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(C) Let $W$ be the weight of the ball,$T$ be the upthrust,and $F$ be the viscous force.When the ball falls at a constant terminal velocity $v_1 = 10\,

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$10$

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(C) Let $W$ be the weight of the ball,$T$ be the upthrust,and $F$ be the viscous force.When the ball falls at a constant terminal velocity $v_1 = 10\, cm/s$,the net force is zero:$W - T - F_1 = 0 \implies F_1 = W - T = W_{eff}$,where $W_{eff}$ is the effective weight.Since $F_1 = 6\pi\eta r v_1$,we have $W_{eff} = 6\pi\eta r v_1$.Now,the ball is pulled upward with an external force $F_{ext} = 2 W_{eff}$.Let the new upward velocity be $v_2$. The forces acting on the ball are the upward force $F_{ext}$,the upthrust $T$ (upward),the weight $W$ (downward),and the new viscous force $F_2$ (downward,as the ball moves up).For constant upward velocity $v_2$,the net force is zero:$F_{ext} + T - W - F_2 = 0$$F_{ext} - (W - T) = F_2$$2 W_{eff} - W_{eff} = F_2$$F_2 = W_{eff}$Since $F_2 = 6\pi\eta r v_2$,we have $6\pi\eta r v_2 = 6\pi\eta r v_1$.Therefore,$v_2 = v_1 = 10\, cm/s$.

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

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