A liquid container, 30 cm tall, can be modelled by a frustum, A, placed below a cylinder B, as shown in Figure 1. Figure 2 shows a sketch of the cone that the frustum A is a part of. The diameter QRS of the top of the frustum, is 12 cm and the diameter VUT of the bottom is 8 cm. The height of the frustum, RU is 9 cm. R and U are the midpoints of QS and VT respectively. A liquid is poured into the container at a constant rate of cm^3 per second. It takes 50 seconds to fill up A completely and it takes another 30 seconds to fill up B completely.

- (a)
Calculate the volume of the frustum, A.
Answer ... cm^3
- (b)
Sketch the height of the liquid in the container with respect to time.
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Worked solution
26(a)
1. Find the height of the smaller cone below the frustum using similar triangles.
Let the height of the smaller cone be cm.
The radii of the bottom and top of the frustum are
By similarity,
2. Calculate the volume of the large cone and the smaller cone.
Height of the large cone
Volume of large cone:
Volume of smaller cone:
3. Subtract the smaller cone from the larger cone to get the frustum volume.
Volume of frustum A
So the volume is approximately
26(b)
1. Mark the key points where each section is filled.
The frustum has height and is filled in .
So the graph passes through
The whole container is tall and cylinder takes another to fill.
So the graph also passes through
2. Decide the shape of the graph in the frustum and cylinder sections.
In the frustum, the diameter increases from to as the liquid rises, so the cross-sectional area increases. Since liquid is poured in at a constant rate, the height rises more slowly as time increases.
Therefore, from to , draw an increasing curve that becomes less steep.
In the cylinder, the cross-sectional area is constant, so the height rises at a constant rate.
Therefore, from to , draw a straight line.
Sketch an increasing curve from to , concave down, then a straight line from to .
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