Match the LIST-I with LIST-II Choose the correct answer from the options given below :

Match the LIST-I withLIST-II

Choose the correct answer from the options given below :
  1. A-III, B-IV, C-I, D-II
  2. A-II, B-III, C-IV, D-I
  3. A-III, B-II, C-I, D-IV
  4. A-III, B-IV, C-II, D-I

Solution

(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) $\frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{k} \frac{\mathrm{AdT}}{\mathrm{dx}}$
$\mathrm{k}=\frac{\left(\mathrm{ML}^2 \mathrm{~T}^{-3}\right) \mathrm{L}}{\mathrm{L}^2 \cdot \mathrm{~K}}=\mathrm{MLT}^{-3} \mathrm{~K}^{-1}$

Asked in: JEE Main 2025 (03 Apr Shift 2)

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