$\begin{array}{1 1}(A)\;5\times 10^3\\(B)\;5\times 10^2\\(C)\;5\times 10^{-4}\\(D)\;5\times 10^{-3}\end{array} $

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$\large\frac{k_2}{k_1}=\frac{R_1}{R_2}$

$\Rightarrow k_2=\large\frac{R_1}{R_2}$$\times k_1$

$\Rightarrow \large\frac{50}{280}$$\times 1.4Sm^{-1}$

$\lambda_m=\large\frac{k_2}{C}=\frac{1/4Sm^{-1}}{0.50mol L^{-1}}$

$\Rightarrow \large\frac{1}{4}\times \frac{10^{-3}}{0.50}$

$\Rightarrow 5\times 10^{-4}Sm^2mol^{-1}$

Hence (C) is the correct answer.

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