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 \[=0.04\text{ }m=4\text{ }cm\]
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