Question

A marble of mass $x$ and diameter $2r$ is gently released in 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. $\frac{x+y}{r}$
• D. $\frac{x-y}{r}$

Answer 1

The correct option is D $\frac{x-y}{r}$

We know that, terminal velocity $v_t=\frac{2}{9}\frac{r^2(\rho -{\rho }^{\prime })g}{\eta }........\left(1\right)$
where $\rho$ and ${\rho }^{\prime }$ be the densities of marble and the liquid, $\eta$ is the viscosity of the liquid.

Given, $x$ is the mass of the marble and $y$ is the mass of liquid displaced.
Now, $x=\frac{4}{3}\pi r^3\rho$
So, we can write $\rho \propto \frac{x}{r^3}$
Similarly, ${\rho }^{\prime }\propto \frac{y}{r^3}$
$\therefore$ From$\left(1\right)$, we can say that,
Terminal velocity $\left(v_t\right)\propto \frac{x-y}{r}$
[$g$ and $\eta$ are constants]
Thus, option (d) is the correct answer.