The equivalent conductance of $0.1\text{N}$ solution of $MgC{l}_2$ is $25{\Omega }^{-1}{\text{cm}}^2{\text{eq}}^{-1}$ at $25{}^{\circ }\text{C}$. A cell with electrodes that are $2{\text{cm}}^2$ in surface area and $0.5\text{cm}$ apart is filled with $0.1\text{N}MgC{l}_2$ solution. How much current $\text{(in A)}$ will flow when the potential difference between the electrodes is $5$ Volts?

The correct option is A $0.05$
Cell constant $\left(\frac{l}{a}\right)$$=\frac{0.5}{2}=\frac{1}{4}$
Specific conductance $\left(K\right)$ $=\frac{\text{Equivalent conductance}}{\text{Volume (cc) containing 1 eq.}}=\frac{25}{10000}(\because \text{For 0.1 N solution volume =}10000\text{cc})=0.0025{\Omega }^{-1}{\text{cm}}^{-1}$

Conductance $=K\times \frac{a}{l}=0.0025\times 4=0.01{\Omega }^{-1}$

$\therefore$ Resistance $\left(R\right)=\frac{1}{0.01}=100\Omega$
$\therefore$ Current $\text{(in A)}$ $=\frac{\text{Potential diffrence (V)}}{\text{Resistance}\left(\Omega \right)}=\frac{5}{100}=0.05\text{A}$