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A water flows through a cylindrical pipe, whose inner radius is 1 cm, at the rate of `80 cm^(-1)` in an empty cylindrical tank, the radius of whose base is 40 cm. What is the rise of water level in tank in half an hour? |
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Answer» Given , radius of tank , `r_(1) = 40 cm` Let hight of water level in tank in half an hour =` h_(1)` Also , given internal radius of cylindrical pipe, `r_(2) = 1cm` and speed of water `= 80 cm//s ` i.e., in 1 water flow = 80 cm `therefore` In 30 (min) water flow `= 80 xx 60 xx 30 = 144000 cm` According to the question, Volume of water in cylindrical tank = Volume of water flow from the circular pipe in half an hour `rArr " " pir_(1)^(2) h_(1) = pi r_(2)^(2) h_(2)` `rArr " " 40 xx 40 xx h_(1) = 1 xx 1xx 144000` `therefore " " h_(1) = (144000)/(40 xx 40) = 90` cm in half an hour. |
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