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.

\frac{3 h}{4}

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\frac{h^{2}}{4 R}

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\frac{5 h}{8}

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\frac{3 h^{2}}{8 R}

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Solution

The correct option is A $\frac{3 h}{4}$
Position of center of mass of a solid cone from the base is $h / 4$ .
Thus, position of centre of mass of a solid cone from the vertex $= h - \frac{h}{4} = \frac{3 h}{4}$
So, $z_{o} = \frac{3 h}{4}$

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Distance of the centre of mass of a solid uniform cone from its vertex is z0. If the radius of its base is R and its height is h, then z0 is equal to.

The correct option is A 3h4Position of center of mass of a solid cone from the base is h/4. Thus, position of centre of mass of a solid cone from the vertex =h−h4=3h4 So, zo=3h4

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.

The correct option is A $\frac{3 h}{4}$
Position of center of mass of a solid cone from the base is $h / 4$ .
Thus, position of centre of mass of a solid cone from the vertex $= h - \frac{h}{4} = \frac{3 h}{4}$
So, $z_{o} = \frac{3 h}{4}$