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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.
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 $=616:2464$
$=1:4$.
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