Physics · Class 11

Drop a sphere through viscous liquid

Weight, buoyancy and viscous drag determine a falling sphere’s speed. As speed grows, drag grows until the net force vanishes and the sphere reaches terminal speed. Viscosity lowers this limiting speed.

Open the interactive experiment →

Drop a sphere through viscous liquid

விழும் கோளின் வேகம் அதிகரிக்கும்போது பாகுநிலை எதிர்ப்பு அதிகரிக்கும். நிகர விசை பூஜ்யமானால் முனைய வேகம் கிடைக்கும்.

Key notes, relationships and model limits

Key relationship

vt=2r²(ρsphere−ρfluid)g/(9η); v=vt(1−e^(−t/τ)); τ=2ρspherer²/(9η).

Remember

Common trap

Terminal speed means zero acceleration, not zero force on each individual interaction.

Model limits

Small isolated sphere in creeping flow with Stokes drag 6πηrv; buoyancy included; no wall or added-mass correction. Check Reynolds number ≪1 before using the result quantitatively.

What you will learn

Relate viscous drag to terminal speed and check the creeping-flow condition.

Try the experiment

Try both scenarios, change one control, and predict the result before checking the graph and readouts.

Think of it this way

Change one input while keeping the others fixed, as you would in a controlled experiment.

Worked example

For r=0.5 mm, density difference=6800 kg/m³ and η=1 Pa·s, vt=0.003706 m/s; the Reynolds number is about 0.0037.

Why this matters

The model connects drop a sphere through viscous liquid with an observable reading. Check the equation and assumptions before applying it.

Before you start

SI units · algebra · graph reading

Check your understanding

In the Stokes regime, doubling radius changes terminal speed by what factor?

A factor of two
A factor of four
A factor of one half

Read the explanation

Terminal speed is proportional to r² when the densities and viscosity are fixed.

Course connection: Class XI Chapter 9. Consult the roadmap for CBSE, NEET, JEE Main and Advanced scope.