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:
A. A
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: n_1 u_1 = n_2 u_2 1 [M_1 L_1^2 T_1^ -3 A_1^ -2 ] = n [M_2 L_2^2 T_2^ -3 A_2^ -2 ] 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. n = ( M_1 M_2 ) ( L_1 L_2 )^2 ( T_1 T_2 )^ -3 ( A_1 A_2 )^ -2 n = ( 1 5 ) ( 1 5 )^2 ( 1 5 )^ -3 ( 1 5 )^ -2 n = 5^ -1 5^ -2 5^3 5^2 n = 5^ -1 - 2 + 3 + 2 = 5^2 = 25 Thus, the magnitude of one SI unit in the new system of units is 25. Answer: 25
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