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(C) Let the height of the cylinder be $H$ and its radius be $r$. Since the cone fits exactly,its radius is also $r$.The volume of the cylinder is $V_{

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$\frac{2h}{3}$

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(C) Let the height of the cylinder be $H$ and its radius be $r$. Since the cone fits exactly,its radius is also $r$.The volume of the cylinder is $V_{cyl} = \pi r^2 H$.The volume of the cone is $V_{cone} = \frac{1}{3} \pi r^2 h$.The total volume of the solid is $V_{total} = V_{cyl} + V_{cone} = \pi r^2 H + \frac{1}{3} \pi r^2 h$.According to the problem,$V_{total} = 3 	imes V_{cone}$.Therefore,$\pi r^2 H + \frac{1}{3} \pi r^2 h = 3 	imes (\frac{1}{3} \pi r^2 h)$.Simplifying the equation:$\pi r^2 H + \frac{1}{3} \pi r^2 h = \pi r^2 h$.Subtract $\frac{1}{3} \pi r^2 h$ from both sides:$\pi r^2 H = \pi r^2 h - \frac{1}{3} \pi r^2 h$.$\pi r^2 H = \frac{2}{3} \pi r^2 h$.Dividing both sides by $\pi r^2$,we get $H = \frac{2}{3} h$.

$A$ solid consists of a circular cylinder with an exact fitting right circular cone placed on the top. The height of the cone is $h$. If the total volume of the solid is three times the volume of the cone,then the height of the cylinder is

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