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Water flows in a streamline motion through a horizontal pipe of circular cross-section as shown in the figure. The pressure difference of water between P and Q is $15\text{ Nm}^{-2}$. The area of cross-section at P and Q are $40\text{ cm}^2$ and $20\text{ cm}^2$, respectively. The rate of flow of water through the pipe, in $\text{cm}^3\text{s}^{-1}$, is :
[Take density of water $= 1000\text{ kg m}^{-3}$]

This question was previously asked in
NEET UG Re-Exam 2026 Question Paper (21-Jun-2026)
The correct answer is
400

To find the rate of flow of water through the pipe, we will use the principle of continuity and Bernoulli's equation for fluid flow.

Step 1: Continuity Equation

The equation of continuity states that the product of the cross-sectional area and the fluid velocity is constant along the pipe for incompressible fluid flow. Mathematically, this is expressed as:

\(A_P v_P = A_Q v_Q\)

Where:

  • \(A_P = 40\text{ cm}^2 = 40 \times 10^{-4} \text{ m}^2\)
  • \(A_Q = 20\text{ cm}^2 = 20 \times 10^{-4} \text{ m}^2\)
  • \(v_P\) and \(v_Q\) are the velocities at points P and Q, respectively.

Step 2: Bernoulli’s Equation

According to Bernoulli’s principle, for streamline flow, the total mechanical energy is constant, i.e.,

\(P_P + \frac{1}{2} \rho v_P^2 = P_Q + \frac{1}{2} \rho v_Q^2\)

Given:

  • \(P_P - P_Q = 15 \text{ Nm}^{-2}\)
  • \(\rho = 1000 \text{ kg m}^{-3}\)

Rearrange the equation to find the relation between \(v_P\) and \(v_Q\):

\(\frac{1}{2} \rho v_P^2 - \frac{1}{2} \rho v_Q^2 = P_P - P_Q = 15\)

Step 3: Solve for Velocities

From the continuity equation, we know:

\(v_Q = \frac{A_P}{A_Q} \cdot v_P = 2v_P\)

Substitute \(v_Q = 2v_P\) into Bernoulli's equation:

\(\frac{1}{2} \times 1000 \times v_P^2 - \frac{1}{2} \times 1000 \times (2v_P)^2 = 15\)

Simplify and solve for \(v_P\):

\(500v_P^2 - 2000v_P^2 = 15 \implies -1500v_P^2 = 15\)

\(v_P^2 = \frac{15}{1500} = \frac{1}{100} \implies v_P = \frac{1}{10} \text{ m/s}\)

Step 4: Calculate the Rate of Flow

The rate of flow (Q) is given by:

\(Q = A_P \cdot v_P\)

\(Q = 40 \times 10^{-4} \times \frac{1}{10} = 400 \text{ cm}^3/\text{s}\)

Thus, the rate of flow of water through the pipe is 400 cm3/s.

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