A solid iron pole consists of a cylinder of h | vedclass

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(B) From the figure,it can be observed that:Height $(h_1)$ of the larger cylinder $= 220 \,cm$Radius $(r_1)$ of the larger cylinder $= \frac{24}{2} =

Vedclass · Accepted Answer

$892.262$

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(B) From the figure,it can be observed that:Height $(h_1)$ of the larger cylinder $= 220 \,cm$Radius $(r_1)$ of the larger cylinder $= \frac{24}{2} = 12 \,cm$Height $(h_2)$ of the smaller cylinder $= 60 \,cm$Radius $(r_2)$ of the smaller cylinder $= 8 \,cm$Total volume of the pole $=$ Volume of the larger cylinder $+$ Volume of the smaller cylinder$= \pi r_1^2 h_1 + \pi r_2^2 h_2$$= \pi(12)^2 	imes 220 + \pi(8)^2 	imes 60$$= \pi(144 	imes 220 + 64 	imes 60)$$= 3.14 	imes (31680 + 3840)$$= 3.14 	imes 35520 = 111532.8 \,cm^3$Mass of $1 \,cm^3$ of iron $= 8 \,g$Total mass of the pole $= 111532.8 	imes 8 = 892262.4 \,g$Since $1 \,kg = 1000 \,g$,the mass in $kg$ is $\frac{892262.4}{1000} = 892.2624 \,kg \approx 892.262 \,kg$.

$A$ solid iron pole consists of a cylinder of height $220 \,cm$ and base diameter $24 \,cm$,which is surmounted by another cylinder of height $60 \,cm$ and radius $8 \,cm$. Find the mass of the pole,given that $1 \,cm^3$ of iron has approximately $8 \,g$ mass. (in $kg$) (Use $\pi = 3.14$)

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