Determine the pressure difference in tube of non$-$uniform cross sectional area as shown in figure. $\Delta P =?$ (in $pa$)

$d_{1}=5\, cm , V_{1}=4\, cm , d_{2}=2\, cm , V_{2}=?$

Question

Determine the pressure difference in tube of non$-$uniform cross sectional area as shown in figure. $\Delta P =?$ (in $pa$)

$d_{1}=5\, cm , V_{1}=4\, cm , d_{2}=2\, cm , V_{2}=?$

AIIMS 2019, Medium

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Solution

Apply the equation of continuity.

$A _{1} v _{1}= A _{2} v _{2}$

$5^{2} \times 4=2^{2} \times v _{2}$

$v _{2}=25$

Apply the energy equation at both the ends.

$P _{1}+\frac{1}{2}\rho v _{1}^{2}= P _{2}+\frac{1}{2}\rho v _{2}^{2}$

$P_{1}-P_{2}=\frac{1}{2}\rho \left(v_{2}^{2}-v_{1}^{2}\right)$

$=\frac{1}{2} \times 10^{3}\left(25^{2}-4^{2}\right)$

$=304500 Pa$

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