A drop of water of radius 0.0015 mm is falling in air. If the coefficient of viscosity of air is 2 × 10–5 kgm–1 s–1, the terminal velocity of the drop will be :- (The density of the water = 103 kgm–3 and g = 10 ms–2

The terminal velocity of a small spherical object falling in a viscous medium is given by Stokes' Law:

v = 2/9 * (r^2) * (ρ - σ) * g / η

where:

• v is the terminal velocity,
• r is the radius of the spherical object,
• ρ is the density of the spherical object,
• σ is the density of the medium,
• g is the acceleration due to gravity, and
• η is the viscosity of the medium.

In this case, we are given:

• r = 0.0015 mm = 0.0015 * 10^-3 m (converted from mm to m),
• ρ = 10^3 kg/m^3 (density of water),
• σ = 1.225 kg/m^3 (density of air at sea level),
• g = 10 m/s^2, and
• η = 2 * 10^-5 kg/m/s.

Substituting these values into the equation, we get:

v = 2/9 * (0.0015 * 10^-3)^2 * (10^3 - 1.225) * 10 / (2 * 10^-5) = 2/9 * 2.25 * 10^-9 * 998.775 * 10 / 2 * 10^-5 = 0.05 m/s.

So, the terminal velocity of the water drop is approximately 0.05 m/s.