Question

Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1 cm If the amount of glycerine collected per second at one end is $4.0 \times 10^{-3} \,kgs ^{-1}$, what is the pressure difference between the two ends of the tube? $\left(\rho_{\text {Glycerine }}=1.3 \times 10^{3} kg m ^{-3}\right)\left(\eta_{\text {Glycerine }}=0.83 Pa . s \right)$

• A. 975 Pa
• B. 1025 Pa
• C. 1050 Pa
• D. 1200 Pa

Answer 1

Answer: (A)

Using Poisuelle's formula, $\frac{V}{t}=\frac{\pi P r^{4}}{8 \cdot \eta \cdot l}$
$\frac{V}{t}=\frac{M}{\rho \cdot t}=\frac{4 \times 10^{-3} kg s ^{-1}}{1.3 \times 10^{3} kg m ^{-3}}$
Now, $P=\left(\frac{V}{t}\right) \cdot\left(\frac{8 \eta l}{\pi r^{4}}\right)$
$=\left(\frac{4 \times 10^{-3} kg s ^{-1}}{1.3 \times 10^{3} kg m ^{-3}}\right) \cdot\left(\frac{8 \times 0.83 Pas (1.5 m )}{3.14 \times\left(10^{-2} m \right)^{4}}\right)$
$=975 \,Pa$