Question

A liquid flows through a horizontal pipe of varying radius. When the ratio of cross-sectional
areas of the pipe is 1:1.5 in its narrower and broader segment, the ratio of liquid kinetic energies in
these two segments has to be: a) 2:3 b) 1.25:1 c) 2:1 d) 1:1.5 e) No answer is correct.l​

Answer 1

Given :

The ratio of cross-sectional areas of the pipe = 1:1.5

To Find :

The ratio of the liquid kinetic energies

Solution :

• The equation of continuity states that for a liquid flowing in a pipe of different cross sections

               $A_{1} v_{1} =A_{2} v_{2}$

                $\frac{v_{1} }{v_{2} } =\frac{A_{2} }{A_{1} }$

                $\frac{v_{1} }{v_{2} } =\frac{1.5}{1}$

• The kinetic energy of the fluid = $\frac{1}{2}mv^{2}$

• The ratio of their kinetic energies is

                         $\frac{K.E_{1}}{K.E_{2}}= \frac{\frac{1}{2} mv_{1} ^{2}}{\frac{1}{2} mv_{2} ^{2}}=\frac{v_{1} }{v_{2} } ^{2}=\frac{2.25}{1}$

The ratio of the kinetic energies of the liquidis 2.25:1