Q. A fluid flows through a horizontal pipe having two different cross-sections of area $A$ and$2\ A$. If the pressure at the thin cross-section is $p$ and fluid velocity is $v$, the velocity and pressure at the thicker cross-section is (take the density of fluid as $\rho$)
A) $\frac{v}{2},\ p + \frac{1}{2}\rho\ v^{2}$
B) $\frac{v}{4},\ p + \frac{3}{8}\rho\ v^{2}$
C) $\frac{v}{2},\ p + \frac{3}{8}\rho\ v^{2}$
D) $v,\ p + \frac{3}{4}\rho\ v^{2}$
Explanation
(c) $A\ v = 2\ A\ v'$ or $v^{'} = v/2$
For a horizontal pipe, according to Bernoull’s theorem
$p + \frac{1}{2}\rho\ v^{2} = {P'}^{'} + \frac{1}{2}\rho\left( \frac{v}{2} \right)^{2}$
Or $p^{'} = p + \frac{1}{2}\rho v^{2}\left( 1 - \frac{1}{4} \right)$
$= p + \frac{3}{8}\rho\ v^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.