A body of mass 0.01 kg starts to move in a viscous liquid with initial velocity 5 m/s. If drag force is 10^-3N then find time after which velocity will become $\frac{1}{10} m / s$

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49 sec

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Detailed Solution

u= 5m/s, v = 1/10 m/s $a = - \frac{10^{- 3}}{0.01} = \frac{- 1}{10} m / s^{2}$
$\Rightarrow t = \frac{v - u}{a} = \frac{\frac{1}{10} - 5}{- \frac{1}{10}} = 49 \sec$

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