Question

The glycerin of density $1.25\times 10^{3}kg{m}^{-3}$ is flowing through a conical tube with end radii $0.1m$ and $0.04m$ respectively. The pressure difference across the ends is $10N{m}^{-2}$ . The rate of flow of glycerine through the tube is

• A. $6.4\times 10^{-2}{m}^2{s}^{-1}$
• B. $6.4\times 10^{-4}{m}^3{s}^{-1}$
• C. $12.8\times 10^{-2}{m}^3{s}^{-1}$
• D. $12.8\times 10^3{m}^3{s}^{-1}$

Answer 1

Answer: (B)

$V=a_1{a}_2\sqrt{\frac{2(p_1-p_2)}{\rho (a_1^2-a_2^2)}}$
$=\pi r_1^2\times \pi r_2^2\sqrt{\frac{2(p_1-p_2)}{\rho \left[{\left(\pi r_1^2\right)}^2-{\left(\pi r_2^2\right)}^2\right]}}$
$=\pi r_1^2{r}_2^2\sqrt{\frac{2(p_1-p_2)}{\rho (r_1^4-r_2^4)}}$
$=\frac{22}{7}\times {\left(0.1\right)}^2\times {\left(0.04\right)}^2\sqrt{\frac{2\times 10}{\left(1.25\times {\left(10\right)}^3\right)[{\left(0.1\right)}^4-{\left(0.04\right)}^4]}}$
$=6.4\times 10^{-4}{m}^3{s}^{-1}$