Q. The resistance of $\frac{N}{10}$ solution is found to be $2.5\ \times 10^{3}\mathrm{\Omega}\ .$ The equivalent conductance of the solution is (cell constant = 1.25${cm}^{- 1}$)
A) ${2.5\mathrm{\Omega}}^{- 1}\ {cm}^{2}\ {equiv}^{- 1}$
B) ${5.0\mathrm{\Omega}}^{- 1}\ {cm}^{2}\ {equiv}^{- 1}$
C) ${2.5\mathrm{\Omega}}^{- 1}\ {cm}^{- 2}\ {equiv}^{- 1}$
D) ${5.0\mathrm{\Omega}}^{- 1}\ {cm}^{- 2}\ {equiv}^{- 1}$
Explanation
(b)
Resistance of $\frac{N}{10}$ solution $= \ 2.5\ \times \ 10^{3}\ \mathrm{\Omega}$
$\kappa\ = \ \frac{1}{resistance}\ \times cell\ constant$
$= \ \frac{1}{2.5\ \times 10^{3}}\ \times 1.25\$
$= \ \frac{1.25\ \times 10^{- 3}}{2.5}\ \ = \ \ 5\ \times \ 10^{- 4}\ \mathrm{\Omega}^{- 1}\ {cm}^{- 1}\$
Equivalent conductance
$= \frac{\kappa\ \times 100}{M}$
${= \ \frac{5\ \times \ 10^{- 4}\ \times \ 1000}{1/10}
}{= {5\mathrm{\Omega}}^{- 1}\ {cm}^{2}\ {equiv}^{- 1}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.