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NEET 2010 (AIPMT, Prelims) · Applications of Dimensional Analysis

Applications of Dimensional Analysis question from NEET 2010 (AIPMT, Prelims)

Q78NEET 2010 · AIPMT · PRELIMSEasy

Applications of Dimensional Analysis

The dimension of 12ε0E2\frac{1}{2}\varepsilon_0 E^2, where ε0\varepsilon_0 is permittivity of free space and E is electric field, is

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Answer: (b)

  1. 12ε0E2\frac{1}{2}\varepsilon_0 E^2 is the energy density of an electric field (energy per volume).
  2. [energy/volume]=[ML2T−2]/[L3]=[ML−1T−2][\text{energy}/\text{volume}] = [ML^2T^{-2}]/[L^3] = [ML^{-1}T^{-2}].
  3. So the answer is (b) [ML−1T−2][ML^{-1}T^{-2}].
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