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JEE Mains · Physics · STD 11 - Units,dimensions and measurement
  • JEE Mains 2025
  • Medium
  • MCQ

Match the LIST-I with LIST-II
LIST-ILIST-II
(A) Boltzmann constantI. \(\mathrm{ML}^2 \mathrm{~T}^{-1}\)
(B) Coefficient of viscosityII. \(\mathrm{MLT}^{-3} \mathrm{~K}^{-1}\)
(C) Planck's constantIII. \(\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}\)
(D) Thermal conductivityIV. \(\mathrm{ML}^{-1} \mathrm{~T}^{-1}\)

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Answer & Solution

Correct Answer

(A) A-III, B-IV, C-I, D-II

Step-by-step Solution

Detailed explanation

(A) \([\mathrm{k}]=\frac{\mathrm{PV}}{\mathrm{NT}}=\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{~K}}=\mathrm{ML}^2 \mathrm{~T}^{-2} \mathrm{~K}^{-1}\)
(B) \([\eta]=\frac{\mathrm{F}}{6 \pi \mathrm{rv}}=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2 \mathrm{~T}^{-1}}=\mathrm{ML}^{-1} \mathrm{~T}^{-1}\)
(C) \([\mathrm{h}]=\frac{\mathrm{E}}{\mathrm{f}}=\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{~T}^{-1}}=\mathrm{ML}^2 \mathrm{~T}^{-1}\)
(D) \(\frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{k} \frac{\mathrm{AdT}}{\mathrm{dx}}\)
\(\mathrm{k}=\frac{\left(\mathrm{ML}^2 \mathrm{~T}^{-3}\right) \mathrm{L}}{\mathrm{L}^2 \cdot \mathrm{~K}}=\mathrm{MLT}^{-3} \mathrm{~K}^{-1}\)
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