Question

A solid sphere moves through a liquid. The liquid opposes the motion of the sphere with a force $F$ which depends on the radius $r$ of the sphere, speed $v$ of the sphere and coefficient of viscosity $\eta$ of the liquid. Find the correct relationship between these physical quantities.

A) $F \propto \eta rv$

B) $F \propto {\eta ^{ - 1}}rv$

C) $F \propto \eta {r^{ - 1}}v$

D) $F \propto \eta {r^{ - 1}}{v^2}$

Answer 1

Here it is mentioned that the opposing force depends on the radius of the sphere, the speed of the sphere and the coefficient of viscosity of the liquid in which the sphere moves. So we can assume that the force is proportional to some power of these quantities and the powers can be determined using the method of dimensions

Complete step by step solution:

Step 1: Describe the problem at hand.

The problem at hand involves a sphere moving through a liquid. As the sphere moves, a force opposing the motion of the sphere comes into play. We assume that this force $F$ is proportional to some powers of the radius $r$ of the sphere, the speed $v$ of the sphere and the coefficient of viscosity $\eta$ of the liquid.

i.e., $F \propto {\eta ^a}{r^b}{v^c}$ -------- (1)

Step 2: Using the method of dimension, obtain the values of the powers $a$, $b$, $c$ .

The dimension of force can be expressed as $F \to \left[ {ML{T^{ - 2}}} \right]$

The corresponding dimension of the coefficient of viscosity of the liquid can be expressed as $\eta \to \left[ {M{L^{ - 1}}{T^{ - 1}}} \right]$

The corresponding dimension of the radius of the sphere can be expressed as $r \to \left[ L \right]$

The corresponding dimension of the speed of the sphere can be expressed as $v \to \left[ {L{T^{ - 1}}} \right]$

Now we can rewrite equation (1) in terms of the dimensions of force, radius, speed and coefficient of viscosity as $\left[ {ML{T^{ - 2}}} \right] = {\left[ {M{L^{ - 1}}{T^{ - 1}}} \right]^a}{\left[ L \right]^b}{\left[ {L{T^{ - 1}}} \right]^c}$

$\Rightarrow {M^1}{L^1}{T^{ - 2}} = {M^1}{L^{\left( { - a + b + c} \right)}}{T^{\left( { - a - c} \right)}}$ ------- (2)

Now we can equate the powers of $M$, $L$ and $T$ on either side of equation (2).

For $M$, we get, $1 = a$

So we have $a = 1$ .

For $T$, we get, $- 2 = - a - c$

$\Rightarrow c = 2 - a = 2 - 1 = 1$

So we have $c = 1$ .

For $L$, we get, $1 = - a + b + c$

$\Rightarrow b = a + 1 - c = 1 + 1 - 1 = 1$

So we have $b = 1$.

Then substituting for $a = 1$, $b = 1$ and $c = 1$ in equation (1) we get, $F \propto {\eta ^1}{r^1}{v^1}$

Thus the required relationship is $F \propto \eta rv$ .

So, the correct option is (A).

Note: The S.I. unit of force is ${\text{N}}$ or ${\text{kgm}}{{\text{s}}^{ - 2}}$ and so we have the dimension of force as $F \to ML{T^{ - 2}}$. The S.I. the unit of the radius of the sphere is meter and meter is a unit of length. So we represent the dimension of radius as $r \to L$. The unit of speed is ${\text{m}}{{\text{s}}^{ - 1}}$ and so its dimension will be $v \to L{T^{ - 1}}$. The unit of the coefficient of viscosity is ${\text{Ns}}{{\text{m}}^{ - 2}}$ or ${\text{kg}}{{\text{m}}^{ - 1}}{{\text{s}}^{ - 1}}$ and hence it has the dimension $\eta \to M{L^{ - 1}}{T^{ - 1}}$.