A spherical ball of density $\rho$ and radius $0.003$ $m$ is dropped into a tube containing a viscous fluid filled up to the $0$ $ cm$ mark as shown in the figure. Viscosity of the fluid $=$ $1.260$ $N.m^{-2}$ and its density $\rho_L=\rho/2$ $=$ $1260$ $kg.m^{-3}$. Assume the ball reaches a terminal speed by the $10$ $cm$ mark. The time taken by the ball to traverse the distance between the $10$ $cm$ and $20$ $cm$ mark is
( $g$ $ =$ acceleration due to gravity $= 10$ $ ms^{^{-2}} )$
Diffcult
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$v_{T}=\frac{2}{9} \frac{r^{2}(\rho-\sigma) g}{\eta}$
$=\frac{2 \times(0.003)^{2}(1260 \times 2-1260) \times 10}{9 \times 1.26}$
$=0.02 \mathrm{m} / \mathrm{s} $
$t=\frac{d}{v_{T}}=\frac{10 \times 10^{-2}}{0.02}=5 s$
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