Question

When a conductance cell was filled with $0.0025M$ solution of $K_{2}S{O}_4$, its resistance was $326ohm$. What will be the specific resistance of $K_{2}S{O}_4$?

Given:
$\text{Length of cell}=100cm$
$\text{Cross-sectional area}=2.14\times {10}^{-4}c{m}^2$.

• A. $4.08\times {10}^{-4}$
• B. $5.73\times {10}^{-4}$
• C. $6.97\times {10}^{-4}$
• D. $8.9\times {10}^{-4}$

Answer 1

The correct option is C $6.97\times {10}^{-4}$

From ohm's law
$V=I\times R$
$R=\frac{V}{I}$

$V:\text{Potential difference}R:\text{Electrical resistance}I:\text{Current}$​

The electrical resistance​ of a conductor is directly proportional ​to its length (l)​ and inversely proportional to its area of cross section (A).
$R\propto \frac{l}{A}$
$R=\rho \times \frac{l}{A}$
Where,
$\rho$ is constant of proportionality
This constant of proportionality is called resistivity or specific resistance.
$\text{Specific resistance,}\rho =R\times \frac{A}{l}$

$R=\frac{\text{Specific resistance}\times l}{A}$
$326=\frac{\text{Specific resistance}\times 100}{2.14\times {10}^{-4}}$
$\text{Specific resistance,}\rho =6.97\times {10}^{-4}$
Hence, option C is correct