JEE Advanced 2026 Paper 2, Physics Section 3 Q5: Units And Dimensions

JEE Advanced2026Paper 2Physics Section 3

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 $μ_{0}$ and $ϵ_{0}$ are the permeability and permittivity of free space, respectively, then in this new system of units, the magnitude of one SI unit of $μ_{0} / ϵ_{0}$ is:

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Solution

Dimensions of impedance. $μ_{0} / ϵ_{0}$ has the dimensions of resistance (impedance of free space), so

$[μ_{0} / ϵ_{0}] = [M^{1} L^{2} T^{- 3} A^{- 2}]$ .

Convert one SI unit to new units. Using $n_{1} u_{1} = n_{2} u_{2}$ with $n_{1} = 1$ :

$n_{2} = \frac{1 \cdot kg \cdot m ^{2} \cdot s ^{- 3} \cdot A ^{- 2}}{( 5 kg ) ( 5 m ) ^{2} ( 5 s ) ^{- 3} ( 5 A ) ^{- 2}} = (\frac{1}{5}) (\frac{1}{5})^{2} (\frac{1}{5})^{- 3} (\frac{1}{5})^{- 2} = 5^{- 1 - 2 + 3 + 2} = 5^{2} = 25$ .

$∴$ The magnitude of one SI unit is $25$ in the new system.

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JEE Advanced 2026 Paper 2, Physics Section 3 Q5: Units And Dimensions

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