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A metal ball of radius 'r' and density 'd' travels with a terminal velocity 'v' in a liquid of density d/4. The terminal velocity of another ball of radius '2r' and density '3d' in the same liquid is

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$\frac{22 v}{3}$

$\frac{11 v}{3}$

$\frac{3 v}{44}$

$\frac{44 v}{3}$

answer is A.

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Detailed Solution

Given,

$r_{1} = r, r_{2} = 2 r$

$ρ_{m} = \frac{d}{4}, ρ_{b_{1}} = d, ρ_{b_{2}} = 3 d$

$V_{t_{1}} = v, V_{t_{2}} = ?$

From the question,

$V_{t} = \frac{2}{9} r^{2} \frac{g}{η} (ρ_{b} - ρ_{m})$

$\Rightarrow V_{t} \propto (ρ_{b} - ρ_{m})$

$\Rightarrow \frac{V_{t_{1}}}{V_{t_{2}}} = {(\frac{r_{1}}{r_{2}})}^{2} (\frac{ρ_{b_{1}} - ρ_{m_{1}}}{ρ_{b_{2}} - ρ_{m_{2}}})$

$\Rightarrow {(\frac{r}{2 r})}^{2} (\frac{d - \frac{d}{4}}{3 d - \frac{d}{4}}) = \frac{3}{44} \Rightarrow V_{t_{2}} = \frac{44}{3} V$

Hence the correct answer is $\frac{44}{3} V .$

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