Question

A water drop of radius 1 μm falls in a situation where the effect of buoyant force is negligible. Co-efficient of viscosity of air is 1.8 × 10-5 Nsm-2 and its density is negligible as compared to that of water (106 gm-3). Terminal velocity of the water drop is
(Take acceleration due to gravity = 10 ms-2)

• A. 145.4 × 10–6 ms–1
• B. 118.0 × 10–6 ms–1
• C. 132.6 × 10–6 ms–1
• D. 123.4 × 10–6 ms–1

Answer 1

Answer: (D)

123.4 × 10–6 ms–1

Answer 2

The correct answer is (D) : 123.4 × 10–6 ms–1
$6πηrv=mg$
$6πηrv=\frac{4}{3}πr^3ρg$
or
$v=\frac{2}{9}\frac{ρr^2g}{η}$
$=\frac{2}{9}×\frac{10^3×(10^{−6})^2×10}{1.8×10^{−5}}$
= 123.4×10–6 m/s