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(D) Let $BPC$ be the hemisphere and $ABC$ be the cone standing on the base of the hemisphere.The radius $BO$ of the hemisphere (as well as of the cone

Vedclass · Accepted Answer

$25.12$

Vedclass · Answer

(D) Let $BPC$ be the hemisphere and $ABC$ be the cone standing on the base of the hemisphere.The radius $BO$ of the hemisphere (as well as of the cone) $= \frac{1}{2} 	imes 4 \, cm = 2 \, cm$.So,the volume of the toy $= \frac{2}{3} \pi r^3 + \frac{1}{3} \pi r^2 h$.$= \left[ \frac{2}{3} 	imes 3.14 	imes (2)^3 + \frac{1}{3} 	imes 3.14 	imes (2)^2 	imes 2 ight] \, cm^3 = 25.12 \, cm^3$.Now,let the right circular cylinder $EFGH$ circumscribe the given solid. The radius of the base of the right circular cylinder $= HP = BO = 2 \, cm$,and its height is $EH = AO + OP = (2 + 2) \, cm = 4 \, cm$.So,the required volume difference $=$ volume of the right circular cylinder $-$ volume of the toy.$= (3.14 	imes 2^2 	imes 4 - 25.12) \, cm^3$.$= (50.24 - 25.12) \, cm^3 = 25.12 \, cm^3$.Hence,the required difference of the two volumes $= 25.12 \, cm^3$.

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