Question

If the radius of a sphere is increased by 2.5 decimetre (dm), then its surface area increases by 110 dm². What is the volume (in dm³) of the sphere?

(Take π = \(\frac{22}{7}\))

1. \(\frac{4}{7}\)
2. \(\frac{3}{7}\)
3. \(\frac{13}{21}\)
4. \(\frac{11}{21}\)

Answer 1

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 dm²

Concept used:

The surface area of sphere = 4πr²

The volume of sphere = (4/3) × πr³

Where r = radius and π = 22/7

a² - b² = (a + b)(a - b)

Calculation:

according to the question,

4π(r + 2.5)² - 4πr² = 110

⇒ 4π[(r + 2.5)² - r²] = 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 dm³

∴ The volume of the sphere is 11/21 dm³.