Question

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

$V=a_{1}a_{2}\sqrt{\frac{2 \left(\right. p_{1} - p_{2} \left.\right)}{\rho \left(\right. a_{1}^{2} - a_{2}^{2} \left.\right)}}$
$=\pi r_{1}^{2}\times \pi r_{2}^{2} \, \sqrt{\frac{2 \left(\right. p_{1} - p_{2} \left.\right)}{\rho \left[\left(\pi r_{1}^{2}\right)^{2} - \left(\pi r_{2}^{2}\right)^{2}\right]}}$
$=\pi r_{1}^{2}r_{2}^{2} \, \sqrt{\frac{2 \left(\right. p_{1} - p_{2} \left.\right)}{\rho \left(\right. r_{1}^{4} - r_{2}^{4} \left.\right)}}$
$=\frac{22}{7}\times \left(0.1\right)^{2}\times \left(0.04\right)^{2} \, \sqrt{\frac{2 \times 10}{\left(1.25 \times \left(10\right)^{3}\right) \left[\right. \left(0.1\right)^{4} - \left(0.04\right)^{4} \left]\right.}}$
$=6.4\times 10^{- 4} \, m^{3 \, }s^{- 1}$