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Question

Question: The unit \(\rightarrow\) is used for....

The unit →\rightarrow is used for.

A

Molar conductivity

B

Equivalent conductivity

C

Specific conductivity

D

Conductivity

Answer

Conductivity

Explanation

Solution

l/a=0.5cm−1,R=50ohml/a = 0.5cm^{- 1},R = 50ohm

p=Ral=500.5=100p = \frac{Ra}{l} = \frac{50}{0.5} = 100

Λ=k×1000N=1p×1000N=1100×10001\Lambda = k \times \frac{1000}{N} = \frac{1}{p} \times \frac{1000}{N} = \frac{1}{100} \times \frac{1000}{1}

10ohm−1cm2gmeq−110ohm^{- 1}cm^{2}gmeq^{- 1}