Problem 40 - Master Resource Book in JEE Main Physics

Question

Work done in splotting a drop of water of \(1 \mathrm{~mm}\) radius into \(10^{6}\) droplets is (surface tension of water \(\left.72 \times 10^{-3} \mathrm{~J} / \mathrm{m}^{2}\right)\) (a) \(\left.9.8 \times 10^{-5}\right\rfloor\) (b) \(895 \times 10^{-5} \mathrm{~J}\) (c) \(5.89 \times 10^{-5} \mathrm{~J}\) (d) \(5.98 \times 10^{-6} \mathrm{~J}\)

Step-by-Step Solution

Answer

Correct answer is (c) 5.98 \times 10^{-5} \text{ J}.

Step 1: Understand the Problem Statement

You need to calculate the work done in splitting a single drop of water with a given radius into a specified number of smaller droplets. The relevant physical parameter here is the surface tension of water, which is given as \(72 \times 10^{-3} \text{ J/m}^2\).

Step 2: Calculate the Initial Surface Area

Start by determining the surface area of the original water drop. Given its radius \(R = 1 \text{ mm} = 0.001 \text{ m}\), use the formula for the surface area of a sphere: \(A_1 = 4\pi R^2\). Plug in the radius to get:\[ A_1 = 4\pi (0.001)^2 = 4\pi \times 10^{-6} \text{ m}^2 \]

Step 3: Calculate the Total Volume of the Drop

Compute the volume of the original drop using the volume formula for a sphere: \(V = \frac{4}{3}\pi R^3\). Substitute the given radius:\[ V = \frac{4}{3}\pi (0.001)^3 = \frac{4}{3}\pi \times 10^{-9} \text{ m}^3 \]

Step 4: Calculate the Radius of Each Smaller Droplet

Since the total volume remains constant during splitting, each of the \(10^6\) droplets has a smaller radius \(r\). Given that the total volume is preserved, the volume of a single small droplet is:\[ V_{small} = \frac{V}{10^6} = \frac{4\pi \times 10^{-9}}{3 \times 10^6} \text{ m}^3 \]Equating this to \(\frac{4}{3}\pi r^3\):\[ \frac{4}{3}\pi r^3 = \frac{4 \pi \times 10^{-9}}{3 \times 10^6} \]Solving this gives:\[ r = \left(\frac{10^{-9}}{10^6}\right)^{1/3} = 10^{-5} \text{ m} \]

Step 5: Calculate the Surface Area of Each Droplet

Now, calculate the surface area of one small droplet using \(r = 10^{-5} \text{ m}\):\[ A_{small} = 4\pi (10^{-5})^2 = 4\pi \times 10^{-10} \text{ m}^2 \]

Step 6: Calculate the Total New Surface Area

Multiply the surface area of one small droplet by the number of droplets:\[ A_2 = 4\pi \times 10^{-10} \times 10^6 = 4\pi \times 10^{-4} \text{ m}^2 \]

Step 7: Calculate the Increase in Surface Area

Subtract the initial surface area from the new total surface area:\[ \Delta A = A_2 - A_1 = 4\pi \times 10^{-4} - 4\pi \times 10^{-6} = 3.96\pi \times 10^{-4} \text{ m}^2 \]

Step 8: Calculate Work Done Using Surface Tension

Work done is given by the increase in surface area multiplied by the surface tension \(\gamma\). Therefore:\[ W = \gamma \Delta A = 72 \times 10^{-3} \times 3.96\pi \times 10^{-4} \]Upon calculating the numerical value, this is:\[ W \approx 5.98 \times 10^{-5} \text{ J} \]

Step 9: Conclusion

After calculating the work done, the answer matches option \((c) 5.98 \times 10^{-5} \text{ J}\).

Key Concepts

Surface Tension

Surface tension is a fascinating physical phenomenon that occurs at the interface of a liquid and a gas, such as the surface of water in contact with air. It refers to the cohesive forces between liquid molecules at the surface, which draw the surface molecules together. This creates a sort of 'film' that can make the surface behave like an elastic sheet. A common everyday example is the way small insects can walk on water without sinking.

In physics problems like the one we're discussing, surface tension is quantified as force per unit length, or energy per unit area—showing how much work is needed to increase the surface area of a liquid. This energy cost is what must be accounted for when splitting a large water droplet into many smaller ones, as more smaller droplets mean more surface area and therefore, more energy is required.

Sphere Volume and Surface Area

Calculating the volume and surface area of spheres is a crucial step in many physics problems, including the one involving droplet splitting. For a sphere, the formulas are:

• Volume (\( V \) ): given by \( V = \frac{4}{3}\pi r^3 \)
• Surface area (\( A \) ): given by \( A = 4\pi r^2 \)

These expressions help determine the amount of water in a droplet and how much surface area it possesses. When dealing with many droplets, like in our problem, these formulas are essential; knowing the radius allows us to compute both the initial surface area and the total for all resulting droplets.

Understanding these formulas enables students to comprehend how changing a sphere's size impacts these two critical properties, which is pivotal when analyzing problems involving merging or splitting droplets.

Physics Problem Solving Steps

Solving physics problems effectively involves a structured approach. Let’s break down the process that was used in the droplet splitting problem:

• Begin by understanding the problem statement and identifying all relevant information and physical parameters, such as surface tension.
• Determine necessary calculations. For example, calculate the initial and new surface areas using known geometric formulas.
• Use physical principles, like the conservation of volume in this case, to derive relationships between quantities, such as the radii of the new droplets.
• Calculate any changes or differences resulting from these relationships, such as the change in surface area that influences work done.

These steps help guide through the complexities of a problem while ensuring all necessary components are addressed, leading to a logical and accurate solution.

Droplet Splitting

Droplet splitting involves dividing a larger body of water into several smaller drops. This process is significant in understanding phenomena in nature and technology, ranging from the formation of raindrops to inkjet printing.

In our exercise, the challenge was to solve the physics behind splitting a single large water droplet into several million smaller ones. The primary focus here is on the conservation of volume, which dictates that the total volume of the smaller droplets equals the original droplet's volume. Once this principle is applied, calculating new parameters like the radius of smaller droplets becomes manageable.

An interesting outcome of this process is the increase in total surface area, leading to more work done due to surface tension. This insight moves students beyond mere number-crunching and invites them to appreciate the broader implications and applications of physics theories.