Ratio of Terminal Velocities for Splitting Liquid Drop

To find the ratio of the terminal velocities $\frac{v_1}{v_2}$, we analyze the relationship between the terminal velocity of a falling sphere and its radius.

1. Expression for Terminal Velocity

The terminal velocity $v$ of a spherical drop of radius $r$ falling through a viscous fluid is given by Stokes' Law: $v = \frac{2}{9} \frac{r^2 (\rho - \sigma) g}{\eta}$

where:

• $r$ is the radius of the drop,
• $\rho$ is the density of the liquid drop,
• $\sigma$ is the density of the surrounding gas,
• $g$ is the acceleration due to gravity,
• $\eta$ is the coefficient of viscosity of the gas.

Since the liquid drop, the gas, and the ambient conditions remain the same, all terms except the radius $r$ are constant. Thus, the terminal velocity is directly proportional to the square of the radius: $v \propto r^2$

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2. Radius of the Smaller Droplets

Let $R$ be the radius of the original big drop and $r$ be the radius of each of the $64$ identical smaller droplets.

By conservation of volume: $\text{Volume of original drop} = 64 \times \text{Volume of one small droplet}$

$\frac{4}{3} \pi R^3 = 64 \times \frac{4}{3} \pi r^3$

$R^3 = 64 r^3$

Taking the cube root on both sides: $R = 4r \implies r = \frac{R}{4}$

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3. Ratio of Terminal Velocities

The terminal velocity $v_1$ of the original drop of radius $R$ is: $v_1 \propto R^2$

The terminal velocity $v_2$ of a smaller droplet of radius $r$ is: $v_2 \propto r^2$

Taking the ratio of the two velocities: $\frac{v_1}{v_2} = \left(\frac{R}{r}\right)^2$

Substituting $R = 4r$: $\frac{v_1}{v_2} = \left(\frac{4r}{r}\right)^2 = 4^2 = 16$

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Conclusion

The ratio of terminal velocities $\frac{v_1}{v_2}$ is $16$.

Correct Option: D