The density of material in CGS system of mass is `4g /cm^(3)` in a system of unit in which unit of length is `10 cm` and unit of mass is `100g` the value of density of material will be

Options

A. 0.04
B. 0.4
C. 40
D. 400

Explanation

To solve the problem, we need to convert the given density from the CGS system to the new system of units where the unit of length is 10 cm and the unit of mass is 100 g.

Given:
- Density in CGS system: \( \rho = 4 \, \text{g/cm}^3 \)
- New unit of length: \( 1 \, \text{new length unit} = 10 \, \text{cm} \)
- New unit of mass: \( 1 \, \text{new mass unit} = 100 \, \text{g} \)

### Step-by-Step Solution:

1. **Convert the mass to the new unit:**
\[ \text{Density} = \frac{\text{Mass}}{\text{Volume}} \]
In the new system, the mass unit is 100 g. Therefore, 4 g in the new units is:
\[ 4 \, \text{g} = \frac{4}{100} \, \text{new mass units} = 0.04 \, \text{new mass units} \]

2. **Convert the length to the new unit:**
\[ 1 \, \text{cm} = \frac{1}{10} \, \text{new length units} \]
Therefore, the volume in the new units is:
\[ 1 \, \text{cm}^3 = \left(\frac{1}{10} \, \text{new length units}\right)^3 = \frac{1}{1000} \, \text{new length units}^3 \]

3. **Calculate the density in the new units:**
\[ \rho_{\text{new}} = \frac{\text{Mass in new units}}{\text{Volume in new units}} \]
Substituting the values:
\[ \rho_{\text{new}} = \frac{0.04 \, \text{new mass units}}{\frac{1}{1000} \, \text{new length units}^3} \]
Simplifying the fraction:
\[ \rho_{\text{new}} = 0.04 \times 1000 \, \frac{\text{new mass units}}{\text{new length units}^3} = 40 \, \frac{\text{new mass units}}{\text{new length units}^3} \]

Thus, the density of the material in the new system of units is \( 40 \, \text{new units} \).