The pressure of water in a water pipe when tap is opened and closed is respectively $3 \times 10^5\,N/m^2$ and $3.5 \times 10^5\,N/m^2$ . With open tap, the velocity of water flowing is ........... $m/s$
, Medium
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$\mathrm{P}_{1}+\frac{1}{2} \rho \mathrm{v}_{1}^{2}=\mathrm{P}_{2}+\frac{1}{2} \rho \mathrm{v}_{2}^{2}$
$\left(\mathrm{P}_{1}-\mathrm{P}_{2}\right)=\frac{1}{2} \rho \mathrm{v}_{2}^{2}$
since tap is closed, $\mathrm{v}_{1}=0$
$\left(3.5 \times 10^{5}-3 \times 10^{5}\right)=\frac{1}{2} \times 10^{3} \times v^{2}$
$\mathrm{v}=10 \mathrm{m} / \mathrm{s}$
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