Question

1. Resistance of decimolar solution is 50 ohm. if electrodes of surface area 0.0004 m^2 each are placed at a distance of 0.02 m then conductivity of solution is......?? a. 1 S cm^-1
b. 0.01 S cm^-1
c. 0.001 S cm^-1
d. 10 S cm^-1

Answer 1

Given : R= 50 ohm

l= 0.02 m

A=0.0004 m^2

R=ρl/A

therefore ρ = RA/l

= 50× 0.0004/0.02

= 1

conductivity (κ) is the inverse of resistivity ,

κ=1/ρ

= 1/1

=1

Answer 2

From the given question the correct option is a.

As we know that,

Given:

$R=$$50$ ohm

$l= 0.02 m$

$A=0.0004$$m^{2}$

Find out:

The conductivity of the solution.

Let us:

R=ρl/A

so,

ρ =$\frac{RA}{I}$

$= 50$× 0.0004/0.02

$= 1$$\frac{0.0004}{0.02}$

The conductivity (κ) is the inverse of resistivity

κ=1/ρ

$=\frac{1}{1}$

$=1$

Hence, the conductivity of the given solution is 1 S cm^-1.