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 $60\,cm / s$ then $\left( P _{ A }- P _{ B }\right)$ is :(Given $P _{ A }$ and $P _{ B }$ are liquid pressures at $A$ and $B$ points.Density $\rho=1000\,kg\,m ^{-3}$
$A$ and $B$ are on the axis of tube $............\,Pa$
JEE MAIN 2023, Diffcult
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From continuity theorem $A_1 V_1=A_2 V_2$
$1.5 \times V_1=25 \times 10^{-2} \times 60$
$V_1=\frac{25 \times 60 \times 10^{-2} \times 10}{1.5}$
$V_1=10\,cm / s$
By Bernoulli's theorem
$P_1+\frac{1}{2} \times 1000 \times(0.1)^2=P_2+\frac{1}{2} \times 1000 \times(0.6)^2$
$P_1+5=P_2+\frac{1}{2} \times 1000 \times 36 \times 10^{-2}$
$P_1+5=P_2+180$
$P_1-P_2=175\,Pa$
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