Obtain Bernoulli’s equation of rest fluid.
Obtain Bernoulli’s equation of rest fluid.
Bernoulli's equation
$\mathrm{P}_{1}+\frac{1}{2} \rho v_{1}^{2}+\rho g h_{1}=\mathrm{P}_{2}+\frac{1}{2} \rho v_{2}^{2}+rho g h_{2}$
When fluid is rest, its velocity at every point is zero. In above equation $v_{1}=0, v_{2}=0$,
$\mathrm{P}_{1}+\rho g h_{1}=\mathrm{P}_{2}+\rho g h_{2}$
$\therefore \mathrm{P}_{1}-\mathrm{P}_{2}=\rho g\left(h_{2}-h_{1}\right)$
Similar Questions
At what speed the velocity head of a stream of water be equal to $40 cm $ of $Hg$ ........ $cm/sec$
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Statement $I$ : When speed of liquid is zero everywhere, pressure difference at any two points depends on equation $\mathrm{P}_1-\mathrm{P}_2=\rho \mathrm{g}\left(\mathrm{h}_2-\mathrm{h}_1\right)$
Statement $II$ : In ventury tube shown $2 \mathrm{gh}=v_1^2-v_2^2$
In the light of the above statements, choose the most appropriate answer from the options given below.
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