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A balloon’s radius increases 3 times when the air is pumped in it. If the surface area of the balloon after pumping the air is A2 and its surface area before pumping the air is A1, then what is the value of \(\frac{A_1}{A_2}\) ?(a) \(\frac{9}{1}\)(b) \(\frac{1}{3}\)(c) \(\frac{3}{1}\)(d) \(\frac{1}{9}\)I got this question during an interview.This key question is from Surface Area of a Sphere topic in division Surface Areas and Volumes of Mathematics – Class 9

Answer»

Right answer is (d) \(\frac{1}{9}\)

The explanation: Let the radius of the balloon before PUMPING the air be “R”, then

A1 = 4πr^2

The radius of the balloon after pumping the air = 3r (given)

HENCE, A2 = 4π(3r)^2

Therefore, \(\frac{A_1}{A_2} = \frac{4πr^2}{4π(3r)^2} = \frac{1}{9}\).



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