The electrical resistance of a column of $0.04MNaOH$ solution of diameter $1.2cm$ and length $50cm$ is $5.55\times {10}^{3}Ohm,$ the resistivity of the column would be

When 1 cc (1 ml) of soln of an electrolyte is placed between two electrodes $1cm$ part . The cross sectional area of solution is thus $1c{m}^2$

$\therefore$  Specific Conductance$\left(k\right)$ $k$ = Conductance$\times \frac{l}{a}$

                                             =  $\frac{1}{R}$ $\times \frac{l}{a}$               $\because$ (Reciprocal of Resistance=Conductance

$$$\frac{l}{a}\left(G^{*}\right)$  is Cell constant

$\text{cell}\text{constant}{G}^{*}=\frac{l}{a}$

                        $G^{*}=\frac{l}{\pi r^2}$

                              $=\frac{50}{3.14\times {\left(0.6\right)}^2}$              $\pi$ = 3.14    r= 0.6

                              $=44.23c{m}^{-1}$

 $\therefore$ $\left(k\right)=\frac{G^{*}}{R}$

$\text{Resistivity}\text{=}\frac{\text{R}}{G^{*}}$                            $\because$ (Reciprocal of Resistance=Conductance

                      $=\frac{5.55\times {10}^3}{44.23}$

                     $=125.48Ohm.cm$