Question
A liquid of density $600\text{ }kg/m^{3}$ flowing steadily in a tube of varying cross-section. The cross-section at a point A is $1.0{\text{ }cm}^{2}$ and that at B is $20{\text{ }mm}^{2}$. Both the points A and B are in same horizontal plane, the speed of the liquid at A is $10\text{ }cm/s$. The difference is pressures at A and B points is
$\_\_\_\_$ Pa.
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Solution
Sol. [Key Concept: Continuity $A_{1}v_{1} = A_{2}v_{2}$ gives $v_{2}$, then Bernoulli $\Delta\ P = \frac{1}{2}\rho(v_{2}^{2} - v_{1}^{2}) = 72$ Pa.]
Using continuity equation,
$$\begin{matrix} & A_{1}V_{1} = A_{2}V_{2} \\ & 1 \times 10 = \frac{1}{5} \times V_{2} \\ & \ \Rightarrow V_{2} = 50\text{ }cm/s \end{matrix}$$
Using Bernoulli's theorem,
$$\begin{matrix} & P_{1} + \frac{1}{2}\rho V_{1}^{2} = P_{2} + \frac{1}{2}\rho V_{2}^{2} \\ & P_{1} - P_{2} = \frac{1}{2}\rho\left\lbrack V_{2}^{2} - V_{1}^{2} \right\rbrack \\ & \ = \frac{1}{2} \times 600\left\lbrack (50)^{2} - (10)^{2} \right\rbrack \times 10^{- 4} = 72\text{ }Pa. \end{matrix}$$
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