Manipal Engineering Manipal Engineering Solved Paper-2015

  • question_answer
    If \[x=\frac{r}{3},\frac{{{d}^{2}}y}{d{{x}^{2}}}=\]then \[x=\frac{r}{3},y\]is equal to

    A)  36                                         

    B) \[\frac{{{N}_{2}}}{{{N}_{1}}}\]

    C) \[\frac{{{N}_{2}}}{{{N}_{1}}}=\exp \left( -\frac{{{E}_{2}}-{{E}_{1}}}{kT} \right)\]                 

    D) \[\lambda =550nm\]

    Correct Answer: C

    Solution :

    We have, \[a.c=\left| c \right|,\left| c-a \right|=2\sqrt{2}\] \[a\times b\]\[\left| (a\times b)\times c \right|\] \[\frac{2}{3}\]\[\frac{3}{2}\] \[\frac{{{d}^{2}}y}{d{{x}^{2}}}={{\left\{ y+{{\left( \frac{dy}{dx} \right)}^{2}} \right\}}^{1/4}}\]\[1+\frac{2}{3}+\frac{6}{{{3}^{2}}}+\frac{10}{{{3}^{3}}}+\frac{14}{{{3}^{4}}}+...\] \[\frac{x}{2}-\frac{y}{3}=1\]\[\frac{x}{-2}+\frac{y}{1}=1\] \[\frac{x}{2}-\frac{y}{3}=-1\]\[\frac{x}{-2}+\frac{y}{1}=-1\]


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