A cylindrical tank is closed at both the ends | vedclass

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(C) The tank consists of a cylinder and two hemispheres at the ends.Radius of the cylinder and hemispheres, $r = 24 / 2 = 12 \, cm$.Height of the cyli

Vedclass · Accepted Answer

$31.68$

Vedclass · Answer

(C) The tank consists of a cylinder and two hemispheres at the ends.Radius of the cylinder and hemispheres, $r = 24 / 2 = 12 \, cm$.Height of the cylindrical part, $h = 78 - (12 + 12) = 78 - 24 = 54 \, cm$.Volume of the tank = Volume of cylinder + $2 	imes$ Volume of hemisphere.Volume = $\pi r^2 h + 2 	imes (2/3 \pi r^3) = \pi r^2 (h + 4/3 r)$.Volume = $3.14159 	imes (12)^2 	imes (54 + 4/3 	imes 12) = 3.14159 	imes 144 	imes 70$.Volume = $31667.2$ approx $31667 \, cm^3$.Since $1000 \, cm^3 = 1 \, litre$, the volume in litres is $31667 / 1000 = 31.667 \, litres \approx 31.68 \, litres$ (using $\pi \approx 22/7$ gives $31.68 \, litres$).

$A$ cylindrical tank is closed at both the ends by hemispheres. The diameter of the tank is $24 \, cm$ and the total height is $78 \, cm$. How many litres of kerosene can it hold?

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