Question

# A solid iron pole consists of a cylinder of height $220cm$ and base diameter$24cm,$ which is surmounted by another cylinder of height$60cm$ and radius $8cm.$ Find the mass of the pole, given that$1c{m}^{3}$ of iron has approximately $8g$ mass.

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Solution

## Height of large cylinder $\left(H\right)=220cm$Radius of large cylinder $\left(R\right)=\frac{24}{2}=12cm$Height of small cylinder $\left(h\right)=60cm$Radius of small cylinder $\left(r\right)=8cm$Step1: We have to find the mass of the pole, given that $1c{m}^{3}$ of iron has approximately $8g$ mass.∴ Volume of iron $=$Volume of the big cylinder $+$Volume of the small cylinder$={\mathrm{\pi R}}^{2}\mathrm{H}+{\mathrm{\pi r}}^{2}\mathrm{h}\phantom{\rule{0ex}{0ex}}=\mathrm{\pi }\left({\mathrm{R}}^{2}\mathrm{H}+{\mathrm{r}}^{2}\mathrm{h}\right)\phantom{\rule{0ex}{0ex}}=3.14×\left(12×12×220+8×8×60\right)$$=111532.8c{m}^{3}$As given $,$ Mass of $1c{m}^{3}=8g$Mass for $111532.8c{m}^{3}=8×111532.8g$ $=892,262.4grams$$=892.2624kg$

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