Question

Eight equal drops of water are falling through air with a steady speed of 10 cm/s. If the drops coalesce, the new velocity is :

• A. 5 cm/s
• B. 40 cm/s
• C. 16 cm/s
• D. 10 cm/s

Answer 1

The Correct Option is B

Step 1: Volume Relation 

The volume of a sphere is given by:

\[ V = \frac{4}{3} \pi r^3. \]

For eight smaller spheres, the total volume is:

\[ 8 \times \frac{4}{3} \pi r^3 = \frac{4}{3} \pi R^3, \]

where \( R \) is the radius of the larger sphere.

Step 2: Radius Relation

Equating the volumes:

\[ 8 \times r^3 = R^3. \]

Taking the cube root of both sides:

\[ R = 2r. \]

Step 3: Terminal Velocity Relation

The terminal velocity of a sphere moving through a fluid is given by:

\[ V = \frac{2r^2}{9\eta} (\rho_b - \rho_{\text{air}}), \]

where:

• \( r \): Radius of the sphere
• \( \eta \): Fluid viscosity
• \( \rho_b \) and \( \rho_{\text{air}} \): Densities of the sphere and the air, respectively.

From the equation, terminal velocity is proportional to the square of the radius:

\[ V \propto r^2. \]

Step 4: Ratio of Velocities

Using the proportionality:

\[ \frac{V_1}{V_2} = \left( \frac{r}{R} \right)^2. \]

Substituting \( R = 2r \):

\[ \frac{V_1}{V_2} = \left( \frac{r}{2r} \right)^2 = \frac{1}{4}. \]

Step 5: Calculate \( V_2 \)

Rewriting the ratio:

\[ V_2 = V_1 \times 4. \]

Given \( V_1 = 10 \, \text{km/s} \):

\[ V_2 = 10 \times 4 = 40 \, \text{km/s}. \]

Final Answer:

The velocity \( V_2 \) is \( 40 \, \text{km/s} \).