# The radius of a spherical balloon increases from $7 \mathrm{~cm}$ to $14 \mathrm{~cm}$ as air is being pumped into it. Find the ratio of surface areas of the balloon in the two cases.

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Given:

The radius of a spherical balloon increases from $7 \mathrm{~cm}$ to $14 \mathrm{~cm}$ as air is being pumped into it.

To do:

We have to find the ratio of surface areas of the balloon in the two cases.

Solution:

Let the initial radius of the spherical balloon be $r_1$ and the final radius be $r_2$

This implies,

Initial radius $r_1=7\ cm$

Final radius $r_2=14\ cm$

Initial surface area of the balloon $=4 \pi r_1^2$

$=4\times\frac{22}{7}\times7^2$

$=88\times7$

$=616\ cm^2$

Final surface area of the balloon $=4 \pi r_2^2$

$=4\times\frac{22}{7}\times(14)^2$

$=88\times28$

$=2464\ cm^2$

Therefore,

The ratio of surface areas of the balloon in the two cases $=616:2464$

$=1:4$.

Updated on 10-Oct-2022 13:46:37