Question

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$r_{1}=r_{2}=8$ cm

Let $h_{1}$ and $h_{2}$ cm be the heights of the cylinder and the cone respectively. Then

$h_{1}=240cm;h_{2}=36cm$

volume of Cylinder $=πr_{1}h_{1}36cm_{3}=(π×8×8×240)cm_{3}=(π×64×240)cm_{3}$

volume of Cone $=31 πr_{2}h_{2}cm_{3}=(31 π×8×8×36)cm_{3}$

Total volume of the iron = volume of the cylinder + Volume of the cone

$=(π×64×240+31 π×64×36)cm_{3}=722 ×64×252cm_{3}=22×64×36cm_{3}$

total weight of the pillar = volume × Weight per $cm_{3}$

$(22×64×36)×7.8gms=395.3664kg$

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