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Q. A solid hemisphere is attached to the top of a cylinder, having the same radius as that of the cylinder. If the height of the cylinder were doubled (keeping both radii fixed), the volume of the entire system would have increased by $50\%$. By what percentage would the volume have increased if the radii of the hemisphere and the cylinder were doubled (keeping the height fixed)?

KVPYKVPY 2018

Solution:

Let the height and radius of cylinder are $h$ are $r$, respectively.
$\therefore $ Volume of cylinder $= \pi r^2h$
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and volume of hemisphere $ = \frac{2}{3} \pi r^3$
$\therefore $ Volume of solid $ = \pi r^2 h + \frac{2}{3} \pi r^3$
When height of cylinder is doubled, then
volume of solid $= 2\pi r^2 h + \frac{2}{3} \pi r^3$
$\therefore \frac{V_2}{V_1} = \frac{3}{2} = \frac{2\pi r^2 h + \frac{2}{3} \pi r^3}{\pi r^2 h + \frac{2}{3} \pi r^3}$
$\Rightarrow \frac{2h + \frac{2}{3} r }{h + \frac{2}{3} r} = \frac{3}{2}$
$\Rightarrow \frac{h}{2} = \frac{r}{3}$
When the radius is doubled, then volume of solid
$ = 4\pi r^2 h + \frac{16 \pi r^3}{3}$
$\therefore \frac{V'_{2}}{V_{1} } = \frac{4h + \frac{16}{3}r}{h + \frac{2}{3}r} $
$ = \frac{4h+8h}{h + h} = 6\left[\because \frac{r}{3} =\frac{h}{3}\right]$
Hence, volume is increased by $500\%$.