Water from a tap emerges vertically downwards | vedclass

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

Question

(B) According to the equation of continuity,the product of cross-sectional area and velocity remains constant for an incompressible fluid: $A_1 v_1 =

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) According to the equation of continuity,the product of cross-sectional area and velocity remains constant for an incompressible fluid: $A_1 v_1 = A_2 v_2$.Given $A_1 = A$,$v_1 = 4 \,m/s$,and $A_2 = \frac{2}{3} A$.Substituting these values: $A 	imes 4 = \frac{2}{3} A 	imes v_2$.Solving for $v_2$: $v_2 = 4 	imes \frac{3}{2} = 6 \,m/s$.Now,using the principle of conservation of energy (Bernoulli's equation) for the falling stream: $P + ho g h_1 + \frac{1}{2} ho v_1^2 = P + ho g h_2 + \frac{1}{2} ho v_2^2$.Since the pressure $P$ is constant (atmospheric pressure) and $h_1 - h_2 = h$,the equation simplifies to: $g h = \frac{1}{2} (v_2^2 - v_1^2)$.Substituting the known values: $10 	imes h = \frac{1}{2} (6^2 - 4^2)$.$10 h = \frac{1}{2} (36 - 16) = \frac{1}{2} (20) = 10$.Therefore,$h = 1 \,m$.

Similar Questions

$A$ streamline flow of a liquid of density $\rho$ is passing through a horizontal pipe of cross-sectional area $A_1$ and $A_2$ at two ends. If the pressure of the liquid is $P$ at a point where the flow speed is $v$,then the pressure at another point where the flow speed becomes $3v$ is:

$A$ manometer connected to a closed tap reads $4.5 \times 10^5 \text{ Pa}$. When the tap is opened,the reading of the manometer falls to $4 \times 10^5 \text{ Pa}$. The velocity of the flow of water is ........ $m/s$.

An $L$-shaped glass tube is just immersed in flowing water such that its opening is pointing against the flowing water. If the speed of the water current is $v$,then

$A$ horizontal pipe of non-uniform cross-section allows water to flow through it with a velocity $1 \ m/s$ when the pressure is $50 \ kPa$ at a point. If the velocity of flow has to be $2 \ m/s$ at some other point,the pressure at that point should be: (in $kPa$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module