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JEE Main 2013 (9 Apr, Shift 1) · Applications of Dimensional Analysis

Applications of Dimensional Analysis question from JEE Main 2013 (9 Apr, Shift 1)

Q74JEE Main 2013 · 9 APR · SHIFT 1Medium

Applications of Dimensional Analysis

From the following, the quantity (constructed from the basic constants of nature), that has the dimensions, as well as correct order of magnitude, vis-a-vis typical atomic size, is:

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Answer: (a)

  1. e24πϵ0r\dfrac{e^2}{4\pi\epsilon_0 r} is an energy, so e24πϵ0mec2\dfrac{e^2}{4\pi\epsilon_0 m_ec^2} is (energy ×\times length)/energy == length.
  2. The other options do not have the dimension of length.
  3. Its value ≈2.8×10−15\approx 2.8\times10^{-15} m, the classical electron radius, which is a sub-atomic length scale.
  4. So the answer is (a).
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