A hollow hemispherical bowl is made of silver | vedclass

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(D) The volume of a hollow hemispherical bowl is given by the formula $V = \frac{2}{3} \pi (R^3 - r^3)$,where $R$ is the outer radius and $r$ is the i

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$14$

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(D) The volume of a hollow hemispherical bowl is given by the formula $V = \frac{2}{3} \pi (R^3 - r^3)$,where $R$ is the outer radius and $r$ is the inner radius.Given $R = 8\, cm$ and $r = 4\, cm$,the volume is $V = \frac{2}{3} \pi (8^3 - 4^3) = \frac{2}{3} \pi (512 - 64) = \frac{2}{3} \pi (448) = \frac{896}{3} \pi\, cm^3$.When the bowl is melted to form a solid right circular cone,the volume remains constant.The volume of a cone is $V = \frac{1}{3} \pi r_c^2 h$,where $r_c = 8\, cm$ is the radius of the cone and $h$ is its height.Equating the volumes: $\frac{896}{3} \pi = \frac{1}{3} \pi (8^2) h$.$\frac{896}{3} = \frac{64}{3} h$.$896 = 64h$.$h = \frac{896}{64} = 14\, cm$.

$A$ hollow hemispherical bowl is made of silver with its outer radius $8\, cm$ and inner radius $4\, cm$. The bowl is melted to form a solid right circular cone of radius $8\, cm$. The height of the cone formed is......$cm$

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