The base of a cone with radius 14 , cm and he | vedclass

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(D) The volume of the solid is the sum of the volume of the cone and the volume of the hemisphere.Given: Radius $r = 14 \, cm$,Height of cone $h = 32

Vedclass · Accepted Answer

$12320$

Vedclass · Answer

(D) The volume of the solid is the sum of the volume of the cone and the volume of the hemisphere.Given: Radius $r = 14 \, cm$,Height of cone $h = 32 \, cm$.Volume of cone $V_1 = \frac{1}{3} \pi r^2 h = \frac{1}{3} 	imes \frac{22}{7} 	imes 14 	imes 14 	imes 32 = \frac{1}{3} 	imes 22 	imes 2 	imes 14 	imes 32 = \frac{19712}{3} \approx 6570.67 \, cm^3$.Volume of hemisphere $V_2 = \frac{2}{3} \pi r^3 = \frac{2}{3} 	imes \frac{22}{7} 	imes 14 	imes 14 	imes 14 = \frac{2}{3} 	imes 22 	imes 2 	imes 14 	imes 14 = \frac{17248}{3} \approx 5749.33 \, cm^3$.Total Volume $V = V_1 + V_2 = \frac{19712 + 17248}{3} = \frac{36960}{3} = 12320 \, cm^3$.

The base of a cone with radius $14 \, cm$ and height $32 \, cm$ is hemispherical. Find its volume in $cm^3$.

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