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Mathematics

The radius of a hemispherical balloon increases from 6 cm to 12 cm as air is being pumped into it. The ratio of the surface areas of the balloon in the two cases is

  1. 1 : 4

  2. 1 : 3

  3. 2 : 3

  4. 2 : 1

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Answer

Radius of balloon in original position (r) = 6 cm,

In pumped position radius (R) = 12 cm.

Ratio of surface areas in two situations = 4πr24πR2\dfrac{4πr^2}{4πR^2}

=r2R2=62122=36144=14=1:4.= \dfrac{r^2}{R^2} \\[1em] = \dfrac{6^2}{12^2} \\[1em] = \dfrac{36}{144} \\[1em] = \dfrac{1}{4} \\[1em] = 1 : 4.

Hence, Option 1 is the correct option.

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