JEE Main & Advanced AIEEE Solved Paper-2007

  • question_answer
    If a line makes an angle of \[C{{H}_{3}}C{{H}_{2}}OH\xrightarrow[{}]{P+{{I}_{2}}}A\xrightarrow[Ether]{Mg}B\xrightarrow[{}]{HCHO}\]with the positive directions of each of X-axis and Y-axis, then the angle that the line makes with the positive direction of the Z-axis is       AIEEE  Solved  Paper-2007

    A)  \[C\xrightarrow[{}]{{{H}_{2}}O}D\]                        

    B)  \[n=3,\text{ }l=1,\text{ }m=1,\text{ }s=+\text{ }1/2\]  

    C)         \[n=3,l=2,m=1,s=+1/2\]              

    D)         \[n=4,\text{ }l=0,\text{ }m=0,\text{ }s=+\text{ }1/2\]

    Correct Answer: D

    Solution :

    Since, a line makes an angle of\[\frac{\pi }{4}\]with positive direction of each of X-axis and Y-axis, therefore \[\alpha =\frac{\pi }{4},\beta =\frac{\pi }{4}\] We know that, \[co{{s}^{2}}\alpha +co{{s}^{2}}\beta +co{{s}^{2}}\gamma =1\] \[\Rightarrow \]\[{{\cos }^{2}}\frac{\pi }{4}+{{\cos }^{2}}\frac{\pi }{4}+{{\cos }^{2}}\gamma =1\] \[\Rightarrow \]\[\frac{1}{2}+\frac{1}{2}+{{\cos }^{2}}\gamma =1\] \[\Rightarrow \]\[{{\cos }^{2}}\gamma =0\]\[\Rightarrow \]\[\gamma ={{90}^{o}}\]


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