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 $${\displaystyle \frac{2849}{3}}$$​ $c{m}^3$ and $${\displaystyle \frac{2618}{3}}$$ $c{m}^3$ 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 $$={\displaystyle \frac{2849}{3}}$$ $c{m}^3$,
the volume of water-filled upto 2cm is $$={\displaystyle \frac{2618}{3}}$$ $c{m}^3$
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

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

    $${\displaystyle \frac{2618}{3}}$$ = $\pi r^2(2r+3h-6)/3$
$\pi r^2(2r+3h-6)$ = $2618$..........(2)
Subtracting the equation $$\left(1\right)-\left(2\right)$$ we get,
$$6\pi r^2=2849-2618$$
 $$6\pi r^2=231$$
$$r^2={\displaystyle \frac{231}{6\pi }}$$
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
$$h={\displaystyle \frac{67}{3}}$$
$$h=22.33$$ $cm$
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.