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

Step-by-step solution

The dimensional formula for permeability of free space ₀ is [M L T⁻² A⁻²] . The dimensional formula for permittivity of free space ₀ is [M⁻¹ L⁻³ T^4 A^2] . The dimensional formula for ₀ ₀ is: ( [M L T⁻² A⁻²] [M⁻¹ L⁻³ T^4 A^2] )^ 1/2 = ( [M^2 L^4 T⁻⁶ A⁻⁴] )^ 1/2 = [M L^2 T⁻³ A⁻²] Let the magnitude of 1 SI unit in the new system be n . Using the principle of dimensional homogeneity: n₁ u₁ = n₂ u₂ 1 [M₁ L₁^2 T₁⁻³ A₁⁻²] = n [M₂ L₂^2 T₂⁻³ A₂⁻²] Given the new units are M₂ = 5 kg, L₂ = 5 m, T₂ = 5 s, and A₂ = 5 A, while t