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NEET 1993 (AIPMT) · Applications of Dimensional Analysis

Applications of Dimensional Analysis question from NEET 1993 (AIPMT)

Q81NEET 1993 · AIPMTEasy

Applications of Dimensional Analysis

The time dependence of a physical quantity pp is given by p=p0exp⁡(−at2)p = p_0 \exp(-at^2), where aa is a constant and tt is the time. The constant aa

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Answer: (b)

  1. The exponent of ee must be dimensionless: [at2]=1[a t^2] = 1.
  2. [a]=[T−2][a] = [T^{-2}].
  3. So the answer is (b).
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