Question : The radius of a spherical balloon is inflated from 7 cm to 10.5 cm. The percentage increase in its surface area is:
Option 1: 150%
Option 2: 125%
Option 3: 120%
Option 4: 135%
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Correct Answer: 125%
Solution :
Given: Radius = 7 cm
New radius = 10.5 cm
Surface area of sphere = $4\pi r^2$
Surface area of old radius = $4\pi r^2$
= $4\pi \times7^2$
= $4\pi \times49$
= $196\pi$
Surface area of new radius = $4\pi r^2$
= $4\pi \times(10.5)^2$
= $4\pi \times110.25$
= $441\pi$
Percentage increase = $\frac{\text{Surface area of new radius - surface area of old Radius}}{\text{surface area of old Radius}}\times100$
= $\frac{441\pi - 196\pi}{196\pi}\times100$
= $\frac{245}{196}\times100$
= $125\%$
Hence, the correct answer is 125%.
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