Soumya took two right circular cones of the same vertical height. One of the two cones is a solid one, while the other is hollow from inside. By measuring the cross-sectional areas of the cones and from the knowledge of symmetry, by using suitable formulas, he found the positions of the centre of gravity of both cones. He found that the difference is about 1.5 cm. What is the vertical height of the two cones?

Let same vertical height of cone be h.

In a right circular cone, the center of gravity lies at a distance of $\dfrac{h}{4}$ from the base for a solid cone, and at a distance of $\dfrac{h}{3}$ from the base for a hollow cone.

Given that the difference between these positions is 1.5 cm, we can set up the following equation:

$\dfrac{h}{3} - \dfrac{h}{4} = 1.5 \\[1em] \dfrac{4h-3h}{12} = \dfrac{3}{2}\\[1em] \dfrac{h}{12}= \dfrac{3}{2}\\[1em] h= \dfrac{36}{2}\\[1em] h= 18 \text{ cm} \\[1em]$

∴ The vertical height of each cone = 18 cm