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  4. Determine the pressure difference in tube of non-uniform cross sectional area as shown in figure. ΔP = ? d1 = 5 cm, V1 = 4, d2 = 2 cm, V2 = ?

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Q. Determine the pressure difference in tube of non-uniform cross sectional area as shown in figure. $\Delta$P = ?
d1 = 5 cm, V1 = 4, d2 = 2 cm, V2 = ?Physics Question Image

AIIMSAIIMS 2019

Solution:

$A_{1}V_{1}=A_{2}V_{2}$
$5^{2}\times4=2^{2}\times V_{2}$
$V_{2}=25$
$P_{1}+\frac{1}{2}+V^{2}_{1}=P_{2}+\frac{1}{2}+V^{2}_{2}$
$P_{1}-P_{2}=\frac{1}{2}+\left(V^{2}_{2}-V^{2}_{1}\right)$
$=\frac{1}{2}\times10^{3}\left(25^{2}-4^{2}\right)$
$=500\left(625-16\right)$
$P_{1}-P_{2}$ =500 × 609 = 304500 Pa