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NEET 2009
Exam
NEET
Year
2009
Subject
Chemistry
Chapter
Solid State

NEET 2009 · Chemistry · Solid State

The figure shows elliptical orbit of a planet m about the sun S ....

The figure shows elliptical orbit of a planet mm about the sun SS. The shaded area SCDS C D is twice the shaded area SABS A B. If t1\mathrm{t}_1 is the time for the planet to move from CC to DD and t2\mathrm{t}_2 is the time to move from AA to BB, then The figure shows elliptical orbit of a planet m about the sun S . The shaded area S C D is twice the shaded area S A B

Options

  1. At1>t2t_1>t_2
  2. Bt1=4t2t_1=4 t_2
  3. Ct1=2t2\mathrm{t}_1=2 \mathrm{t}_2
  4. Dt1=t2\mathrm{t}_1=\mathrm{t}_2

Correct answer

C. t1=2t2\mathrm{t}_1=2 \mathrm{t}_2

Step-by-step 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 iei e, areal velocity is constant. dAdt= constant \frac{\mathrm{dA}}{\mathrm{dt}}=\text{ constant }or A1t1=A2t2\frac{\mathrm{A}_1}{\mathrm{t}_1}=\frac{\mathrm{A}_2}{\mathrm{t}_2}\begin{array}{llrl} \Rightarrow& \mathrm{t}}_{1} & =\frac{\mathrm{A}_1}{\mathrm{~A}_2} \mathrm{t}}_{2} \\ \text{ Given }& \mathrm{A}}_{1} & =2 \mathrm{~A}}_{2} \\ \therefore & \mathrm{t}}_{1} & =2 \mathrm{t}}_{2} \end{array}$
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