A cylinder, which has a diameter of 27 and a height of 30 , contains two lead spheres with radii 6 and 9 , with the larger sphere sitting on the bottom of the cylinder, as shown. Water is poured into the cylinder so that it just covers both spheres. The volume of water required is
(A) $3672 \pi$
(D) $3114 \pi$
(B) $3660 \pi$
(C) $3375 \pi$
(E) $4374 \pi$

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-131.jpg?height=401&width=242&top_left_y=195&top_left_x=1451)
We examine a vertical cross-section of the cylinder and the spheres that passes through the vertical axis of the cylinder and the centres of the spheres. (There is such a cross-section since the spheres will be pulled into this position by gravity.)

Let the centres of the spheres be $O_{1}$ and $O_{2}$, as shown.

Join the centres of the spheres to each other and to the respective points of tangency of the the spheres to the walls of the cylinder.

Then $O_{1} O_{2}=6+9=15$ ( $O_{1} O_{2}$ passes through the point of tangency of the two spheres), $O_{1} P=6$ and $O_{2} Q=9$, using the radii of the spheres.

Next, draw $\triangle O_{1} R O_{2}$ so that $R O_{1}$ is perpendicular to $O_{1} P$ and $\angle O_{1} R O_{2}=90^{\circ}$. Then looking at the width of the cylinder, since $P O_{1}=6$ and $O_{2} Q=9$, we have

$$
\begin{aligned}
P O_{1}+R O_{2}+O_{2} Q & =27 \\
R O_{2} & =12
\end{aligned}
$$

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-236.jpg?height=374&width=390&top_left_y=756&top_left_x=1301)

By Pythagoras in $\triangle O_{1} R O_{2}$, we see that $O_{1} R=9$.

Then the depth of the water will be

Radius of lower sphere $+R O_{1}+$ Radius of higher sphere $=9+9+6=24$. So

Volume of water $=($ Volume of cylinder to height of 24 $)-($ Volume of spheres $)$

$$
\begin{aligned}
& =\pi\left(\frac{27}{2}\right)^{2}(24)-\frac{4}{3} \pi(6)^{3}-\frac{4}{3} \pi(9)^{3} \\
& =4374 \pi-288 \pi-972 \pi \\
& =3114 \pi
\end{aligned}
$$

Therefore, the volume of water required is $3114 \pi$ cubic units.

ANSWER: (D)