JEE Challenger
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PhyMedAdv 2026 • Paper 2

Magnitude of Impedance of Free Space in New Unit System

In a new system of units, the units of mass, length, time and current are 5 kg5\ \text{kg}, 5 m5\ \text{m}, 5 s5\ \text{s} and 5 A5\ \text{A}, respectively. If μ0\mu_0 and ϵ0\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 μ0/ϵ0\sqrt{\mu_0/\epsilon_0}, is:

Official Numerical Answer25

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Step-by-Step Solution

The quantity Z0=μ0ϵ0Z_0 = \sqrt{\frac{\mu_0}{\epsilon_0}} represents the characteristic impedance of free space, which has the dimension of electrical resistance.

To find the dimensional formula of resistance RR: Resistance R=Potential DifferenceCurrent=WorkCharge×Current=WorkCurrent2×Time\text{Resistance } R = \frac{\text{Potential Difference}}{\text{Current}} = \frac{\text{Work}}{\text{Charge} \times \text{Current}} = \frac{\text{Work}}{\text{Current}^2 \times \text{Time}}

Therefore, the dimensions of Z0Z_0 are given by: [Z0]=[ML2T−2][I]2[T]=[M1L2T−3I−2][Z_0] = \frac{[\text{M} \text{L}^2 \text{T}^{-2}]}{[\text{I}]^2 [\text{T}]} = [\text{M}^1 \text{L}^2 \text{T}^{-3} \text{I}^{-2}]

Let n1n_1 and n2n_2 be the numerical values of 1 SI unit1\text{ SI unit} of impedance in the SI system and the new unit system, respectively. Using the principle of homogeneity of physical quantities: n1U1=n2U2n_1 U_1 = n_2 U_2

Where the base units in the SI system (U1U_1) are: M1=1 kg,L1=1 m,T1=1 s,I1=1 AM_1 = 1\text{ kg}, \quad L_1 = 1\text{ m}, \quad T_1 = 1\text{ s}, \quad I_1 = 1\text{ A}

And the base units in the new system (U2U_2) are: M2=5 kg,L2=5 m,T2=5 s,I2=5 AM_2 = 5\text{ kg}, \quad L_2 = 5\text{ m}, \quad T_2 = 5\text{ s}, \quad I_2 = 5\text{ A}

For n1=1n_1 = 1 (one SI unit): n2=n1(M1M2)1(L1L2)2(T1T2)−3(I1I2)−2n_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{I_1}{I_2}\right)^{-2}

Substituting the base values: n2=1×(15)1×(15)2×(15)−3×(15)−2n_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}

n2=15×125×125×25=25n_2 = \frac{1}{5} \times \frac{1}{25} \times 125 \times 25 = 25

Thus, the magnitude of one SI unit of μ0/ϵ0\sqrt{\mu_0/\epsilon_0} in this new system of units is 2525.

Magnitude of Impedance of Free Space in New Unit System | Physics PYQ Solution - JEE Challenger