Problem 46 - Master Resource Book in JEE Main Physics

Question

What is the ratio of surface energy of 1 small drop and 1 large drop if 1000 drops combined to form 1 large drop? (a) \(100: 1\) (b) \(1000: 1\) (c) \(10: 1\) (d) \(1: 100\)

Step-by-Step Solution

Answer

The ratio of surface energy is \(1:100\).

Step 1: Understanding the Problem

We are given that the volume of 1000 small drops is equal to the volume of 1 large drop. We need to find the ratio of surface energy between a small drop and a large drop.

Step 2: Establish Relationship Between Volumes

Let's assume the radius of one small drop is \( r \) and the radius of the large drop is \( R \). Since volume is conserved, the volume of 1000 small drops is equal to the volume of the large drop: \( 1000 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3 \).

Step 3: Simplify the Volume Equation

Simplifying the volume equation from Step 2, we find \( 1000r^3 = R^3 \). Therefore, the radius of the large drop is \( R = 10r \).

Step 4: Calculate Surface Areas

The surface area of the small drop is \( 4\pi r^2 \). The surface area of the large drop is \( 4\pi R^2 = 4\pi (10r)^2 = 400\pi r^2 \).

Step 5: Calculate Surface Energy

If the surface tension of the liquid is \( \sigma \), then the surface energy for a small drop is \( 4\pi r^2 \sigma \) and for the large drop is \( 400\pi r^2 \sigma \).

Step 6: Find the Ratio of Surface Energies

The ratio of the surface energy of a small drop to that of a large drop is \( \frac{4\pi r^2 \sigma}{400\pi r^2 \sigma} = \frac{1}{100} \).

Key Concepts

Understanding Surface Tension

Surface tension is a property of liquids that describes the elastic tendency of a fluid surface. This phenomenon allows for the formation of droplets and is responsible for the cohesive forces at the surface of a liquid, which pull the surface molecules down and towards each other. These forces are a result of the intermolecular attractions, leading to a minimized surface area.
When solving physics problems like the one in the exercise, surface tension is crucial because it determines how energy is distributed across the surface of a liquid.

• Surface tension is measured in force per unit length, typically in Newtons per meter (N/m).
• It is the reason why small droplets can form spherical shapes, minimizing the surface area for a given volume.
• In the context of this problem, understanding that surface tension affects the energy at the surface helps explain why surface energy calculations are relevant in determining energy ratios between different sized droplets.

Principle of Volume Conservation

Volume conservation is a fundamental principle applicable in fluid dynamics. In the context of forming droplets, the total volume of smaller droplets should equal the volume of the larger droplet they form when combined. This conservation of volume is essential in the exercise solution, as it allows for calculating the radii of droplets.
For the problem:

• The volume of a sphere is calculated using the formula: \( V = \frac{4}{3} \pi r^3 \).
• From the solution, the equation for conservation of volume is \( 1000 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3 \).
• By simplifying, the relationship \( 1000r^3 = R^3 \) is derived, which helps determine that the radius of the large drop is ten times that of a small one.

Calculating Sphere Surface Area

The surface area of a sphere is a pivotal concept in understanding surface energy problems. The calculation is fairly straightforward but is essential for applying surface tension principles.

• The formula for the surface area of a sphere is \( A = 4 \pi r^2 \).
• In this problem, we used it to first find the surface area of a single small drop.
• For the large drop, we then determined that \( R = 10r \), leading to its surface area being \( 4 \pi (10r)^2 = 400 \pi r^2 \).
• These calculations reflect how increasing the radius of a sphere affects its surface area, which in turn influences the surface energy, as it is directly proportional to surface tension and surface area.

JEE Physics Problem-Solving Strategy

The Joint Entrance Examination or JEE is known for testing students on their conceptual understanding and problem-solving abilities, especially in physics. Tackling problems like this one effectively involves understanding key concepts, setting up equations accurately, and bridging different areas of physics together.
To solve JEE problems successfully:

• Start by understanding the problem statement clearly and identifying what physical principles are involved, such as conservation laws or equations of motion.
• Break the problem into steps, as demonstrated in the provided solution. This approach helps in systematically addressing each part of the problem.
• Make use of assumptions and approximations only when appropriate, as these can aid in simplifying complex problems.
• Always cross-verify your final answer by analyzing whether it makes physical sense, checking units, and reviewing key equations in the solution.

This structured method ensures that students not only find the correct solution but also deeply understand the underlying physical principles.