A force F is applied on a body (which moves on a straight line) for a duration of 3 s. The momentum of the body changes from 10 g cm/s to 40 g cm/s. The magnitude of the force F is

Options

A. 10 dyne
B. 10 newton
C. 120 dyne
D. 12 dyne

Explanation

To solve the problem, we need to find the magnitude of the force \( F \) applied to a body whose momentum changes over a specific time interval. We will use the relationship between force, change in momentum, and time.

### Step-by-Step Solution:

1. **Identify the initial and final momentum**:
- Initial momentum \( p_i = 10 \, \text{g cm/s} \)
- Final momentum \( p_f = 40 \, \text{g cm/s} \)

2. **Calculate the change in momentum**:
\[ \Delta p = p_f - p_i = 40 \, \text{g cm/s} - 10 \, \text{g cm/s} = 30 \, \text{g cm/s} \]

3. **Determine the time duration**:
- The time duration \( t = 3 \, \text{s} \)

4. **Use the formula for force**:
According to Newton's second law, the force \( F \) is equal to the rate of change of momentum:
\[ F = \frac{\Delta p}{t} \]

5. **Substitute the values into the equation**:
\[ F = \frac{30 \, \text{g cm/s}}{3 \, \text{s}} = 10 \, \text{g cm/s}^2 \]

6. **Convert the units to dynes**:
In the CGS system, \( 1 \, \text{g cm/s}^2 = 1 \, \text{dyne} \). Therefore:
\[ F = 10 \, \text{dyne} \]

### Final Answer:
The magnitude of the force \( F \) is \( 10 \, \text{dyne} \).

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