A cylindrical tank is closed at both the ends | vedclass

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(D) The tank consists of one cylinder and two identical cones.Radius of the cylinder and cones $(r)$ = $14 / 2 = 7 \, cm$.Height of the cylinder $(h_1

Vedclass · Accepted Answer

$4312$

Vedclass · Answer

(D) The tank consists of one cylinder and two identical cones.Radius of the cylinder and cones $(r)$ = $14 / 2 = 7 \, cm$.Height of the cylinder $(h_1)$ = $20 \, cm$.Height of each cone $(h_2)$ = $12 \, cm$.Volume of the tank = (Volume of cylinder) + $2$ $	imes$ (Volume of cone).Volume of cylinder = $\pi r^2 h_1 = (22/7) 	imes 7^2 	imes 20 = 22 	imes 7 	imes 20 = 3080 \, cm^3$.Volume of one cone = $(1/3) \pi r^2 h_2 = (1/3) 	imes (22/7) 	imes 7^2 	imes 12 = 22 	imes 7 	imes 4 = 616 \, cm^3$.Total volume = $3080 + 2 	imes 616 = 3080 + 1232 = 4312 \, cm^3$.

$A$ cylindrical tank is closed at both the ends by cones of height $12 \, cm$. The diameter of the cylindrical part is $14 \, cm$ and its height is $20 \, cm$. Find the volume of this tank (in $cm^3$).

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