A A-III, B-IV, C-I, D-II B A-II, B-III, C-IV, D-I C A-III, B-II, C-I, D-IV D A-III, B-IV, C-II, D-I
Show answer and solution ▾ Answer: (A) A-III, B-IV, C-I, D-II
(A) [ k ] = P V N T = M L 2 T − 2 K = M L 2 T − 2 K − 1 [\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} [ k ] = NT PV = K ML 2 T − 2 = ML 2 T − 2 K − 1 [ η ] = F 6 π r v = M L T − 2 L 2 T − 1 = M L − 1 T − 1 [\eta]=\frac{\mathrm{F}}{6 \pi \mathrm{rv}}=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2 \mathrm{~T}^{-1}}=\mathrm{ML}^{-1} \mathrm{~T}^{-1} [ η ] = 6 π rv F = L 2 T − 1 MLT − 2 = ML − 1 T − 1 [ h ] = E f = M L 2 T − 2 T − 1 = M L 2 T − 1 [\mathrm{h}]=\frac{\mathrm{E}}{\mathrm{f}}=\frac{\mathrm{ML}^2 \mathrm{~T}^{-2}}{\mathrm{~T}^{-1}}=\mathrm{ML}^2 \mathrm{~T}^{-1} [ h ] = f E = T − 1 ML 2 T − 2 = ML 2 T − 1 d Q d t = k A d T d x \frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{k} \frac{\mathrm{AdT}}{\mathrm{dx}} dt dQ = k dx AdT k = ( M L 2 T − 3 ) L L 2 ⋅ K = M L T − 3 K − 1 \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} k = L 2 ⋅ K ( ML 2 T − 3 ) L = MLT − 3 K − 1 
%20A-III%2C%20B-IV%2C%20C-I%2C%20D-II%0A(B)%20A-II%2C%20B-III%2C%20C-IV%2C%20D-I%0A(C)%20A-III%2C%20B-II%2C%20C-I%2C%20D-IV%0A(D)%20A-III%2C%20B-IV%2C%20C-II%2C%20D-I){kind=link}