Question

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)?

• A. 300%
• B. 400%
• C. 500%
• D. 600%

Answer 1

The correct option is C 500%

Let the radius of the cylinder be ${}^{\prime }{r}^{\prime }$ and its height be ${}^{\prime }{h}^{\prime }$
So the radius of hemisphere will be ${}^{\prime }{r}^{\prime }$.
Total volume, $V=\pi r^{2}h+\frac{2}{3}\pi r^3$
Now the height of cylinder is doubled. So, the volume will be,
$V_1=2\pi r^{2}h+\frac{2}{3}\pi r^3$
we know,
$\frac{V_1-V}{V}\times 100=50$%$=\frac{\pi r^{2}h}{\pi r^{2}h+\frac{2}{3}\pi r^3}$
$\Rightarrow \frac{h}{h+\frac{2}{3}r}=0.5\Rightarrow h=0.5(h+\frac{2}{3}r)\Rightarrow 0.5h=\frac{1}{3}r\Rightarrow h=\frac{2}{3}r.....\left(i\right)$

When the radii is doubled, the volume will be
$V_2=\pi (2r{)}^{2}h+\frac{2}{3}\pi (2r{)}^3=4\pi r^{2}h+\frac{16}{3}\pi r^3$
The percentage increase in the volume will be,
$\frac{V_2-V}{V}\times 100=\frac{3\pi r^{2}h+\frac{14}{3}\pi r^3}{\pi r^{2}h+\frac{2}{3}\pi r^3}\times 100=\frac{3h+\frac{14}{3}r}{h+\frac{2}{3}r}\times 100$
using equation (i),
$=\frac{\frac{6}{3}r+\frac{14}{3}r}{\frac{2}{3}r+\frac{2}{3}r}\times 100=\frac{20}{4}\times 100=500\%$

Hence the percent increase when radii is doubled is 500%.