Velocity of Efflux at The Bottom of A Tank
Let's discuss a problem based on the velocity of efflux at the bottom of a tank. Try the problem yourself and read the solution here.
The Problem:
A large tank is filled with water. The total pressure at the bottom is (3.0atm). If a small hole is punched at the bottom, what is the velocity of efflux?
Solution:
A large tank is filled with water. The total pressure at the bottom is (3.0atm). A small hole is punched at the bottom.
Pressure at the bottom due to water coloumn $$ (3-1)atm$$ $$=2atm$$ $$ =2\times 10^5 Pa$$
The equation for pressure is $$ P=h\rho g$$
Hence, $$ h=\frac{P}{\rho g}$$ $$=\frac{2\times 10^5}{1000g}$$ $$=\frac{200}{g}$$
Hence, velocity $$ v=\sqrt{2gh}$$ $$=\sqrt{\frac{200}{2g}}$$ $$=20m/s$$
Maximum Height of Water in A Tank With A Hole
Let's discuss a problem based on the maximum height of water in a tank with a hole. Try the problem yourself first and then read the solution.
The Problem: Maximum Height of Water
A large tank is filled with water of (70 cm^3/s). A hole of cross-section (0.25cm^2) is punched at the bottom of the tank. Find the maximum height to which the tank can be filled.
Solution:
For the water level to remain stationary volume efflux= rate of filling = (x)
The velocity $$ v=\sqrt{2gh}$$ where (g) is the acceleration due to gravity.
Hence,$$ vA=\sqrt{2gh}A$$ $$ =x$$ $$=70cm^3//s$$
The maximum height $$ h=\frac{x^2}{2gA^2}$$ $$=\frac{70^2}{2\times 980\times(0.25)^2}$$ $$=40cm$$
Work Done By A Force Acting on The Piston
Let's discuss a problem and find out the work done by a force acting on the piston. First, try the problem and then read the solution.
The Problem:
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:
The 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.
$$ W=\frac{1}{2}\frac{\rho V^3}{s^2t^2}$$
More Problem: https://cheenta.com/velocity-of-efflux-at-the-bottom-of-a-tank/