Question

An incompressible non viscous fluid flows steadily through a cylindrical pipe which has radius $2R$ at point $A$ and radius $R$ at point $B$ farther along the flow direction. If the velocity of the fluid at point $A$ is $V$, its velocity at the point $B$ will be :

• A. $P-\frac{1}{2}\delta V^2$
• B. $P-\delta V^2$
• C. $P+\delta V^2$
• D. $P-7.5\delta V^2$

Answer 1

Answer: (A), (B)

$P-\delta V^2$
$P-\frac{1}{2}\delta V^2$

We know that, from continuity equation,
$A_1{V}_2=A_2{V}_2$
(Symbols have usual meanings)
$\pi (2R{)}^{2}V=\pi R^2{V}_B$
$(V_B=$ Velocity of the fluid at point $B)$
$\Rightarrow \pi (2R{)}^{2}V=\pi R^2{V}_B$
$\Rightarrow \pi 4R^{2}V=\pi R^2{V}_B$
$4V=V_B$
$\therefore V_B=4V$