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JEE Main 2019 (10 Jan, Shift 1) · Applications of Dimensional Analysis

Applications of Dimensional Analysis question from JEE Main 2019 (10 Jan, Shift 1)

Q62JEE Main 2019 · 10 JAN · SHIFT 1Medium

Applications of Dimensional Analysis

The density of a material in SI units is 128 kg m–3. In certain units in which the unit of length is 25 cm and the unit of mass is 50 g, the numerical value of density of the material is -

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Answer: (a)

  1. New unit of density =0.05 kg(0.25 m)3=0.050.015625=3.2= \dfrac{0.05\text{ kg}}{(0.25\text{ m})^3} = \dfrac{0.05}{0.015625} = 3.2 kg m−3^{-3}.
  2. Value =1283.2=40= \dfrac{128}{3.2} = 40.
  3. So the answer is (a) 40.
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