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JEE Advanced 2016Paper 1
Exam
JEE Advanced
Year
2016
Paper
Paper 1
Subject
Physics
Chapter
Units and Dimensions

JEE Advanced 2016 · Paper 1 · Physics · Units and Dimensions

A length - scale (l) depends on the permittivity (ε) of a...

A length - scale (l)\left(l\right) depends on the permittivity (ε)\left(ε\right) of a dielectric material, Boltzmann constant kB{k}_{B} , the absolute temperature T, the number per unit volume (n) of certain charged particles, and the charge (q) carried by each of the particles, Which of the following expressions (s) for ll is (are) dimensionally correct?

Options

  1. Al = \sqrt{\left(\dfrac{n{q}^{2}}{ε{k}_{B}T}\right)}
  2. Bl = \sqrt{\left(\dfrac{ε{k}_{B}T}{n{q}^{2}}\right)}
  3. Cl = \sqrt{\left(\dfrac{{q}^{2}}{ε{n}^{\dfrac{2}{3}}{K}_{B}T}\right)}
  4. Dl = \sqrt{\left(\dfrac{{q}^{2}}{ε{n}^{\dfrac{1}{3}}{K}_{B}T}\right)}

Correct answer

B. l = \sqrt{\left(\dfrac{ε{k}_{B}T}{n{q}^{2}}\right)}

Step-by-step solution

We know,
F=14πE0q2r2⇒q2E0=(Fr2)4πF = \dfrac{1}{4\pi{E}_{0}}\dfrac{{q}^{2}}{{r}^{2}} ⇒ \dfrac{{q}^{2}}{{E}_{0}} = \left(F{r}^{2}\right)4\pi
So, dimension
q2E0=dim(Fr2)=MLT−2×L2=ML3T−2\dfrac{{q}^{2}}{{E}_{0}} = dim \left(F{r}^{2}\right) = ML{T}^{- 2} \times {L}^{2} = M{L}^{3}{T}^{- 2}
Similarly; E=32KBT⇒dim(KBT)=dim(Energy)=ML2T−2E = \dfrac{3}{2}{K}_{B}T ⇒ dim \left({K}_{B}T\right) = dim \left(Energy\right) = M{L}^{2}{T}^{- 2}
(i) nq2EkBT=L−3×ML3T−2ML2T−2=1L\sqrt{\dfrac{n{q}^{2}}{E{k}_{B}T}} = \sqrt{\dfrac{{L}^{- 3} \times M{L}^{3}{T}^{- 2}}{M{L}^{2}{T}^{- 2}} = \dfrac{1}{L}}
(ii) (E)×volFr2=MLT−2×L3MLT−2×L2=L\sqrt{\dfrac{\left(E\right) \times vol}{F{r}^{2}}} = \sqrt{\dfrac{ML{T}^{- 2} \times {L}^{3}}{ML{T}^{- 2} \times {L}^{2}}} = L
(iii) Fr2(vol)23(Kℇ)=MLT−2×L2×L2ML2T−2=L3=L32\sqrt{\dfrac{F{r}^{2}{\left(vol\right)}^{\dfrac{2}{3}}}{( Kℇ )}} = \sqrt{\dfrac{ML{T}^{- 2} \times {L}^{2} \times {L}^{2}}{M{L}^{2}{T}^{- 2}}} = \sqrt{{L}^{3}} = {L}^{\dfrac{3}{2}}
(iv) Fr2(vol)13Energy=MLT−2L2×LML2T−2=L\sqrt{\dfrac{F{r}^{2}{\left(vol\right)}^{\dfrac{1}{3}}}{Energy}} = \sqrt{\dfrac{ML{T}^{- 2}{L}^{2} \times L}{M{L}^{2}{T}^{- 2}}} = L

[ ∴∴ dimension n = dim \left(\dfrac{1}{vol}\right) = {L}^{- 3} ]
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