2. Sample Papers
  3. NEET

NEET Sample Paper NEET Sample Test Paper-37

  • 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 \[{{t}_{1}}\]is the time for the planet to move from C to D and\[{{t}_{2}}\] is the time to move from A to B, then

    A) \[{{t}_{1}}>{{t}_{2}}\]

    B) \[{{t}_{1}}=4{{t}_{2}}\]

    C) \[{{t}_{1}}=2{{t}_{2}}\]

    D) \[{{t}_{1}}={{t}_{2}}\]  

    Correct Answer: C

    Solution :

     Applying Kepler's law of area of planetary motion. The line joining the sun to the planet sweeps out equal areas in equal time interval i.e., areal velocity is constant. \[\frac{dA}{dt}=\]constant or \[\frac{{{A}_{1}}}{{{t}_{1}}}=\frac{{{A}_{2}}}{{{t}_{2}}}\Rightarrow {{t}_{1}}=\frac{{{A}_{1}}}{{{A}_{2}}}{{t}_{2}}\] Given \[{{A}_{1}}=2{{A}_{2}}\] \[\therefore \] \[{{t}_{1}}=2{{t}_{2}}\]


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