Find the volume of water in the container.

Find the value of $h$.

$\pi  \times {8^2} \times 12$     (M1)

Note:     Award (M1) for correct substitution into the volume of a cylinder formula.

$2410{\text{ c}}{{\text{m}}^3}{\text{ }}(2412.74 \ldots {\text{ c}}{{\text{m}}^3},{\text{ }}768\pi {\text{ c}}{{\text{m}}^3})$     (A1)     (C2)

[2 marks]

$\frac{4}{3}\pi  \times {2.9^3} + 768\pi  = \pi  \times {8^2}h$     (M1)(M1)(M1)

Note:     Award (M1) for correct substitution into the volume of a sphere formula (this may be implied by seeing 102.160…), (M1) for adding their volume of the ball to their part (a), (M1) for equating a volume to the volume of a cylinder with a height of $h$.

OR

$\frac{4}{3}\pi  \times {2.9^3} = \pi  \times {8^2}(h - 12)$     (M1)(M1)(M1)

Note:     Award (M1) for correct substitution into the volume of a sphere formula (this may be implied by seeing 102.160…), (M1) for equating to the volume of a cylinder, (M1) for the height of the water level increase, $h - 12$. Accept $h$ for $h - 12$ if adding 12 is implied by their answer.

$(h = ){\text{ }}12.5{\text{ (cm) }}\left( {12.5081 \ldots {\text{ (cm)}}} \right)$     (A1)(ft)     (C4)

Note:     If 3 sf answer used, answer is 12.5 (12.4944…). Follow through from part (a) if first method is used.

[4 marks]