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) $400$
Continuity: $A_P v_P = A_Q v_Q$, so $v_Q = \dfrac{40}{20}v_P = 2v_P$.
Bernoulli (horizontal pipe):
$$P_P - P_Q = \frac{1}{2}\rho\left(v_Q^2 - v_P^2\right) = \frac{1}{2}\rho\left(3v_P^2\right)$$
$$15 = \frac{1}{2}(1000)(3v_P^2) \;\Rightarrow\; v_P^2 = 0.01 \;\Rightarrow\; v_P = 0.1\ \text{m s}^{-1} = 10\ \text{cm s}^{-1}$$
Rate of flow:
$$Q = A_P v_P = 40 \times 10 = 400\ \text{cm}^3\text{s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics