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Chapter 15: Problem 67

A can of height \(h\) and cross-sectional area \(A_{0}\) is initially full of water. A small hole of area \(A_{1} \ll A_{0}\) is cut in the bottom of the can. Find an expression for the time it takes all the water to drain from the can. (Hint: Call the water depth \(y\), use the continuity equation, and integrate.)

Short Answer

Expert verified
The expression for the time it takes all the water to drain from the can is given by \(t = 2\sqrt{h} A_{0} / (A_{1} \sqrt{2g})\).

Step by step solution

01

Determine the Exit Velocity

Using Torricelli's theorem, which says the speed \(v\) of outflow through a hole depends on the height \(y\) of the fluid above the hole, the exit velocity is given by \(v = \sqrt{2gy}\), with \(g\) being the acceleration due to gravity.
02

Establish the Rate of Change of Volume

The volume \(V\) of the water reducing in the can is given by \(V = A_{0}y\), so the rate of change of volume \(dV/dt = A_{0}dy/dt\). The rate of volume decrease equals to the volume flow out of the hole, therefore \(A_{0}dy/dt = -A_{1}v\). Substitute \(v\) with \(\sqrt{2gy}\) from step 1 to get \(-A_{0}dy/dt = A_{1}\sqrt{2gy}\).
03

Integrate To Find Time

Rearrange the equation as \(dy/\sqrt{y} = -A_{1}/A_{0} * \sqrt{2g} dt\). Then integrate both sides from \(h\) to \(0\) for the left side (considering \(y\) decreases from \(h\) to \(0\)), and from \(0\) to \(t\) for the right side. On performing the integration, we get \(-2\sqrt{h} = -tA_{1}/A_{0} * \sqrt{2g}\). Solve this equation for \(t\) to get the expression for time.

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