Distance of the centre of mass of a solid uniform cone from its vertex is $$z_0$$. If the radius of its base is $$R$$ and its height is $$h$$ then $$z_0$$ is equal to

We have a solid, right-circular cone of uniform (constant) density. Its geometrical dimensions are the height $$h$$ and the radius of the base $$R$$. We want the distance $$z_0$$ of the centre of mass from the vertex (the sharp tip). By symmetry, the centre of mass must lie on the axis of the cone, so we only need the coordinate along this axis.

To find $$z_0$$ we will use the definition of the centre of mass for a continuous body:

$$z_0=\dfrac{\displaystyle\int z \, dm}{\displaystyle\int dm}$$

Here the element of mass $$dm$$ is the mass of a thin disc cut perpendicular to the axis at a distance $$z$$ from the vertex. Every algebraic step follows.

Let the constant volume density be $$\rho$$ (mass per unit volume). First we relate the radius $$r$$ of a disc taken at distance $$z$$ to the fixed dimensions $$R$$ and $$h$$. Because the cone narrows linearly, the ratio of any disc’s radius to its distance from the vertex equals the ratio of the base radius to the full height:

$$\dfrac{r}{z}=\dfrac{R}{h}\qquad\Longrightarrow\qquad r=\dfrac{R}{h}\,z.$$

The disc has thickness $$dz$$, so its volume is the area $$\pi r^{2}$$ multiplied by $$dz$$. Substituting the expression for $$r$$:

$$dV = \pi r^{2}\,dz = \pi\left(\dfrac{R}{h}z\right)^{2}dz = \pi\dfrac{R^{2}}{h^{2}}\,z^{2}\,dz.$$

Because density is uniform, the mass element is

$$dm = \rho\,dV = \rho\,\pi\dfrac{R^{2}}{h^{2}}\,z^{2}\,dz.$$

Next, we calculate the total mass $$M$$ of the cone by integrating $$dm$$ from the vertex $$z=0$$ to the base $$z=h$$:

$$M = \int_{0}^{h} dm = \rho\pi\dfrac{R^{2}}{h^{2}}\int_{0}^{h} z^{2}\,dz.$$

We evaluate the integral $$\int z^{2}dz$$ first:

$$\int_{0}^{h} z^{2}\,dz = \left[\dfrac{z^{3}}{3}\right]_{0}^{h} = \dfrac{h^{3}}{3}.$$

Substituting this back, we get

$$M = \rho\pi\dfrac{R^{2}}{h^{2}}\cdot\dfrac{h^{3}}{3} = \rho\pi R^{2}\dfrac{h}{3}.$$

Now we need the numerator of the centre-of-mass formula, namely $$\int z\,dm$$:

$$\int_{0}^{h} z\,dm = \rho\pi\dfrac{R^{2}}{h^{2}}\int_{0}^{h} z^{3}\,dz.$$

The integral $$\int z^{3}dz$$ is

$$\int_{0}^{h} z^{3}\,dz = \left[\dfrac{z^{4}}{4}\right]_{0}^{h} = \dfrac{h^{4}}{4}.$$

Hence

$$\int z\,dm = \rho\pi\dfrac{R^{2}}{h^{2}}\cdot\dfrac{h^{4}}{4} = \rho\pi R^{2}\dfrac{h^{2}}{4}.$$

We now form the ratio for $$z_0$$:

$$z_0 = \dfrac{\displaystyle\rho\pi R^{2}\dfrac{h^{2}}{4}} {\displaystyle\rho\pi R^{2}\dfrac{h}{3}} = \dfrac{h^{2}}{4}\,\cdot\,\dfrac{3}{h} = \dfrac{3h}{4}.$$

The density $$\rho$$, the factor $$\pi R^{2}$$, and all other common terms cancel in the ratio, leaving a simple fraction of the height.

So, the centre of mass lies on the axis at a distance $$\dfrac{3h}{4}$$ from the vertex.

Hence, the correct answer is Option C.