A tank 'X' can be filled 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 among these is four times as much in cross section as the inlet pipe of the tank 'X', find the time taken to fill the second tank. [Speed of water flowing into the second tank is half the speed of the water flowing into the first tank]
A. 5/14 min
B. 5/7 min
C. 5/6 min
D. 6/5 min
Answer: Option B
Solution(By Examveda Team)
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 is k. Then the cross sectional areas of 3 pipes of second tank will be 4k, 16k and 36k respectively. Total cross sectional area of inlet pipes of second tank is 56k.Time taken in filling the second tank is = 10* k/56k *2 *2 = 5/7 min.
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