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:
Two identical spheres fall through the same viscous liquid; given densities and v_A = 0.4 m/s, find v_B.

Step 2: Key Formula or Approach:
For identical sizes and same liquid, terminal velocity v ∝ (ρ_s – ρ_l).

Step 3: Detailed Explanation:
v_A/v_B = (7.5–1.5)/(3–1.5) = 6.0/1.5 = 4. 0.4/v_B = 4 → v_B = 0.1 m/s.

Step 4: Final Answer:
Terminal velocity of B is 0.1 m/s, matching option (B).