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 determine the terminal velocity of a water drop falling through the air, we can consider the forces acting on it. As the question hints, we are neglecting buoyant force, so the primary forces at play are the gravitational force and the viscous drag force.

The formula for terminal velocity (\(v_t\)) in a viscous medium is given by Stokes' law:

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

Where:

• \(r\) is the radius of the drop (1 μm = 1 × 10^–6 m)
• \(\rho\) is the density of water (10⁶ gm^–3 = 10³ kg/m³)
• \(g\) is the acceleration due to gravity (10 m/s²)
• \(\eta\) is the coefficient of viscosity (1.8 × 10^–5 Ns/m²)

Step-by-Step Calculation:

1. Convert the given radius to meters: \(r = 1 \times 10^{-6}\) m.
2. Substitute the known values into the terminal velocity formula:

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

1. Calculate the numerator:

\(2 \times (1 \times 10^{-6})^2 = 2 \times 10^{-12}\)

\(2 \times 10^{-12} \times 10^3 \times 10 = 2 \times 10^{-8}\)

1. Calculate the denominator:

\(9 \times 1.8 \times 10^{-5} = 16.2 \times 10^{-5}\)

1. Divide the numerator by the denominator to find \(v_t\):

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

Or, equivalently: \(123.4 \times 10^{-6} \, \text{m/s}\)

Thus, the terminal velocity of the water drop is 123.4 × 10^–6 ms^–1, which means the correct option is:

123.4 × 10^–6 ms^–1