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CBSE Class 12-science Answered

3.13
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Asked by lovemaan5500 | 13 Aug, 2019, 05:50: PM
answered-by-expert Expert Answer
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
 
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
 
 
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
 
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
Answered by Varsha | 14 Aug, 2019, 12:12: PM

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