A tank can be filled in by an inlet pipe of circular cross section in 10 min. Another tank twice as big is filled in by 3 inlet pipes whose diameters are in the ratio 1 : 2 : 3 . If the smallest pipe of these is four times as much in cross section as the pipe of the first tank, find the time taken to fill the second tank. [Speed of water in the second tank is half the speed of the water in the first tank.]

• If the ratio of diameters of 3 inlet pipes of second tank is 1 : 2 : 3, the ratio of their cross-sectional areas will be 1 : 4 :9. Suppose cross-sectional area of inlet pipe of first tank be k. Then the cross-sectional areas of three pipes of second tank will be 4k, 16k and 36k respectively. Total cross-sectional area of pipes of second tank is 56k.
  
  Time taken in filling the second tank is = 10 x k/56k x 2 x 2 = 5/7 min.
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