VIT Engineering VIT Engineering Solved Paper-2014

  • question_answer
    When a resistor of 11\[\Omega \] is connected in series with an electric cell. The current flowing in it is 0.5 A. Instead when a resistor of 5\[\Omega \] is connected to the same electric cell in series, the current increases by 0.4 A. The internal resistance of the cell is

    A) 1.5\[\Omega \]                

    B) 2\[\Omega \]

    C) 2.5\[\Omega \] 

    D) 3.5\[\Omega \]

    Correct Answer: C

    Solution :

    \[N{{i}^{3+}}=\left[ Ar \right]3{{d}^{7}}4{{s}^{0}}\Rightarrow 3\text{unpaired electrons}\]                      \[CuS{{O}_{4}}+2Kl\to Cu{{l}_{2}}+{{K}_{2}}S{{O}_{4}}\]                      \[2Cu{{l}_{2}}\to \underset{\text{Cuprous iodide}}{\mathop{2Cul}}\,+{{l}_{2}}\]                       \[S{{c}^{3+}}>C{{r}^{3+}}>F{{e}^{3+}}>M{{n}^{3+}}\]                    \[C{{r}^{3+}}>M{{n}^{3+}}>F{{e}^{3+}}>S{{c}^{3+}}.\] On solving these equation,  \[\text{ }\!\!\Delta\!\!\text{ }{{\text{T}}_{\text{b}}}\text{=}\frac{\text{1000 }\!\!\times\!\!\text{ }{{\text{k}}_{\text{b}}}\text{ }\!\!\times\!\!\text{ w}}{\text{W }\!\!\times\!\!\text{ M}}\]                


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