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(D) The terminal velocity $v_{T}$ of a spherical ball of radius $r$ and density $\sigma$ falling in a medium of density $\rho$ and viscosity $\eta$ is

Vedclass · Accepted Answer

$\frac{79}{36}$

Vedclass · Answer

(D) The terminal velocity $v_{T}$ of a spherical ball of radius $r$ and density $\sigma$ falling in a medium of density $ho$ and viscosity $\eta$ is given by $v_{T} = \frac{2 r^{2}(\sigma - ho) g}{9 \eta}$.Given that the masses of the two balls are equal,$m_{1} = m_{2}$.Since $m = 	ext{volume} 	imes 	ext{density} = \frac{4}{3} \pi r^{3} \sigma$,we have $\frac{4}{3} \pi r_{1}^{3} ho_{1} = \frac{4}{3} \pi r_{2}^{3} ho_{2}$.Given $r_{1} = 1 \; mm$,$r_{2} = 2 \; mm$,and $ho_{1} = 8 ho_{2}$,we check the mass condition: $1^{3} 	imes 8 ho_{2} = 8 ho_{2}$ and $2^{3} 	imes ho_{2} = 8 ho_{2}$. The condition holds.The ratio of terminal velocities is $\frac{v_{1}}{v_{2}} = \left(\frac{r_{1}}{r_{2}}ight)^{2} \frac{(ho_{1} - ho_{medium})}{(ho_{2} - ho_{medium})}$.Here,$ho_{medium} = 0.1 ho_{2}$.Substituting the values: $\frac{v_{1}}{v_{2}} = \left(\frac{1}{2}ight)^{2} \frac{(8 ho_{2} - 0.1 ho_{2})}{(ho_{2} - 0.1 ho_{2})} = \frac{1}{4} 	imes \frac{7.9 ho_{2}}{0.9 ho_{2}} = \frac{1}{4} 	imes \frac{79}{9} = \frac{79}{36}$.

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