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MHT CET 2019 (2 May, Shift 2) · Applications of Dimensional Analysis

Applications of Dimensional Analysis question from MHT CET 2019 (2 May, Shift 2)

Q56MHT CET 2019 · 2 MAY · SHIFT 2Easy

Applications of Dimensional Analysis

The force ' FF ' acting on a body of density ' dd ' are related by the relation F=ydF = \frac{y}{\sqrt{d}}. The dimensions of ' yy ' are

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Answer: (a)

  1. y=Fdy = F\sqrt{d}.
  2. [F]=[MLT−2][F] = [MLT^{-2}], [d]=[ML−3]1/2=[M1/2L−3/2][\sqrt{d}] = [ML^{-3}]^{1/2} = [M^{1/2}L^{-3/2}].
  3. [y]=[M3/2L−1/2T−2][y] = [M^{3/2}L^{-1/2}T^{-2}].
  4. So the answer is (a).
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