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Fluid Mechanics

JEE Main level, single correct, medium

56 of 192

Question 56 of 192

The figure shows a liquid of given density flowing steadily in horizontal tube of varying cross-section. Cross-sectional areas at $A$ is $1.5{cm}^{2},$ and $B$ is $25{mm}^{2},$ if the speed of liquid at $B$ is $60cm{s}^{-1}$ then $({P}_{A}–{P}_{B})$ is
(Given ${P}_{A}$ and ${P}_{B}$ are liquid pressures at $A$ and $B$ points.
Density $\rho =1000kg{m}^{-3}$
$A$ and $B$ are on the axis of tube)

Show answer and solution

Answer: (C) $175Pa$

Using Bernoulli's theorem,

{P}_{A}+\dfrac{1}{2}\rho {V}_{A}^{2}={P}_{B}+\dfrac{1}{2}\rho {V}_{B}^{2}...(i)

Using equation of continuity,

{A}_{A}{V}_{A}={A}_{B}{V}_{B}

So,

1.5\times {10}^{-4}\times {V}_{A}=25\times {10}^{-6}\times {V}_{B}

{V}_{A}=\dfrac{25\times {10}^{-6}}{1.5\times {10}^{-4}}\times 0.6

=\dfrac{1}{10}m{s}^{-1}

Substituting in equation (i)

\Rightarrow {P}_{A}-{P}_{B}=\dfrac{1000}{2}\left[0.{6}^{2}-0.{1}^{2}\right]=\dfrac{1000}{2}\times 0.7\times 0.5

=175Pa

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