Haryana PMT Haryana PMT Solved Paper-2010

  • question_answer 24.      Total surface area of a cathode is \[0.05\text{ }{{m}^{2}}\] and 1    A current passes through it for 1 h. Thickness of            nickel deposited on the cathode is (given that    density of nickel = 9 g/cc and its \[ECE=3.04\times {{10}^{-~~4}}g/C\])

    A)  2.4m                                    

    B)  2.4 \[\mu \]m

    C)  2.4 mm                               

    D)  None of these

    Correct Answer: B

    Solution :

                    Mass deposited = density \[\times \] volume of the metal \[m=\rho \times A\times x\]                      ?...(i) Hence, from Faradays first law m = Zit   ... (ii) So, from Eqs. (i) and (ii)                 \[Zit=\rho Ax\] \[\Rightarrow \]               \[x=\frac{Zit}{\rho A}\]                 \[=\frac{3.04\times {{10}^{-4}}\times {{10}^{-3}}\times 1\times 3600}{9000\times 0.05}\]                 \[=2.4\times {{10}^{-6}}m=2.4\mu m\]

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