Two solid metallic right circular cones have same height $h$ . The radii of their bases are $r_{1}$ and $r_{2}$ . The two cones are melted together and recast into a right circular cylinder of height $h$ . Show that radius of the base of the cylinder is $\sqrt{\frac{1}{3} (r_{1}^{2} + r_{2}^{2})}$ .

True

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False

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Solution

The correct option is A True
Volume of the cylinder $=$ Volume of Cone 1 $+$ Volume of Cone 2.
Since height is the same for the two cones, let $r_{1}$ and $h$ be the radius and height of the first cone and $r_{2}$ and $h$ be the radius and height of the second cone.
Volume of a cone $= \frac{1}{3} π r^{2} h$ , where $r$ is the radius of the base of the cone and $h$ is the height.
Volume of a Cylinder of Radius " $R$ " and height " $h$ " $= π R^{2} h$
So, $π R^{2} h = \frac{1}{3} π {r_{1}}^{2} h + \frac{1}{3} π {r_{2}}^{2} h$
$\Rightarrow R^{2} = \frac{1}{3} ({r_{1}}^{2} + {r_{2}}^{2})$
$\Rightarrow R = \sqrt{\frac{1}{3} (r_{1}^{2} + {r_{2}}^{2})}$
Hence, the radius of the base of the cylinder $= \sqrt{\frac{1}{3} (r_{1}^{2} + {r_{2}}^{2})}$

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