Question 14

NUMERICALMEDIUM

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:

Correct Answer: 25

Detailed Solution

The quantity μ0/ϵ0\sqrt{\mu_0/\epsilon_0} is the impedance of free space, which has dimensions of resistance ([R][R]). The dimensions of resistance are [R]=[ML2T−3I−2][R] = [ML^2 T^{-3} I^{-2}]. Let the magnitude in the new system be n2n_2. Using n1u1=n2u2n_1 u_1 = n_2 u_2: 1⋅[M1L12T1−3I1−2]=n2⋅[M2L22T2−3I2−2]1 \cdot [M_1 L_1^2 T_1^{-3} I_1^{-2}] = n_2 \cdot [M_2 L_2^2 T_2^{-3} I_2^{-2}] n2=1⋅(M1M2)1⋅(L1L2)2⋅(T1T2)−3⋅(I1I2)−2n_2 = 1 \cdot \left(\frac{M_1}{M_2}\right)^1 \cdot \left(\frac{L_1}{L_2}\right)^2 \cdot \left(\frac{T_1}{T_2}\right)^{-3} \cdot \left(\frac{I_1}{I_2}\right)^{-2} Substituting the given units (M2=5M1M_2 = 5 M_1, L2=5L1L_2 = 5 L_1, T2=5T1T_2 = 5 T_1, I2=5I1I_2 = 5 I_1): n2=1⋅(15)⋅(15)2⋅(15)−3⋅(15)−2n_2 = 1 \cdot \left(\frac{1}{5}\right) \cdot \left(\frac{1}{5}\right)^2 \cdot \left(\frac{1}{5}\right)^{-3} \cdot \left(\frac{1}{5}\right)^{-2} n2=5−1⋅5−2⋅53⋅52=52=25n_2 = 5^{-1} \cdot 5^{-2} \cdot 5^3 \cdot 5^2 = 5^2 = 25.

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