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A cylinder filled with water of volume $$V$$ is fitted with a piston and placed horizontally. There is a hole of cross sectional area $$s$$ at the other end of the cylinder, $$s$$ being much smaller than the cross sectional area of the piston. Show that the work to be done by a constant force acting on the piston to squeeze all water from the cylinder in time $$t$$ is given by $$W=\frac{\rho V^3}{2s^2t^2}$$ where $$\rho$$ is the density of water.

Solution:

Volume of water flowing out per second $$Q=sv$$
where $$v$$ is the speed of sound and $$s$$ is the cross-sectional area.
Volume flowing out $$V=Qt=svt$$ $$\frac{ \rho v^2}{2}=P$$$$= \frac{F}{A}$$ $$=\frac{FL}{AL}$$ $$=\frac{W}{V}$$
where $$L$$ is the length of the cylinder and $$W$$ is the work done.
$$W=\frac{1}{2}\frac{\rho V^3}{s^2t^2}$$