Question

A test-tube consists of a hemisphere and a cylinder of the same radius. The volume of water required to fill the whole tube is \[\dfrac{{2849}}{3}\]​ $cm^3$ and \[\dfrac{{2618}}{3}\] $cm^3$ of water are required to fill the tube to a level which is 2 $cm$ below the top of the tube. Find the radius and volume of the whole tube.

Answer 1

Hint: First we consider a test tube of a hemisphere and a cylinder of the same radius.
We use the formula for volume of the hemisphere and volume of the cylinder then we add these volumes and find the volume of a test tube.
Finally we subtract both the volume to get the radius of these test tube

Complete step-by-step answer:
It is given that the volume of water required to fill the entire tube is \[ = \dfrac{{2849}}{3}\] $cm^3$,
the volume of water-filled upto 2cm is \[ = \dfrac{{2618}}{3}\] $cm^3$
Let \[r\] be the radius of the test tube and
Let \[h\] be the height of the test tube.
Volume of the test tube = volume of the hemisphere + volume of the cylinder

  \[\dfrac{{2849}}{3}\] = ${\pi {r^2}(2r + 3h)}/{3}$
${\pi {r^2}(2r + 3h)}$ = $2849$.........(1)
water is filled \[2\] $cm$ below the top of tube, \[H = h - 2\] $cm$
Now, volume

    \[\dfrac{{2618}}{3}\] = ${\pi {r^2}(2r + 3h - 6)}/{3}$
${\pi {r^2}(2r + 3h - 6)}$ = $2618$..........(2)
Subtracting the equation \[\left( 1 \right) - \left( 2 \right)\] we get,
\[6\pi {r^2} = 2849 - 2618\]
 \[6\pi {r^2} = 231\]
\[{r^2} = \dfrac{{231}}{{6\pi }}\]
Putting $\pi = 3.14$ and dividing the values we get
\[{r^2} = 12.25\]
Taking root on both sides we get,
\[r = 3.5\] $cm$
Substituting the value of r in equation (1) we get,
which implies that,
\[\pi {(3.5)^2}(2(3.5) + 3h) = 2849\]
\[7 + 3h = 74\]
\[3h = 74 - 7\]
\[3h = 67\]
Divide 67 by 3 we get the value of height
\[h = \dfrac{{67}}{3}\]
\[h = 22.33\] $cm$
Therefore, the radius of the test tube is \[r = 3.5\] $cm$ and the height \[h = 22\] $cm$

Note: Students make common mistakes i.e. they have not considered the marked point which is $2$ $cm$ from the top of the tube.