Question

Two metal spheres of densities in the ratio $3:2$ and diameter in the ratio $1:2$ is released from rest in two vertical liquid columns of coefficients of viscosity in the ratio $4:3$. If the viscous force on them is same, then the ratio of their instantaneous velocities is
$\begin{aligned} & A)1:2 \\\ & B)3:2 \\\ & C)4:3 \\\ & D)8:3 \\\ \end{aligned}$

Answer 1

According to Stokes law, viscous force on a spherical body is proportional to viscosity of the liquid, into which the body is falling and radius of the spherical body. Moreover, it is also proportional to the velocity of the spherical body. Thus, from the formula for viscous force acting on a spherical body, we can arrive at the ratio of instantaneous velocities of two such spherical bodies.

Formula used: $F=6\pi \eta rv$

Complete step by step answer:
According to Stokes law, viscous force or viscous drag acting on a spherical body is proportional to the coefficient of viscosity of the liquid, into which the body is falling as well as the radius of the spherical body. Moreover, it is also proportional to the velocity of the spherical body. Mathematically, viscous force acting on a spherical body is given by
$F=6\pi \eta rv$
where
$F$ is the viscous force acting on a spherical body
$\eta$ is the coefficient of viscosity of the liquid, into which the spherical body is falling
$r$ is the radius of the spherical body
$v$ is the instantaneous velocity of the spherical body
Let this be equation 1.
Coming to our question, we are given that two metal spheres of densities in the ratio $3:2$ and diameter in the ratio $1:2$ are released from rest in two vertical liquid columns of coefficients of viscosity in the ratio 4:3. If the viscous force on them is the same, we are required to determine the ratio of their instantaneous velocities of the given spheres.
Using equation 1, if ${{F}_{1}}=6\pi {{\eta }_{1}}{{r}_{1}}{{v}_{1}}$ and ${{F}_{2}}=6\pi {{\eta }_{2}}{{r}_{2}}{{v}_{2}}$ represent the viscous forces on the two given spherical bodies, we have
${{F}_{1}}={{F}_{2}}\Rightarrow 6\pi {{\eta }_{1}}{{r}_{1}}{{v}_{1}}=6\pi {{\eta }_{2}}{{r}_{2}}{{v}_{2}}$
as provided in the question
Let this be equation 2.
Solving equation 2, we have
$\dfrac{{{v}_{1}}}{{{v}_{2}}}=\dfrac{{{\eta }_{2}}}{{{\eta }_{1}}}\dfrac{{{r}_{2}}}{{{r}_{1}}}\Rightarrow \dfrac{{{v}_{1}}}{{{v}_{2}}}=\dfrac{3}{4}\times \dfrac{2}{1}\Rightarrow \dfrac{{{v}_{1}}}{{{v}_{2}}}=\dfrac{3}{2}$
since we are given that
$\dfrac{{{\eta }_{1}}}{{{\eta }_{2}}}=\dfrac{4}{3}$ and $\dfrac{{{r}_{1}}}{{{r}_{2}}}=\dfrac{1}{2}$
Let this be equation 3.
From the above equation, we can conclude that the ratio of instantaneous velocities of the two given metal spheres is $3:2$.

So, the correct answer is “Option B”.

Note: In the question we have the ratio of $\dfrac{{{\eta }_{1}}}{{{\eta }_{2}}}$, but in the calculation part, we have $\dfrac{{{\eta }_{2}}}{{{\eta }_{1}}}$and it is for this reason that we have substituted $3:4$ instead of $4:3$. Similarly, in the question, we have the ratio of$\dfrac{{{r}_{1}}}{{{r}_{2}}}$, but in the calculation part, we have $\dfrac{{{r}_{2}}}{{{r}_{1}}}$and it is for this reason that we have substituted $2:1$ instead of $1:2$. It is important to be careful while making these substitutions since carelessness shown here can cause major mistakes in calculations. Also, the given ratio of densities of the metal spheres is irrelevant for solving the given question.