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water flows out through a circular pipe whose internal diameter is 4/3 cm at the rate of 0.63m per sec into a cylindrical tank the radius of whose base is 0.2 m by how much will the level of water rise in one hour
Water flow rate = cross section area of pipe × velocity of flow

Let d be the diameter of pipe.

Cross section area of pipe = ( π/4) ×d2 = ( π/4) × (4/3)2 × 10-4 m3 = 1.396 × 10-4 m2

velocity of flow = 0.63 m/s

Water flow rate = 1.396 × 10-4 × 0.63  m3 / s = 0.880 × 10-4 m3 / s

Volume of water collected in cylindrical tank for 1 hour = 0.880 × 10-4 × 3600  m3 = 0.3168 m3

if the base radius r of cylindrical tank is 0.2 m , then height h of water in cylinder is calculated as

m3

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