A manometer connected to a closed tap reads $4.5 \times {10^5}$ pascal. When the tap is opened the reading of the manometer falls to $4 \times {10^5}$ pascal. Then the velocity of flow of water is ........  $m{s^{ - 1}}$
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(d)$\frac{{{P_1} - {P_2}}}{{\rho g}} = \frac{{{v^2}}}{{2g}}$ ==> $\frac{{4.5 \times {{10}^5} - 4 \times {{10}^5}}}{{{{10}^3} \times g}} = \frac{{{v^2}}}{{2g}}$

$v=10m/s$ 

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