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(A) According to Kepler's second law of planetary motion,the radius vector joining the sun and the planet sweeps out equal areas in equal intervals of

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The time taken in travelling $DAB$ is less than that for $BCD$.

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(A) According to Kepler's second law of planetary motion,the radius vector joining the sun and the planet sweeps out equal areas in equal intervals of time.In the given figure,the sun $S$ is at one of the foci of the elliptical orbit.The area swept by the planet in path $DAB$ is less than the area swept in path $BCD$ because the sun is closer to the arc $DAB$.Since the area swept in path $DAB$ is smaller than the area swept in path $BCD$,the time taken to travel $DAB$ must be less than the time taken to travel $BCD$ to satisfy the condition of equal area in equal time.Alternatively,according to the law of conservation of angular momentum,the orbital velocity of the planet is higher when it is closer to the sun (near $A$) and lower when it is farther away (near $C$).Thus,the planet travels faster along the path $DAB$ and slower along the path $BCD$,confirming that the time taken for $DAB$ is less than that for $BCD$.

$A$ planet is revolving around the sun in an elliptical path as shown in the figure. Which of the following is the correct option?

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