A steady flow of a liquid of density rho is s | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) According to the equation of continuity,$A_1 v_1 = A_2 v_2$.Given $A_1 = 2A$,$v_1 = \sqrt{2} \ m \ s^{-1}$,and $A_2 = A$.Substituting these values

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(C) According to the equation of continuity,$A_1 v_1 = A_2 v_2$.Given $A_1 = 2A$,$v_1 = \sqrt{2} \ m \ s^{-1}$,and $A_2 = A$.Substituting these values: $(2A)(\sqrt{2}) = (A)v_2 \Rightarrow v_2 = 2\sqrt{2} \ m \ s^{-1}$.For a steady flow,applying Bernoulli's equation at points $1$ and $2$:$P_1 + \rho g h_1 + \frac{1}{2} \rho v_1^2 = P_2 + \rho g h_2 + \frac{1}{2} \rho v_2^2$.Rearranging the terms: $P_1 - P_2 = \rho g (h_2 - h_1) + \frac{1}{2} \rho (v_2^2 - v_1^2)$.Given $P_1 - P_2 = 100 \ N \ m^{-2}$ (since pressure is higher at the wider section $1$ where speed is lower),$h_1 - h_2 = 10 \ cm = 0.1 \ m$,so $h_2 - h_1 = -0.1 \ m$.$100 = \rho [10(-0.1) + \frac{1}{2}((2\sqrt{2})^2 - (\sqrt{2})^2)]$.$100 = \rho [-1 + \frac{1}{2}(8 - 2)]$.$100 = \rho [-1 + 3] = 2\rho$.$\rho = 50 \ kg \ m^{-3}$.

Similar Questions

What is an aerofoil? Explain.

The flow speeds of air on the lower and upper surfaces of the wing of an aeroplane are $v$ and $2v$ respectively. The density of air is $\rho$ and the surface area of the wing is $A$. The dynamic lift on the wing is:

$A$ steady flow of water passes along a horizontal tube from a wide section $X$ to the narrower section $Y$,see figure. Manometers are placed at $P$ and $Q$ at the sections. Which of the statements $A, B, C, D$ is most correct?

Write the limitations of Bernoulli's equation.

Vedclass Test Series

Exam Paper Generator

Online Exam Module