hey guys. help me with this.

two vertical tank in conical shape. both inverted. have their vertices connected by a short horizontal pipe. the larger tank is full of water, has a diameter base of 5 m. the smaller tank is empty, has an altitude of 4 m, has an diameter base of 5 m. if the water is allowed to flow through the connecting pipe until the smaller tank is full. find the total volume of the water in the two tanks and find the altitude of the larger tank.

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22r^2*h/21Log in to reply

yes

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can you give me the solution on how you solve this?

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the hieght to which water rise will be same i.e 4 m thus the hieght of large tank be consider to be 4m then volume of water will be in 2 tank =2

22r^2*h/21 =52.38m^3Log in to reply

so it means that larger tank and smaller tank is equal?

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as the tank connected by pipe so water rise will be same so altitude of large tank be 4m thus applying volume of cone i.e 22

r^2h/21, as hieght and diameter is same so ,volume=44r^2h/21 r=d/2,thus r=2.5m ,h=4m , putting values in above equation we get 52.38m^3 if asked in litter then 1m^3=1000ltrs , so 52380ltrs : )Log in to reply

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there diameter is also same

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