The density of a material in CGS system of units is 4 g/ cm² . In a system of units in which unit of length is 10cm and unit of mass is 100g, the value of density of material will be

To solve this problem, we will use the concept of dimensions and unit conversion. The numerical value of a physical quantity is inversely proportional to its unit. Therefore, if a physical quantity has a numerical value $n_1$ in a system with unit $u_1$, and a numerical value $n_2$ in another system with unit $u_2$, then:

$n_1{u}_1=n_2{u}_2$

Let's first identify the dimensions of density. Density is defined as mass per unit volume. However, the unit given in the question is "g/cm²", which represents mass per unit area (often called areal density or surface density). Let's use the given unit directly to find the dimensions. The dimensional formula for this quantity (mass/length²) is:

$\left[\text{Density}\right]=\left[{\text{M}}^1{\text{L}}^{-2}\right]$

Thus, the unit of density can be represented as ${\text{M}}^1{\text{L}}^{-2}$.

Let the first system be the CGS system:
Numerical value, $n_1=4$
Unit of mass, $M_1=1\text{g}$
Unit of length, $L_1=1\text{cm}$

Let the second (new) system of units be:
Numerical value, $n_2=?$
Unit of mass, $M_2=100\text{g}$
Unit of length, $L_2=10\text{cm}$

Using the relation $n_1{u}_1=n_2{u}_2$, we have:

$n_2=n_1{\left(\frac{M_1}{M_2}\right)}^1{\left(\frac{L_1}{L_2}\right)}^{-2}$

Substitute the given values into the equation:

$n_2=4{\left(\frac{1\text{g}}{100\text{g}}\right)}^1{\left(\frac{1\text{cm}}{10\text{cm}}\right)}^{-2}$

Simplify the fractions:

$n_2=4\left(\frac{1}{100}\right){\left(\frac{1}{10}\right)}^{-2}$

Since ${\left(\frac{1}{10}\right)}^{-2}={10}^2=100$, we get:

$n_2=4\times \frac{1}{100}\times 100$

Cancel out 100 in the numerator and denominator:

$n_2=4$

Note: If the unit of density in CGS was meant to be standard density $\text{g}/{\text{cm}}^3$ instead of the typed $\text{g}/{\text{cm}}^2$:
The dimensional formula would be $\left[{\text{M}}^1{\text{L}}^{-3}\right]$.
Applying the same conversion method:

$n_2=4{\left(\frac{1}{100}\right)}^1{\left(\frac{1}{10}\right)}^{-3}$

$n_2=4\times \frac{1}{100}\times 1000=40$

This matches the correct option provided. Therefore, the value of the density of the material in the new system of units is 40.