VIT Engineering VIT Engineering Solved Paper-2014

  • question_answer
    If a magnet is suspended at angle 30° to the magnet meridian, the dip of needle makes angle of 45° with the horizontal, the real dip is

    A) \[{{\tan }^{-1}}\left( \frac{\sqrt{3}}{2} \right)\]                 

    B) \[{{\tan }^{-1}}\left( \sqrt{3} \right)\]

    C) \[{{\tan }^{-1}}\left( \sqrt{\frac{3}{2}} \right)\]                 

    D)  \[{{\tan }^{-1}}\left( \frac{2}{\sqrt{3}} \right)\]

    Correct Answer: D

    Solution :

    \[L=\frac{{{\mu }_{r}}.{{\mu }_{0}}{{N}^{2}}A}{l}\]                     \[=\frac{600\times 4\pi \times {{10}^{-7}}\times {{\left( 2000 \right)}^{2}}\times \left( 1.5 \right)\times {{10}^{-4}}}{0.3}\]                      \[2C+3{{H}_{2}}\to {{C}_{2}}{{H}_{6}};\Delta H=-21.1\]                 \[C+OL2\to C{{O}_{2}};\Delta H=-94.1\]


You need to login to perform this action.
You will be redirected in 3 sec spinner