Question

A solid sphere of radius R acquires a terminal velocity ${v_1}$ when falling (due to gravity) through a viscous fluid having a coefficient of viscosity $\eta$. The sphere is broken into 27 identical solid spheres. If each of these spheres acquires a terminal velocity ${v_2}$ when falling through the same fluid, the ratio $\left( {{v_1}/{v_2}} \right)$ equals:
A. $\dfrac{1}{{27}}$
B. $\dfrac{1}{{9}}$
C. 27
D. 9

Answer 1

The terminal velocity of a sphere falling through a liquid of certain viscosity is directly proportional to the square of the radius of the sphere. By comparing the radii of the solid sphere and the 27 identical solid spheres, we can obtain the ratio of their terminal velocities.

Formula used:
The formula for the terminal velocity of a sphere falling down a liquid is given as follows:
${V_T} = \dfrac{2}{9}\dfrac{{{r^2}}}{\eta }\left( {{\rho _0} - {\rho _l}} \right)g$

Complete step by step answer:
We are given a solid sphere which has radius R and it acquires a terminal velocity ${v_1}$ when falling (due to gravity) through a viscous fluid which has a coefficient of viscosity $\eta$.
Then this sphere is broken into 27 identical solid spheres which have radius r and same density as the original sphere and each of these spheres acquires a terminal velocity ${v_2}$ when falling through the same fluid of same density and viscosity.
We know that the terminal velocity of a sphere falling down a liquid is given as follows:
${V_T} = \dfrac{2}{9}\dfrac{{{r^2}}}{\eta }\left( {{\rho _0} - {\rho _l}} \right)g$
Here r signifies the radius of the sphere, ${\rho _0}$ is the density of the sphere while ${\rho _l}$ is the density of the liquid through which it is falling; g signifies the acceleration due to gravity.
For a given liquid and a given sphere of fixed density, all the parameters on the right hand side are constant except the radius of the sphere. We can say that the terminal velocity of a given sphere of radius r is directly proportional to the square of the radius.
Therefore, for the initial solid sphere, we can write that
${v_1} \propto {R^2}$
While for each of the 27 identical spheres, we can write
${v_2} \propto {r^2}$
Dividing them we get
$\dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{{R^2}}}{{{r^2}}}$ …(i)
Now we need to find out the relation between the radii of the original sphere and the 27 identical spheres. The volume of the initial sphere will be equal to the sum of the volumes of the 27 identical spheres. Therefore, we can equate their volumes in the following way.
$\dfrac{4}{3}\pi {R^3} = 27 \times \dfrac{4}{3}\pi {r^3} \\\ {R^3} = 27 \times {r^3} \\\ \dfrac{R}{r} = 3 \\\$
Using this relation in equation (i), we get
$\dfrac{{{v_1}}}{{{v_2}}} = \dfrac{{{R^2}}}{{{r^2}}} = 9$
This is the required value.

So, the correct answer is “Option D”.

Note:
It should be noted that the terminal velocity of the original solid sphere having radius R is nine times larger than the smaller sphere which has a radius three times less than that of the original sphere. Hence, larger the radius of the sphere, greater is its terminal velocity while falling through a liquid.