Question
Water flows through a cylindrical pipe, whose inner radius is 1cm, at the rate of 80cm/sec in an empty cylindrical tank, the radius of whose base is 40cm. What is the rise of water level in tank in half an hour?
✓
Answer
Given, radius of tank, $r_1=40 cm$
Let height of water level in tank in half an hour $=h_1$
Also, given internal radius of cylindrical pipe, $r_1=1 cm$
and speed of water $=80 cm / s$ in 1 water flow $=80 cm$
$\therefore \ln 30(min)$ water flow $=80 \times 60 \times 30=144000 cm$
According to the question,
Volume of water in cylindrical tank = Volume of water flow the circular pipe in half an hour
$\Rightarrow\pi\text{r}^2_1\text{h}_1=\pi\text{r}^2_1\text{h}_2$
$\Rightarrow40\times40\times\text{h}_1=1\times144000$
$\therefore\text{h}_1=\Big(\frac{144000}{40\text{x}40}\Big)=90\text{cm}$
Hence, the level of water in cylindrical tank rises 90cm in half an hour.
Let height of water level in tank in half an hour $=h_1$
Also, given internal radius of cylindrical pipe, $r_1=1 cm$
and speed of water $=80 cm / s$ in 1 water flow $=80 cm$
$\therefore \ln 30(min)$ water flow $=80 \times 60 \times 30=144000 cm$
According to the question,
Volume of water in cylindrical tank = Volume of water flow the circular pipe in half an hour
$\Rightarrow\pi\text{r}^2_1\text{h}_1=\pi\text{r}^2_1\text{h}_2$
$\Rightarrow40\times40\times\text{h}_1=1\times144000$
$\therefore\text{h}_1=\Big(\frac{144000}{40\text{x}40}\Big)=90\text{cm}$
Hence, the level of water in cylindrical tank rises 90cm in half an hour.
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