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JEE Main 2025 (2 Apr, Shift 2) · Units & Dimensions

Units & Dimensions question from JEE Main 2025 (2 Apr, Shift 2)

Q13JEE Main 2025 · 2 APR · SHIFT 2Easy

Units & Dimensions

If μ0\mu_0 and ϵ0\epsilon_0 are the permeability and permittivity of free space, respectively, then the dimension of (1μ0ϵ0)\left(\frac{1}{\mu_0\epsilon_0}\right) is:

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Answer: (b)

  1. c=1/μ0ϵ0c = 1/\sqrt{\mu_0\epsilon_0}, so 1/(μ0ϵ0)=c21/(\mu_0\epsilon_0) = c^2.
  2. [c2]=L2T−2[c^2] = \mathrm{L^2T^{-2}}
  3. So the answer is (b) L2/T2\mathrm{L^2/T^2}.
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