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 Nm − 2 . The area of cross-section

Question

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 $15$ Nm $^{- 2}$ . The area of cross-section at $P$ and $Q$ are $40$ cm $^{2}$ and $20$ cm $^{2}$ , respectively. The rate of flow of water through the pipe, in cm $^{3}$ s $^{- 1}$ , is :
[Take density of water $= 1000$ kg m $^{- 3}$ ]
NEET 2026 (Re-NEET)

TestHub · Accepted Answer

From the equation of continuity, the rate of flow is constant:

$A_{P} v_{P} = A_{Q} v_{Q}$

Given $A_{P} = 40 cm^{2}$ and $A_{Q} = 20 cm^{2}$ , we have:

$40 v_{P} = 20 v_{Q} \Rightarrow v_{Q} = 2 v_{P}$

Applying Bernoulli's equation for a horizontal pipe:

$P_{P} + \frac{1}{2} ρ v_{P}^{2} = P_{Q} + \frac{1}{2} ρ v_{Q}^{2}$

$P_{P} - P_{Q} = \frac{1}{2} ρ (v_{Q}^{2} - v_{P}^{2})$

Substitute the given values ( $P_{P} - P_{Q} = 15 N m^{- 2}$ , $ρ = 1000 kg m^{- 3}$ ) and $v_{Q} = 2 v_{P}$ :

$15 = \frac{1}{2} \times 1000 \times ((2 v_{P})^{2} - v_{P}^{2})$

$15 = 500 \times (4 v_{P}^{2} - v_{P}^{2})$

$15 = 500 \times 3 v_{P}^{2}$

$15 = 1500 v_{P}^{2}$

$v_{P}^{2} = \frac{1}{100} \Rightarrow v_{P} = 0.1 m s^{- 1} = 10 cm s^{- 1}$

The rate of flow of water is:

$Q = A_{P} v_{P} = 40 cm^{2} \times 10 cm s^{- 1} = 400 cm^{3} s^{- 1}$

Answer: $400$

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