A liquid of density $900 {kgm}^{- 3}$ is filled in a cylinder drical tank of upper radius $0.9 m$ and lower radius us $0.3 m$ A capillary tube of length l is attached at the bottom of the tank as shown in the figure. The capillary has outer radius 0.002 m and inner radius r. When pressure P is applied at the top of the tank, the rate of flow of the liquid out of the capillary tube is $8 \times 10^{- 6} m^{3} s^{- 1}$ , but if the capillary tube is removed, the liquid comes out of the tank with a speed of $10 {ms}^{- 1}$ . Determine the coefficient of viscosity of the liquid. Given $p r^{2} = 10^{- 6} m^{2} and r^{2} / l = 2 \times 10^{- 6} m$ .

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2.23 $\times 10^{- 4} N s m^{- 2}$

3.12 $\times 10^{- 3} N s m^{- 2}$

$1.25 \times 10^{- 3} {Nsm}^{- 2}$

4.45 $\times 10^{- 2} N s m^{- 2}$

answer is A.

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Detailed Solution

Let $P_{0}$ be the atmospheric pressure. When the capillary tube is not attached,

$\begin{array}{l} P + P_{0} + ρ g H + \frac{1}{2} ρ v_{1}^{2} = P_{0} + \frac{1}{2} ρ v_{2}^{2} \\ \Rightarrow P + ρ g H = \frac{1}{2} ρ (v_{2}^{2} - v_{1}^{2}) \\ = \frac{1}{2} ρ v_{2}^{2} [1 - {(\frac{v_{1}}{v_{2}})}^{2}] \\ = \frac{1}{2} ρ v_{2}^{2} [1 - \frac{A_{2}}{A_{1}}] (∵ A_{1} v_{1} = A_{2} v_{2}) \end{array}$

$\begin{array}{l} = \frac{1}{2} \times 900 \times (10)^{2} \times (1 - \frac{1}{9}) \\ = 4 \times 10^{4} Pa \end{array}$

$∴$ Pressure difference between the ends of the capillary tube is $△ P = 4 \times 10^{4} Pa$

Volume of liquid flowing per second in the capillary tube is

$\begin{array}{l} Q = \frac{π r^{4} Δ P}{8 η l} \\ \Rightarrow η = \frac{π r^{4} Δ P}{8 Q l} = (π r^{2}) \times \frac{r^{2}}{l} \times \frac{Δ P}{8 Q} \\ = \frac{10^{- 6} \times (2 \times 10^{- 6}) \times 4 \times 10^{4}}{8 \times 8 \times 10^{- 6}} \\ = 1.25 \times 10^{- 3} {Nsm}^{- 2} \end{array}$