Water is flowing steadily through a horizontal tube of non uniform cross-section. If the pressure of water is $4$ $\times $ $10^4$ $N/m^2$ at a point where cross-section is $0.02$ $m^2$ and velocity of flow is $2$ $m/s$, what is pressure at a point where cross-section reduces to $0.01$ $m^2$.
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As we know
$A_{1} V_{1}=A_{2} V_{2}$
Where $A_{1}=0.02 \mathrm{m}^{2}$
$V_{1}=2 m / s$
$A_{2}=0.01 \mathrm{m}^{2}$
$\Longrightarrow 0.02 \times 2=0.01 \times V_{2}$
$\Longrightarrow V_{2}=4 m / s$
Now using Bernoulli's theorem
$\frac{P_{1}}{\rho}+\frac{1}{2} V_{1}^{2}=\frac{P_{2}}{\rho}+\frac{1}{2} v_{2}^{2}$
$\Longrightarrow P_{2}=P_{1}+\frac{\rho_{1}}{2}\left[V_{1}^{2}-V_{1}^{2}\right]$
$\Longrightarrow P_{2}=4 \times 10^{10}+\frac{1000}{2}\left[2^{2}-4^{2}\right]-4 \times 10^{10}=3.4 \times 10^{4} N / m^{2}$
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