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Question

When a conductance cell was filled with 0.0025 M solution of K2SO4, its resistance was 326 ohm. What will be the specific resistance of K2SO4?

Given:
Length of cell =100 cm
Cross-sectional area =2.14×104 cm2.

A
4.08×104
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B
5.73×104
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C
6.97×104
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D
8.9×104
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Solution

The correct option is C 6.97×104
From ohm's law
V=I×R
R=VI

V:Potential differenceR:Electrical resistanceI:Current

The electrical resistance​ of a conductor is directly proportional ​to its length (l)​ and inversely proportional to its area of cross section (A).
RlA
R=ρ×lA
Where,
ρ is constant of proportionality
This constant of proportionality is called resistivity or specific resistance.
Specific resistance, ρ=R×Al

R=Specific resistance×lA
326=Specific resistance×1002.14×104
Specific resistance, ρ=6.97×104
Hence, option C is correct

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