Q. The glycerin of density $1.25 \times 10^{3}\ kgm^{- 3}$ is flowing through a conical tube with end radii 0.1 m and 0.04 m respectively. The pressure difference across the ends is $10\ Nm^{- 2}$. 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) $12.8 \times 10^{3}\ m^{3}s^{- 1}$
Explanation
(b) $V = a_{1}a_{2}\sqrt{\frac{2(p_{1} - p_{2})}{\rho(a_{1}^{2} - a_{2}^{2})}}$
$= \pi r_{1}^{2} \times \pi r_{2}^{2}\ \sqrt{\frac{2(p_{1} - p_{2})}{\rho\left\lbrack \left( \pi r_{1}^{2} \right)^{2} - \left( \pi r_{2}^{2} \right)^{2} \right\rbrack}}$
$= \pi r_{1}^{2}r_{2}^{2}\ \sqrt{\frac{2(p_{1} - p_{2})}{\rho(r_{1}^{4} - r_{2}^{4})}}$
$= \frac{22}{7} \times (0.1)^{2} \times (0.04)^{2}\ \sqrt{\frac{2 \times 10}{\left( 1.25 \times 10^{3} \right)\lbrack(0.1)^{4} - (0.04)^{4}\rbrack}}$
$= 6.4 \times 10^{- 4}\ m^{3}s^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.