Problem 44 - Master Resource Book in JEE Main Physics

Question

What change in surface energy will be noticed when a drop of radius \(R\) splits up inte 1000 droplets of radius \(\boldsymbol{r}\), surface tension \(\boldsymbol{T}\) ? (a) \(4 \pi R^{2} T\) (b) \(7 \pi R^{2} T\) (c) \(16 \pi R^{2} T\) (d) \(36 \pi R^{2} T\)

Step-by-Step Solution

Answer

The surface energy change is \(36\pi R^2 T\), so the answer is (d).

Step 1: Determine Change in Surface Area

The surface area of a sphere is given by the formula \( A = 4\pi r^2 \). Initially, we have one large drop with radius \(R\), so its surface area is \(4\pi R^2\). After splitting, we have 1000 smaller droplets each with radius \(r\), so the total surface area becomes \(1000 \times 4\pi r^2\).

Step 2: Relate Volumes Before and After

The volume remains constant before and after the split. The volume of one drop with radius \(R\) is \(\frac{4}{3}\pi R^3\). The volume of one smaller droplet with radius \(r\) is \(\frac{4}{3}\pi r^3\). Since there are 1000 droplets, the total volume is \(1000 \times \frac{4}{3}\pi r^3\). By setting these equal, we get \(\frac{4}{3}\pi R^3 = 1000 \times \frac{4}{3}\pi r^3\).

Step 3: Solve for Smaller Radius \(r\)

Cancel the common terms and solve for \(r\): \(R^3 = 1000r^3\). Taking the cube root of both sides gives \(r = \frac{R}{10}\).

Step 4: Calculate the Initial and Final Surface Areas

Initially, the surface area was \(4\pi R^2\). The new total surface area after the droplets are formed is \(1000 \times 4\pi r^2 = 1000 \times 4\pi \left( \frac{R}{10} \right)^2 = 1000 \times 4\pi \times \frac{R^2}{100} = 40\pi R^2\).

Step 5: Calculate Change in Surface Area

The change in surface area is the new surface area minus the initial surface area: \(40\pi R^2 - 4\pi R^2 = 36\pi R^2\).

Step 6: Calculate Change in Surface Energy

Surface energy is given by the product of surface area and surface tension \(T\). Thus, the change in surface energy is \(36\pi R^2 T\).

Key Concepts

Understanding Surface Area

The concept of surface area is crucial when dealing with phenomena such as the splitting of droplets. Surface area refers to the total area that the surface of a solid object occupies. For spheres, the formula used is

• The initial surface area of a single larger droplet is calculated with the formula: \(A = 4\pi R^2\), where \(R\) is the radius of the sphere.
• When the droplet divides into smaller ones, say 1000 droplets, each with radius \(r\), the total new surface area becomes: \(1000 \times 4\pi r^2\).
• This formula helps in understanding how surface area changes when objects are resized or split.

These changes in surface area relate directly to changes in physical properties like surface energy. Thus, grasping the calculations of surface area is key to solving such problems.

The Role of Surface Tension

Surface tension is a critical concept in understanding how liquids behave, especially in phenomena like droplets maintaining shape. It refers to the cohesive force at the surface of a liquid that makes it behave like a stretched elastic membrane.

• The measure of surface tension is crucial as it determines how much energy is stored at the surface.
• In our problem, the surface energy change is calculated by multiplying the change in surface area by the surface tension \(T\).
• The equation reflects this calculation: \(36\pi R^2 T\), illustrating the importance of both surface area and surface tension in energy calculations.

Imagine a stretched balloon. The air pressure inside causes it to expand, similar to how surface tension keeps a droplet intact. Understanding surface tension helps one predict not just shape but also energy changes during transformations like droplet division.

Comprehending Sphere Volume

Spheres are fascinating shapes, especially in geometry, and play a substantial role in applications like droplet calculations. The volume of a sphere is calculated using the formula:

• \(V = \frac{4}{3}\pi R^3\), where \(R\) is the sphere's radius.
• When a large droplet splits into smaller ones, the total volume remains the same.
• For our problem, initially, the single droplet had a volume of \(\frac{4}{3}\pi R^3\). After splitting, the total volume of 1000 droplets is \(1000 \times \frac{4}{3}\pi r^3\).
• Volume conservation allows us to equate these and find that \(r = \frac{R}{10}\), illustrating how volume ensures the total mass or substance quantity remains unchanged despite shape transformations.

Volume is a fundamental concept when dissecting geometric problems, especially in maintaining consistency in measurements and predictions of physical phenomena.