The radii of the base of two right circular solid cones of same height are $r_{1}$ and $r_{2}$ respectively. The cones are melted and recast into a solid sphere of radius $R$ . Show that the height of each cone is given by $h = \frac{4 R^{3}}{r_{1}^{2} + r_{2}^{2}}$

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Solution

Let $h$ be the height of each cone.
Then sum of the volumes of two cones = Volume of the sphere

$\Rightarrow \frac{1}{3} π r_{1}^{2} h + \frac{1}{3} π r_{2}^{2} h = \frac{4}{3} π R^{3}$

$\Rightarrow$ $(r_{1}^{2} + r_{2}^{2}) h = 4 R^{3}$

$\Rightarrow h = \frac{4 R^{3}}{(r_{1}^{2} + r_{2}^{2})}$ $[H e n c e p r o v e d]$

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The radii of the base of two right circular solid cones of same height are r1 and r2 respectively. The cones are melted and recast into a solid sphere of radius R. Show that the height of each cone is given by h=4R3r21+r22

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