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 $15 \mathrm{Nm}^{-2}$. The area of cross-section at $P$ and $Q$ are $40 \mathrm{~cm}^2$ and $20 \mathrm{~cm}^2$, respectively. The rate of flow of water through the pipe, in $\mathrm{cm}^3 \mathrm{~s}^{-1}$, is:

[Take density of water $=1000 \mathrm{~kg} \mathrm{~m}^{-3}$]

![](https://res.cloudinary.com/ddr1kuyb3/image/upload/v1786187138/questions/m8tvyzqsxjlhdimwf7jd.png)

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 $15 \mathrm{Nm}^{-2}$ . The area of cross-section at $P$ and $Q$ are $40 \mathrm{~cm}^2$ and $20 \mathrm{~cm}^2$ , respectively. The rate of flow of water through the pipe, in $\mathrm{cm}^3 \mathrm{~s}^{-1}$ , is:

[Take density of water $=1000 \mathrm{~kg} \mathrm{~m}^{-3}$ ]

[Take density of water $=1000 \mathrm{~kg} \mathrm{~m}^{-3}$]

![](https://res.cloudinary.com/ddr1kuyb3/image/upload/v1786187138/questions/m8tvyzqsxjlhdimwf7jd.png)

Accepted Answer

A. 1. Since the fluid is incompressible and flows through a pipe of varying cross-section, the **continuity equation** applies, meaning the volume flow rate stays the same at $P$ and $Q$.
$$
A_P V_P = A_Q V_Q
$$
Given the areas, so
$$
40 V_P = 20 V_Q \implies V_Q = 2V_P
$$

2. Since the pipe is horizontal, **Bernoulli's equation** relates the pressure difference to the change in velocity.
$$
P_P - P_Q = \frac{1}{2}ho(V_Q^2 - V_P^2)
$$
Substituting the given values,
$$
15 = \frac{1}{2}	imes 1000 	imes (4V_P^2 - V_P^2)
$$

3. Solving this equation gives the velocity at $P$.
$$
V_P = \sqrt{\frac{15}{1500}} = 0.1\ \mathrm{m/s}
$$
Therefore
$$
V_Q = 0.2\ \mathrm{m/s}
$$

4. The **rate of flow** through the pipe equals $A_P V_P$.
$$
	ext{Rate} = 40	imes10^{-4}	imes0.1 = 4	imes10^{-4}\ \mathrm{m^3/s}
$$
Converting to $\mathrm{cm^3/s}$,
$$
4	imes10^{-4}	imes10^6 = 400\ \mathrm{cm^3/s}
$$

Hence, the answer is **A. 400**.