Question

A solid iron cuboidal block of dimensions $4.4m,2.6m,1m$ is cast into a hollow cylindrical pipe of internal radius $30cm$and thickness $5cm.$. Find length of the pipe.

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Solution

Step 1: Find the volume of cylindrical pipeDimensions of cuboidal blockLength $\left(l\right)=4m$Breadth $\left(b\right)=2.6m$Height $\left(h\right)=1m$Dimensions of cylindrical pipe:Internal radius $\left({r}_{2}\right)$ $=30cm$ $=0.3m$Thickness$=5cm=0.05m$So the outer radius $\left({r}_{1}\right)$ $=30+5cm=35cm=0.35m$Let the length of the pipe $=hmetre$We know, the volume of a hollow cylinder $=\mathrm{\pi h}\left({{\mathrm{r}}_{1}}^{2}-{{\mathrm{r}}_{2}}^{2}\right)$, where $\mathrm{\pi }=\frac{22}{7},{r}_{1}=\mathrm{external}\mathrm{radius},{r}_{2}=\mathrm{internal}\mathrm{radius}$We know, when a shape is recast into a different shape, then the volume stays the same.Here, Cuboidal block is recast into cylindrical pipe. So, volume of the cuboidal block equals the volume of hollow cylinder.We know, Volume of a cuboidal tank $=l×b×h$ $=4.4×2.6×1{m}^{3}=11.44{m}^{3}$Then, Volume of cylindrical pipe will also be$=11.44{m}^{3}$ [ Because the cuboidal tank is melt and recast into the cylindrical pipe]Step 2: Find the height of the pipeAlso, we have volume of a hollow cylinder $=\mathrm{\pi h}\left({{\mathrm{r}}_{1}}^{2}-{{\mathrm{r}}_{2}}^{2}\right)$So equating the above two, we get:$⇒\mathrm{\pi h}\left({{\mathrm{r}}_{1}}^{2}-{{\mathrm{r}}_{2}}^{2}\right)=11.44$Now, putting the values of ${r}_{1},{r}_{2},\mathrm{\pi }$ in the equation above, we get:$⇒11.44{m}^{3}=\frac{22}{7}×h×\left(0.{35}^{2}-0.{3}^{2}\right)\phantom{\rule{0ex}{0ex}}⇒11.44=\frac{22}{7}×h×\left(0.1225-0.09\right)\phantom{\rule{0ex}{0ex}}⇒11.44=\frac{22}{7}×h×\left(0.0325\right)\phantom{\rule{0ex}{0ex}}⇒11.44=\frac{0.715}{7}×h\phantom{\rule{0ex}{0ex}}⇒\mathrm{h}=\frac{11.44×7}{0.715}\phantom{\rule{0ex}{0ex}}⇒\mathrm{h}=112\mathrm{m}$Thus, the length of the pipe is $112m$

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