Let A be area swept out by the line joining the earth and the sun during Feb. 1993. The area swept out by the line during a typical week in Feb. 1993 is.

A

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2 A

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4 A

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\frac{A}{4}

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Solution

The correct option is D $\frac{A}{4}$
Kepler's 2nd law; Equal area is swept in equal interval of time by the planet revolving around the sun i-e, $\frac{A}{t} = c o n s t a n t$
Thus $\frac{A_{2}}{t_{2}} = \frac{A_{1}}{t_{1}}$
For $A_{1} = A t_{1} = t t_{2} = \frac{t}{4}$
$\frac{A_{2}}{\frac{t}{4}} = \frac{A}{t} ⟹ A_{2} = \frac{A}{4}$

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Let A be area swept out by the line joining the earth and the sun during Feb. 1993. The area swept out by the line during a typical week in Feb. 1993 is.

The correct option is D A4Kepler's 2nd law; Equal area is swept in equal interval of time by the planet revolving around the sun i-e, At=constantThus A2t2=A1t1For A1=At1=tt2=t4A2t4=At⟹A2=A4