question_answer

Water is flowing at the rate of 2.52 km/h through a cylindrical pipe into a cylindrical tank, the radius of whose base is 40 cm. If the increase in the level of water in the tank, in half an hour is 3/15 m, find the internal diameter of the pipe.

Answer:

Let the internal radius of the pipe be x m.

Radius of base of tank \[=40\text{ }cm=\frac{2}{5}m\]

Speed of water flowing through the pipe

\[=2.52\,\,km/hr.\]

\[=\frac{2.52}{2}\times 1000\]

\[=1260\,m\] in half an hour

Volume of water flown in half an hour

\[=\pi {{r}^{2}}h\]

\[=\frac{22}{7}\times x\times x\times 1260\]

\[=3960\,\,{{x}^{2}}\]

Level of water raised in the tank

\[=3.15\text{ }m\]

\[=\frac{315}{100}m\]

Now, \[\pi \times \frac{2}{5}\times \frac{2}{5}\times \frac{315}{100}=3960{{x}^{2}}\]

\[{{x}^{2}}=\frac{22\times 2\times 2\times 315}{7\times 5\times 5\times 100\times 3960}\]

\[{{x}^{2}}=\frac{4}{10000}\]

\[{{x}^{2}}=\frac{2}{100}=0.02\,m\]

Internal diameter of the pipe \[=0.04\text{ }m=4\text{ }cm\]