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

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

Q51JEE Main 2020 · 2 SEP · SHIFT 2Easy

Applications of Dimensional Analysis

If momentum (P), area (A) and time (T) are taken to be the fundamental quantities then the dimensional formula for energy is

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Answer: (c)

  1. Let [E]=[PaAbTc][E] = [P^aA^bT^c] with [P]=[MLT−1][P] = [MLT^{-1}], [A]=[L2][A] = [L^2].
  2. [ML2T−2]=[MaLa+2bT−a+c][ML^2T^{-2}] = [M^aL^{a+2b}T^{-a+c}].
  3. a=1a = 1; 1+2b=2⇒b=1/21 + 2b = 2 \Rightarrow b = 1/2; −1+c=−2⇒c=−1-1 + c = -2 \Rightarrow c = -1.
  4. So the answer is (c) [PA1/2T−1][PA^{1/2}T^{-1}].
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