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?

In the experiment, the steel ball falls downward through the liquid at a constant speed of 10 cm/s. Since the speed is constant, the acceleration is zero, meaning the net force acting on the ball is zero. Let's denote the mass of the ball as $$ m $$ and the acceleration due to gravity as $$ g $$. The weight of the ball acting downward is $$ mg $$. There is an upward buoyant force, $$ F_b $$, due to the liquid. Additionally, when the ball moves, a viscous force opposes the motion. For downward motion, the viscous force acts upward.

At constant downward speed of 10 cm/s, the net force is zero. The effective weight (the net gravitational force) is $$ mg - F_b $$, which acts downward. This is balanced by the upward viscous force, $$ F_v $$. Therefore:

$$ mg - F_b = F_v $$

Let $$ W_{\text{eff}} = mg - F_b $$, so:

$$ W_{\text{eff}} = F_v $$

Given that the downward speed is 10 cm/s, we have $$ F_v = k \times 10 $$, where $$ k $$ is the constant of proportionality for viscous force (assuming Stokes' law, where viscous force is proportional to speed). Thus:

$$ W_{\text{eff}} = k \times 10 $$

Now, the ball is pulled upward with a force equal to twice its effective weight, so the applied force upward is $$ 2W_{\text{eff}} $$. When moving upward, the forces acting on the ball are:

• Upward forces: Applied force $$ 2W_{\text{eff}} $$ and buoyant force $$ F_b $$
• Downward forces: Weight $$ mg $$ and viscous force $$ F_v' $$ (which now acts downward because it opposes the upward motion)

The net force upward is:

$$ F_{\text{net, upward}} = (2W_{\text{eff}} + F_b) - (mg + F_v') $$

Substitute $$ W_{\text{eff}} = mg - F_b $$:

$$ F_{\text{net, upward}} = 2(mg - F_b) + F_b - mg - F_v' $$

Simplify step by step:

$$ F_{\text{net, upward}} = 2mg - 2F_b + F_b - mg - F_v' $$

$$ F_{\text{net, upward}} = (2mg - mg) + (-2F_b + F_b) - F_v' $$

$$ F_{\text{net, upward}} = mg - F_b - F_v' $$

$$ F_{\text{net, upward}} = W_{\text{eff}} - F_v' $$

When the ball moves upward at a constant speed, the net force becomes zero (no acceleration). Therefore:

$$ W_{\text{eff}} - F_v' = 0 $$

$$ F_v' = W_{\text{eff}} $$

The viscous force $$ F_v' $$ is proportional to the upward speed $$ v' $$, so $$ F_v' = k v' $$. From earlier, $$ W_{\text{eff}} = k \times 10 $$. Substituting:

$$ k v' = k \times 10 $$

Divide both sides by $$ k $$ (assuming $$ k \neq 0 $$):

$$ v' = 10 \text{ cm/s} $$

Thus, the ball moves upward at a constant speed of 10 cm/s.

Hence, the correct answer is Option C.