If the radius of a sphere is increased by 2.5 decimeter (dm), then its surface area increases by 110 dm². What is 50% of the volume (in dm³) of the sphere? (Take \(\pi \) = \(\frac{22}{7}\))

\(\frac{11}{42}\) dm³

Surface area of sphere = 4\(\pi \)r²

4\(\pi \) (r + 2.5)² - 4\(\pi \)r² = 110

4\(\pi \) (r² + 6.25 + 5r) - 4\(\pi \)r² = 110

4\(\pi \)r² + 25\(\pi \) + 20\(\pi \)r - 4\(\pi \)r² = 110

20r = \(\frac{110 × 7}{22}\) - 25

r = \(\frac{1}{2}\) decimeter

volume of sphere = \(\frac{4}{3}\)\(\pi \)r³ = \(\frac{4}{3}\) × \(\frac{22}{7}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) × \(\frac{1}{2}\) = \(\frac{11}{21}\)

50% of the volume = \(\frac{1}{2}\) × \(\frac{11}{21}\) = \(\frac{11}{42}\) dm³