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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When the radius of a sphere is increased by 5 cm, its surface area increases by 704 cm$$^2$$. The diameter of the originalsphere, is (Take $$\pi = \frac{22}{7}$$)