To begin, we can work out the volume of
each solid:
Solid 1) A sphere of radius $1 \ \mathrm{cm}$ :
$$\textrm{[Volume]} = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi \
\mathrm{cm^3}$$
Solid 2) A solid cylinder of height $\frac{4}{3} \ \mathrm{cm}$ and
radius $1 \ \mathrm{cm}$:
$$\textrm{[Volume]} = \pi r^2 h = \frac{4}{3}\pi \
\mathrm{cm^3}$$
Solid 3) A solid circular cone of base radius $1 \ \mathrm{cm}$ and
height $4 \ \mathrm{cm}$.
$$\textrm{[Volume]} = \frac{1}{3} \times \textrm{[Volume of a
cylinder]} = \frac{1}{3}\pi r^2 h = \frac{4}{3}\pi \
\mathrm{cm^3}$$
Solid 4) A solid cylinder of height $\frac{4}{9}\ \mathrm{cm}$ with
a hole drilled through it, leaving an annular cross-section with
interal and external radii $2 \ \mathrm{cm}$ and $1 \
\mathrm{cm}$.
$$
\begin{align}
\textrm{[Volume]} & = \textrm{[Volume of the outer cylinder]} -
\textrm{[Volume of the inner cylinder]} \\
& = \pi (r_{outer}^{2} - r_{inner}^{2}) h = \frac{4}{3}\pi
\mathrm{\ cm^3}\\
\end{align}
$$
We can now work out what the axes
represent:
$y$-axis:
The maximum $y$ value reached by all curves is identical at around
4.2, it is in fact equal to $\frac{4}{3}\pi$. This suggests the y
axis is a measure of volume, all solids will eventually displace a
fluid of $\frac{4}{3}\pi \mathrm{\ cm^3}$.
The $y$ axis represents the volume of fluid displaced (or the
volume of the solid immersed) in $\mathrm{cm^3}$.
$x$-axis:
The $x$-axis represents the time elapsed in minutes since lowering
began.
Simon now works out which curve matches
which solid, with some clear reasoning:
Curve 1:
The time taken to fully immerse the object $\approx 1.3 \mathrm{\
minutes}$
The volume displaced varies linearly with time, this must therefore
represent:
Solid 2 (cylinder of height $\frac{4}{3} \mathrm{\ cm}$ and radius
$1 \mathrm{\ cm}$) lowered vertically
or
Solid 4 (A solid cylinder of height $\frac{4}{9} \mathrm{\ cm}$
with a hole drilled through it) lowered vertically.
The time taken to fully immerse the object = $\frac{4}{3}\pi
\mathrm{\ minutes}$
Curve 1 is therefore a cylinder of height $\frac{4}{3} \mathrm{\
cm}$ and radius $1 \mathrm{\ cm}$ lowered vertically
Curve 2 and Curve 3:
The time taken to fully immerse the object = $2 \mathrm{\
minutes}$
This could therefore be:
Solid 2 lowered sideways or solid 1 lowered in any
orientation.
Solid 2 lowered sideways would initially displace a greater fluid
than solid 1, it can therefore be seen that curve 2 corresponds to
solid 2 lowered sideways and curve 3 to solid 1.
Curve 4 and Curve 5:
The time taken to fully immerse the object = $4 \mathrm{\
minutes}$
This could therefore be:
Solid 3 lowered vertically or solid 4 lowered sideways
Consider solid 3 lowered vertically:
Volume immersed as a function of height ($h$)
$$V(h) = \frac{1}{3}\pi r^2 h$$
If we immerse this cone point first the radius varies with the
height of object immersed as:
The volume immersed as a function of h is equal to the volume
generated when we rotate the equation of a circle about the
$x$-axis by 360 degrees and evaluate this integral between the
limits $r$ and $(r-h)$.
A solid cylinder of height $\frac{4}{9} \mathrm{\ cm}$ with a hole
drilled through it, leaving an annular cross-section with interal
and external radii $2 \mathrm{\ cm}$ and $1 \mathrm{\ cm}$ lowered
sideways.
When $h$ is less than $1 \mathrm{\ cm}$ the volume immersed takes
the same form as curve 2 but simply changing the radius from $1
\mathrm{\ cm}$ to $2 \mathrm{\ cm}$ and changing the length of the
solid from $\frac{4}{3} \mathrm{\ cm}$ to $\frac{4}{9} \mathrm{\
cm}$.
When h is greater than $1\mathrm{\ cm}$ the volume immersed can be
found by subtracting the area of the inner segment from the outer
segment and then multiplying by the length of the cylinder.