In a particular system the units of length, mass and time are chosen to be `10 cm, 10 g` and `0.1 s` respectively. The unit of force in this system will be equal to

Options

A. `0.1 N`
B. `1 N`
C. `10 N`
D. `100 N`

Explanation

To find the unit of force in the given system, we will use the formula for force derived from Newton's second law of motion, which states:

\[ F = m \cdot a \]

where:
- \( F \) is the force,
- \( m \) is the mass,
- \( a \) is the acceleration.

Acceleration can be expressed in terms of length and time as:

\[ a = \frac{L}{t^2} \]

where:
- \( L \) is the length,
- \( t \) is the time.

Thus, we can rewrite the force formula as:

\[ F = m \cdot \frac{L}{t^2} \]

Now we will substitute the units given in the problem into this formula.

1. **Identify the units:**
- Mass \( m = 10 \, \text{g} \)
- Length \( L = 10 \, \text{cm} \)
- Time \( t = 0.1 \, \text{s} \)

2. **Substituting the units into the force formula:**
\[ F = (10 \, \text{g}) \cdot \left(\frac{10 \, \text{cm}}{(0.1 \, \text{s})^2}\right) \]

3. **Calculating the acceleration:**
\[ (0.1 \, \text{s})^2 = 0.01 \, \text{s}^2 \]
Therefore,
\[ a = \frac{10 \, \text{cm}}{0.01 \, \text{s}^2} = 1000 \, \text{cm/s}^2 \]

4. **Now substituting back to find force:**
\[ F = 10 \, \text{g} \cdot 1000 \, \text{cm/s}^2 \]
\[ F = 10000 \, \text{g cm/s}^2 \]

5. **Converting the force to standard units:**
We know that \( 1 \, \text{g cm/s}^2 = 1 \, \text{dyne} \), so:
\[ F = 10000 \, \text{dyne} \]

Thus, the unit of force in this system is \( 10000 \, \text{dyne} \).