Question : If the radius of a sphere is increased by 2.5 decimetres (dm), then its surface area increases by 110 dm$^2$. What is the volume (in dm$^3$ ) of the sphere? (Take $\pi=\frac{22}{7}$)
Option 1: $\frac{13}{21}$
Option 2: $\frac{3}{7}$
Option 3: $\frac{4}{7}$
Option 4: $\frac{11}{21}$
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Correct Answer: $\frac{11}{21}$
Solution : The radius of a sphere is increased by 2.5 decimeters (dm), then its surface area increases by 110 dm 2 We know that, The surface area of the sphere = $4πr^2$ The volume of sphere = $\frac{4}{3} × πr^3$ Where r = radius and $π = \frac{22}{7}$ a 2 – b 2 = (a + b)(a - b) According to the question, $4π(r + 2.5)^2 - 4πr^2 = 110$ ⇒ $4π[(r + 2.5)^2 - r^2] = 110$ ⇒ $4π[(r + 2.5 + r)(r + 2.5 - r)] = 110$ ⇒ $4 × \frac{22}{7} × (2r + 2.5) × 2.5 = 110$ ⇒ $10 × (2r + 2.5) = 35$ ⇒ $20r + 25 = 35$ ⇒ $r = \frac{1}{2}$ dm Now, Volume = $\frac{4}{3} × π × (\frac{1}{2})^3$ = $\frac{4}{3} × \frac{22}{7} × \frac{1}{8}$ = $\frac{88}{168}$ = $\frac{11}{21}$ dm$^3$ Hence, the correct answer is $\frac{11}{21}$.
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