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JEE Main 2024 (31 Jan, Shift 1) · Applications of Dimensional Analysis

Applications of Dimensional Analysis question from JEE Main 2024 (31 Jan, Shift 1)

Q23JEE Main 2024 · 31 JAN · SHIFT 1Easy

Applications of Dimensional Analysis

A force is represented by F=ax2+bt12F = ax^2 + bt^{\frac{1}{2}}

where x=x = distance and t=t = time. The dimensions of b2/ab^2/a are:

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Answer: (b)

  1. [a]=[F]/[L2]=[ML−1T−2][a] = [F]/[L^2] = [ML^{-1}T^{-2}].
  2. [b]=[F]/[T1/2]=[MLT−5/2][b] = [F]/[T^{1/2}] = [MLT^{-5/2}], so [b2]=[M2L2T−5][b^2] = [M^2L^2T^{-5}].
  3. [b2/a]=[M2L2T−5]/[ML−1T−2]=[ML3T−3][b^2/a] = [M^2L^2T^{-5}]/[ML^{-1}T^{-2}] = [ML^3T^{-3}].
  4. So the answer is (b).
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