A liquid flows through a horizontal tube. The | vedclass

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Question

Rate of flow $=\frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{A}_{1} \mathrm{V}_{1}=\mathrm{A}_{2} \mathrm{V}_{2}$

Apply energy conservation

$\mathrm{P

Vedclass · Accepted Answer

${v_2} - {v_1} = \sqrt {2gh} $

Vedclass · Answer

Rate of flow $=\frac{\mathrm{dQ}}{\mathrm{dt}}=\mathrm{A}_{1} \mathrm{V}_{1}=\mathrm{A}_{2} \mathrm{V}_{2}$

Apply energy conservation

$\mathrm{P}_{1}+\frac{1}{2} \rho \mathrm{V}_{1}^{2}+\rho \mathrm{gh}_{1}=\mathrm{P}_{2}+\frac{1}{2} \rho \mathrm{V}_{2}^{2}+\rho \mathrm{gh}_{2} ; \mathrm{h}_{1}=\mathrm{h}_{2}$

$\mathrm{P}_{1}+\frac{1}{2} \rho \mathrm{V}_{1}^{2}=\mathrm{P}_{2}+\frac{1}{2} \rho \mathrm{V}_{2}^{2}$

$\mathrm{P}_{1}-\mathrm{P}_{2}=\frac{1}{2} \rho\left(\mathrm{V}_{2}^{2}-\mathrm{V}_{1}^{2}\right)$

$\mathrm{V}_{2}^{2}-\mathrm{V}_{1}^{2}=2 \mathrm{gh} ;\left\{\mathrm{P}_{1}-\mathrm{P}_{2}=\rho \mathrm{gh}\right\}$

A liquid flows through a horizontal tube. The velocities of the liquid in the two sections which have areas of cross-section $A_1$ and $A_2$ are $v_1$ and $v_2$ respectively. The difference in the levels of the liquid in the two vertical tubes is $h$ . The incorrect statement is

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