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CBSE Class 10 Answered

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
question image
Asked by advikanagare29 | 13 Nov, 2023, 09:07: PM
answered-by-expert Expert Answer
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
 
begin mathsize 14px style straight pi space straight r squared space straight h space equals space 0.3168 end style m3 
begin mathsize 14px style h space equals space fraction numerator 0.3168 over denominator straight pi cross times 0.2 cross times 0.2 end fraction space m space equals space 2.521 space m end style
Answered by Thiyagarajan K | 13 Nov, 2023, 10:42: PM
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