A solid metallic hemisphere of radius 8 , cm | vedclass

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Question

(A) Let the height of the cone be $h \, cm$.Given,the radius of the base of the cone $r = 6 \, cm$.The volume of the cone is given by $V_{cone} = \fra

Vedclass · Accepted Answer

$28.44$

Vedclass · Answer

(A) Let the height of the cone be $h \, cm$.Given,the radius of the base of the cone $r = 6 \, cm$.The volume of the cone is given by $V_{cone} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi (6)^2 h = \frac{36 \pi h}{3} = 12 \pi h \, cm^3$.Given,the radius of the hemisphere $R = 8 \, cm$.The volume of the hemisphere is given by $V_{hemisphere} = \frac{2}{3} \pi R^3 = \frac{2}{3} \pi (8)^3 = \frac{2}{3} \pi (512) = \frac{1024 \pi}{3} \, cm^3$.Since the hemisphere is melted and recast into a cone,their volumes must be equal:$12 \pi h = \frac{1024 \pi}{3}$$h = \frac{1024 \pi}{3 	imes 12 \pi} = \frac{1024}{36} = \frac{256}{9} \approx 28.44 \, cm$.Thus,the height of the cone is $28.44 \, cm$.

$A$ solid metallic hemisphere of radius $8 \, cm$ is melted and recast into a right circular cone of base radius $6 \, cm$. Determine the height of the cone (in $cm$).

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