A cylinder is closed at both the ends by hemi | vedclass

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Question

(B) The solid consists of a cylinder and two hemispheres at its ends.Radius of the cylinder $(r)$ = $4.2 \, cm$.Total height of the solid = $15 \, cm$

Vedclass · Accepted Answer

$676.368$

Vedclass · Answer

(B) The solid consists of a cylinder and two hemispheres at its ends.Radius of the cylinder $(r)$ = $4.2 \, cm$.Total height of the solid = $15 \, cm$.Height of the cylindrical part $(h)$ = Total height - $2 	imes$ radius of hemisphere = $15 - 2(4.2) = 15 - 8.4 = 6.6 \, cm$.Volume of the solid = Volume of cylinder + $2 	imes$ Volume of hemisphere.Volume of solid = $\pi r^2 h + 2 	imes (\frac{2}{3} \pi r^3) = \pi r^2 (h + \frac{4}{3} r)$.Volume = $\frac{22}{7} 	imes (4.2)^2 	imes (6.6 + \frac{4}{3} 	imes 4.2)$.Volume = $\frac{22}{7} 	imes 17.64 	imes (6.6 + 5.6) = 22 	imes 2.52 	imes 12.2$.Volume = $55.44 	imes 12.2 = 676.368 \, cm^3$.

$A$ cylinder is closed at both the ends by hemispheres. The radius of the cylinder is $4.2 \, cm$ and the total height of the solid is $15 \, cm$. Find the volume of this solid (in $cm^3$).

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