Problem 41 - Master Resource Book in JEE Main Physics

Question

A drop of liquid of diameter \(2.8 \mathrm{~mm}\) breaks up into 125 identical drops. The change in energy is nearly \(\left(S=75\right.\) dyne \(\left.\mathrm{cm}^{-1}\right)\) (a) zero (b) \(19 \mathrm{erg}\) (c) 46 erg (d) \(74 \mathrm{erg}\)

Step-by-Step Solution

Answer

The change in energy is nearly 46 erg.

Step 1: Calculate Original Volume of the Drop

The volume of a sphere is given by the formula \(V = \frac{4}{3} \pi r^3\). The original drop has a diameter of \(2.8 \text{ mm}\), so the radius is \(1.4 \text{ mm}\). So, the volume is \(V_{original} = \frac{4}{3} \pi (1.4)^3\).

Step 2: Calculate Volume of Each Smaller Drop

The drop breaks into 125 identical smaller drops. Each of these smaller drops has a total volume equal to the original drop divided by 125. So, \(V_{small} = \frac{V_{original}}{125}\).

Step 3: Calculate Radius of Smaller Drops

For each smaller drop, use the formula for the volume of a sphere to find the radius: \(r_{small} = \left( \frac{3V_{small}}{4\pi} \right)^{1/3}\).

Step 4: Calculate Surface Area of Original and Smaller Drops

The surface area of a sphere is given by \(A = 4\pi r^2\). First, calculate the surface area of the original drop \(A_{original}\), and then calculate the total surface area of all smaller drops \(A_{small} = 125 \times 4\pi r_{small}^2\).

Step 5: Calculate Change in Surface Energy

The change in surface energy is given by \(\Delta E = S \times (A_{small} - A_{original})\). Calculate this change using the provided surface tension \(S = 75 \text{ dyne/cm}\).

Step 6: Unit Conversion

Convert the energy from dyne-cm to erg by noting that \(1 \text{ dyne cm} = 1 \text{ erg}\). No conversion is necessary because they are the same.

Step 7: Conclusion

Choose the closest value that matches the calculated change in energy.

Key Concepts

Surface Tension

Surface tension is a fascinating property of liquids where the surface behaves like a stretched elastic membrane. This force is what allows small insects to walk on water or why water beads up on surfaces. It's all about the cohesive forces between liquid molecules that are stronger at the surface. In this exercise, the surface tension of the liquid is given as 75 dyne/cm, reflecting its ability to resist an external force. This characteristic is crucial when calculating the change in energy as a liquid splinters into smaller droplets. Surface tension plays a key role in determining the amount of energy required to change a liquid's shape. This energy is proportional to the change in the surface area when droplets combine or break apart. In simpler terms, when the drop of liquid breaks into smaller drops, the surface area increases, which involves energy changes influenced by the liquid's surface tension.

Volume and Surface Area Calculations

Understanding the relationship between volume and surface area is vital in physics exercises dealing with liquids. In this case, the original liquid drop is spherical with a given diameter of 2.8 mm, allowing us to find its volume using the sphere formula:

• Volume of a Sphere: \( V = \frac{4}{3} \pi r^3 \)

where \(r\) is the radius (half the diameter).
For the original drop, \(r = 1.4 \text{ mm}\), and you can compute its volume. When the drop divides into 125 smaller ones, each new drop's volume is one 125th that of the initial.
The more challenging task is calculating each new drop's radius from its volume:

• Radius from Volume: \( r_{small} = \left( \frac{3V_{small}}{4\pi} \right)^{1/3} \)

Next, we need to work out the surface areas.

• Surface Area of a Sphere: \( A = 4\pi r^2 \)

By comparing the original drop's surface area with the total surface area of the smaller drops, we gain insights into how surface area influences energy across physical transformations.

Energy Conversion in Physics

Energy conversion in physics is about transforming energy from one form to another. In our scenario, it's specifically about surface energy change with a physical transformation: a liquid drop fragmenting into smaller parts. This process affects energy because increased surface areas due to the breakup define how surface energy varies.
The change in surface energy when converting from a large drop to many smaller ones can be determined using the formula:

• Change in Surface Energy: \( \Delta E = S \times (A_{small} - A_{original}) \)

In simpler terms, this equation captures how much the new surface area exceeds the old one, multiplied by the surface tension \(S\).
Once you calculate the surface energies for both the large drop and the collection of smaller drops, you subtract the original from the new total. The resulting energy difference reflects how much energy the system must gain or lose due to this transformation.
When considering energy units, it's helpful to know that here, dyne-cm equals erg, simplifying calculations since no conversion factors are needed. Understanding energy conversion in this context emphasizes the principles of conservation of energy while demonstrating transformations affecting the energetic state of physical systems.