Using Stokes' Law for terminal velocity:
vt=9η2r2g(ρs−ρf).
Given: r = 0.1 cm, g = 1000 cm/s², ρs=10.5 g/cm³, ρf=1.5 g/cm³, η = 10 Poise = 10 g/(cm⋅s).
Substituting:
vt=9×102×(0.1)2×1000×(10.5−1.5)
=902×0.01×1000×9
=90180=2.0 cm/s.
JEE Main 2026 — Physics Mechanics
A small metallic sphere of diameter 2 mm and density 10.5 g/cm3 is dropped in glycerine having viscosity 10 Poise and density 1.5 g/cm3 respectively. The terminal velocity attained by the sphere is ____ cm/s. (π=722 and g=10 m/s2)
Held on 23 Jan 2026 · Verified 6 Jul 2026.
3.0
1.0
2.0
1.5
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