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A wide vessel has a small hole at the bottom. It holds a layer of water (density rho1) of thickness h1 lying below a layer of kerosene (density rho2) of thickness h2 floating on top. Ignoring viscosity, find the speed v with which water flows out of the bottom hole.
- v = sqrt(2*g*(h1 + h2*rho2/rho1))
- v = sqrt(2*g*(h1 + h2))
- v = sqrt(2*g*h1)
- v = sqrt(2*g*(h1*rho1/rho2 + h2))
Correct answer: v = sqrt(2*g*(h1 + h2*rho2/rho1))
Solution
The gauge pressure at the hole comes from both layers: P = rho1*g*h1 + rho2*g*h2. Applying Bernoulli (the water leaving has density rho1), this pressure converts to kinetic energy: P = (1/2)*rho1*v². Solving, v = sqrt(2*g*(h1 + (rho2/rho1)*h2)).
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