A vessel, in the form of an inverted cone, is filled with water to the brim. Its height is 32 cm and diameter of the base is 25.2 cm. Six equal solid cones are dropped in it, so that they are fully submerged. As a result one-fourth of water in the original cone overflows. What is the volume of each of the solid cones submerged?

Radius of vessel (R) = $\dfrac{25.2}{2}$ = 12.6 cm.

Total volume of water in vessel = $\dfrac{1}{3}πR^2H$

On submerging six equal cones in vessel, one-fourth of water in the original cone overflows.

Let radius of small cones be r and height be h,

$\therefore 6 \times \text{Vol. of each cone} = \dfrac{1}{4} \times \dfrac{1}{3}πR^2H \\[1em] \Rightarrow \text{Vol. of each cone} = \dfrac{1}{72}πR^2H \\[1em] \Rightarrow \text{Vol. of each cone} = \dfrac{1}{72} \times \dfrac{22}{7} \times (12.6)^2 \times 32 \\[1em] \Rightarrow \text{Vol. of each cone} = 221.76 \text{ cm}^3.$

Hence, volume of each cone = 221.76 cm³.