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: $\frac{11}{21}$
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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