An ideal fluid flows (laminar flow) through a pipe of non-uniform diameter. The maximum and minimum diameters of the pipes are $6.4\, cm$ and $4.8\, cm$, respectively. The ratio of the minimum and the maximum velocities of fluid in this pipe is :

Question

Given below are two statements:
Statement I: Pressure of a fluid is exerted only on a solid surface in contact as the fluid-pressure does not exist everywhere in a still fluid.
Statement II: Excess potential energy of the molecules on the surface of a liquid, when compared to interior, results in surface tension.
In the light of the above statements, choose the correct answer from the options given below

Answer 1

To solve this problem, we need to apply the principle of continuity for fluid flow. The principle of continuity states that for an incompressible fluid flowing through a pipe of varying diameter, the product of the cross-sectional area and the velocity of the fluid at any point in the pipe is constant. Mathematically, this is represented as:

A_1V_1 = A_2V_2

where A_1 and A_2 are the cross-sectional areas at points 1 and 2, and V_1 and V_2 are the velocities at those points.

The areas A_1 and A_2 can be calculated using the formula for the area of a circle:

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

Given:

Maximum diameter, d_1 = 6.4\ \text{cm}
Minimum diameter, d_2 = 4.8\ \text{cm}

First, calculate A_1 and A_2:

A_1 = \pi \left(\frac{6.4}{2}\right)^2 = \pi \cdot 3.2^2 = 10.24\pi\ \text{cm}^2
A_2 = \pi \left(\frac{4.8}{2}\right)^2 = \pi \cdot 2.4^2 = 5.76\pi\ \text{cm}^2

Using the continuity equation A_1V_1 = A_2V_2, we can express the velocity ratio \frac{V_2}{V_1} as:

\frac{V_2}{V_1} = \frac{A_1}{A_2} = \frac{10.24\pi}{5.76\pi}

Simplifying gives:

\frac{V_2}{V_1} = \frac{10.24}{5.76} = \frac{64}{36} = \frac{16}{9}

Therefore, the inverse ratio, or the ratio of the minimum to the maximum velocity, is:

\frac{V_1}{V_2} = \frac{9}{16}

Thus, the correct answer is \frac{9}{16}, which matches the given correct answer.

Answer 2