Category: Physics Concept Cards

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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
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  • 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:
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Concave and Convex Mirrors

Worksheet

SaitechAI — Objective Questions (Focal Length & Radius, f = R/2)

SaitechAI — Objective Type Questions

Topic: Relation between focal length and radius of curvature, \( f=\dfrac{R}{2} \). Assume magnitudes (ignore sign convention unless asked); write units.

Questions

  1. A spherical mirror has \( R=40\cm \). Find \( f \).
  2. Given \( R=1.2\m \). Find \( f \).
  3. The focal length is \( f=15\cm \). Find \( R \).
  4. For \( R=80\cm \), compute \( f \).
  5. If \( f=0.75\m \), compute \( R \).
  6. For \( R=24\cm \), compute \( f \).
  7. Given \( f=25\cm \), compute \( R \).
  8. State the relation between \( f \) and \( R \) for a spherical mirror (paraxial).
  9. If radius of curvature doubles, how does focal length (magnitude) change?
  10. For \( R=100\cm \), compute \( f \).
  11. If \( f=2\m \), compute \( R \).
  12. For \( R=30\cm \), compute \( f \).

Answer Key

  1. \(20\cm\)
  2. \(0.6\m\)
  3. \(30\cm\)
  4. \(40\cm\)
  5. \(1.5\m\)
  6. \(12\cm\)
  7. \(50\cm\)
  8. \( f=\dfrac{R}{2} \) (equivalently \( R=2f \))
  9. It doubles.
  10. \(50\cm\)
  11. \(4\m\)
  12. \(15\cm\)

Tip: Always keep units consistent; convert \( \mathrm{cm} \leftrightarrow \mathrm{m} \) when needed.

© SaitechAI • Physics — Objective Questions