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

The Correct Option is D

To find the terminal velocity of the water drop, we can use Stokes' Law. According to Stokes' Law, the terminal velocity \(v_t\) of a spherical object moving under the influence of gravity through a fluid is given by:

\(v_t = \frac{{2r^2g(\rho - \sigma)}}{{9\eta}}\) 

where:

• \(r\) is the radius of the drop.
• \(g\) is the acceleration due to gravity.
• \(\rho\) is the density of the drop (water).
• \(\sigma\) is the density of the fluid (air) - negligible in this case.
• \(\eta\) is the coefficient of viscosity of the fluid (air).

Given data:

• \(r = 1 \, \mu m = 1 \times 10^{-6} \, m\)
• \(g = 10 \, \text{m/s}^2\)
• \(\rho = 10^6 \, \text{gm/m}^3 = 10^3 \, \text{kg/m}^3\)
• \(\eta = 1.8 \times 10^{-5} \, \text{Ns/m}^2\)
• \(\sigma = 0 \, \text{kg/m}^3\) (negligible)

Substituting these values into the formula:

\(v_t = \frac{{2 \times (1 \times 10^{-6})^2 \times 10 \times (10^3 - 0)}}{{9 \times 1.8 \times 10^{-5}}}\)

Simplifying further:

\(v_t = \frac{{2 \times 10^{-12} \times 10^4}}{{9 \times 1.8 \times 10^{-5}}}\)

\(v_t = \frac{{2 \times 10^{-8}}}{{16.2 \times 10^{-5}}}\)

\(v_t = \frac{{2 \times 10^{-8}}}{{16.2 \times 10^{-5}}} \, \approx 1.234 \times 10^{-4} \, \text{m/s}\)

Converting it to the format given in the options:

\(v_t = 123.4 \times 10^{-6} \, \text{ms}^{-1}\)

Therefore, the correct answer is 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