JEE Main & Advanced Chemistry Electrochemistry / विद्युत् रसायन Question Bank Conductor and Conductance

  • question_answer
    Given \[l/a=0.5c{{m}^{-1}},\,R=50ohm,\,N=1.0\]. The equivalent conductance of the electrolytic cell is [Orissa JEE 2005]

    A)                 \[10\,oh{{m}^{-1}}c{{m}^{2}}gm\,e{{q}^{-1}}\] 

    B)                 \[20\,oh{{m}^{-1}}c{{m}^{2}}gm\,e{{q}^{-1}}\]

    C)                 \[300\,oh{{m}^{-1}}c{{m}^{2}}gme{{q}^{-1}}\]  

    D)                 \[100oh{{m}^{-1}}c{{m}^{2}}\,gme{{q}^{-1}}\]

    Correct Answer: A

    Solution :

               \[l/a=0.5\,c{{m}^{-1}},\,R=50\,ohm\]                    \[p=\frac{Ra}{l}=\frac{50}{0.5}=100\]                    \[\Lambda =k\times \frac{1000}{N}=\frac{1}{p}\times \frac{1000}{N}=\frac{1}{100}\times \frac{1000}{1}\]                                 \[10\,oh{{m}^{-1}}c{{m}^{2}}\,gm\,e{{q}^{-1}}\]


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