Question

Glycerine flows steadily through a horizontal tube of length $1.5m$ and radius $1cm$. If the amount of glycerin collected per second at one end is $4.0\times {10}^{3}kg{s}^{-1}$, The pressure difference between the two ends of the tube is . Given that density of glycerin $=1.3\times {10}^{3}kg{m}^{-3}$ and viscosity of glycerin $=0.83Pa-s.$

• A. $9.75Pa$
• B. $975.52Pa$
• C. $97.5Pa$
• D. $975.52kPa$

Answer 1

The correct option is B $975.52Pa$

Amount of glycerin at one end $=4\times {10}^{3}kg/s$
Flow rate of glycerin $=\frac{4\times {10}^3}{1.3\times {10}^3}{m}^3/s$

$\frac{m}{v}=d$$=\frac{40}{13}{m}^3/s$

From poiseuille equ,
$Q=\frac{\pi r^4\Delta p}{8\eta L}$

$\frac{40}{13}=\frac{\pi {10}^{-8}}{8\times 0.83\times 1.5}\times \Delta p$

$\frac{40\times {10}^8\times 8\times 0.83\times 1.5}{\pi \times 13}=\Delta p$

$\Delta p=975.49\times {10}^{6}pa$