Question

A sphere of brass released in a long liquid column attains a terminal speed ${v_o}$. If the terminal speed attained by sphere of marble of the same radius and released in the same liquid is $n{v_o}$, then the value of $n$ will be (Given : the specific gravities of brass, marble and the liquid are $8.5,\,2.5$ and $0.8$ respectively):
A. $\dfrac{5}{{17}}$
B. $\dfrac{{17}}{{77}}$
C. $\dfrac{{11}}{{31}}$
D. $\dfrac{{17}}{5}$

Answer 1

To find the value of $n$, you will need to recall the formula for terminal velocity. Here, there are two cases, in one brass sphere is used and in the other marble sphere is used. Write down the terminal speed or velocity for each sphere one by one. Then divide the two expressions to get the value of $n$.

Formula used:
The formula for terminal velocity is given by,
$v = \dfrac{2}{9}{r^2}\left( {\dfrac{{\rho - {\rho _l}}}{\eta }} \right)g$ (i)
where $r$ is the radius of the sphere, $\eta$ is the viscosity, $g$ is acceleration due to gravity, $\rho$ is the relative density of the object and ${\rho _l}$ is the relative density of the fluid.

Complete step by step answer:
Given, the terminal velocity of brass sphere, ${v_o}$.Terminal velocity of marble sphere , $n{v_o}$.The radii of both the spheres are the same and they are released in the same liquid.Relative density of brass, ${\rho _b} = 8.5$.Relative density of marble, ${\rho _m} = 2.5$.Relative density of liquid, ${\rho _l} = 0.8$ .Let the radius of brass sphere and marble sphere be $r$.First, we will take the case of brass sphere,
Here, $\rho = {\rho _b} = 8.5$ , ${\rho _l} = 0.8$ and $v = {v_o}$
Putting these values in equation (i) we get,
${v_o} = \dfrac{2}{9}{r^2}\left( {\dfrac{{8.5 - 0.8}}{\eta }} \right)g$
$\Rightarrow {v_o} = \dfrac{2}{9}{r^2}\left( {\dfrac{{7.7}}{\eta }} \right)g$ (ii)

In the case of marble sphere,
Here, $\rho = {\rho _m} = 2.5$
Since the liquid is the same, the values $r$, ${\rho _l}$ and $\eta$ will remain the same.
Putting these values in equation (ii) we get,
$n{v_o} = \dfrac{2}{9}{r^2}\left( {\dfrac{{2.5 - 0.8}}{\eta }} \right)g$
$\Rightarrow n{v_o} = \dfrac{2}{9}{r^2}\left( {\dfrac{{1.7}}{\eta }} \right)g$ (iii)

Now, dividing equation (iii) by (ii), we get
$\dfrac{{n{v_o}}}{{{v_o}}} = \dfrac{{\left( {\dfrac{2}{9}{r^2}\left( {\dfrac{{1.7}}{\eta }} \right)g} \right)}}{{\left( {\dfrac{2}{9}{r^2}\left( {\dfrac{{7.7}}{\eta }} \right)g} \right)}}$
$\Rightarrow \dfrac{{n{v_o}}}{{{v_o}}} = \dfrac{{1.7}}{{7.7}}$
$\therefore n = \dfrac{{17}}{{77}}$
Therefore the value of $n$ is $\dfrac{{17}}{{77}}$.

Hence, the correct answer is option B.

Note: Terminal velocity can be defined as the highest speed achieved by an object falling through a liquid. Terminal velocity is achieved when the downward gravitational force is equal to the sum of the object’s buoyancy and the drag force. At this point the net force is zero so the acceleration is zero and the object falls at constant speed.