JEE Main & Advanced Sample Paper JEE Main Sample Paper-13

  • question_answer
    If the inclination of the diameter \[PP'\] of the ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] to the major axis is \[\theta \] and \[PP{{'}^{2}}\] is the AM of squares of major and minor axis, then tan 9 is equal to

    A)  \[\frac{b}{a}\]                                 

    B)  \[\frac{a}{b}\]

    C)  \[\frac{\pi }{4}\]                                             

    D)  \[\frac{\pi }{6}\]

    Correct Answer: A

    Solution :

     Let any two points on the ellipse be \[{{c}_{2}}\] and        \[{{c}_{1}}\] \[={{E}_{i}}\]      \[={{E}_{f}}\]     \[{{E}_{i}}=\frac{-GMm}{R}:\]    \[{{E}^{f}}=-\frac{GMm}{2(2R+R)}\] \[[\because \,\,\text{height}\,=2R)\] \[={{E}_{f}}-{{E}_{i}}\]    \[\Rightarrow \] \[\Delta E=\frac{-GMm}{6R}+\frac{GMm}{R}=\frac{5}{6}\,\frac{GMm}{R}\]        \[\Rightarrow \] \[\Delta E=\frac{5}{6}\,mgR\]     \[mg={{V}_{w}}{{\rho }_{w}}\,g+{{V}_{oil}}\,g\] \[{{V}_{w}}\]     \[{{V}_{oil}}\] and \[{{\rho }_{w}}\] \[{{\rho }_{oil}}\] \[\Rightarrow \] Slope of \[m=(2\times 10\times 10\times 1)\,+(8\times 10\times 10\times 0.6)\] \[m=200+480=680\,g\]


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