Question

A marble of mass x and diameter 2r is gently released a tall cylinder containing honey. If the marble displaces mass y( < x) of the liquid, then the terminal velocity is proportional to
A. $$x+y$$
B. $x-y$
C. ${\displaystyle \frac{x+y}{r}}$
D. ${\displaystyle \frac{x-y}{r}}$

Answer 1

Hint: The formula for terminal velocity for a solid sphere is given as:
$$v={\displaystyle \frac{2}{9}}.{\displaystyle \frac{({\rho }_S-{\rho }_L)g{r}^2}{\mu }}$$
Where,
$\mu$- Viscosity of liquid
${\rho }_S$- Density of Solid sphere
${\rho }_L$- Density of Liquid
$r$- Radius of solid sphere
To calculate the terminal velocity, only one quantity is unknown which is the density of liquid. This can be calculated by using the given mass of displaced liquid.

Complete step by step answer:
Formula to be used for the problem is given as:
$$v={\displaystyle \frac{2}{9}}.{\displaystyle \frac{({\rho }_S-{\rho }_L)g{r}^2}{\mu }}$$
All the parameters are given in the question except for the densities of solid and liquid.
For the density of solid sphere, mass and radius is given i.e.
$${\rho }_S={\displaystyle \frac{mass}{volume}}={\displaystyle \frac{x}{{\displaystyle \frac{4}{3}}\pi r^3}}$$
For the density of liquid, the mass of displaced liquid is given. Since this liquid is being displaced the solid sphere in a long cylinder so the volume of liquid displaced will be equal to the solid sphere. So,
$$\begin{gathered} V_L=V_S\Rightarrow V_L={\displaystyle \frac{4}{3}}\pi r^3 \\ {\rho }_L={\displaystyle \frac{mass}{volume}}={\displaystyle \frac{y}{{\displaystyle \frac{4}{3}}\pi r^3}} \end{gathered}$$
Therefore, substituting the expression of density of solid sphere and liquid in the terminal velocity expression, we have,
$$\begin{gathered} v={\displaystyle \frac{2}{9}}.{\displaystyle \frac{({\rho }_S-{\rho }_L)g{r}^2}{\mu }}={\displaystyle \frac{2}{9}}.{\displaystyle \frac{({\displaystyle \frac{x}{{\displaystyle \frac{4}{3}}\pi r^3}}-{\displaystyle \frac{y}{{\displaystyle \frac{4}{3}}\pi r^3}})g{r}^2}{\mu }} \\ \Rightarrow v={\displaystyle \frac{2}{9}}.{\displaystyle \frac{(x-y)g}{({\displaystyle \frac{4}{3}}\pi r)\mu }}\Rightarrow v={\displaystyle \frac{1}{6}}.{\displaystyle \frac{(x-y)g}{(r)\pi \mu }} \\ \Rightarrow v\propto {\displaystyle \frac{(x-y)}{r}} \end{gathered}$$
The terminal velocity is therefore proportional to ${\displaystyle \frac{(x-y)}{r}}$

So, the correct answer is “Option d”.

Additional Information:
The terminal velocity is the maximum constant velocity which can be attained by the object falling in a fluid. As the object falls the drag force exerted by the fluid increases until it becomes equal to the weight of the object. Hence the net force is zero which means zero acceleration therefore, constant velocity. In reality, the object approaches the terminal velocity asymptotically.

Note:
The formula used for the terminal formula in the problem is only for spherical objects falling in a liquid. This formula is derived from Stokes’ Law. Stokes worked mainly on spheres moving in viscous fluids and came up with the formula $F_d=6\pi \mu Rv$.