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 $1.8 \times 10^{–5} Nsm^{–2}$ and its density is negligible as compared to that of water $(10^6 gm^{–3})$. Terminal velocity of the water drop is

• A. $145.4 \times 10^{–6} \text{ms}^{–1}$
• B. $118.0 \times 10^{–6} \text{ms}^{–1}$
• C. $132.6 \times 10^{–6} \text{ms}^{–1}$
• D. $123.4 \times 10^{–6} \text{ms}^{–1}$

Answer 1

Answer: (D)

$123.4 \times 10^{–6} \text{ms}^{–1}$

Answer 2

$6\pi\eta rv=mg$
$6\pi\eta rv= \frac{4}{3} πr3ρg$

or

$v= \frac{2}{9} \text{ρr}^2 \frac{g}{\eta}$

$= \frac{2}{9} \times \frac{10^3×(10^-6)^{-2}×10 }{1.8×10^{-5}}$

$=123.4 \text 10^{–6} m/s$