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JEE Main 2020 (9 Jan, Shift 1) · Applications of Dimensional Analysis

Applications of Dimensional Analysis question from JEE Main 2020 (9 Jan, Shift 1)

Q55JEE Main 2020 · 9 JAN · SHIFT 1Easy

Applications of Dimensional Analysis

A quantity f is given by f=hc5Gf = \sqrt{\frac{hc^5}{G}} where c is speed of light, G universal gravitational constant and h is the Planck's constant. Dimension of f is that of:

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Answer: (a)

  1. [hc5]=ML2T−1⋅L5T−5=ML7T−6[hc^5] = ML^2T^{-1}\cdot L^5T^{-5} = ML^7T^{-6}.
  2. Dividing by [G]=M−1L3T−2[G] = M^{-1}L^3T^{-2}: M2L4T−4M^2L^4T^{-4}.
  3. Square root: ML2T−2ML^2T^{-2}, the dimension of energy.
  4. So the answer is (a) energy.
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