Question

When a solid ball of volume $V$ is dropped into a viscous liquid, then a viscous force $F$ acts on it. If another ball of volume $2V$ of the same material is dropped in the same liquid then the viscous force experienced by it will be
(A) $2nF$
(B) $\dfrac{{nF}}{2}$
(C) $2F$
(D) $\dfrac{F}{2}$

Answer 1

Hint: Viscous force is the resistant force provided by the fluid against the direction of the object. To find the viscous force of the solid ball, first use the formula of viscous force, substitute the stroke's law in it. Simplify the obtained formula and get the final relation of the viscous force and the volume. Substitute the volume of the solid ball to find viscous force.

Formula used:
(1) The viscous force is given by
${F_v} = 6\pi \eta rv$
Where ${F_v}$ is the viscous force acts on the second dropped ball, $\eta $ is the coefficient of viscosity, $r$ is the radius of the solid ball and $v$ is the velocity of the solid ball in the viscous liquid.
(2) The stoke’s law is given by
$v = \dfrac{{2{r^2}\left( {\rho - {\rho _0}} \right)g}}{{9\eta }}$
Where $\rho $ is the density of the solid ball and ${\rho _0}$ is the density of the medium of the liquid.

Complete step by step answer:
Given: First, the volume of the solid ball dropped is $V$
The viscous force of the liquid is $F$
Second, the volume of the solid ball dropped is $2V$
The formula of the viscous force is considered.
${F_v} = 6\pi \eta rv$
Substituting the stroke's law in it.
${F_v} = 6\pi \eta r\dfrac{{2{r^2}\left( {\rho - {\rho _0}} \right)g}}{{9\eta }}$
By simplifying the above equation, we get
${F_v} = \dfrac{{4\pi {r^3}\left( {\rho - {\rho _0}} \right)g}}{3}$
From the above equation, it is clear that the ${F_v}\alpha {r^3}$. Hence it is also written as the
${F_v}\alpha V$
Thus the viscous force is directly proportional to that of the volume of the material.
Hence if the volume of the solid ball doubles, then the viscous force also becomes double.

Hence, the correct answer is option (C).

Note: In the above calculation the cube of the radius is considered as the volume. This is because the radius is the single dimensional parameter of the sphere and the volume is the three dimensional parameter of the sphere, hence the cube of the radius provides volume.