Correct Answer - Option 4 :

\(\frac{11}{21}\)
**Given:**

The radius of a sphere is increased by 2.5 decimeter(dm),

then its surface area increases by 110 dm^{2}

**Concept used:**

The surface area of sphere = 4πr^{2}

The volume of sphere = (4/3) × πr^{3}

Where r = radius and π = 22/7

a^{2} - b^{2} = (a + b)(a - b)

**Calculation:**

according to the question,

4π(r + 2.5)^{2} - 4πr^{2} = 110

⇒ 4π[(r + 2.5)^{2} - r^{2}] = 110

⇒ 4π[(r + 2.5 + r)(r + 2.5 - r)] = 110

⇒ 4 × 22/7 × (2r + 2.5) × 2.5 = 110

⇒ 10 × (2r + 2.5) = 35

⇒ 20r + 25 = 35

⇒ r = 1/2 dm

now,

Volume = 4/3 × π × (1/2)^{3}

⇒ 4/3 × 22/7 × 1/8

⇒ 88/168 = 11/21 dm^{3}

**∴ The volume of the sphere is 11/21 dm**^{3}.