The generatrix of a cone forms an angle $\alpha$ with the plane of its base, $\cos \alpha=\frac{1}{4}$. A sphere is inscribed in the cone, and a plane is drawn through the circle of contact between the sphere and the lateral surface of the cone. The volume of the part of the cone enclosed between this plane and the base plane of the cone is 37. Find the volume of the remaining part of the cone.
Let $P$ be the vertex of the cone, $PH$ its height, and $O$ the center of the inscribed sphere. Consider the axial section of the cone - an isosceles triangle $PAB$ with angle $\alpha$ at the base $AB$. The section of the sphere is a circle with center $O$, inscribed in the triangle $PAB$ and touching the base $AB$ at point $H$, and the lateral sides $PA$ and $PB$ at points $M$ and $N$ respectively. Let $K$ be the midpoint of $MN$, $R$ the radius of the base of the cone, $V$ its volume, and $v$ the sought volume. Then $V-v=37$. Since the cone cut off from the given cone by the specified plane is similar to the given cone, $v=k^3 V$, where $k$ is the similarity coefficient. Further, we have:

\[
\begin{aligned}
& k=\frac{PM}{PA}=\frac{PA-AM}{PA}=\frac{PA-AH}{PA}=1-\frac{AH}{PA}=1-\frac{R}{PM}=1-\cos \alpha=1-\frac{1}{4}=\frac{3}{4}, \\
& v=\left(\frac{3}{4}\right)^3 V=\frac{27}{64} V, V-v=V-\frac{27}{64} V=\frac{37}{64} V=37, V=64
\end{aligned}
\]

Therefore, $v=\frac{27}{64} V=27$.

## Answer

27.00