Q. Given $l/a = 0.5{cm}^{- 1},\ R\ = \ 50$ ohm, $N\ = \ 1.0.$ The equivalent conductance of the electrolytic cell is
A) $10\mathrm{\Omega}^{- 1}\ {cm}^{2}\ g\ {equiv}^{- 1}$
B) $20\mathrm{\Omega}^{- 1}\ {cm}^{2}\ g\ {equiv}^{- 1}$
C) $300\mathrm{\Omega}^{- 1}\ {cm}^{2}\ g\ {equiv}^{- 1}$
D) $100\mathrm{\Omega}^{- 1}\ {cm}^{2}\ g\ {equiv}^{- 1}$
Explanation
(a)
Given, $\frac{l}{a} = 0.5\ {cm}^{- 1}$
$R\ = \ 50\mathrm{\Omega}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ N\ = \ 1.0$
Specific conductance $(\kappa)\ = \ \frac{1}{\rho} = \ \frac{l}{R.a} = \ \frac{0.5}{50}\$
$\Lambda\ \ \ \ = \ \kappa\ \times \ \frac{1000}{N}\$
= $\frac{0.5}{50}\ \times \ \frac{1000}{0.1}$
$= \ 10\ \mathrm{\Omega}^{- 1}\ {cm}^{2}\ g\ {eq}^{- 1}\$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.