JEE PYQ: Mechanical Properties Of Fluids Question 17
Question 17 - 2020 (06 Sep 2020 Shift 2)
A fluid is flowing through a horizontal pipe of varying cross-section, with speed $v$ ms$^{-1}$ at a point where the pressure is $P$ Pascal. At another point where pressure is $\frac{P}{2}$ Pascal its speed is $V$ ms$^{-1}$. If the density of the fluid is $\rho$ kg m$^{-3}$ and the flow is streamline, then $V$ is equal to:
(1) $\sqrt{\frac{P}{\rho} + v}$
(2) $\sqrt{\frac{2P}{\rho} + v^2}$
(3) $\sqrt{\frac{P}{2\rho} + v^2}$
(4) $\sqrt{\frac{P}{\rho} + v^2}$
Show Answer
Answer: (4) $\sqrt{\frac{P}{\rho} + v^2}$
Solution
Using Bernoulli’s equation: $P_1 + \frac{1}{2}\rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2}\rho v_2^2 + \rho g h_2$ For horizontal pipe, $h_1 = 0$ and $h_2 = 0$ and taking $P_1 = P$, $P_2 = \frac{P}{2}$: $\Rightarrow P + \frac{1}{2}\rho v^2 = \frac{P}{2} + \frac{1}{2}\rho V^2$ $\Rightarrow \frac{P}{2} + \frac{1}{2}\rho v^2 = \frac{1}{2}\rho V^2 \Rightarrow V = \sqrt{v^2 + \frac{P}{\rho}}$
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