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](http://images.topperlearning.com/topper/new-ate/top_mob169988983265778223993873529-7961-4ede-beb7-54f6fc823033.jpg)
Asked by advikanagare29 | 13 Nov, 2023, 21:07: PM
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](https://images.topperlearning.com/topper/tinymce/cache/3f267fa5c1a4c2e013ca6bf262d200f3.png)
![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](https://images.topperlearning.com/topper/tinymce/cache/051a0b697935c7ced58b7e8801316e6e.png)
Answered by Thiyagarajan K | 13 Nov, 2023, 22:42: PM
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