Question

Water flows through a cylindrical pipe whose inner radius is $1\mathrm{cm}$, at the rate of $80\mathrm{cm}/\mathrm{s}$ in an empty cylindrical tank, the radius of whose base is $40\mathrm{cm}$. What is the rise of water level in tank in half an hour?

Answer 1

We have,

Radius of tank, ${\mathrm{r}}_1=40\mathrm{cm}$

Also, the internal radius of cylindrical pipe,${\mathrm{r}}_2=1\mathrm{cm}$

Speed of water $=80\mathrm{cm}/\mathrm{s}$ that means in $1\mathrm{second}$ water flow $=80\mathrm{cm}$

Therefore,

$\mathrm{In}30\mathrm{minutes}\mathrm{water}\mathrm{flow}=80\mathrm{x}60\mathrm{x}30=144000\mathrm{cm}$

As per the question,

$$\begin{gathered} \mathrm{Volume}\mathrm{of}\mathrm{water}\mathrm{in}\mathrm{cylindrical}\mathrm{tank}=\mathrm{Volume}\mathrm{of}\mathrm{water}\mathrm{flow}\mathrm{from}\mathrm{the}\mathrm{circular}\mathrm{pipe}\mathrm{in}\mathrm{half}\mathrm{an}\mathrm{hour} \\ {{\mathrm{\pi r}}^2}_1{\mathrm{h}}_1={{\mathrm{\pi r}}_2}^2{\mathrm{h}}_2 \\ \mathrm{On}\mathrm{substituting},\mathrm{we}\mathrm{get}, \\ \Rightarrow \mathrm{\pi }\times 40\times 40\times {\mathrm{h}}_1=\mathrm{\pi }\times 1\times 1\times 144000 \\ \Rightarrow {\mathrm{h}}_1=\frac{144000}{40\times 40} \\ \Rightarrow {\mathrm{h}}_1=\frac{1440}{16} \\ \Rightarrow {\mathrm{h}}_1=90\mathrm{cm} \end{gathered}$$

Therefore, in cylindrical tank, the level of water rises $90\mathrm{cm}$ in half an hour.