Question

If the terminal speed of a sphere A [density $\rho_A = 7.5\text{ kg}\cdot\text{m}^{-3}$] is $0.4\text{ m/s}$ in a viscous liquid [density $\rho_L = 1.5\text{ kg}\cdot\text{m}^{-3}$], then the terminal speed of a sphere B [density $\rho_B = 3\text{ kg}\cdot\text{m}^{-3}$] of the same size in the same liquid is

• A. $0.3\text{ m/s}$
• B. $0.1\text{ m/s}$
• C. $0.2\text{ m/s}$
• D. $0.4\text{ m/s}$

Hint:

Notice that the effective weight driving the descent of sphere $A$ is proportional to $7.5 - 1.5 = 6$, while for sphere $B$ it is $3 - 1.5 = 1.5$. Since $1.5$ is exactly one-quarter of $6$, sphere $B$ experiences one-quarter of the driving force and must fall at exactly one-quarter of the speed: $\frac{0.4}{4} = 0.1\text{ m/s}$!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are analyzing two identical spheres, $A$ and $B$, falling through the same viscous liquid. We are given the densities of both spheres, the density of the liquid, and the terminal velocity of sphere $A$ ($0.4\text{ m/s}$). We need to determine the terminal velocity of sphere $B$.

Step 2: Key Formula or Approach:
According to Stokes' Law, the terminal velocity $v$ of a spherical body falling through a viscous fluid is given by: $$v = \frac{2}{9} \frac{r^2 g (\rho_s - \rho_l)}{\eta}$$ Since both spheres have the exact same size ($r$ is identical) and are falling through the exact same liquid ($\eta$ and $\rho_l$ are identical), the terminal velocity is directly proportional to the net effective density difference: $$v \propto (\rho_s - \rho_l)$$

Step 3: Detailed Explanation:
Let's set up the ratio of the terminal velocities for spheres $A$ and $B$: $$\frac{v_A}{v_B} = \frac{\rho_A - \rho_L}{\rho_B - \rho_L}$$ Substitute the given values ($\rho_A = 7.5$, $\rho_B = 3$, and $\rho_L = 1.5$) into the ratio: $$\frac{0.4}{v_B} = \frac{7.5 - 1.5}{3 - 1.5}$$ Simplify the terms in the numerator and denominator: $$\frac{0.4}{v_B} = \frac{6.0}{1.5} = 4$$ Now, isolate and solve for $v_B$: $$4 \cdot v_B = 0.4 \implies v_B = \frac{0.4}{4} = 0.1\text{ m/s}$$

Step 4: Final Answer:
The terminal speed of sphere B is $0.1\text{ m/s}$, which corresponds to option (B).