Problem 994 - JEE Physics

Question

A big drop of radius \(R\) is formed by 1000 small droplets of coater then the radius of small drop is (A) \((\mathrm{R} / 2)\) (B) \((\mathrm{R} / 5)\) (C) \((\mathrm{R} / 6)\) (D) \((\mathrm{R} / 10)\)

Step-by-Step Solution

Answer

The radius of the small droplet is \(\frac{R}{10}\). So, the correct option is (D).

Step 1: Write the volume relationship for the big drop and small droplets

Since the big drop is formed by combining 1000 small droplets, the total volume of the small droplets equals the volume of the big drop. The volume of a sphere is given by \(V = \frac{4}{3} \pi r^3\), where V is the volume and r is the radius. Let the radius of the small droplet be r. Then we have: Volume of big drop = 1000 * Volume of small droplet \(\frac{4}{3}\pi R^3 = 1000 * \frac{4}{3} \pi r^3\)

Step 2: Solve for the radius of the small droplet

We can now solve for the radius of the small droplet by dividing both sides of the equation by \(\frac{4}{3}\pi\). \(R^3 = 1000 * r^3\) Now take the cube root of both sides. \(R = 10 * r\) Solve for r. \(r = \frac{R}{10}\) The radius of the small droplet is \(\frac{R}{10}\). So, the correct option is (D).

Key Concepts

Relationships of Radii

Understanding how radii relate in spheres is crucial, especially when you are dealing with volumes. For a sphere, the volume can be defined by the formula \(V = \frac{4}{3} \pi r^3\), where \(r\) is the radius. This formula helps us establish a relationship between the volume of larger spheres and smaller ones.

In the provided exercise, you have a large sphere formed by multiple smaller spheres. Each of these smaller spheres is a "droplet." Therefore, if you know the volume of the larger sphere, you can calculate the volume – and thus the radius – of the smaller spheres by reversing or scaling down the formula.

So, if 1000 small droplets make up the large droplet, then their combined volume equals the volume of the large sphere. This relationship is foundational for solving the problem and answering questions about individual droplet sizes.

Cube Root Calculation

Finding the cube root is a key step in many geometry-related exercises, especially when dealing with sphere volumes. When we are told that the volume of the big droplet equals 1000 times the volume of a small droplet, and each volume is given by \(\frac{4}{3} \pi r^3\), solving for \(r\) involves dividing the volumes and then finding the cube root.

The equation becomes \(R^3 = 1000 \times r^3\). To isolate \(r\), you take the cube root of both sides of the equation. This step simplifies the equation to \(R = 10 \times r\). The cube root of 1000 is 10 because multiplying 10 by itself three times gives 1000.

This easy cube root calculation is what ultimately gives us the proportion \(r = \frac{R}{10}\), showing that the radius of each small droplet is one-tenth of the large droplet's radius.

Volume Conservation

Volume conservation is a concept where the total volume remains constant even as it changes shape. In our exercise, volume conservation is applied to spherical droplets.

When thousand small droplets come together to form one big droplet, the principle of volume conservation informs us that these small droplets' combined volume is exactly the same as the big droplet's volume.

Using the formula for the volume of a sphere \(\frac{4}{3} \pi r^3\), you can see how combining the exact total volume of smaller spheres results in the volume of a larger sphere, and how by maintaining this volume, we maintain a certain balance or proportion with respect to the spheres involved in the transformation. This whole concept is important for understanding not just this problem but other real-world phenomena where volume conservation might apply.