Q 11-09-093JEE MainJEE Main 2023 (13 Apr, Shift 1)Medium
The figure shows a liquid of given density flowing steadily in horizontal tube of varying cross-section. Cross-sectional areas at $A$ is $1.5\ \text{cm}^2$ and $B$ is $25\ \text{mm}^2$. If the speed of liquid at $B$ is $60\ \text{cm s}^{-1}$ then $(P_A-P_B)$ is (Given $P_A$ and $P_B$ are liquid pressures at $A$ and $B$ points. Density $\rho=1000\ \text{kg m}^{-3}$. $A$ and $B$ are on the axis of tube)
Answer: (C) $175\ \text{Pa}$
Continuity: $v_A=\dfrac{25\times60}{150}=10\ \text{cm s}^{-1}$.
$$P_A-P_B=\frac12\rho(v_B^2-v_A^2)=500\times(0.36-0.01)=175\ \text{Pa}$$
Solution by Sreeraj P, M.Sc Physics