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 60

Question

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 $60 cm s^{- 1}$ then $(P_{A} - P_{B})$ is

(Given $P_{A}$ and $P_{B}$ are liquid pressures at $A$ and $B$ points.

Density $ρ = 1000 kg m^{- 3}$

$A$ and $B$ are on the axis of tube)

TestHub · Accepted Answer

Using Bernoulli's theorem, P A + 1 2 ρ V A 2 = P B + 1 2 ρ V B 2 . . . ( i ) Using equation of continuity, A A V A = A B V B So, 1 . 5 × 10 - 4 × V A = 25 × 10 - 6 × V B V A = 25 × 10 - 6 1 . 5 × 10 - 4 × 0 . 6 = 1 10 m s - 1 Substituting in equation (i) ⇒ P A - P B = 1000 2 0 . 6 2 - 0 . 1 2 = 1000 2 × 0 . 7 × 0 . 5 = 175 P a

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