A horizontal pipe line carries water in a streamline flow.At a point along the tube where the cross-sectional area is $10^{-2}$ $m^2$, the water velocity is $2$ $ms^{^{-1}}$ and the pressure is $8000 $ $Pa$. The pressure of water at another point where the cross-sectional area is $0.5$ $\times 10^{-2}$ $m^2$ is ........ $Pa$
Medium
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Area at other point is half. So, speed will be double.
Now, $p_{1}+\frac{1}{2} \rho v_{1}^{2}=p_{2}+\frac{1}{2} \rho v_{2}^{2}$
$\therefore p_{2}=p_{1}+\frac{1}{2} \rho v_{1}^{2}-v_{2}^{2}$
$=(8000)+\frac{1}{2} \times 1000(4-16)$
$=2000 P a$
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