Q.An ideal fluid flows through a pipe of circular cross-section made of two sections with diameters 2.5 cm and 3.75 cm. The ratio of the velocities in the two pipes is

The key idea here is the Equation of Continuity, which states that for an incompressible, non-viscous fluid in steady flow through a pipe, the volume flow rate remains constant.

Here, $A$ is the cross-sectional area and $v$ is the fluid velocity.

For a circular cross-section, the area $A = \pi (D/2)^2 = \frac{\pi D^2}{4}$, where $D$ is the diameter.

Substituting this into the Equation of Continuity:

$\frac{\pi D_1^2}{4} v_1 = \frac{\pi D_2^2}{4} v_2$

$D_1^2 v_1 = D_2^2 v_2$ …

When an ideal fluid flows through a pipe, its mass flow rate must remain constant at every cross-section, assuming no fluid is added or removed along the pipe. An ideal fluid is considered incompressible, meaning its density ($\rho$) does not change. Therefore, if the mass flow rate ($\dot{m} = \rho A v$) is constant and density is constant, the volume flow rate ($Q = A v$) must also be constant. This principle is known as the Equation of Continuity.

Here, $A$ is the cross-sectional area of the pipe and $v$ is the average velocity of the fluid across that cross-section. If the pipe narrows, the fluid must speed up to maintain the same volume flow rate. If the pipe widens, the fluid slows down.

1. Identify the given information and the goal.

   We are given the diameters of two sections of a pipe:

   $d_1 = 2.5 \text{ cm}$

   $d_2 = 3.75 \text{ cm}$

   We need to find the ratio of the velocities, $v_1 : v_2$.

2. Apply the Equation of Continuity.

   For an ideal fluid flowing through a pipe, the volume flow rate ($Q$) is constant.

   $A_1 v_1 = A_2 v_2$

   Here, $A_1$ and $A_2$ are the cross-sectional areas of the two pipe sections, and $v_1$ and $v_2$ are the fluid velocities in those sections, respectively.

3. Express the cross-sectional area in terms of diameter.

   The pipe has a circular cross-section. The area of a circle is given by $A = \pi r^2$, where $r$ is the radius. Since the diameter $d = 2r$, we have $r = d/2$.

   Substituting this into the area formula:

   $A = \pi \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{4}$

4. Substitute the area expressions into the Equation of Continuity.

   Using the formula for area, we can write:

   $\left(\frac{\pi d_1^2}{4}\right) v_1 = \left(\frac{\pi d_2^2}{4}\right) v_2$

5. Simplify the equation and find the ratio of velocities.

   We can cancel out the common term $\frac{\pi}{4}$ from both sides:

   $d_1^2 v_1 = d_2^2 v_2$

   Now, rearrange this equation to find the ratio $v_1 : v_2$: …

Continuity: A1v1=A2v2, A=piD^2/4. D1^2v1=D2^2*v2 => v1/v2=(D2/D1)^2=(3.75/2.5)^2=(1.5)^2=2.2 …