Question : The radius of the ends of a frustum of a solid right-circular cone 45 cm high is 28 cm and 7 cm. If this frustum is melted and reconstructed into a solid right circular cylinder whose radius of base and height are in the ratio 3: 5, find the curved surface area (in ${cm}^2$ ) of this cylinder. [Use $\pi=\frac{22}{7}$.]
Option 1: 4580
Option 2: 4610
Option 3: 4640
Option 4: 4620
Correct Answer: 4620
Solution :
According to the question,
Area of the frustum = Volume of the cylinder
⇒ $\frac{1}{3}\pi$ ((7)
2
+ (28)
2
+ 7×28) 45 = $\pi r^{2} h$ where $r$, $h$ are the radius and heights of the cylinder
⇒ $\frac{1}{3}\pi(49 + 754 + 196)$ 45 = $\pi × 9x^{2} × 5x$
⇒ (49 + 754 + 196) 15 = $9x^{2} × 5x$
⇒ 1029 = $3x^{3}$
⇒ $x^{3}$ = $\frac{1029}{3}$ = 343
⇒ $x$ = $\sqrt{343}$ = 7
Curved surface area = $2\pi r h$
= 2 $\frac{22}{7}$ × 35 × 21
= 2 × 22 × 5 × 21
= 4620
Hence, the correct answer is 4620.
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