imc

First, we draw the vertical cross-section passing through the middle of the frustum. Let the top base have a diameter of $2$ and the bottom base has a diameter of $2r$. Then using the Pythagorean theorem we have: $(r+1{)}^2=(2s{)}^2+(r-1{)}^2$, which is equivalent to: $r^2+2r+1=4s^2+r^2-2r+1$. Subtracting $r^2-2r+1$ from both sides, $4r=4s^2$, and solving for $s$, we end up with $$s=\sqrt{r}.$$ Next, we can find the volume of the frustum (truncated cone) and the sphere. Since we know $V_{\text{frustum}}=2V_{\text{sphere}}$, we can solve for $s$ using $V_{\text{frustum}}=\frac{\pi h}{3}(R^2+r^2+Rr)$ we get: $$V_{\text{frustum}}=\frac{\pi \cdot 2\sqrt{r}}{3}(r^2+r+1)$$ Using $V_{\text{sphere}}=\frac{4s^3\pi }{3}$, we get $$V_{\text{sphere}}=\frac{4(\sqrt{r}{)}^3\pi }{3}$$ so we have: $$\frac{\pi \cdot 2\sqrt{r}}{3}(r^2+r+1)=2\cdot \frac{4(\sqrt{r}{)}^3\pi }{3}.$$ Dividing by $\frac{2\pi \sqrt{r}}{3}$, we get $$r^2+r+1=4r$$ which is equivalent to $$r^2-3r+1=0$$ by the Quadratic Formula, $r=\frac{3\pm \sqrt{(-3{)}^2-4\cdot 1\cdot 1}}{2\cdot 1}$ , so $$r=\frac{3+\sqrt{5}}{2}⟶\boxed{\frac{3+\sqrt{5}}{2}}$$