Question
A solid iron pole having cylindrical portion $110\ cm$ high and of base diameter $12\ cm$ is surmounted by a cone $9\ cm$ high. Find the mass of the pole, given that the mass of $1\ cm^3$ of iron is $8\ gm$.
✓
Answer
Diamter of the base $= 12\ cm$
$\therefore$ Radius (r) $=\Big(\frac{12}{2}\Big)=6\text{cm}$
Height of the cylindrical portion $(h_1) = 110\ cm$
and height of conical portion $(h_2) = 9\ cm$
$\therefore$ Total volume of the pole
$=\pi\text{r}^2\text{h}_1+\frac{1}{3}\pi\text{r}^2\text{h}_2$
$=\pi\text{r}^2\Big(\text{h}_1+\frac{1}{3}\text{h}_2\Big)$
$=\frac{22}{7}\times(6)^2\Big(110+\frac{1}{3}\times9\Big)\text{cm}^2$
$=\frac{22}{7}+36(110+3)\text{cm}^3$
$=\frac{22}{7}\times36\times113\text{cm}^3$
$=\frac{89496}{7}\text{cm}^3$
Weight of $1\ cm^3$ of iron $= 8\ gm$
$\therefore$ Total weight $=\frac{89496}{7}\times8\text{gms}$
$=\frac{89496\times8}{7\times1000}\text{kg}=102.281\text{kg}$
$\therefore$ Mass of the pole $= 102.281\ kg$
$\therefore$ Radius (r) $=\Big(\frac{12}{2}\Big)=6\text{cm}$
Height of the cylindrical portion $(h_1) = 110\ cm$
and height of conical portion $(h_2) = 9\ cm$
$\therefore$ Total volume of the pole
$=\pi\text{r}^2\text{h}_1+\frac{1}{3}\pi\text{r}^2\text{h}_2$
$=\pi\text{r}^2\Big(\text{h}_1+\frac{1}{3}\text{h}_2\Big)$
$=\frac{22}{7}\times(6)^2\Big(110+\frac{1}{3}\times9\Big)\text{cm}^2$
$=\frac{22}{7}+36(110+3)\text{cm}^3$
$=\frac{22}{7}\times36\times113\text{cm}^3$
$=\frac{89496}{7}\text{cm}^3$
Weight of $1\ cm^3$ of iron $= 8\ gm$
$\therefore$ Total weight $=\frac{89496}{7}\times8\text{gms}$
$=\frac{89496\times8}{7\times1000}\text{kg}=102.281\text{kg}$
$\therefore$ Mass of the pole $= 102.281\ kg$
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