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Square of the time period of revolution of a planet is proportional to the cube of the mean distance between sun and planet. If mean distance is increased by 2% then percentage change in its time period is

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2/3

answer is D.

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Detailed Solution

If original mean distance is r then after 2% increase its value will be 1.02 r

$As T^{2} \propto r^{3} or, T \propto r^{\frac{3}{2}}$

$\frac{T_{2}}{T_{1}} = {(\frac{r_{2}}{r_{1}})}^{\frac{3}{2}} = {(\frac{1.02}{1})}^{\frac{3}{2}} = {(1 + 0.02)}^{\frac{3}{2}}$

applying binomial approximation

$\frac{T_{2}}{T_{1}} = 1 + 0.03$

$% change in time period (\frac{T_{2}}{T_{1}} - 1) \times 100 = 3 %$

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