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JEE Main 2020 (4 Sep, Shift 2) · Applications of Dimensional Analysis

Applications of Dimensional Analysis question from JEE Main 2020 (4 Sep, Shift 2)

Q53JEE Main 2020 · 4 SEP · SHIFT 2Easy

Applications of Dimensional Analysis

A quantity x is given by (IFv2WL4)\left(\frac{IFv^2}{WL^4}\right) in terms of moment of inertia I, force F, velocity v, work W and Length L. The dimensional formula for x is same as that of:

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Answer: (c)

  1. [I]=ML2[I] = ML^2, [F]=MLT−2[F] = MLT^{-2}, [v2]=L2T−2[v^2] = L^2T^{-2}, [W]=ML2T−2[W] = ML^2T^{-2}, [L4]=L4[L^4] = L^4.
  2. Numerator: ML2⋅MLT−2⋅L2T−2=M2L5T−4ML^2\cdot MLT^{-2}\cdot L^2T^{-2} = M^2L^5T^{-4}; denominator: ML6T−2ML^6T^{-2}.
  3. [x]=ML−1T−2[x] = ML^{-1}T^{-2}, which is energy per unit volume.
  4. So the answer is (c) energy density.
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