A lead sphere of mass m falls in a viscous li | vedclass

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(C) The terminal velocity $V$ of a spherical body falling through a viscous liquid is given by the formula: $V = \frac{2g(\rho - \sigma)r^2}{9\eta}$.S

Vedclass · Accepted Answer

$4 V_0$

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(C) The terminal velocity $V$ of a spherical body falling through a viscous liquid is given by the formula: $V = \frac{2g(\rho - \sigma)r^2}{9\eta}$.Since the material is the same,the density $\rho$ is constant. The terminal velocity is proportional to the square of the radius: $V \propto r^2$.Given that the mass $m$ is increased to $8m$,and density is constant,the volume $V_{ol} = \frac{m}{\rho}$ also increases by a factor of $8$.Since $V_{ol} = \frac{4}{3}\pi r^3$,if the volume becomes $8$ times,the radius $r$ becomes $2$ times $(r_2 = 2r_1)$.Therefore,the new terminal velocity $V_2$ is given by: $\frac{V_2}{V_0} = \left(\frac{r_2}{r_1}\right)^2 = (2)^2 = 4$.Thus,$V_2 = 4V_0$.

$A$ lead sphere of mass $m$ falls in a viscous liquid with terminal velocity $V_0$. Another lead sphere of mass $8m$ but of the same material will fall through the same liquid with terminal velocity:

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