The density of a material in CGS system is `2g//cm^(3)`. In a system of units in which unit of length is `2` cm and unit of mass is `4` g what is the numerical value of the density of the material?

Explanation

To find the numerical value of the density of the material in a different system of units, we can use the relationship between the densities in two different unit systems. The formula we will use is:

\[ \frac{n_1}{n_2} = \frac{u_1}{u_2} \quad \text{(where } n \text{ is density and } u \text{ is the unit of measurement)} \]

### Step-by-Step Solution:

1. **Identify Given Values:**
- Density in CGS (System 1): \( n_1 = 2 \, \text{g/cm}^3 \)
- Unit of length in System 1: \( u_1 = 1 \, \text{cm} \)
- Unit of mass in System 1: \( m_1 = 1 \, \text{g} \)
- Unit of length in System 2: \( u_2 = 2 \, \text{cm} \)
- Unit of mass in System 2: \( m_2 = 4 \, \text{g} \)

2. **Set Up the Density Conversion Formula:**
\[ n_2 = n_1 \cdot \frac{m_1}{m_2} \cdot \left(\frac{u_1}{u_2}\right)^{-3} \]

3. **Substitute the Known Values:**
\[ n_2 = 2 \, \text{g/cm}^3 \cdot \frac{1 \, \text{g}}{4 \, \text{g}} \cdot \left(\frac{1 \, \text{cm}}{2 \, \text{cm}}\right)^{-3} \]

4. **Calculate the Mass Ratio:**
\[ \frac{1 \, \text{g}}{4 \, \text{g}} = \frac{1}{4} \]

5. **Calculate the Length Ratio:**
\[ \left(\frac{1 \, \text{cm}}{2 \, \text{cm}}\right)^{-3} = \left(\frac{1}{2}\right)^{-3} = 2^3 = 8 \]

6. **Combine the Results:**
\[ n_2 = 2 \cdot \frac{1}{4} \cdot 8 \]

7. **Simplify the Expression:**
\[ n_2 = 2 \cdot 2 = 4 \]

### Final Answer:
The numerical value of the density of the material in the new system of units is \( n_2 = 4 \, \text{(in the new units)} \).

---