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InterviewSolution
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A cylinder containing water up to height 25 cm has a hole of cross-section `0.25 cm^(2)` at its bottom. It is counter possed in a balance. What is the initial change in the balancing weight when water begins to flow out? |
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Answer» Here, `a = 0.25 cm^(2) = 0.25 xx 10^(-4)m^(-2)`, . `h = 25 cm = 0.25 m` Initial velocity of water flowing out of hole `upsilon = sqrt(2gh)` When wate emerges out from the hole. The weight of water decrease. The decrease in weight is equal to the rate of change of linear momentum. Initially velocity of liquid flowing out `upsilon` be taken constant for a small interval of time `Delta t`. then `F = (Delta p)/(Delta t) = (Delta (m upsilon))/(Delta t) = upsilon (Delta m)/(Delta t)` `=(upsilonDelta(V rho))/(Delta t) = upsilon rho (Delta V)/(Delta t)` Where, `(Delta V)/(Delta t)` = volume flowing out per second `=aupsilon` `F = upsilon rho.(a upsilon) = a upsilon^(2) rho` `=a(2gh) rho = 2 a gh rho` `= 2 xx (0.25 xx 10^(-4)) xx 9.8 xx 0.25 xx 10^(3)` ltrgt `=12.5 xx 9.8 xx 10^(-3) N = 12.5 xx 10^(-3)kg f`. |
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