JEE MAIN - Physics (2016 - 9th April Morning Slot - No. 3)
Figure shows elliptical path abcd of a planet around the sun S such that the area of triangle csa is $${1 \over 4}$$ the area of the ellipse. (See figure) With db as the semimajor axis, and ca as the semiminor axis. If t1 is the time taken for planet to go over path abc and t2 for path taken over cda then :
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t1 = t2
t1 = 2t2
t1 = 3t2
t1 = 4t2
Explanation
Let the area of ellipse = A
$$ \therefore $$ Area of abcSa = $${A \over 2} + {A \over 4}$$ = Area of half of the ellipse + Area of the triangle = $${{3A} \over 4}$$
Area of adcSa = $${A \over 2} - {A \over 4}$$ = $${A \over 4}$$
$$ \therefore $$ $${{{t_1}} \over {{t_2}}}$$ = $${{{{3A} \over 4}} \over {{A \over 4}}}$$ = 3
$$ \therefore $$ t1 = 3t2
$$ \therefore $$ Area of abcSa = $${A \over 2} + {A \over 4}$$ = Area of half of the ellipse + Area of the triangle = $${{3A} \over 4}$$
Area of adcSa = $${A \over 2} - {A \over 4}$$ = $${A \over 4}$$
$$ \therefore $$ $${{{t_1}} \over {{t_2}}}$$ = $${{{{3A} \over 4}} \over {{A \over 4}}}$$ = 3
$$ \therefore $$ t1 = 3t2
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