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 :
- 5 cm/s
- 40 cm/s
- 16 cm/s
- 10 cm/s
The Correct Option is B
Solution and Explanation
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 \( 40 \, \text{km/s} \).
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