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: Understanding the Concept:
When a solid spherical object is dropped into a dense viscous fluid medium (such as glycerin), it initially accelerates downward under the action of gravity. As its downward speed climbs, it experiences an opposing upward resistive dragging force called the viscous drag force, governed by Stokes' Law. Eventually, the upward viscous drag plus the upward buoyant force perfectly balance out the downward weight of the sphere. At this exact point of dynamic equilibrium, the net acceleration drops to zero, and the sphere continues falling at a constant, maximum limiting speed called the terminal velocity ($v$).

Step 2: Key Formula or Approach:
The standard equation derived from balancing forces for a sphere of radius $r$ and material density $\rho$ falling through a fluid of density $\sigma$ and viscosity coefficient $\eta$ is: $$ v = \frac{2}{9} \frac{r^2(\rho - \sigma)g}{\eta} $$ Since both spheres in this scenario are made from the same material ($\rho$ is constant) and are dropping through the same fluid ($\sigma$ and $\eta$ are constant), all terms in the equation except for the radius are identical constants. Therefore, the terminal velocity of a falling sphere is directly proportional to the square of its radius: $$ v \propto r^2 $$

Step 3: Detailed Explanation:
Let's set up a proportional comparison between the two spheres: $$ \frac{v_2}{v_1} = \left(\frac{r_2}{r_1}\right)^2 $$ From the given parameters: - Initial sphere radius $r_1 = R$, with terminal velocity $v_1$. - Second sphere radius $r_2 = 3R$, with terminal velocity $v_2$. Substitute these radius values directly into the comparative scaling equation: $$ \frac{v_2}{v_1} = \left(\frac{3R}{R}\right)^2 $$ $$ \frac{v_2}{v_1} = (3)^2 = 9 $$ $$ v_2 = 9v_1 $$ The terminal velocity of the second sphere is exactly $9$ times greater than the first, which directly matches option (B).

Step 4: Final Answer:
The terminal velocity $v_2$ of the second sphere is $9v_1$.