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JEE Main 2021 (27 Jul, Shift 2) · Units & Dimensions

Units & Dimensions question from JEE Main 2021 (27 Jul, Shift 2)

Q64JEE Main 2021 · 27 JUL · SHIFT 2Medium

Units & Dimensions

Match List I with List II.

Choose the correct answer from the options given below

List - IList - II
(a)Capacitance, C(i)M1L1T−3A−1M^{1}L^{1}T^{-3}A^{-1}
(b)Permittivity of free space, ε0\varepsilon_0(ii)M−1L−3T4A2M^{-1}L^{-3}T^{4}A^{2}
(c)Permeability of free space, μ0\mu_0(iii)M−1L−2T4A2M^{-1}L^{-2}T^{4}A^{2}
(d)Electric field, E(iv)M1L1T−2A−2M^{1}L^{1}T^{-2}A^{-2}
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Answer: (a)

  1. Capacitance C=qV=[AT][ML2T−3A−1]=[M−1L−2T4A2]C = \dfrac{q}{V} = \dfrac{[\mathrm{AT}]}{[\mathrm{ML^2T^{-3}A^{-1}}]} = [\mathrm{M^{-1}L^{-2}T^4A^2}] → (iii).
  2. ε0=q24πFr2=[M−1L−3T4A2]\varepsilon_0 = \dfrac{q^2}{4\pi F r^2} = [\mathrm{M^{-1}L^{-3}T^4A^2}] → (ii).
  3. μ0=1ε0c2=[MLT−2A−2]\mu_0 = \dfrac{1}{\varepsilon_0 c^2} = [\mathrm{MLT^{-2}A^{-2}}] → (iv).
  4. E=F/q=[MLT−3A−1]E = F/q = [\mathrm{MLT^{-3}A^{-1}}] → (i).
  5. So the answer is (a) (a)-(iii), (b)-(ii), (c)-(iv), (d)-(i).
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