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 $15\ \text{N m}^{-2}$. 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 $= 1000\ \text{kg m}^{-3}$]
Answer: (D) $400$
Concept. Continuity ($Av$ constant) plus Bernoulli for a horizontal pipe ($\Delta P = \dfrac12\rho(v_2^2 - v_1^2)$).
Step 1. Continuity. $v_Q = \dfrac{40}{20}v_P = 2v_P$.
Step 2. Bernoulli.
$$15 = \frac12(1000)(4v_P^2 - v_P^2) \;\Rightarrow\; v_P^2 = 0.01 \;\Rightarrow\; v_P = 10\ \text{cm/s}$$
Step 3. Flow rate. $Q = 40 \times 10 = 400\ \text{cm}^3\text{/s}$.
Answer. $400$ (Option D).
Solution by Sreeraj P, M.Sc Physics