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

Step 1: Picture the situation.
Water flows through a horizontal pipe whose mouth A is narrow ($4\,cm^2$) and outlet B is wider ($8\,cm^2$). A narrow region forces fluid to move fast, a wide region lets it slow down, and pressure trades off against speed. We are given $P_A=10^5\,Nm^{-2}$ and $v_A=20\,ms^{-1}$.
Step 2: Use continuity to get the speed at B.
The same volume of water per second passes every section, so $A_A v_A = A_B v_B$. Substituting, $4\times 20 = 8\times v_B$, which gives $v_B = \dfrac{80}{8} = 10\,ms^{-1}$.
Step 3: State Bernoulli's principle for a horizontal pipe.
Since both points are at the same height, the gravity terms cancel and we keep \[ P_A + \tfrac12 \rho v_A^2 = P_B + \tfrac12 \rho v_B^2 \] with $\rho = 1000\,kgm^{-3}$ for water.
Step 4: Compute the kinetic pressure terms.
At A, $\tfrac12 \rho v_A^2 = \tfrac12 (1000)(20)^2 = 2.0\times 10^5$. At B, $\tfrac12 \rho v_B^2 = \tfrac12 (1000)(10)^2 = 0.5\times 10^5$.
Step 5: Solve for the pressure at B.
Rearranging, $P_B = P_A + \tfrac12 \rho v_A^2 - \tfrac12 \rho v_B^2 = 10^5 + 2.0\times 10^5 - 0.5\times 10^5$, so $P_B = 2.5\times 10^5\,Nm^{-2}$.
Step 6: Sanity check.
The pipe widened, so the water slowed down, and slower water carries higher pressure. A rise from $10^5$ to $2.5\times 10^5$ is exactly that trend, so the answer is option (2).
\[ \boxed{P_B = 2.5\times 10^5\,Nm^{-2}} \]