5. Given a trapezoid, the ratio of its bases is 2. If the trapezoid is rotated $360^{\circ}$ around the larger base, a certain solid figure $\Phi_{1}$ is obtained. If the trapezoid is rotated $360^{\circ}$ around the smaller base, a certain solid figure $\Phi_{2}$ is obtained. Find the ratio of the volumes of the resulting solid figures $\Phi_{1}$ and $\Phi_{2}$.
Solution. Consider trapezoid $ABCD$ with bases $AD=2a, BC=a$. Draw the heights of the trapezoid $BH_{1}=BH_{2}=h$. Denote $AH_{1}=a_{1}, AH_{2}=a_{2}$, where $a_{1}+a_{2}=a$.

When the trapezoid is rotated around the larger base, a cylinder is formed with a generatrix equal to $a$ and a base radius equal to the height of the trapezoid $h$, from which cones are cut out at the bases, with the radius of the base equal to the height of the trapezoid, and heights equal to $a_{1}$ and $a_{2}$. If the rotation is around the smaller base, a cylinder is formed with a generatrix equal to $b$ and a base radius equal to the height of the trapezoid $h$, to which the cones described above are attached at the bases.

Now let's calculate the volumes

$$
\begin{aligned}
& V_{\Phi_{1}}=2 a \pi h^{2}-\frac{\pi h^{2}}{3}\left(a_{1}+a_{2}\right)=\frac{5}{3} \pi h^{2} a . \\
& V_{\Phi_{1}}=a \pi h^{2}+\frac{\pi h^{2}}{3}\left(a_{1}+a_{2}\right)=\frac{4}{3} \pi h^{2} a .
\end{aligned}
$$

From the last two equalities, we easily obtain the answer.

Answer: $\frac{5}{4}$.