The figure shows a liquid of given density flowing steadily in...

Using Bernoulli's theorem,

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

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)

$⇒ {P}_{A} - {P}_{B} = \dfrac{1000}{2}\left[0 . {6}^{2} - 0 . {1}^{2}\right] = \dfrac{1000}{2} \times 0 . 7 \times 0 . 5$

$= 175Pa$