Question 189 of 192

In an experiment, a small steel ball falls through a liquid at a constant speed of $10 \mathrm{~cm} / \mathrm{s}$. If the steel ball is pulled upward with a force equal to twice its effective weight, how fast will it move upward ?

A$5 \mathrm{~cm} / \mathrm{s}$BZeroC$10 \mathrm{~cm} / \mathrm{s}$D$20 \mathrm{~cm} / \mathrm{s}$

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Answer: (C) $10 \mathrm{~cm} / \mathrm{s}$

Weight of the body
$\mathrm{W}=\mathrm{mg}=\frac{4}{3} \pi \mathrm{r}^3 \rho \mathrm{g}$
$\mathrm{T}=\frac{4}{3} \pi \mathrm{r}^3 \sigma \mathrm{g}$
and $\mathrm{F}=6 \pi \eta \mathrm{vr}$
When the body attains terminal velocity net force acting on the body is zero. i.e.,
$\mathrm{W}-\mathrm{T}-\mathrm{F}=0$
And terminal velocity $\mathrm{v}=\frac{2}{9} \frac{\mathrm{r}^2(\rho-\sigma) \mathrm{g}}{\eta}$
As in case of upward motion upward force is twice its effective weight, therefore, it will move with same speed $10 \mathrm{~cm} / \mathrm{s}$