Given,
internal diameter of pipe = 2 cm
internal radius of pipe = \(\frac{2}{2}\) = 1 cm
rate of flow of water = 6 m/s = 600 cm/s
radius of base of cylindrical tank = 60 cm
so,
rise in height in cylindrical tank = \(\frac{rate\,of\,flow\,of\,water\times total\,time\times volume\,of\,pipe}{volume\,of\,cylinderical\,tank}\)
= \(\frac{600\times 30\times 60\times π\times 1\times 1}{π\times 60\times 60}\) = 300 cm = 3 m