Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{Length}\mathrm{of}\mathrm{the}\mathrm{rectangular}\mathrm{block},l=4.4\mathrm{m}, \\ \mathrm{Breadth}\mathrm{of}\mathrm{the}\mathrm{rectangular}\mathrm{block},\mathrm{b}=2.6\mathrm{m}, \\ \mathrm{Height}\mathrm{of}\mathrm{the}\mathrm{rectangular}\mathrm{block},h=1\mathrm{m}, \\ \mathrm{Internal}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{cylindrical}\mathrm{pipe},r=30\mathrm{cm}=0.3\mathrm{m}\mathrm{and} \\ \mathrm{Thickness}\mathrm{of}\mathrm{the}\mathrm{pipe}=5\mathrm{cm}=0.05\mathrm{m} \\ \mathrm{Also},\mathrm{the}\mathrm{external}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{pipe}=0.3+0.05=0.35\mathrm{m} \\ \mathrm{Let}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{pipe}\mathrm{be}H. \\ \mathrm{Now}, \\ \mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{pipe}=\mathrm{Volume}\mathrm{of}\mathrm{the}\mathrm{block} \\ \Rightarrow \mathrm{\pi }{R}^{2}H-\mathrm{\pi }{r}^{2}H=lbh \\ \Rightarrow \mathrm{\pi }\left(R^2-r^2\right)H=lbh \\ \Rightarrow \frac{22}{7}\times \left(0.{35}^2-0.3^2\right)H=4.4\times 2.6\times 1 \\ \Rightarrow \frac{22}{7}\times \left(0.1225-0.09\right)H=4.4\times 2.6 \\ \Rightarrow \frac{22}{7}\times 0.0325\times H=4.4\times 2.6 \\ \Rightarrow H=\frac{4.4\times 2.6\times 7}{22\times 0.0325} \\ \therefore H=112\mathrm{m} \end{gathered}$$

So, the length of the pipe is 112 m.