The glycerine of density $1 .25 \times 10^{3} {kgm}^{- 3}$ is flowing through a conical tube with end radii 0.1 m and 0.04 m respectively. The pressure difference across the ends is
10 Nm^-2. The rate of flow of glycerine through the tube is

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$12.8 \times 10^{- 2} m^{3} s^{- 1}$

$6.4 \times 10^{- 2} m^{3} s^{- 1}$

$6.4 \times 10^{- 4} m^{3} s^{- 1}$

$12.8 \times 10^{3} m^{3} s^{- 1}$

answer is B.

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Detailed Solution

$\begin{matrix} Q = a_{1} a_{2} \sqrt{\frac{2 (p_{1} - p_{2})}{ρ (a_{1}^{2} - a_{2}^{2})}} \\ = {πr}_{1}^{2} \times {πr}_{2}^{2} \sqrt{\frac{2 (p_{1} - p_{2})}{ρ [{({πr}_{1}^{2})}^{2} - {({πr}_{2}^{2})}^{2}]}} \\ = {πr}_{1}^{2} r_{2}^{2} \sqrt{\frac{2 (p_{1} - p_{2})}{ρ (r_{1}^{4} - r_{2}^{4})}} \\ = \frac{22}{7} \times (0 .1)^{2} \times (0 .04)^{2} \sqrt{\frac{2 \times 10}{(1 .25 \times 10^{3}) [(0 .1)^{4} - (0 .04)^{4}]}} \\ = 6 .4 \times 10^{- 4} m^{3} s^{- 1} \end{matrix}$