Question

A spherical ball of radius r falls through a viscous liquid. If the radius is doubled, the terminal velocity becomes:

• A. Same
• B. 2 times
• C. 4 times
• D. 8 times

Hint:

Remember that terminal velocity depends on the square of the radius ($v_t \propto r^2$).
If the radius increases by a factor of $x$, the terminal velocity will increase by a factor of $x^2$.

Answer 1

The Correct Option is C

Step 1: Write Stokes' law for terminal velocity.
For a sphere falling through a viscous fluid, terminal velocity is given by
\[ v_t = \frac{2r^2(\rho - \sigma)g}{9\eta} \]
which shows that terminal velocity is directly proportional to the square of the radius, all other quantities held fixed.
Step 2: Apply the scaling relationship.
Since \( v_t \propto r^2 \), doubling the radius means the new terminal velocity scales by a factor of \( 2^2 \).
Step 3: Compute the new terminal velocity.
\[ \frac{v_t'}{v_t} = \left(\frac{2r}{r}\right)^2 = 4 \]
So the terminal velocity becomes 4 times its original value.
\[ \boxed{4 \text{ times}} \]