Question

The rate of flow of glycerine of density $1.25\times {10}^{3}kg/m^3$ through a conical section of a pipe, if the radii of its ends are $0.1m$ and $0.04m$ and the pressure drop across its length is $10N/m^2$ is:

• A. $3.14\times {10}^{-3}{m}^3/s$
• B. $6.28\times {10}^3{m}^3/s$
• C. $6.28\times {10}^{-4}{m}^3/s$
• D. $3.14\times {10}^{-4}{m}^3/s$

Answer 1

The correct option is C $6.28\times {10}^{-4}{m}^3/s$

$d=1.25\times {10}^{3}kg/m^3$

$\frac{1}{2}\times 1.25\times {10}^3(V_2^2-V_1^2)=10$

$V_2^2-V_1^2=\frac{8}{5}\times {10}^{-2}$

$A_1{V}_1=A_2{V}_2$

$V_1=4\times {10}^{-2}{V}_2$

from equation (i) and (i)

$V_2=\frac{8\times {10}^{-1}}{2\pi }m/s$

$($because $\sqrt{10}=\pi )$

now rate $=A_2{V}_2=6.28\times {10}^{-4}{m}^3/s$

correct option is (C).