Q. Water flows through a vertical tube of variable cross-section. The area of cross-section at $A$ and $B$ are 6 and 3 $mm^{2}$ respectively. If 12 cc of water enters per second through $A$, find the pressure difference $p_{A} - p_{B}\ (g = 10\ ms^{- 2})$ The separation between cross-section at $A$ and $B$ is 100 cm
A) $1.6 \times 10^{5}\ dyne\ cm^{- 2}$
B) $2.29 \times 10^{5}\ dyne\ cm^{- 2}$
C) $5.9 \times 10^{5}\ dyne\ cm^{- 2}$
D) $3.9 \times 10^{5}\ dyne\ cm^{- 2}$
Explanation
(a) $v_{1} = \frac{V}{A_{1}} = \frac{12 \times 10^{- 6}}{6 \times 10^{- 6}} = 2\ ms^{- 1} = 200\ cms^{- 1}$
$v_{2} = \frac{V}{A_{2}} = \frac{12 \times 10^{- 6}}{3 \times 10^{- 6}} = 4\ ms^{- 1} = 400\ cms^{- 1}$
$p_{A} - p_{B} = \rho g\left( h_{2} - h_{1} \right) + \frac{\rho}{2}(v_{2}^{2} - v_{1}^{2})$
$= 1 \times 1000(100) + \frac{1}{2}(16 \times 10^{4} - 4 \times 10^{4})$
$= 10^{5} + 6 \times 10^{4} = 1.6 \times 10^{5}\ dyne\ cm^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.