Q177 / 180 · KCET 2012 · Physics

A planet moving around sun sweeps area $\mathrm{A}_{1}$ in 2 days, $\mathrm{A}_{2}$ in 3 days and $\mathrm{A}_{3}$ in 6 days. Then, the relation between $\mathrm{A}_{1}, \mathrm{~A}_{2}$ and $\mathrm{A}_{3}$ is
Figure

  • A
    $6 \mathrm{~A}_{1}=3 \mathrm{~A}_{2}=2 \mathrm{~A}_{3}$
  • Correct answer
    B
    $3 A_{1}=2 A_{2}=A_{3}$
  • C
    $2 \mathrm{~A}_{1}=3 \mathrm{~A}_{2}=6 \mathrm{~A}_{3}$
  • D
    $3 A_{1}=2 A_{2}=6 A_{3}$

Solution

By Kepler's second law of motion, $\frac{A_{1}}{t_{1}}=\frac{A_{2}}{t_{2}}=\frac{A_{3}}{t_{3}}$ $\frac{A_{1}}{2}=\frac{A_{2}}{3}=\frac{A_{3}}{6}$ $\frac{3 A_{1}}{6}=\frac{2 A_{2}}{6}=\frac{A_{3}}{6}$ $3 A_{1}=2 A_{2}=A_{3}$

Hence, the correct answer is option: 'B'.

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