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VIT Engineering VIT Engineering Solved Paper-2014

  • question_answer
    If \[\cos \alpha +i\sin \alpha ,b=\cos \beta +i\sin \beta ,\]\[c=\cos \gamma +i\sin \gamma \] and \[\frac{b}{c}+\frac{c}{a}+\frac{a}{b}=1,\]then \[\cos \left( \beta -\gamma  \right)+\cos \left( \gamma -\alpha  \right)+\cos \left( \alpha -\beta  \right)\] is equal to

    A) \[\frac{3}{2}\]                                   

    B) \[-\frac{3}{2}\]

    C) 0                                             

    D) 1

    Correct Answer: D

    Solution :

    We have,  \[=1.25\times {{10}^{7}}kWh\]                    \[i=\frac{5}{20+30}=\frac{5}{50}A\] and             \[=5\times {{10}^{3}}W\] Now,           \[=5000kW\]                      \[=\frac{mgh}{t}\]                                         \[=7.5\times 9.8\times 4.7\] \[=3454.5kW\]     \[\text{=}\frac{\text{Power used}}{\text{Power consumed}}\text{ }\!\!\times\!\!\text{ 100}\]                ...(i) Similarly,   \[=\frac{3454.5}{5000}\times 100=69%\]       ...(ii) and     \[E=V+{{l}_{r}}\]           ...(iii) On adding  Eqs. (i), (ii), and (iii), we get  \[{{\lambda }_{red}}>{{\lambda }_{violet}}\]                                \[\lambda \]                              \[r=\frac{mV}{B}=\frac{V}{\left( \frac{e}{m} \right)B}\] On equating real parts, we get \[r=\frac{6\times {{10}^{7}}}{1.7\times {{10}^{11}}\times 1.5\times {{10}^{-2}}}\] 


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