Question

Water is flowing through a cylindrical pipe of internal diameter 2 cm, into a cylindrical tank of base radius 40 cm, at the rate of 0.4 m per second. Determine the rise in level of water in the tank in half an hour. [CBSE 2013]

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \mathrm{the}\mathrm{internal}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{cylindrical}\mathrm{pipe},r=\frac{2}{2}=1\mathrm{cm}\mathrm{and} \\ \mathrm{the}\mathrm{base}\mathrm{radius}\mathrm{of}\mathrm{cylindrical}\mathrm{tank},R=40\mathrm{cm}, \\ \mathrm{Also},\mathrm{the}\mathrm{rate}\mathrm{of}\mathrm{water}\mathrm{flow},h=0.4\mathrm{m}/\mathrm{s}=40\mathrm{cm}/\mathrm{s} \\ \mathrm{Let}\mathrm{the}\mathrm{rise}\mathrm{in}\mathrm{level}\mathrm{of}\mathrm{water}\mathrm{be}H. \\ \mathrm{Now}, \\ \mathrm{The}\mathrm{volume}\mathrm{of}\mathrm{water}\mathrm{flowing}\mathrm{out}\mathrm{of}\mathrm{the}\mathrm{cylindrical}\mathrm{pipe}\mathrm{in}1\sec =\mathrm{\pi }{r}^{2}h=\mathrm{\pi }\times 1\times 1\times 40=40\mathrm{\pi }{\mathrm{cm}}^3 \\ \mathrm{So},\mathrm{the}\mathrm{volume}\mathrm{of}\mathrm{water}\mathrm{flowing}\mathrm{out}\mathrm{of}\mathrm{the}\mathrm{cylindrical}\mathrm{pipe}\mathrm{in}\mathrm{half}\mathrm{an}\mathrm{hour}(30\min )=40\mathrm{\pi }\times 60\times 30=72000\mathrm{\pi }{\mathrm{cm}}^3 \\ \mathrm{As}, \\ \mathrm{Volume}\mathrm{of}\mathrm{water}\mathrm{in}\mathrm{the}\mathrm{cylindrical}\mathrm{tank}=\mathrm{Volume}\mathrm{of}\mathrm{standing}\mathrm{water}\mathrm{in}\mathrm{cylindrical}\mathrm{pipe}\mathrm{for}\mathrm{half}\mathrm{an}\mathrm{hour} \\ \Rightarrow \mathrm{\pi }{R}^{2}H=72000\mathrm{\pi } \\ \Rightarrow R^{2}H=72000 \\ \Rightarrow 40\times 40\times H=72000 \\ \Rightarrow H=\frac{72000}{40\times 40} \\ \therefore H=45\mathrm{cm} \end{gathered}$$

So, the rise in level of water in the tank in half an hour is 45 cm.

Disclaimer: The answer given in the textbook is incorrect. The same has been corrected above.