JEE Advanced 2013Paper 2
- Exam
- JEE Advanced
- Year
- 2013
- Paper
- Paper 2
- Subject
- Physics
- Chapter
- Units and Dimensions
JEE Advanced 2013 · Paper 2 · Physics · Units and Dimensions
Match List I with List II and select the correct answer using the...
Match List with List and select the correct answer using the codes given below the lists: \begin{cases}\textList-I & \textList-II \\ \text{a.} & \text{Boltzmann constant} & \text{p.} & \left[{\text{ML}^{2}{\text{T}^{-1}\right] \\ \text{b.} & \text{Coefficient of viscosity} & \text{q.} & \left[{\text{ML}^{-1}{\text{T}^{-1}\right] \\ \text{c.} & \text{Planck constant} & \text{r.} & \left[{\text{MLT}^{- 3}{\text{K}^{-1}\right] \\ \text{d.} & \text{Thermal conductivity} & \text{s.} & \left[{\text{ML}^{2}{\text{T}^{- 2}{\text{K}^{-1}\right]\end{cases}
Options
- Aa-s;b-r;c-p;d-q;
- Ba-s;b-r;c-p;d-q;
- Ca-r;b-s;c-q;d-p;
- Da-s;b-q;c-p;d-r;
Correct answer
D. a-s;b-q;c-p;d-r;
Step-by-step solution
(a)
(b) F = ηA\dfrac{dv}{dx} \to \left[η\right] = \dfrac{{\text{MLT}}^{- 2}}{{\text{L}}^{2}{\text{LT}}^{-1}{\text{L}}^{-1}} = {\text{ML}}^{-1}{\text{T}}^{-1}
(c) E = hν \to {\text{ML}}^{2}{\text{T}}^{- 2} = \left[\text{h}\right]{\text{T}}^{-1} \to \left[\text{h}\right] = {\text{ML}}^{2}{\text{T}}^{-1}
(d) \dfrac{dQ}{dt} = \dfrac{kA Δ θ}{ℓ} \to \left[\text{k}\right] = \dfrac{{\text{ML}}^{2}{\text{T}}^{- 3}\text{L}}{{\text{L}}^{2}\text{K}} = {\text{MLT}}^{- 3}{\text{K}}^{-1}
(b) F = ηA\dfrac{dv}{dx} \to \left[η\right] = \dfrac{{\text{MLT}}^{- 2}}{{\text{L}}^{2}{\text{LT}}^{-1}{\text{L}}^{-1}} = {\text{ML}}^{-1}{\text{T}}^{-1}
(c) E = hν \to {\text{ML}}^{2}{\text{T}}^{- 2} = \left[\text{h}\right]{\text{T}}^{-1} \to \left[\text{h}\right] = {\text{ML}}^{2}{\text{T}}^{-1}
(d) \dfrac{dQ}{dt} = \dfrac{kA Δ θ}{ℓ} \to \left[\text{k}\right] = \dfrac{{\text{ML}}^{2}{\text{T}}^{- 3}\text{L}}{{\text{L}}^{2}\text{K}} = {\text{MLT}}^{- 3}{\text{K}}^{-1}