A cylinder is closed at both the ends by 18, | vedclass

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Question

(C) The solid consists of a cylinder and two cones at its ends.Given:Radius of cylinder and cones $(r)$ = $7\, cm$.Height of each cone $(h_1)$ = $18\,

Vedclass · Accepted Answer

$6160$

Vedclass · Answer

(C) The solid consists of a cylinder and two cones at its ends.Given:Radius of cylinder and cones $(r)$ = $7\, cm$.Height of each cone $(h_1)$ = $18\, cm$.Total height of the solid $(H)$ = $64\, cm$.Height of the cylinder $(h_2)$ = $H - 2 	imes h_1 = 64 - 2(18) = 64 - 36 = 28\, cm$.Volume of the solid = Volume of cylinder + $2 	imes$ Volume of cone.Volume = $\pi r^2 h_2 + 2 	imes (\frac{1}{3} \pi r^2 h_1)$.Volume = $\pi r^2 (h_2 + \frac{2}{3} h_1)$.Volume = $\frac{22}{7} 	imes 7^2 	imes (28 + \frac{2}{3} 	imes 18)$.Volume = $22 	imes 7 	imes (28 + 12) = 154 	imes 40 = 6160\, cm^3$.

$A$ cylinder is closed at both the ends by $18\, cm$ high cones. The radius of the cylinder is $7\, cm$ and the total height of the solid is $64\, cm$. Find the volume of this solid (in $cm^3$).

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