Chapter: Mechanical Properties of Fluids· Physics

A spherical drop of oil falls at a constant speed of 9.8 cm/s in steady air. Calculate the radius of the drop. The density of oil is 0.9013 g/cm3, density of air is 0.0013 g/cm3 and the coefficient of viscosity of air is 1.8 × 10–4 poise.

Q.22 A spherical drop of oil falls at a constant speed of 9.8 cm/s in steady air. Calculate the radius of the drop. The density of oil is 0.9013 g/cm³, density of air is 0.0013 g/cm³ and the coefficient of viscosity of air is 1.8 × 10^–4 poise.

Doubtly Asked 29 Oct 2026 Board exam · July 2024 hsc 12th Save WhatsApp Report

1 answer

Best answer

Answer: \(r = 3\times10^{-3}\) cm \(= 3\times10^{-5}\) m

Given (CGS units): \(v = 9.8\) cm/s, \(\rho = 0.9013\) g/cm\(^3\), \(\sigma = 0.0013\) g/cm\(^3\), \(\eta = 1.8\times10^{-4}\) poise, \(g = 980\) cm/s\(^2\).

The drop falls with terminal velocity given by Stokes’ law:

\[v = \frac{2r^2(\rho - \sigma)g}{9\eta} \Rightarrow r^2 = \frac{9\eta v}{2(\rho - \sigma)g}\]\[r^2 = \frac{9\times1.8\times10^{-4}\times9.8}{2\times0.9\times980} = \frac{1.62\times10^{-3}\times9.8}{1.8\times980} = \frac{1.62\times10^{-3}}{180} = 9\times10^{-6}\ \mathrm{cm^2}\]\[r = 3\times10^{-3}\ \mathrm{cm} = 3\times10^{-5}\ \mathrm{m}\]

Doubtly AI · 29 Oct 2026 AI-generated · not yet verified Report

Was this answer helpful?

A spherical drop of oil falls at a constant speed of 9.8 cm/s in steady air. Calculate the radius of the drop. The density of oil is 0.9013 g/cm3, density of air is 0.0013 g/cm3 and the coefficient of viscosity of air is 1.8 × 10–4 poise.

1 answer

Your answer