Question

A sphere of mass $M$ and radius $R$ falls through a glycerin column and attains a terminal velocity $v_1$. Another sphere of same material but radius $3R$ falls through it. Its terminal velocity $v_2$ is:

• A. $3v_1$
• B. $9v_1$
• C. $27v_1$
• D. $v_1 / 3$ Correct Answer: (B) $9v_1$

Hint:

Don't fall into the trap of overthinking the mass parameter ($M$) given in the question! For a sphere dropping through a viscous fluid, the terminal velocity depends exclusively on the square of its linear dimension (radius), not linearly on its mass. If the radius triples ($\times 3$), the velocity jumps by $3^2 = \mathbf{9}$ instantly!

Answer 1

The Correct Option is B

Step 1: Recall what sets the terminal speed.
When a sphere falls steadily in a fluid, its weight is balanced by buoyancy and drag. Working this balance out gives the terminal speed \[ v = \frac{2}{9}\frac{r^2(\rho-\sigma)g}{\eta} \]

Step 2: Spot what stays the same.
Both spheres are the same material in the same glycerin, so density, fluid density, viscosity and $g$ are all fixed. Only the radius differs, so \[ v \propto r^2 \]

Step 3: Compare the two spheres.
\[ \frac{v_2}{v_1} = \left(\frac{r_2}{r_1}\right)^2 = \left(\frac{3R}{R}\right)^2 \]

Step 4: Get the answer.
\[ \frac{v_2}{v_1} = 3^2 = 9 \quad\Rightarrow\quad v_2 = 9v_1 \] The mass $M$ in the question is not needed. So the second sphere falls 9 times faster, which is option (B).
\[ \boxed{v_2 = 9v_1} \]