Tag: surface tension

Posted on by — Leave a comment

Worksheet in Bubbles in Surface Tension

A drop of water of diameter 0.2 cm is broken up into 27,000 droplets of equal volumen. How much work will be done against surface tension in the process? (ST. = 7 x 10^-2 N/m)

Example Problem

SaitechAI Worksheet – Surface Tension Work (Droplet Splitting)

SaitechAI Worksheet

Topic: Work done against surface tension when a single drop splits into many droplets (Equal-volume droplets)
10 Questions × 2 marks = 20
Timer
Start: —
End: —
Duration: —
  • Assume spherical drops. Volume conserved: \( n\frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3 \Rightarrow r=\dfrac{R}{n^{1/3}}\).
  • Surface-area change: \( \Delta A = 4\pi R^2\left(n^{1/3}-1\right)\).
  • Work against surface tension: \( W=\sigma\,\Delta A \). Use SI units (R in m).
  • Show steps in the answer box. Toggle Key to verify.
Score: 0/20

Q1. A water drop of diameter 2 mm is broken into 1000 equal droplets. Surface tension = 0.072 N m−1. Find the work done.

Marks:
Key: \(R=1\times10^{-3}\,\text{m},\ n=10^3\Rightarrow n^{1/3}=10\). \(\Delta A=4\pi R^2(10-1)=36\pi\times10^{-6}\,\text{m}^2\). \(W=\sigma\Delta A=0.072\times36\pi\times10^{-6}\approx \boxed{8.143\times10^{-6}\,\text{J}}\).

Q2. A mercury drop of radius 0.8 cm is sprayed into \(10^5\) equal droplets. Surface tension = 0.465 N m−1. Energy required?

Marks:
Key: \(R=8.0\times10^{-3}\,\text{m},\ n^{1/3}\approx46.416\). \(\Delta A=4\pi R^2(n^{1/3}-1)\). \(W\approx \boxed{1.698\times10^{-2}\,\text{J}}\).

Q3. A water drop of radius 0.2 cm splits into 27,000 equal droplets. Surface tension = 0.070 N m−1. Compute the work.

Marks:
Key: \(R=2.0\times10^{-3}\,\text{m},\ n^{1/3}=30\). \(\Delta A=4\pi R^2(29)\). \(W=\sigma\Delta A\approx \boxed{1.020\times10^{-4}\,\text{J}}\).

Q4. A soap solution (σ = 0.025 N m−1) drop of diameter 1 cm is atomized into 8000 equal droplets. Work done?

Marks:
Key: \(R=5.0\times10^{-3}\,\text{m},\ n^{1/3}=20\). \(\Delta A=4\pi R^2(19)\). \(W\approx \boxed{1.492\times10^{-4}\,\text{J}}\).

Q5. A glycerine drop of radius 0.5 cm is broken into \(10^6\) equal droplets. σ = 0.063 N m−1. Energy?

Marks:
Key: \(R=5.0\times10^{-3}\,\text{m},\ n^{1/3}=100\). \(\Delta A=4\pi R^2(99)\). \(W\approx \boxed{1.959\times10^{-3}\,\text{J}}\).

Q6. A water drop of diameter 3 mm splits into 125 equal droplets. σ = 0.072 N m−1. Work done?

Marks:
Posted on by — Leave a comment

Soap bubbles and Surface Tension

Interactive worksheet

SaitechAI | Coalescing Drops – Energy Release Worksheet & Calculator

SaitechAI Worksheet: Energy Released when Two Liquid Drops Coalesce

Interactive calculator + step-by-step derivation with MathJax. Units are SI by default.

Tip: inputs are in mm; calculator converts to meters automatically.
Show step-by-step solution

Theory (for any two spherical drops)

When two spherical drops of radii \(r_1\) and \(r_2\) coalesce to form a single spherical drop of radius \(R\), the total volume is conserved but the surface area decreases. The decrease in surface energy equals the energy released:

\[ \textbf{Volume conservation:}\quad \frac{4}{3}\pi R^3=\frac{4}{3}\pi\left(r_1^3+r_2^3\right)\Rightarrow R=\left(r_1^3+r_2^3\right)^{1/3} \] \[ \textbf{Surface areas:}\quad A_i=4\pi\left(r_1^2+r_2^2\right),\qquad A_f=4\pi R^2 \] \[ \textbf{Energy released:}\quad \Delta E=S\,(A_i-A_f)=4\pi S\left(r_1^2+r_2^2-R^2\right) \]

All radii must be in meters for \(\Delta E\) in joules.

Create More Practice

© SaitechAI

Posted on by — Leave a comment

Coalescence in liquid drops

Dr E. Ramanathan PhD | Class 11 Physics | Surface Tension

This problem is based on coalescience of liquid drops in the Surface Tension topic of Class 11 Physics.

Concept Explanation — Energy Released on Coalescence of Liquid Drops

When two small liquid drops merge (coalesce) to form a single larger drop, the total surface area decreases because the surface-to-volume ratio reduces. Since surface energy is directly proportional to surface area, this reduction leads to a release of surface energy in the form of heat or kinetic energy.

4. Interpretation

  • The merged drop has less total surface area.
  • The difference in surface energy (due to reduced surface area) appears as released energy.
  • Hence, energy released = decrease in surface area × surface tension.

Key Insight

Smaller drops have higher surface area per unit volume, meaning higher surface energy.
When they combine into a larger drop, the surface area reduces → surface energy decreases → energy is released.

Interactive Worksheet – Click here