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Q. A manometer connected to a closed tap reads $ 3.5\times {{10}^{5}}\,N/{{m}^{2}} $ . When the valve is opened, the reading of manometer falls to $ 3.0\times {{10}^{5}}\,N/{{m}^{2}} $ , then velocity of flow of water is

ManipalManipal 2008Mechanical Properties of Fluids

Solution:

Bernoullis theorem for unit mass of liquid
$ \frac{p}{\rho }+\frac{1}{2}{{v}^{2}}= $ constant
As the liquid starts flowing, it pressure energy decreases
$ \frac{1}{2}{{v}^{2}}=\frac{{{p}_{1}}-{{p}_{2}}}{\rho }\Rightarrow \frac{1}{2}{{v}^{2}}=\frac{3.5\times {{10}^{5}}-3\times {{10}^{5}}}{{{10}^{3}}} $
$ \Rightarrow \,\,{{v}^{2}}=\frac{2\times 0.5\,\times {{10}^{5}}}{{{10}^{3}}}\,\Rightarrow \,{{v}^{2}}=100 $
$ \Rightarrow \,\,v=10\,m/s $