Question

The speed of a ball of radius 2 cm in a viscous liquid is 20 cm/s. What will be the speed of a ball of radius 1 cm in same liquid?

• A. 10 cm/s
• B. 4 cm/s
• C. 5 cm/s
• D. 8 cm/s

Hint:

When the radius is halved ($r \rightarrow \frac{r}{2}$), the square dependency means the terminal velocity drops by a factor of $2^2 = 4$. Simply divide the original speed by 4: $\frac{20}{4} = 5\ \text{cm/s}$.

Answer 1

The Correct Option is C

Step 1: Identify the physics.
A ball moving steadily through a viscous liquid reaches a terminal speed. We are given the terminal speed for a $2\,\text{cm}$ ball and want it for a $1\,\text{cm}$ ball in the same liquid.
Step 2: Recall the terminal velocity law.
By Stokes' law the terminal speed is $v = \dfrac{2}{9}\dfrac{r^2(\rho - \sigma)g}{\eta}$.
Step 3: Decide what stays constant.
The liquid and ball material are unchanged, so $\rho$, $\sigma$, $\eta$ and $g$ are fixed and $v \propto r^2$.
Step 4: Form the ratio.
$\dfrac{v_2}{v_1} = \left(\dfrac{r_2}{r_1}\right)^2$.
Step 5: Insert the radii and speed.
With $r_1 = 2\,\text{cm}$, $r_2 = 1\,\text{cm}$, $v_1 = 20\,\text{cm/s}$: $\dfrac{v_2}{20} = \left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4}$.
Step 6: Solve for $v_2$.
$v_2 = \dfrac{20}{4} = 5\,\text{cm/s}$, which is option (3).
\[ \boxed{v_2 = 5\ \text{cm/s}} \]