NEET AIPMT SOLVED PAPER SCREENING 2009

  • question_answer
    The figure shows elliptical orbit of a planet m about the sun S. The shaded area SCD is twice the shaded area SAB. If \[\frac{{{\varepsilon }^{2}}\sqrt{{{R}^{2}}+{{\left( L\omega -\frac{1}{C\omega } \right)}^{2}}}}{R}\] is the time for the planet to move from C to D and \[\frac{{{\varepsilon }^{2}}\left[ {{R}^{2}}+{{\left( L\omega -\frac{1}{C\omega } \right)}^{2}} \right]}{R}\] is the time to move from A to B, then

    A) \[\frac{{{\varepsilon }^{2}}R}{\sqrt{{{R}^{2}}+{{\left( L\omega -\frac{1}{C\omega } \right)}^{2}}}}\]

    B)                        \[V=-{{x}^{2}}y-x{{z}^{3}}+4\]   

    C)        \[\overrightarrow{E}\] 

    D)        \[\overrightarrow{E}=\hat{i}(2xy+{{z}^{3}})+\hat{j}{{x}^{2}}+\hat{k}3x{{z}^{2}}\]

    Correct Answer: C

    Solution :

    Key Idea Apply Kepler's law of area of planetary motion. The line joining the sun to the planet sweeps out equal areas in equal time interval ie, areal velocity is constant. \[{{Z}_{2}}\]constant or\[{{M}_{2,}}\] \[{{r}_{0.}}\]\[{{M}_{1}}\times {{M}_{2}}\] Given   \[{{Z}_{1}}{{Z}_{2}}\] \[{{Z}_{1}}\]\[{{M}_{1}}\]


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