Question

The pressure exerted by water in a vessel at its bottom is $32\times {10}^{2}N{m}^{-2}$. A capillary tube of length $0.25m$ and radius $1mm$ is fixed near the bottom of the vessel to the side of the vessel. If volume of water flown out of the tube in $70sec$ is $44\times {10}^{-6}{m}^3$, the coefficient of viscosity of water is:

• A. $8\times {10}^{-3}Pa$-$s$
• B. $0.8\times {10}^{-3}Pa$-$s$
• C. $560\times {10}^{-3}Pa$-$s$
• D. $0.56\times {10}^{-3}Pa$-$s$

Answer 1

The correct option is A $8\times {10}^{-3}Pa$-$s$

Pressure difference leading to flow of water $=32\times {10}^{2}Nm$
From Poiseuille's eqn,
$\Rightarrow \frac{44\times {10}^{-6}}{70}{m}^3/s=\frac{\pi \times {10}^{-12}\times 32\times {10}^2}{8\eta 0.25}$
$\Rightarrow \eta =\frac{\pi \times {10}^{-4}\times 32\times 70}{8\times 0.25\times 44}$
$\Rightarrow \eta =7.996\times {10}^{-3}Pa-s$
$\Rightarrow \eta \approx 8\times {10}^{-3}Pa-s$