A spherical liquid drop of radius r acquires | vedclass

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

Vedclass · Accepted Answer

$16$

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(D) The terminal velocity $v_t$ of a spherical drop falling through a viscous medium is given by Stokes' Law: $v_t = \frac{2r^2(ho - \sigma)g}{9\eta}$, where $ho$ is the density of the liquid, $\sigma$ is the density of the gas, and $\eta$ is the viscosity.Since $v_t \propto r^2$, we have $\frac{v_1}{v_2} = \frac{R^2}{r^2}$, where $R$ is the radius of the large drop and $r$ is the radius of the small droplet.When a large drop of radius $R$ is broken into $n = 64$ identical droplets of radius $r$, the volume remains conserved:$\frac{4}{3}\pi R^3 = n 	imes \frac{4}{3}\pi r^3$$R^3 = 64r^3 \Rightarrow R = 4r \Rightarrow \frac{R}{r} = 4$.Substituting this into the ratio of velocities:$\frac{v_1}{v_2} = \left(\frac{R}{r}ight)^2 = (4)^2 = 16$.

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