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Chapter 9: Problem 62

A sphere of radius \(1.0 \mathrm{cm}\) is dropped into a glass cylinder filled with a viscous liquid. The mass of the sphere is \(12.0 \mathrm{g}\) and the density of the liquid is \(1200 \mathrm{kg} / \mathrm{m}^{3} .\) The sphere reaches a terminal speed of \(0.15 \mathrm{m} / \mathrm{s} .\) What is the viscosity of the liquid?

Short Answer

Expert verified
Answer: The viscosity of the liquid is approximately \(1.046\, \text{Pa}\cdot\text{s}\).

Step by step solution

01

Write down the given values

We are given: - Radius of the sphere, \(r = 1.0\,\text{cm} = 0.01\,\text{m}\) - Mass of the sphere, \(m = 12.0\,\text{g} = 0.012\,\text{kg}\) - Density of the liquid, \(\rho_\text{liquid} = 1200\, \text{kg}/\text{m}^3\) - Terminal speed of the sphere, \(v_\text{terminal} = 0.15\,\text{m/s}\)
02

Calculate the gravitational force acting on the sphere

Using Newton's second law, the gravitational force on the sphere is given by: \(F_\text{gravity} = mg\) where \(g = 9.81\, \text{m/s}^2\) is the acceleration due to gravity. Plug in the given values to find the gravitational force: \(F_\text{gravity} = 0.012\,\text{kg} \times 9.81\, \text{m/s}^2 = 0.11772\, \text{N}\)
03

Apply Stoke's Law

At terminal speed, the gravitational force is balanced by the viscous drag force from the liquid. Stoke's Law relates this drag force to the viscosity of the liquid \(\eta\), the radius of the sphere \(r\), and the terminal speed \(v_\text{terminal}\) by the formula: \(F_\text{drag} = 6\pi \eta r v_\text{terminal}\) At terminal speed, \(F_\text{drag} = F_\text{gravity}\), so \(6\pi \eta r v_\text{terminal} = 0.11772\, \text{N}\)
04

Solve for the viscosity of the liquid

We'll now solve for \(\eta\). Rearranging the equation from Step 3: \(\eta = \frac{F_\text{gravity}}{6\pi r v_\text{terminal}}\) Plug in the given values and calculate the viscosity: \(\eta = \frac{0.11772\, \text{N}}{6\pi \times 0.01\, \text{m} \times 0.15\, \text{m/s}} \approx 1.046\, \text{Pa}\cdot\text{s}\) So, the viscosity of the liquid is approximately \(1.046\, \text{Pa}\cdot\text{s}\).

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