The height of a circular cylinder is increased six times and the base area is decreased to one-ninth of its value. The factor by which the lateral surface area of the cylinder increases, is [SSC (CGL) 2012] |
A) \[2\]
B) \[\frac{1}{2}\]
C) \[\frac{2}{5}\]
D) \[\frac{3}{2}\]
Correct Answer: A
Solution :
Curved surface area of cylinder \[=2\pi rh\] |
New height = 6h |
Now, base area is decreased by 9 times. |
\[\therefore \]Radius will be decreased by 3 times. |
New radius \[=\frac{1}{3}r\] |
Curved surface area\[=2\pi \frac{1}{3}r\times 6h=(2\pi rh)\times 2\] |
Hence, lateral surface area is increased by 2 times. |
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