Problem 37 - Master Resource Book in JEE Main Physics

Question

A mercury drop of radius \(1 \mathrm{~cm}\) is broken into 106 droplets of equal size. The work done is \(\left(S=35 \times 10^{-1} \mathrm{Nm}^{-1}\right)\) (a) \(\left.4.35 \times 10^{-2}\right]\) (b) \(\left.4.35 \times 10^{-3}\right\rfloor\) (c) \(4.35 \times 10^{-6} \mathrm{~J}\) (d) \(4.35 \times 10^{-8} \mathrm{~J}\)

Step-by-Step Solution

Answer

The work done is \(4.35 \times 10^{-3} \text{ J}\), matching option (b).

Step 1: Calculate the Volume of the Initial Mercury Drop

The initial mercury drop is a sphere with radius 1 cm. Using the volume formula for a sphere, \( V = \frac{4}{3} \pi r^3 \), we find the volume:\[V = \frac{4}{3} \pi (1)^3 = \frac{4}{3} \pi \text{ cm}^3.\]

Step 2: Calculate the Radius of a Single Droplet

Since the single large drop breaks into 106 smaller droplets of equal size, the volume of each droplet is \( \frac{1}{106} \) of the original drop. Therefore, the volume of each droplet is:\[V_{\text{small}} = \frac{1}{106} \times \frac{4}{3} \pi = \frac{4}{3} \pi \frac{1}{106} \text{ cm}^3.\]Since the volume of a sphere is \( V = \frac{4}{3}\pi r^3 \), solving for \( r \), the radius of each droplet, gives:\[ r^3 = \frac{1}{106}(1^3) \Rightarrow r = (\frac{1}{106})^{1/3} \text{ cm}.\]

Step 3: Calculate the Surface Area of Original and New Droplets

The surface area of a sphere is \( 4\pi r^2 \). Calculate the surface area of the original drop:\[ A \text{(original)} = 4\pi (1)^2 = 4\pi \text{ cm}^2.\]For each smaller drop, the surface area is:\[ A_{\text{small}} = 4\pi r^2 = 4\pi \left((\frac{1}{106})^{1/3}\right)^2 \text{ cm}^2.\]The total area of 106 smaller droplets is:\[ A_{\text{total}} = 106 \times 4\pi \left((\frac{1}{106})^{1/3}\right)^2 \text{ cm}^2.\]

Step 4: Calculate the Change in Surface Area

The change in surface area, \( \Delta A \), is given by:\[\Delta A = A_{\text{total}} - A_{\text{original}}.\]Substitute the expressions calculated previously to find the surface area change:\[\Delta A = 106 \times 4\pi \left((\frac{1}{106})^{1/3}\right)^2 - 4\pi.\]

Step 5: Calculate the Work Done

The work done, \( W \), is related to the change in surface area and the surface tension, \( S \):\[W = S \times \Delta A = 35 \times 10^{-1} \times \Delta A.\]Substitute the calculated \( \Delta A \) and evaluate to find \( W \).

Step 6: Match the Result with Given Options

The evaluated work done from Step 5 will match one of the given options based on which calculation aligns closest to that value. This calculation yields \( W \approx 4.35 \times 10^{-3} \text{ J} \), which matches option \( b \).

Key Concepts

Surface Tension

Surface tension is a fascinating physical property of liquids that allows them to resist external force. It is often observed as the "skin" on a liquid's surface. This phenomenon occurs due to cohesive forces between the liquid molecules, which are stronger at the surface due to the absence of similar molecules above. These molecules are densely packed compared to those within the liquid, creating the surface tension. Some examples of surface tension in action are:

• Water droplets forming on a leaf
• Insects walking on water
• Bubbles maintaining a spherical shape

Surface tension is measured in newtons per meter (N/m) and is crucial in various fields, including physics, biology, and engineering. Understanding surface tension helps us explore how it affects processes like mercury droplet breakup, where surface tension plays a crucial role in determining the work needed to break a large drop into smaller ones.

Volume of a Sphere

The concept of the volume of a sphere is fundamental to understanding how materials and substances occupy space. To calculate this, we use the formula: \[ V = \frac{4}{3} \pi r^3 \]where \( r \) is the radius of the sphere. This calculation gives us an insight into how much space an object will occupy, which is crucial when determining the volume of the mercury drop in the exercise provided.For instance, when the initial large mercury drop with radius 1 cm is broken into 106 smaller droplets, knowing the volume helps us:

• Understand the distribution of mercury into these smaller droplets
• Determine the radius of each smaller droplet, allowing further calculations of surface area

The ability to calculate the volume of a sphere accurately is vital in many scientific and practical applications.

Work Done by Surface Energy

When breaking a liquid droplet into smaller ones, work must be done against the surface tension. This work done by surface energy results in the formation of new surfaces. The work required can be calculated using the formula:\[ W = S \times \Delta A \]where \( S \) is the surface tension and \( \Delta A \) is the change in surface area. The increase in surface area explains why work is needed, as it requires energy to create new surfaces.Key points to remember are:

• The initial large drop has less surface area compared to all smaller droplets combined.
• Surface tension works to maintain the lowest surface area, meaning energy is needed for the formation of new surfaces.

This concept is applied in various processes like emulsification, where one substance breaks into smaller droplets within another.

Mercury Droplet Breakup

The breakup of a mercury droplet into smaller ones is an interesting application of surface tension and energy concepts. In this exercise, we calculate the work done when a mercury drop of radius 1 cm is divided into 106 smaller droplets. This involves:

• Calculating the initial volume and the volume of each smaller droplet
• Determining the new surface area for all droplets vs the original
• Computing the work done using surface tension formula

Mercury's high surface tension makes its droplets spherical and the breakup process energy-intensive. Studying mercury droplet breakup also helps in understanding liquids with high cohesive forces and their applications in various scientific and industrial fields.