
Identifying that $\sqrt{\mu_0/\epsilon_0}$ has units of Ohms (Resistance) simplifies the dimensional analysis significantly.
Be very careful with negative signs in exponents during substitution; for example, $(1/5)^{-3}$ is $5^3$.
Ensure you use $n_2 = n_1 (\text{ratio})^a$ correctly to find the magnitude in the new system.
Be very careful with negative signs in exponents during substitution; for example, $(1/5)^{-3}$ is $5^3$.
Ensure you use $n_2 = n_1 (\text{ratio})^a$ correctly to find the magnitude in the new system.
Updated On: May 20, 2026
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Solution and Explanation
Step 1: Understanding the Question:
The physical quantity $\sqrt{\mu_0/\epsilon_0}$ is known as the intrinsic impedance of free space ($\eta_0$).
It has the dimensions of electrical resistance.
The goal is to determine the numerical value of one SI unit of this quantity in a hypothetical system with different base units.
Step 2: Key Formula or Approach:
The unit conversion formula is $n_1 [U_1] = n_2 [U_2]$, where $n$ is the numerical value and $U$ is the unit.
The dimensions of resistance $R$ (and thus $\sqrt{\mu_0/\epsilon_0}$) are:
$[R] = \frac{[Potential]}{[Current]} = \frac{[Work/Charge]}{[Current]} = \frac{[ML^2 T^{-2}/AT]}{[A]} = [ML^2 T^{-3} A^{-2}]$.
Let $n_1 = 1$ in SI units ($M_1=1\text{kg}, L_1=1\text{m}, T_1=1\text{s}, A_1=1\text{A}$).
In the new system ($M_2=5\text{kg}, L_2=5\text{m}, T_2=5\text{s}, A_2=5\text{A}$), we find $n_2$.
Step 3: Detailed Explanation:
• Dimensional analysis equation:
\[ n_2 = n_1 \times \left( \frac{M_1}{M_2} \right)^1 \times \left( \frac{L_1}{L_2} \right)^2 \times \left( \frac{T_1}{T_2} \right)^{-3} \times \left( \frac{A_1}{A_2} \right)^{-2} \]
• Substituting given values:
\[ n_2 = 1 \times \left( \frac{1}{5} \right)^1 \times \left( \frac{1}{5} \right)^2 \times \left( \frac{1}{5} \right)^{-3} \times \left( \frac{1}{5} \right)^{-2} \]
• Simplifying the exponents:
\[ n_2 = \frac{1}{5} \times \frac{1}{5^2} \times 5^3 \times 5^2 \]
\[ n_2 = 5^{-1} \times 5^{-2} \times 5^3 \times 5^2 = 5^{-1-2+3+2} = 5^2 \]
\[ n_2 = 25. \]
Step 4: Final Answer:
One SI unit of the intrinsic impedance $\sqrt{\mu_0/\epsilon_0}$ corresponds to a magnitude of 25 in the new system of units.
The physical quantity $\sqrt{\mu_0/\epsilon_0}$ is known as the intrinsic impedance of free space ($\eta_0$).
It has the dimensions of electrical resistance.
The goal is to determine the numerical value of one SI unit of this quantity in a hypothetical system with different base units.
Step 2: Key Formula or Approach:
The unit conversion formula is $n_1 [U_1] = n_2 [U_2]$, where $n$ is the numerical value and $U$ is the unit.
The dimensions of resistance $R$ (and thus $\sqrt{\mu_0/\epsilon_0}$) are:
$[R] = \frac{[Potential]}{[Current]} = \frac{[Work/Charge]}{[Current]} = \frac{[ML^2 T^{-2}/AT]}{[A]} = [ML^2 T^{-3} A^{-2}]$.
Let $n_1 = 1$ in SI units ($M_1=1\text{kg}, L_1=1\text{m}, T_1=1\text{s}, A_1=1\text{A}$).
In the new system ($M_2=5\text{kg}, L_2=5\text{m}, T_2=5\text{s}, A_2=5\text{A}$), we find $n_2$.
Step 3: Detailed Explanation:
• Dimensional analysis equation:
\[ n_2 = n_1 \times \left( \frac{M_1}{M_2} \right)^1 \times \left( \frac{L_1}{L_2} \right)^2 \times \left( \frac{T_1}{T_2} \right)^{-3} \times \left( \frac{A_1}{A_2} \right)^{-2} \]
• Substituting given values:
\[ n_2 = 1 \times \left( \frac{1}{5} \right)^1 \times \left( \frac{1}{5} \right)^2 \times \left( \frac{1}{5} \right)^{-3} \times \left( \frac{1}{5} \right)^{-2} \]
• Simplifying the exponents:
\[ n_2 = \frac{1}{5} \times \frac{1}{5^2} \times 5^3 \times 5^2 \]
\[ n_2 = 5^{-1} \times 5^{-2} \times 5^3 \times 5^2 = 5^{-1-2+3+2} = 5^2 \]
\[ n_2 = 25. \]
Step 4: Final Answer:
One SI unit of the intrinsic impedance $\sqrt{\mu_0/\epsilon_0}$ corresponds to a magnitude of 25 in the new system of units.
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