Question

A spherical balloon expands when it is taken from the cold outdoors to the inside of a warm house. If its surface area increases 17.4percent002, by what percentage does the radius of the balloon change?

Answer 1

Answer on Question #73062-Physics-Classical Mechanics

A spherical balloon expands when it is taken from the cold outdoors to the inside of a warm house. If its surface area increases 17.4 percent 002, by what percentage does the radius of the balloon change?

Solution

$$\frac {\Delta A}{A} = \frac {4 \pi r ^ {\prime 2}}{4 \pi r ^ {2}} - 1 = \left(\frac {r ^ {\prime}}{r}\right) ^ {2} - 1$$ $$\left(\frac {r ^ {\prime}}{r}\right) ^ {2} - 1 = 0.174$$ $$\left(\frac {r ^ {\prime}}{r}\right) ^ {2} = 1 + 0.174 = 1.174$$ $$\left(\frac {r ^ {\prime}}{r}\right) = \sqrt {1.174}$$ $$\frac {\Delta r}{r} = \left(\frac {r ^ {\prime}}{r}\right) - 1 = \sqrt {1.174} - 1 = 0.0835.$$

Answer: 8.35%.

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