Question

Water flows through a horizontal pipe AB of non-uniform cross section. Water enters at A of area \(4cm^2\) with pressure \(10^5Nm^{-2}\) and velocity \(20ms^{-1}\). It leaves at B of area \(8cm^2\). The pressure at B is

• A. \(0.5\times10^5Nm^{-2}\)
• B. \(2.5\times10^5Nm^{-2}\)
• C. \(3.5\times10^5Nm^{-2}\)
• D. \(4.5\times10^5Nm^{-2}\)

Hint:

For fluid flow in horizontal pipes: first apply continuity equation to find unknown velocity, then substitute in Bernoulli equation.

Answer 1

The Correct Option is B

Concept: Use equation of continuity \[ A_1v_1=A_2v_2 \] And Bernoulli equation \[ P_1+\frac12\rho v_1^2=P_2+\frac12\rho v_2^2 \] Since pipe is horizontal, gravitational term cancels.

Step 1: Find velocity at B \[ A_1=4,\qquad A_2=8 \] \[ 4(20)=8v_2 \] \[ 80=8v_2 \] \[ v_2=10ms^{-1} \]

Step 2: Apply Bernoulli theorem Density of water \[ \rho=1000kgm^{-3} \] \[ P_1+\frac12\rho v_1^2=P_2+\frac12\rho v_2^2 \] \[ 10^5+\frac12(1000)(20)^2=P_2+\frac12(1000)(10)^2 \] \[ 10^5+200000=P_2+50000 \] \[ 300000=P_2+50000 \] \[ P_2=250000 \] \[ P_2=2.5\times10^5 \] Thus \[ \boxed{2.5\times10^5Nm^{-2}} \]