Mathematics · Oct / Nov 2024 · Paper 41 · Question 5
YMA01
As printed in the paper
The question as text
5. 12 cm r cm 30 cm h cm Figure 3 Figure 3 shows a container in the shape of a hollow, inverted, right circular cone. The height of the container is 30 cm and the radius is 12 cm, as shown in Figure 3. The container is initially empty when water starts flowing into it. When the height of water is h cm, the surface of the water has radius r cm and the volume of water is V cm 3
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5(a)[2]
2
r
12
12
1
1
π
2
e.g.
=
r
=
h
V
=
π
r h
2
=
h
h
h
30
30
3
3
5
4
π
h
3
*
V
=
75
5(b)[4]
4
π
h
3
d
V
12
π
h
2
4
π
h
2
V
=
=
=
75
d
h
75
25
V
1.5 60 2
=
×
×
π
4
π
h
3
(
)
180
π=
h
3
=
3375
h
=
... 15
75
d
h
d
h
d
V
75
=
=
×
2
π
=
...
d
t
d
V
d
t
12
π
×
15
2
1
18 (cm s−1)
(a)
M1:
Uses the 12 and the 30 to find the ratio between h and r and substitutes into the volume
formula to find V in terms of h. Condone poor bracketing / invisible brackets for this mark.
2
May just state
r
=
h
5
2
2
This mark cannot be for substituting in e.g.
r =
instead of
r
=
h
5
5
A1*: Correct proof with no errors including brackets. There must be at least one intermediate
12
stage of working between establishing
and proceeding to the given answer.
r
=
h
30
(b)
Note that substituting
h =
1.5
or
h =
90
will score M0M0 in (b)
Correct expression for d
V
B1:
h in any form
d
M1:
Attempts their volume (“2
π×”) after 1.5 minutes and uses this with the given result from
t
(a) to find h. They may find h by equating their expression for the volume to the given result
from (a) before substituting
and rearranging to find h. Do not be concerned by the
t =
90
mechanics of the rearrangement. Condone slips.
Correct chain rule attempt using their h to find d
h
M1:
t . It is dependent on having used t = 90
d
(or condone t = 1.5 ) when finding h.
A1:
Correct value. Allow awrt 0.0556 (cm s−1) Units not required for this mark.
d
h
75
Alt(b) I Solving the differential equation
=
and then using
90
t =
d
t
6
h
2
d
V
12
π
h
2
4
π
h
2
B1:
=
=
d
h
75
25
d
h
d
h
d
V
75
M1:
Attempts
=
×
=
to achieve an expression in h, attempts to solve their
d
t
d
V
d
t
6
h
2
differential equation
6
h
2
d
h
=
75
dt
2
h
3
=
75
t
+
c
o.e. and uses
t
=
0,
h
=
0
c
=
1
(
)
(you may not see the working for c) to find h = f(t) (
h
=
37.5
t
). Do not be concerned by
3
the mechanics of the rearrangement.
d
h
1 37.5
2
d
h
−
M1:
Attempts
=
3
t
3
and substitutes
t
=
90
=
...
d
t
3
d
t
A1:
As above in main scheme
Alt(b) II Equating volumes, rearranging to
h =
...
and then using
t =
90
B1:
V
=
2
π
t
4
π
h
3
1
M1:
Equates their 2 t
π equal to
and rearranges to
h
=
f( )
t
(
h
(
37.5
t
)
). Do not be
=
3
75
concerned by the mechanics of the rearrangement.
d
h
1 37.5
2
d
h
−
M1:
Attempts
=
3
t
3
and substitutes
t
=
90
=
...
d
t
3
d
t
A1:
As above in main scheme
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