Problem 1002 - JEE Physics

Question

A spherical drop of coater has radius \(1 \mathrm{~mm}\) if surface tension of contex is \(70 \times 10^{-3} \mathrm{~N} / \mathrm{m}\) difference of pressures between inside and outside of the spherical drop is (A) \(35 \mathrm{~N} / \mathrm{m}^{2}\) (B) \(70 \mathrm{~N} / \mathrm{m}^{2}\) (C) \(140 \mathrm{~N} / \mathrm{m}^{2}\) (D) zero

Step-by-Step Solution

Answer

The pressure difference between inside and outside of the spherical drop is \(140\) N/\(m^2\), which corresponds to option (C).

Step 1: Identify the given values

We are given the following values in the problem: - Radius of the spherical drop (r) = 1 mm = \(1 \times 10^{-3}\) m (converted to meters) - Surface tension of coater (T) = \(70 \times 10^{-3}\) N/m

Step 2: Apply the formula for pressure difference

The formula for pressure difference (ΔP) between the inside and outside of a spherical drop due to surface tension is given by: ΔP = \(\frac{2T}{r}\) Now we can plug in the given values for surface tension and radius: ΔP = \(\frac{2 (70 \times 10^{-3})}{1 \times 10^{-3}}\)

Step 3: Calculate the pressure difference

Simplify the expression: ΔP = \(\frac{140}{1}\) N/\(m^2\) ΔP = \(140\) N/\(m^2\) So, the pressure difference between inside and outside of the spherical drop is \(140\) N/\(m^2\), which corresponds to option (C).

Key Concepts

Understanding Spherical Drops

A spherical drop is a perfect little ball of liquid held together by the forces of surface tension. Think of how small drops of water bead up on a leaf or countertop. This spherical shape minimizes the surface area for a given volume, which is an energy-efficient configuration for the drop. Surface tension acts along the surface of the liquid, pulling it tightly and causing the formation of a sphere.

Here are a few things to keep in mind about spherical drops:

• The shape allows for minimal surface tension energy.
• Gravity and air resistance can also affect a drop's shape, but in very tiny droplets, these influences are negligible compared to surface tension.
• Spherical drops can be observed in various scenarios, including raindrops, dew, and small liquid beads.

Understanding spherical drops offers insight into the fascinating characteristics of liquids and how they interact with their surroundings.

Pressure Difference Explained

Pressure difference in a spherical drop is an interesting topic primarily governed by the principles of fluid mechanics and surface tension. Inside any drop, the fluid molecules interact differently than those on the surface because of surface tension. This produces a difference in pressure inside compared to the outside atmosphere.

To calculate this pressure difference, the crucial formula is:\[ΔP = \frac{2T}{r}\]where:

• \(ΔP\) is the pressure difference across the surface.
• \(T\) is the surface tension of the liquid.
• \(r\) is the radius of the spherical drop.

Using this formula, you can find how much more pressure is inside the drop than outside. This is not only significant in understanding physical phenomena but also has real-world applications in fields like medicine and environmental science.

The formula tells us that pressure difference is directly proportional to surface tension and inversely proportional to the radius, leading to higher pressure in smaller droplets if the surface tension remains constant.

Relevance to JEE Physics

The concepts of surface tension and pressure difference in spherical drops are vital components of the JEE Physics syllabus. This topic tests the understanding of theoretical principles and their mathematical applications, which are crucial for exams like JEE.

Students preparing for JEE exams should focus on:

• Understanding the formula \(ΔP = \frac{2T}{r}\) and its derivation.
• Familiarity with unit conversions, as seen with converting radius units from millimeters to meters.
• The ability to solve problems related to spherical drops and surface tension by applying relevant formulas.

Gaining proficiency in this topic provides a gateway to mastering broader concepts in physical chemistry and advanced fluid dynamics. Additionally, developing a strong grasp of pressure-related equations will be beneficial when tackling questions on energy conservation, mechanics of fluids, and other related areas in physics.