Question 137
When the radius of a sphere is increased by 5 cm, its surface area increases by 704 cm$$^2$$. The diameter of the original
sphere, is (Take $$\pi = \frac{22}{7}$$)
Solution
Surface area of the sphere = 4$$ \times \pi \times r^2$$
Let the radius of the sphere in the initial be r.
According to the question,
4$$ \times \pi \times (r + 5)^2 - 4 \times \pi \times r^2 = 704$$
$$\frac{22}{7} \times (r + 5)^2 - \frac{22}{7} \times r^2 = 176$$
$$r^2 + 25 + 10r - r^2 = 56$$
r = 3.1
Diameter of the original sphere = 2r = 2 $$\times$$ 3.1 = 6.2 cm
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