Chapter 9, Problem 72P

To determine

The viscosity of the liquid.

Answer to Problem 72P

Viscosity is $2.4\text{Pa}\cdot \text{s}$

Explanation of Solution

Radius of sphere is $1.0\text{cm}$, mass of sphere is $12.0\text{g}$, density of liquid is $1200\text{kg}/{\text{m}}^{\text{3}}$, and the terminal speed is $0.15\text{m}/\text{s}$.

There will be no acceleration for sphere moving with terminal velocity. That means, the net force on sphere in vertical direction is zero.

The free body diagram is shown below in figure 1.

Write the equation for net force on sphere in vertical direction.

$F_D-F_B-m_{s}g=0$. (I)

Here, $F_D$ is the viscous drag force on the sphere,$F_B$ is the buoyant force, $m_s$ is the mass of sphere, and $g$ is the gravitational acceleration.

Write the equation for $F_B$.

$F_B=m_{l}g$                                                                                                                   (II)

Here, $m_l$ is the mass of liquid displaced.

Write the equation for $F_D$.

$F_D=6\pi \eta r{v}_t$                                                                                                                   (III)

Here, $\eta$ is the coefficient of viscosity of the fluid, $r$ is the radius of sphere, and $v_t$ is the terminal velocity of the sphere.

Rewrite equation (I) by substituting equations (II) and (III).

$\begin{array}{c}6\pi \eta r{v}_t-m_{l}g-m_{s}g=0\\ 6\pi \eta r{v}_t-g\left(m_s-m_l\right)=0\end{array}$ (IV)

Write the equation for $m_l$ in terms of volume and density.

$m_l=\frac{4}{3}\pi r^3{\rho }_l$

Here, ${\rho }_l$ is the density of liquid.

Rewrite equation (IV) by substituting the above relation for $m_l$.

$6\pi \eta r{v}_t-g\left(m_s-\frac{4}{3}\pi r^3{\rho }_l\right)=0$

Rewrite the above relation in terms of $\eta$.

$\eta =\frac{g\left(m_s-\frac{4}{3}\pi r^3{\rho }_l\right)}{6\pi r{v}_t}$

Conclusion:

Substitute $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $12\text{g}$ for $m_s$, $3.14$ for $\pi$, $1.0\text{cm}$ for $r$, $1200\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_l$, and $0.15\text{m}/\text{s}$ for $v_t$ in the above equation to find $\eta$.

$\begin{array}{c}\eta =-\frac{\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\left[\left(12\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\right)-\left(\frac{4}{3}\left(3.14\right){\left(\left(1.0\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)\right)}^3\left(1200\text{kg}/{\text{m}}^{\text{3}}\right)\right)\right]}{6\left(3.14\right)\left(1.0\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)\left(0.15\text{m}/\text{s}\right)}\\ =\frac{0.0679}{0.0283}\text{Pa}\cdot \text{s}\\ =2.4\text{Pa}\cdot \text{s}\end{array}$

Therefore, the viscosity is $2.4\text{Pa}\cdot \text{s}$