Distance of the center 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 :

Center of Mass PYQ — JEE Main 2015 · Shift 0 · Q67 with Solution

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Correct answer: D

Take the vertex of the uniform solid cone at $z=0$ and the base at $z=h$ .

We need the $z$ -coordinate of the center of mass:

z_0 = \frac{\int z\,dm}{\int dm}

Because the cone is uniform, density is constant, so we can use volume elements:

z_0 = \frac{\int z\,dV}{\int dV}

At a distance $z$ from the vertex, take a thin disc of thickness $dz$ .

Its radius $r$ varies linearly with $z$ :

\frac{r}{z} = \frac{R}{h} \quad \Rightarrow \quad r = \frac{R}{h}z

So the disc volume is

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

V = \int_0^h dV = \pi \frac{R^2}{h^2} \int_0^h z^2 \, dz

V = \pi \frac{R^2}{h^2} \left[\frac{z^3}{3}\right]_0^h = \pi \frac{R^2}{h^2} \cdot \frac{h^3}{3} = \frac{1}{3}\pi R^2 h

\int_0^h z\, dV = \pi \frac{R^2}{h^2} \int_0^h z \cdot z^2 \, dz = \pi \frac{R^2}{h^2} \int_0^h z^3 \, dz

\int_0^h z\, dV = \pi \frac{R^2}{h^2} \left[\frac{z^4}{4}\right]_0^h = \pi \frac{R^2}{h^2} \cdot \frac{h^4}{4} = \frac{\pi R^2 h^2}{4}

z_0 = \frac{\int z\,dV}{V} = \frac{\frac{\pi R^2 h^2}{4}}{\frac{1}{3}\pi R^2 h}

z_0 = \frac{h^2}{4} \cdot \frac{3}{h} = \frac{3h}{4}

z_0 = \frac{3h}{4}

So the correct option is:

D. $\dfrac{3h}{4}$

This is the distance from the vertex.
(Equivalently, the center of mass is at a distance $h/4$ from the base.)