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JEE Main 2022 (29 Jul, Shift 1) · Units & Dimensions

Units & Dimensions question from JEE Main 2022 (29 Jul, Shift 1)

Q55JEE Main 2022 · 29 JUL · SHIFT 1Medium

Units & Dimensions

Given below are two statements: One is labelled as Assertion (A) and other is labelled as Reason (R).

Assertion (A): Time period of oscillation of a liquid drop depends on surface tension (S), if density of the liquid is ρ\rho and radius of the drop is r, then T=Kρr3/S3/2\mathrm{T} = \mathrm{K}\sqrt{\rho \mathrm{r}^3/\mathrm{S}^{3/2}} is dimensionally correct, where K is dimensionless.

Reason (R): Using dimensional analysis we get R.H.S. having different dimension than that of time period.

In the light of above statements, choose the correct answer from the options given below.

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Answer: (d)

  1. [ρr3]=[ML−3][L3]=[M][\rho r^3] = [\mathrm{ML^{-3}}][\mathrm{L^3}] = [\mathrm{M}]; [S]=[MT−2]⇒[S3/2]=[M3/2T−3][S] = [\mathrm{MT^{-2}}] \Rightarrow [S^{3/2}] = [\mathrm{M^{3/2}T^{-3}}].
  2. ρr3S3/2=[M−1/2T3]=[M−1/4T3/2]≠[T]\sqrt{\dfrac{\rho r^3}{S^{3/2}}} = \sqrt{[\mathrm{M^{-1/2}T^{3}}]} = [\mathrm{M^{-1/4}T^{3/2}}] \neq [\mathrm{T}].
  3. So the formula is dimensionally wrong: (A) false, (R) true. (Correct form: T=Kρr3/ST = K\sqrt{\rho r^3/S}.)
  4. So the answer is (d).
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