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: Understanding the Question:
The question asks how the terminal velocity of a falling spherical ball changes if its radius is doubled while falling through a viscous fluid.
This is a standard fluid mechanics problem governed by Stokes' Law.

Step 2: Key Formula or Approach:
The terminal velocity ($v_t$) of a spherical body falling through a viscous medium is given by the formula:
\[ v_t = \frac{2}{9} \frac{r^2 (\rho - \sigma) g}{\eta} \]
Where:
- $r$ is the radius of the spherical ball.
- $\rho$ is the density of the ball.
- $\sigma$ is the density of the liquid.
- $\eta$ is the coefficient of viscosity of the liquid.
- $g$ is the acceleration due to gravity.

Step 3: Detailed Explanation:

• From the terminal velocity formula, we establish that if the densities of the ball and the fluid, the gravity, and the viscosity remain constant, terminal velocity is directly proportional to the square of the radius:
\[ v_t \propto r^2 \]

• Let the initial radius of the ball be $r_1 = r$, and the corresponding initial terminal velocity be $v_{t1} = v$.

• The final radius of the ball is doubled: $r_2 = 2r$.

• Let the new terminal velocity be $v_{t2}$.

• We take the ratio of the terminal velocities:
\[ \frac{v_{t2}}{v_{t1}} = \left(\frac{r_2}{r_1}\right)^2 \]

• Substitute $r_1$ and $r_2$ into the ratio:
\[ \frac{v_{t2}}{v} = \left(\frac{2r}{r}\right)^2 = 2^2 = 4 \]
\[ v_{t2} = 4v \]

• This shows that doubling the radius results in a fourfold increase in the terminal velocity.


Step 4: Final Answer:
The terminal velocity becomes 4 times the original value.