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

Explanation

To solve the problem, we need to analyze the forces acting on the steel ball when it is pulled upwards with a force equal to twice its effective weight. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the forces acting on the ball
When the steel ball is falling through the syrup at a constant speed of 10 cm/s, it experiences three main forces:
1. **Weight of the ball (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 (B)**: This is the upward force exerted by the syrup, which acts against the weight of the ball.
3. **Viscous drag force (F_v)**: This is the resistance force due to the syrup, which also acts upwards.

At terminal velocity (constant speed), the net force acting on the ball is zero. Therefore, we can write:
\[ W - B - F_v = 0 \]
This implies:
\[ F_v = W - B \]
The effective weight of the ball is defined as:
\[ W_{\text{effective}} = W - B \]

### Step 2: Determine the effective weight
Since the ball is falling at a constant speed, the effective weight can be expressed as:
\[ W_{\text{effective}} = F_v \]
Given that the ball is falling at a constant speed of 10 cm/s, we can denote this speed as \( v = 10 \, \text{cm/s} \).

### Step 3: Calculate the upward force applied
According to the problem, the steel ball is pulled upwards with a force equal to twice its effective weight:
\[ F_{\text{up}} = 2 W_{\text{effective}} \]

### Step 4: Analyze the net force when pulling the ball upwards
When the ball is pulled upwards, the forces acting on it are:
- The upward force \( F_{\text{up}} = 2 W_{\text{effective}} \)
- The downward forces, which include the weight of the ball \( W \) and the viscous drag force \( F_v \).

The net force \( F_{\text{net}} \) acting on the ball when it is pulled upwards can be expressed as:
\[ F_{\text{net}} = F_{\text{up}} - (W + F_v) \]

Substituting \( F_{\text{up}} \) and \( F_v \):
\[ F_{\text{net}} = 2 W_{\text{effective}} - (W + W_{\text{effective}}) \]
\[ F_{\text{net}} = 2 W_{\text{effective}} - W - W_{\text{effective}} \]
\[ F_{\text{net}} = W_{\text{effective}} - W \]

### Step 5: Determine the acceleration
The net force will cause the ball to accelerate upwards. According to Newton's second law:
\[ F_{\text{net}} = ma \]
Where \( a \) is the acceleration of the ball. Therefore:
\[ W_{\text{effective}} - W = ma \]

### Step 6: Calculate the upward speed
Since the ball was initially moving downwards at a constant speed of 10 cm/s, when it is pulled upwards with a force equal to twice its effective weight, it will start moving upwards. The new speed will be equal to the initial speed because the forces are balanced in a similar manner. Thus, the speed of the ball moving upwards will also be:
\[ v = 10 \, \text{cm/s} \]

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

---