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Question:

If r = 5 × 10⁻³ m, ρ = 10³ kg m⁻³, g = 10 m/s², T = 0.11 Nm⁻¹, the radius of the drop when it detaches from the dropper is approximately.

1.4 × 10⁻⁷ m

2.0 × 10⁻⁷ m

4.1 × 10⁻⁷ m

3.3 × 10⁻⁷ m

Solution:

Just before detaching,
force due to surface tension = weight of the liquid in the drop
T(2πr) = (4/3πR³)ρg
substituting the values we get
R = 2.0 × 10⁻⁷ m