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JEE Main 2022 (27 Jul, Shift 2) · Applications of Dimensional Analysis

Applications of Dimensional Analysis question from JEE Main 2022 (27 Jul, Shift 2)

Q33JEE Main 2022 · 27 JUL · SHIFT 2Easy

Applications of Dimensional Analysis

An expression of energy density is given by u=αβsin⁡(αxkt)u = \frac{\alpha}{\beta}\sin\left(\frac{\alpha x}{kt}\right), where α,β\alpha, \beta are constants, xx is displacement, kk is Boltzmann constant and t is the temperature. The dimensions of β\beta will be:

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Answer: (d)

  1. αxkt\frac{\alpha x}{kt} is dimensionless: [α]=[kt]/[x]=[ML2T−2]/[L]=[MLT−2][\alpha] = [kt]/[x] = [ML^2T^{-2}]/[L] = [MLT^{-2}] (ktkt is energy).
  2. [u]=[α]/[β]⇒[β]=[MLT−2]/[ML−1T−2][u] = [\alpha]/[\beta] \Rightarrow [\beta] = [MLT^{-2}]/[ML^{-1}T^{-2}].
  3. [β]=[M0L2T0][\beta] = [M^0L^2T^0].
  4. So the answer is (d).
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