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

The Correct Option is D

 To find the terminal velocity of a water drop of radius \(1 \, \mu m\) falling under the influence of a viscous force in a medium where the buoyant force can be neglected, we use Stokes' Law.

Stokes' Law for the drag force (\(F_d\)) experienced by a spherical object moving through a viscous fluid is given by: 

\[F_d = 6 \pi \eta r v\]

where:

• \(\eta = 1.8 \times 10^{-5} \, \text{Ns/m}^2\) is the coefficient of viscosity of air.
• \(r = 1 \, \mu m = 1 \times 10^{-6} \, \text{m}\) is the radius of the drop.
• \(v\) is the terminal velocity.

 

At terminal velocity, the gravitational force acting on the drop is balanced by the drag force: 

\[F_g = F_d\]

The gravitational force (\(F_g\)) on the water drop is given by: 

\[F_g = \frac{4}{3} \pi r^3 \rho g\]

where:

• \(\rho = 10^6 \, \text{g/m}^3 = 10^3 \, \text{kg/m}^3\) is the density of water.
• \(g = 9.8 \, \text{m/s}^2\) is the acceleration due to gravity.

 

Equating \(F_g = F_d\) gives: 

\[\frac{4}{3} \pi r^3 \rho g = 6 \pi \eta r v\]

Solving for \(v\), we get: 

\[v = \frac{2 r^2 \rho g}{9 \eta}\]

Substituting the given values: 

\[v = \frac{2 \times (1 \times 10^{-6})^2 \times 10^3 \times 9.8}{9 \times 1.8 \times 10^{-5}}\]

Calculating the above expression: 

\[v = \frac{2 \times 1 \times 10^{-12} \times 10^3 \times 9.8}{1.62 \times 10^{-4}}\]\[v = \frac{19.6 \times 10^{-9}}{1.62 \times 10^{-4}}\]\[v = \frac{19.6}{1.62} \times 10^{-5}\]\[v \approx 12.34 \times 10^{-5} \, \text{m/s}\]\[v \approx 123.4 \times 10^{-6} \, \text{ms}^{-1}\]

Therefore, the terminal velocity of the water drop is \(123.4 \times 10^{-6} \, \text{ms}^{-1}\), which is the correct answer.