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(A) Given: Height of the cone $h = 12\, cm$ and radius of the cone $R = 6\, cm$.Let the radius of the hemisphere (which is the same as the radius of t

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$3$

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(A) Given: Height of the cone $h = 12\, cm$ and radius of the cone $R = 6\, cm$.Let the radius of the hemisphere (which is the same as the radius of the sphere) be $r\, cm$.The volume of a hemisphere is given by $V_h = \frac{2}{3} \pi r^3$.The volume of a cone is given by $V_c = \frac{1}{3} \pi R^2 h$.According to the problem,$8$ bowlfuls of the hemisphere fill the conical vessel:$8 	imes V_h = V_c$$8 	imes \left( \frac{2}{3} \pi r^3 ight) = \frac{1}{3} \pi R^2 h$Substitute the values:$8 	imes \frac{2}{3} \pi r^3 = \frac{1}{3} \pi (6)^2 (12)$Divide both sides by $\frac{1}{3} \pi$:$16 r^3 = 36 	imes 12$$16 r^3 = 432$$r^3 = \frac{432}{16} = 27$$r = \sqrt[3]{27} = 3\, cm$.Thus,the radius of the sphere is $3\, cm$.

$A$ sphere is cut into two hemispheres. One of them is used as a bowl. It takes $8$ bowlfuls of this to fill a conical vessel of height $12\, cm$ and radius $6\, cm$. The radius of the sphere (in centimeter) will be

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