CBSE Class 12-science Answered
3.13
![question image](https://images.topperlearning.com/topper/new-ate/topr_411144269491227309rps20190813132610.jpeg)
Asked by lovemaan5500 | 13 Aug, 2019, 17:50: PM
Given:
V = 20 V
I = 1.198 ampere
R =?
R
![equals straight V over straight I
equals fraction numerator 20 over denominator 1.198 end fraction
equals 16.69 space straight capital omega](https://images.topperlearning.com/topper/tinymce/cache/6cfb71d798317e856452864a65f409a9.png)
Conductance
![equals 1 over straight R
equals fraction numerator 1 over denominator 16.69 end fraction
equals 0.0599 space ohm to the power of negative 1 end exponent](https://images.topperlearning.com/topper/tinymce/cache/bafa08d1c7bdce1f1fd5d3fc5731ef4c.png)
Radius of column = 4/2 = 2 cm
Area of cross section = πr2
=3.14× 4 = 12.56 cm2
Cell constant
![equals straight l over straight A
equals fraction numerator 12 over denominator 12.56 end fraction
equals 0.955 space cm to the power of negative 1 end exponent](https://images.topperlearning.com/topper/tinymce/cache/2fbae6851376f4107a755d2540c4eb44.png)
Specific conductance, K= Conductance×Cell constant
= 0.0599×0.955
=0.0572 ohm−1 cm−1
Equivalent conductance,Λ
![straight capital lambda space equals fraction numerator straight K cross times 1000 over denominator straight C end fraction
space space space equals fraction numerator 0.0572 cross times 1000 over denominator 0.1 end fraction
straight capital lambda space equals 570 space ohm to the power of negative 1 end exponent eq to the power of negative 1 end exponent cm squared](https://images.topperlearning.com/topper/tinymce/cache/8368b5311adb622e436d4045731e6ea7.png)
Answered by Varsha | 14 Aug, 2019, 12:12: PM
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