VIT Engineering VIT Engineering Solved Paper-2014

  • question_answer
    The equivalent conductivity of a solution containing 2.54 g of \[CuS{{O}_{4}}\] per L is \[91.0{{W}^{-1}}\]\[c{{m}^{2}}e{{q}^{-1}}.\] Its conductivity would be

    A) \[2.9\times {{10}^{-3}}{{\Omega }^{-1}}c{{m}^{-1}}\]

    B) \[1.8\times {{10}^{-2}}{{\Omega }^{-1}}c{{m}^{-1}}\]

    C) \[2.4\times {{10}^{-4}}{{\Omega }^{-1}}c{{m}^{-1}}\]

    D) \[3.6\times {{10}^{-3}}{{\Omega }^{-1}}c{{m}^{-1}}\]

    Correct Answer: A

    Solution :

    \[=\frac{10}{100}\times 6+\frac{10}{100}\times 4\] \[=0.6+0.4=1\] \[\frac{\Delta R}{R}=\frac{1}{10}\]


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