A lead sphere of mass m falls with a terminal | vedclass

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(D) The formula for terminal velocity is given by:$V_{T} = \frac{2}{9} r^2 \frac{(\rho-\sigma)g}{\eta}$Here,$V_{T} \propto r^2$.The relationship betwe

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$4V$

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(D) The formula for terminal velocity is given by:$V_{T} = \frac{2}{9} r^2 \frac{(ho-\sigma)g}{\eta}$Here,$V_{T} \propto r^2$.The relationship between mass and radius is:$m \propto 	ext{volume} \propto r^3$Therefore,$\frac{m_1}{m_2} = \left(\frac{r_1}{r_2}ight)^3$.Given $m_1 = m$ and $m_2 = 8m$,we have:$\frac{m}{8m} = \left(\frac{r_1}{r_2}ight)^3 \Rightarrow \frac{1}{8} = \left(\frac{r_1}{r_2}ight)^3$$\frac{r_1}{r_2} = \frac{1}{2} \Rightarrow r_2 = 2r_1$.Now,the ratio of terminal velocities is:$\frac{V_{T2}}{V_{T1}} = \left(\frac{r_2}{r_1}ight)^2 = (2)^2 = 4$.Thus,$V_{T2} = 4V_{T1} = 4V$.

$A$ lead sphere of mass $m$ falls with a terminal velocity $V$ in a viscous liquid. With what terminal velocity will another lead sphere of mass $8m$ fall in the same liquid?

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