A small steel ball falls through a syrup at a constant speed of `10cms^-1`. If the steel ball is pulled upwards with a force equal to twice its effective weight, how fast will it move upwards?

Options

A. `10cms^-1`
B. `20cms^-1`
C. `5cms^-1`
D. zero

Answer

To solve the problem step by step, we will analyze the forces acting on the steel ball and how they change when the ball is pulled upwards.

### Step 1: Understand the forces acting on the ball when it falls
When the steel ball falls through the syrup at a constant speed of 10 cm/s, it experiences three forces:
1. **Weight (W)**: This is the gravitational force acting downwards, given by \( W = mg \) where \( m \) is the mass of the ball and \( g \) is the acceleration due to gravity.
2. **Buoyant Force (F_b)**: This is the upward force exerted by the syrup, which opposes the weight of the ball.
3. **Viscous Force (F_v)**: This is the drag force acting downwards due to the viscosity of the syrup.

Since the ball is falling at a constant speed, the net force acting on it is zero. Thus, we can write:
\[ W - F_b - F_v = 0 \]

### Step 2: Determine the effective weight of the ball
The effective weight of the ball (W_eff) is defined as the weight minus the buoyant force:
\[ W_{eff} = W - F_b = mg - F_b \]

### Step 3: Analyze the situation when the ball is pulled upwards
When the steel ball is pulled upwards with a force equal to twice its effective weight, the forces acting on it change:
- The upward force applied (F_applied) is \( 2W_{eff} \).
- The downward forces are the effective weight \( W_{eff} \) and the viscous force \( F_v \).

### Step 4: Set up the equation for the upward motion
The net force acting on the ball when it is pulled upwards can be expressed as:
\[ F_{net} = F_{applied} - (W_{eff} + F_v) \]
Substituting \( F_{applied} = 2W_{eff} \):
\[ F_{net} = 2W_{eff} - (W_{eff} + F_v) \]
\[ F_{net} = 2W_{eff} - W_{eff} - F_v \]
\[ F_{net} = W_{eff} - F_v \]

### Step 5: Compare the forces
Now, notice that when the ball was falling, the viscous force \( F_v \) was equal to the effective weight \( W_{eff} \) (since the ball was falling at a constant speed). Thus, we can say:
\[ W_{eff} = F_v \]

### Step 6: Determine the upward speed
Now substituting \( F_v \) with \( W_{eff} \):
\[ F_{net} = W_{eff} - W_{eff} = 0 \]

This means that when the ball is pulled upwards with a force equal to twice its effective weight, it will move upwards at the same speed it was falling, which is 10 cm/s.

### Final Answer
The steel ball will move upwards at a speed of **10 cm/s**.

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