Question

Consider two solid spheres $P$ and $Q$ each of density $8\,{\text{gc}}{{\text{m}}^{ - 3}}$ and diameters $1\,{\text{cm}}$ and ${\text{0}}{\text{.5}}\,{\text{cm}}$ respectively. Sphere $P$ is dropped into a liquid of density ${\text{0}}{\text{.8}}\,{\text{gc}}{{\text{m}}^{ - 3}}$ and viscosity $\eta = 3$ poise Sphere $Q$ is dropped into a liquid of density $1.6\,{\text{g}}/{\text{c}}{{\text{m}}^3}$ and viscosity $\eta = 2$ poise. The ratio of the terminal velocities of $P$ and $Q$ is:

Answer 1

First of all, we will find the terminal velocities of the two spheres separately. After dropping the spheres into the liquid, they will have different terminal velocities for the differences in the viscosities of the two liquids. We will divide terminal velocity of one sphere by the velocity of the other.

Formula used:
Using terminal velocity formula,
$v = \dfrac{{2g{r^2}\left( {\rho - \sigma } \right)}}{{9\eta }}$ …… (1)
Where,
$\rho$ is density
$\sigma$ is density of liquid
$\eta$ is viscosity.

Complete step by step answer:
Given,
Diameter for $P$, ${d_p} = 1\,{\text{cm}}$.
Radius for $P$, ${r_p} = 0.5\,{\text{cm}}$.
Diameter for $Q$, ${d_q} = 0.5\,{\text{cm}}$.
Radius for $Q$,${r_q} = 0.25\,{\text{cm}}$.
Viscosity for P, ${\eta _p} = 3\,{\text{P}}$.
Viscosity for Q, ${\eta _q} = 2\,{\text{P}}$.
Density for both P and Q is, $\rho p = \rho q = 0.8\,{\text{g}}/{\text{c}}{{\text{m}}^3}$.

We have to find the terminal velocity ratio,
So,
${v_p} = \dfrac{{2g{r_p}^2\left( {{\rho _p} - {\sigma _p}} \right)}}{{9{\eta _p}}}$ and ${v_q} = \dfrac{{2g{r_q}^2\left( {{\rho _q} - {\sigma _q}} \right)}}{{9{\eta _q}}}$.
Now,
\dfrac{{{v_p}}}{{{v_q}}} = \dfrac{{2g{r_p}^2\left( {{\rho _p} - {\sigma _p}} \right)/9{\eta _p}}}{{2g{r_q}^2\left( {{\rho _q} - {\sigma _q}} \right)/9{\eta _q}}} \\\ \Rightarrow\dfrac{{{v_p}}}{{{v_q}}} = \dfrac{{{r_p}^2\left( {{\rho _p} - {\sigma _p}} \right)/{\eta _p}}}{{{r_q}^2\left( {{\rho _q} - {\sigma _q}} \right)/{\eta _q}}} \\\ …… ($1$)
So, put all the value in equation $1$ we get,
\dfrac{{{v_p}}}{{{v_q}}} = \dfrac{{{r_p}^2\left( {{\rho _p} - {\sigma _p}} \right)/{\eta _p}}}{{{r_q}^2\left( {{\rho _q} - {\sigma _q}} \right)/{\eta _q}}} \\\ \Rightarrow \dfrac{{{v_p}}}{{{v_q}}} = \dfrac{{{{\left( {0.5} \right)}^2}\left( {8 - 0.8} \right)/3}}{{{{\left( {0.25} \right)}^2}\left( {8 - 1.6} \right)/2}} \\\ \Rightarrow \dfrac{{{v_p}}}{{{v_q}}} = \dfrac{{{{\left( {0.5} \right)}^2}\left( {7.2} \right) \times 2}}{{{{\left( {0.25} \right)}^2}\left( {6.4} \right) \times 3}} \\\ \Rightarrow \dfrac{{{v_p}}}{{{v_q}}} = \dfrac{{3.6}}{{1.2}} \\\
$\therefore \dfrac{{{v_p}}}{{{v_q}}} = 3$

Hence, the required answer is $3$.

Additional information:
Terminal velocity: Terminal velocity is the highest velocity an object will reach as it falls through a fluid (air is the most common example). It happens when the drag force (${F_d}$) and buoyancy total is equal to the gravity (${F_G}$) downward force acting on the object. Since the net force is zero on the object, there is zero acceleration on the object.

Note: Remember that there are a maximum of two forces acting on a dropping object in the air which is not affected by wind or other sideways forces: weight and air resistance (also known as drag). The weight does not move. While the object is stationary, the air resistance is zero, but it increases as the object speeds up.