Two solid spheres S_{1} and S_{2} of same uni | vedclass

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(B) The terminal velocity $v_{T}$ of a sphere of radius $r$ and density $d$ falling in a medium of density $\rho$ and viscosity $\eta$ is given by $v_

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$4$

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(B) The terminal velocity $v_{T}$ of a sphere of radius $r$ and density $d$ falling in a medium of density $\rho$ and viscosity $\eta$ is given by $v_{T} = \frac{2r^{2}g(d-\rho)}{9\eta}$.Since $g, d, \rho,$ and $\eta$ are constant for both spheres, $v_{T} \propto r^{2}$.The mass of a sphere is $m = \frac{4}{3}\pi r^{3}d$, which implies $r \propto m^{1/3}$.Substituting this into the proportionality for terminal velocity, we get $v_{T} \propto (m^{1/3})^{2} = m^{2/3}$.Therefore, the ratio of terminal velocities is $\frac{v_{1}}{v_{2}} = \left(\frac{m_{1}}{m_{2}}\right)^{2/3}$.Given $\frac{m_{1}}{m_{2}} = 8$, we have $\frac{v_{1}}{v_{2}} = (8)^{2/3} = (2^{3})^{2/3} = 2^{2} = 4$.

Two solid spheres $S_{1}$ and $S_{2}$ of same uniform density fall from rest under gravity in a viscous medium and after some time, reach terminal velocities $v_{1}$ and $v_{2}$ respectively. If the ratio of masses $\frac{m_{1}}{m_{2}}=8$, then $\frac{v_{1}}{v_{2}}$ will be equal to

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