The radius of a spherical gas balloon increases from 7 cm to 21 cm as air is being pumped into it. The ratio of the surface areas of the balloon in th - Quantitative Aptitude MCQ Competitive Examination 2026 | ExamTest.live

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The radius of a spherical gas balloon increases from 7 cm to 21 cm as air is being pumped into it. The ratio of the surface areas of the balloon in the two cases is:

Solution & Step-by-step Explanation

The surface area (S) of a sphere of radius r is given by S = 4 π r^{2} . Let the initial radius be r_{1} = 7 cm and final radius be r_{2} = 21 cm . The ratio of their surface areas is: \frac{S _{1}}{S _{2}} = \frac{4 π r _{1}^{2}}{4 π r _{2}^{2}} = (\frac{r _{1}}{r _{2}})^{2} \frac{S _{1}}{S _{2}} = (\frac{7}{21})^{2} = (\frac{1}{3})^{2} = \frac{1}{9} Thus, the ratio of the surface areas is 1 : 9 .

Try it yourself before checking the explanation above.

The radius of a spherical gas balloon increases from 7 cm to 21 cm as air is being pumped into it. The ratio of the surface areas of the balloon in the two cases is:

9 ∶ 1

49 ∶ 21

1 ∶ 9

7 ∶ 441

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