Question

If the terminal velocity of a metal sphere of mass 8 g falling through a liquid is 3 \( \text{cms}^{-1} \), then the terminal velocity of another sphere of mass 64 g made of the same metal falling through same liquid is

• A. \( 6 \, \text{cms}^{-1} \)
• B. \( 3 \, \text{cms}^{-1} \)
• C. \( 12 \, \text{cms}^{-1} \)
• D. \( 18 \, \text{cms}^{-1} \)

Hint:

Scaling laws are very useful here. \( v_t \propto r^2 \) and \( m \propto r^3 \), so \( v_t \propto m^{2/3} \). Memorizing this saves time on finding radius explicitly.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:

Terminal velocity \( v_t \) is proportional to the square of the radius (\( r^2 \)). Since mass \( m \propto r^3 \) (for same density), we can relate velocity to mass.
Step 2: Key Formula or Approach:

1. Mass \( m = \frac{4}{3} \pi r^3 \rho \implies r \propto m^{1/3} \). 2. Terminal Velocity \( v_t \propto r^2 \). 3. Combine: \( v_t \propto (m^{1/3})^2 \propto m^{2/3} \).
Step 3: Detailed Explanation:

Let sphere 1 have mass \( m_1 = 8 \) g and velocity \( v_1 = 3 \) cm/s. Let sphere 2 have mass \( m_2 = 64 \) g and velocity \( v_2 \). Ratio: \[ \frac{v_2}{v_1} = \left( \frac{m_2}{m_1} \right)^{2/3} \] Substitute values: \[ \frac{v_2}{3} = \left( \frac{64}{8} \right)^{2/3} \] \[ \frac{v_2}{3} = (8)^{2/3} = (2^3)^{2/3} = 2^2 = 4 \] \[ v_2 = 3 \times 4 = 12 \, \text{cm/s} \]
Step 4: Final Answer:

The terminal velocity is \( 12 \, \text{cms}^{-1} \).